Class 8

Go through BSE Odisha Class 8 Science Solutions Chapter 9 The Amazing World of Solutes, Solvents, and Solutions Question Answer to understand textbook questions more clearly.

Class 8 Science Curiosity Chapter 9 Question Answer

Class 8 Science Ch 9 The Amazing World of Solutes, Solvents, and Solutions Question Answer

Class 8 Science Chapter 9 The Amazing World of Solutes, Solvents, and Solutions Question Answer

Probe and Ponder Questions

Question 1.
What do you think is happening in the picture given below?

Answer:

• The picture shows Mahatma Gandhi obtaining salt from the sea during the Salt March, accompained by followers.
• This illustrates the historical process of extracting salt from seawater through evaporation, where seawater acts as a natural solution with salt as the solute and water as the solvan, highlighting concepts of solubility and traditional salt production.

Question 2.
What happens when you add too much sugar to your tea and it stops dissolving? How can you solve this problem?
Answer:

• When too much sugar is added, the tea becomes a saturated solution and excess sugar settles at the bottom as it can no longer dissolve at that temperature.
• To solve this, heat the tea to increase the solubility of sugar, allowing more to dissolve, as solubility generally increases with temperature for solids in liquids.

Question 3.
Why do sugar and salt dissolve in water but not in oil? Why is water considered a good solvent?
Answer:
Sugar and salt dissolve in water because water mixes evenly with many with many substances and can break them down into smaller particles forming uniform solution. They do not dissolve in oil because oil cannot mix well with them. Water is considered a good solvent because it can dissolve a large number of substances, so it is often called a universal solvent.

Question 4.
Why are water bottles usually tall and cylindrical in shape instead of spherical?
Answer:

• Water bottles are tall and cylindrical as they are easy to hold, store and use efficiently.
• They provide better grip, stability and use less material for the same volume compared to spheres, which would roll and be harder to handle.

Question 5.
Share your questions …………
Answer:
Based on the chapter, potential questions could include:

• How does temperature affect the solubility of gases differently from solids?
• Why does ice float on water despite being a solid?
• What role does density play in everyday phenomena like hot air balloons?

InText Questions

Question 1.
We know air is a mixture. Would a mixture of gases also be considered a solution? (Page 135)
Answer:
Yes, just like liquid solution where water act as a solvent, gases can also form solution – with air being a common example. Air is a gaseous solution. Since nitrogen is present in the largest amount in the air, it is considered as the solvent, while oxygen, argon, carbon dioxide, and other gases are considered as solutes.

Question 2.
What will happen if we keep on adding more salt in a given amount of water? (Page 136)
Answer:
A stage comes when the added salt does not dissolve completely and undissolved salt settles at the bottom.

Question 3.
Do gases also dissolve in water ? (Page 139)
Answer:
Yes, many gases, including oxygen dissolve in water. All aquatic life like fishes, even plants utilises these dissolved oxygen to sustain.

Question 4.
How many types of mixture are there? What special name is given to uniform mixture? How would you able to see their components?
Answer:
Now I understand that the mixtures we use can be of two types-uniform and nonuniform. Uniform mixtures are called solutions, and their components are not visible separately. In non-uniform mixtures, the components can be seen either with the naked eye or with a magnifying device.

Question 5.
I observed that in some nonuniform mixtures, such as sawdust in water, the sawdust floats, whereas in the mixture of sand and water, the sand sinks. I wonder why that happens? (Page 139)
Answer:
This happens because sawdust is lighter than water but sand is heavier than water. In other words, density of sawdust is less than water but density of sand is more than water.

Question 6.
Why are measuring cylinders always designed narrow and tall instead of wider and short like a beaker? (Page 144)
Answer:
A measuring cylinder is designed to be narrow and tall to improve the accuracy and precision of liquid volume measurements. It also minimizes the distortion of the liquid’s curved surface (the meniscus), making it easier for a person to read the volume consistently at eye level.

Question 7.
I wonder how the level of a coloured liquid is measured? (Page 145)
Answer:
For coloured liquids take reading from the top of the meniscus.

Question 8.
What is the maximum amount of solute which a fixed amount of solvent can dissolve ?
Answer:
It is called the solubility.

The Amazing World of Solutes, Solvents, and Solutions Class 8 Questions and Answers

Keep the Curiosity Alive (Pages 149-151)

Question 1.
State whether the statements given below are True [T] or False [F]. Correct the false statement(s).

(i) Oxygen gas is more soluble in hot water rather than in cold water.
Answer:
False: Oxygen is more soluble in cold water.

(ii) A mixture of sand and water is a solution.
Answer:
False: A Mixture of sand and water is not a solution. Sand does not dissolve in water but settles down.

(iii) The amount of space occupied by any object is called its mass.
Answer:
False: The amount of space occupied by any object is called its volume

(iv) An unsaturated solution has more solute dissolved than a saturated solution.
Answer:
False: A saturated solution has more solute dissolved than an unsaturated solution.

(v) The mixture of different gases in the atmosphere is also a solution.
Answer:
True.

Question 2.
Fill in the blanks:

(i) The volume of a solid can be measured by the method of displacement, where the solid is ……… in water and the ………… in water level is measured.
Answer:
placed; rise

(ii) The maximum amount………… dissolved in ………… of……….. at a particular temperature is called solubility at that temperature.
Answer:
solute, solvent

(iii) Generally, the density ………… with increase in temperature.
Answer:
decrease

(iv) The solution in which glucose has completely dissolved in water, and no more glucose can dissolve at a given temperature, is called a ………… solution of glucose.
Answer:
saturated

Question 3.
You pour oil into a glass containing some water. The oil floats on top. What does this tell you?
(i) Oil is denser than water.
(ii) Water is denser than oil.
(iii) Oil and water have the same density.
(iv) Oil dissolves in water.
Answer:
(ii) Water is denser than oil.
Oil floats because its density is lower than water’s, causing less dense substances to float on denser ones.

Question 4.
A stone sculpture weighs 225 g and has a volume of 90cm³. Calculate its density and predict whether it will float or sink in water.
Answer:
Density of stone =\(\frac{225}{90}\)=2.5 g/cm³
Since this is greater than water’s density ( 1g/cm³ )the sculpture will sink in water.

Question 5.
Which one of the following is the most appropriate statement, and why are the other statements not appropriate?
(i) A saturated solution can still dissolve more solute at a given temperature.
(ii) An unsaturated solution has dissolved the maximum amount of solute possible at a given temperature.
(iii) No more solute can be dissolved into the saturated solution at that temperature.
(iv) A saturated solution forms only at high temperatures.
Answer:
Statement (iii) is most appropriate.
(i) It is not appropriate as a saturated solution cannot dissolve more solute at a given temperature.
(ii) An unsaturated solution can have more solute dissolved at a given temperature.
(iii) Correct.
(iv) A Saturated solution can be formed at all temperatures.

Question 6.
You have a bottle with a volume of 2 litres. You pour 500 mL of water into it. How much more water can the bottle hold?
Answer:
The bottle of 2 litres capacity can hold 1500 mL more of water besides 500 mL.

Question 7.
An object has a mass of 400 g and a volume of 40cm³. What is its density?
Answer:
Density = \(\frac{\text { mass }}{\text { volume }}\)=\(\frac{400}{40}\)=10 g/cm³

Question 8.
Analyse Figures (a) and (b). Why does the unpeeled orange float, while the peeled one sinks? Explain.

Answer:
An unpeeled orange displaces more water, and so it floats. Peeled orange displaces less water than its weight, so it sinks.

Question 9.
Object A has a mass of 200 g and a volume of 40 cm³. Object B has a mass of 240 g and a volume of 60 cm³. Which object is denser?
Answer:
Density of object A=\(\frac{200}{40}\)= 5g/cm³
Density of object B= \(\frac{240}{60}\)=4g /cm³
Conclusion: Object A is denser, having more density than B.

Question 10.
Reema has a piece of modeling clay that weighs 120 g. She first moulds it into a compact cube that has a volume of 60 cm³. Later, she flattens it into a thin sheet. Predict what will happen to its density.
Answer:
The density of flattened clay will decrease as it displaces more liquid.

Question 11.
A block of iron has a mass of 600 g and a density of 7.9g/cm³. What is its volume?
Answer:
We know density =\(\frac{\text { mass }}{\text { volume }}\)
∴ Volume = \(\frac{\text { mass }}{\text { density }} \)
= \(\frac{600 \mathrm{~g}}{7.9 \mathrm{~g} / \mathrm{cm}^3}\)=75.94 cm}³

Question 12.
You are provided with an experimental setup as shown in Figures (a) and (b). On keeping the test tube (Figure b) in a beaker containing hot water ∼70°C, the water level in the glass tube rises. How does it affect the density?

Answer:
The density of water in setup (b) will decrease.

Class 8 Science Chapter 9 Question Answer

Activity 1

Let us investigate
Aim: To find the capacity of water to dissolve solutes.

Materials Required: A glass tumbler, salt.

Procedure:

• Take a clean glass tumbler and half filled with water.
• Now add one spoonful of salt into it and stir it untill it dissolves completely (see figure).
• Continue adding a spoonful of salt into the glass tumbler and stir. Observe how many spoons of salt you can add before it stops dissolving completely.
• Note down your observations in table given below.

Observations:
Table: Dissolution of salt in water

Answer:

Amount of salt taken (teaspoon)	Observation (salt dissolves/salt does not dissolve)
One	Salt dissolves
Two	Salt dissolves
Three	Salt dissolves
Four	Salt dissolves with difficulty
Five	Salt does not dissolve and settle at the bottom

Inferences:

• After adding a few more spoons of salt, a stage comes when the added salt does not dissolve completely and the undissolved salt settles at the bottom.
• This indicates water reaches a limit and cannot dissolve more salt. This is called saturated solution.

Activity 2.

Let us experiment (Demonstration activity)
Aim: To show that solubility of substances increases with increase in temperature.
Materials Required: Baking soda (sodium hydrogen carbonate), a glass beaker.

Procedure:

• First of all we will take about 50 mL of water in a glass beaker and measure its temperature using a laboratory thermometer, say 20°C.
• Now, add a spoonful of baking soda (sodium hydrogen carbonate) to the water and stir until it dissolves. Continue adding small amounts of baking soda while stirring, till some solid baking soda is left undissolved at the bottom of the beaker.
• Heat the mixture to 50°C while stirring.
• Observe the undissolved baking soda dissolving.
• Add more baking soda until undissolved solid remains again.
• Heat further to 70°C while stirring and observe again.

Observations:

• At 20°C: Limited baking soda dissolves; excess remains undissolved.
• At 50°C: Previously undissolved baking soda dissolves; more can be added before saturation.
• At 70°C: Even more baking soda dissolves, showing increased capacity.

Inference: Water at 70°C dissolves more baking soda than at 50°C and much more than at 20°C.

Activity 3.

Let us measure
Aim: To measure the mass of objects.
Materials Required: Digital weighing balance, a watch glass, stone or solid objects.

Procedure:

• Switch ON the digital weighing balance.
• Observe the initial reading on the digital weighing balance display.
• The balance should show a zero reading initially. If not, then we must bring it to zero by pressing the tare or reset button (Figure (a).
• Now, put a dry and clean watch glass or butter paper on the pan.
• Note down the reading on the digital weighing balance.
• Reset the digital weighing balance reading to zero by pressing the tare or reset button as shown in figure (b).
• Now, carefully place the solid object, such as stone, on the watch glass [Figure (c)].
• Note the reading displayed on the balance, which gives the mass of the stone, say 15.400 g.
• Repeat the experiment with different objects like an apple, orange etc.
• You can use any other type of balance available in your school.

Observations: Mass of different objects are different.
Inference: A digital weighing balance or a balance give the measurement of mass. e.g., Mass of stone =15.400 g
Mass of an apple =150 g

Activity 4.

Let us observe and calculate
Aim: (i) To measure the maximum volume of liquid using a measuring cylinder.
(ii) To find the smallest value that a measuring cylinder can read.
Materials Required: A measuring cylinder.
Procedure: Take a measuring cylinder of 100 mL.

Observations:

• There are 10 smaller divisions between 10 mL and 20 mL or between 40 mL and 50 mL.
• 10 small divisions =10 mL
  So, one small division = \(\frac{10}{10}\)=1 mL
• It can measure volume upto 100 mL.

Inference:

1. What is the maximum volume it can measure?
Answer: The cylinder is marked as 100 mL; therefore, it can measure volume up to 100 mL.

2. What is the smallest volume it can measure?

3. How much is the volume difference indicated between the two bigger marks (for example, between 10 mL and 20 mL )?
Answer: 10 mL

4. How many smaller divisions are there between the two bigger marks?
Answer: 10 smaller divisions
Fig: Measuring cylinder of 100 m L

5. How much volume does one small division indicate?
Answer: 1mL

Activity 5.

Let us measure 50 mL of water
Aim: To measure 50 mL of water.
Materials Required: A dry measuring cylinder, a droper.

Procedure:

• Take a clean and dry measuring cylinder on a flat surface and pour water slowly to the mark. [see figure (a)]
• Use a dropper to add or remove water for exact level.
• If you observe carefully then you will findthat the water inside the measuring cylinder forms a curved surface. This curved surface is called the meniscus [see figure (b)].
• Keep eyes at level with the bottom of the meniscus for accurate reading.
• As soon as it reaches the required level-that is, 50 mL – transfer this water to the required container.
• For coloured liquids read the top of the meniscus.

Observations: Reading of the bottom of the meniscus is observed 50 mL.
Inference: The volume of water is 50 mL.

Determining Volume of Solid Objects with Regular Shapes

• For cuboid shapes (e.g., notebook, shoe box, dice), measure length (l), width (w), height (h) with a scale.
• Formula: Volume = l ×w ×h.

Activity 6.

Let us calculate
Aim: To determine volume of solid objects with regular shapes (like cube, cuboid).
Materials Required: For cuboid shapes (e.g., notebook, shoe box), For cube: a dice, scale.

Procedure:

• Take various objects with a cuboid shapes, such as a notebook, a shoe box, or a dice.
• Now, measure the length (l), width (w), and height (h) of the objects with the help of a scale. Suppose the length of the notebook is 25 cm, the width is 18 cm, and the height is 2 cm.

Observations:

	Note book	dice	shoe box
length (l) width or breadth (w) height	25 cm 18 cm 2 cm	1 cm 1 cm 1 cm	10 cm 5 cm 3 cm

Demonstration:
Volume of cuboid =l × w × h
Volume of cube = a³ or (side)³
∴ Volume of Note book =25 × 18 × 2= 900cm³
Volume of a dice =1cm × 1 cm × 1 cm = 1 cm³
Volume of a shoe box =10 × 5 × 3 = 150 cm³

Inference:

Volume of a cuboid = l times w times h
Volume of cube = side × side × side
Note: The values of volume are obtained in units of mL, which can be written in the equivalent unit cm³ for solids.

Activity 7.

Let us measure
Aim: To determine the volume of objects with irregular shapes.
Materials Required: A measuring cylinder, various objects such as a stone, metal keys etc.

Procedure:

• Take some objects from your surroundings, like stone, metal keys, and so on.
• Now, pour water in a measuring cylinder up to any desired volume, say 50 mL [Figure (a)] and record the initial volume taken in table.
• Now, tie the object, say a stone, with the help of a thread and slowly lower it down into the measuring cylinder.
• Note down your observation.
• Now, record the final volume after the level rises, say 55 mL, as shown in [Figure (b)].
• Subtract the initial volume from the final volume after the object is put into the measuring cylinder. This is the volume of the object.

Observations:

Table: Volume of irregular solids

S.No.	Object	Initial volume of water in the measuring cylinder (mL) (A)	Final volume of water in the measuring cylinder (mL) (B)	Volume of water displaced in the measuring cylinder (mL) (B-A)	Volume of the object ( cm3)
1. 2. 3.	Stone Metal key Any other	50 mL	55 mL	5 mL	5cm3

Inference:
Volume of object = Final volume of water – Initial volume of water
∴ Volume of stone =55 mL-50 mL
= 5mL = 5 cm³

The Amazing World of Solutes, Solvents, and Solutions Class 8 Extra Questions and Answers

Short Answer Type Questions

Question 1.
How do you decide which liquid is the solute and which is the solvent when two liquids mix ?
Answer:
The component present in the smaller amount is the solute and the component in the larger amount is the solvent.

Question 2.
Why does sugar dissolve in water but sand does not form a solution?
Answer:
Sugar particles interact with water and disperse evenly to form a clear solution, while sand does not dissolve and settles at the bottom.

Question 3.
What does it mean when a solution becomes saturated ?
Answer:
A saturated solution has dissolved the maximum amount of solute at that temperature, so any extra solute remains undissolved.

Question 4.
How can heating turn a saturated solution into an unsaturated one ?
Answer:
Heating usually increases wi bility for many solids, so the solvent can cassolve more solute and the previously saturated solution becomes unsaturated.

Question 5.
Floating of ice is an important phenomenon. How does it helps animals living in lakes and oceans?
Answer:
This is important for animals living in lakes and oceans because ice floats, it forms a layer on top, keeping the water underneath warm enough for fish and other creatures to survive, even in extremely cold weather.

Question 6.
If mass of an object is 16.400 g and its volume is 5cm³, then calculate the density of object.
Answer:
Mass of object =16.400 g
Volume of object =5 cm³
∴ Density = \(\frac{\text { Mass }}{\text { Volume }}\)= \(\frac{16.4 \mathrm{~g}}{5 \mathrm{~cm}^3}\) 5cm³ =3.28 g/cm³

Question 7.
Why hot air balloons rises in the sky ?
Answer:
As temperature increases the volume of gases filled inside the balloons increases and its density decreases. So, density of gases inside the balloon is less than the cool air around it and hence it rises.

Long Answer Type Questions

Question 1.
Explain with an example how the same substances can form different types of mixtures depending on proportion and how to identify them.
Answer:
(a) Oil and water usually form a non-uniform mixture with separate layers when oil is added in ordinary amounts, so it is not a solution.
(b) However, a very small amount of acetic acid in water forms vinegar, which is a true solution because it is uniform and clear.
(c) To identify them, look for clarity, absence of layers, and particles that do not settle on standing. If the mixture is cloudy or separates, it is not a true solution. If it remains clear and uniform, it is a solution.

Question 2.
Explain how relative density helps predict floating and sinking better than mass alone, using two same-sized objects made of different materials.
Answer:
(a) Mass alone can be misleading because it does not consider volume, but relative density compares a material’s density to water. If a same-sized wooden block and an iron block are placed in water, the wood floats and iron sinks because wood’s density is less than water’s, while iron’s is greater.

(b) Relative density less than 1 means it will float in water; greater than 1 means it will sink. Thus, relative density accurately predicts behavior in a liquid

Case-Study Based Questions

1. Read the following passage carefully and answer the questions that follow : At a picnic, friends make lemonade by stirring sugar into water. After adding too much sugar, they notice some grains settle at the bottom of the glass. They realise no more sugar can dissolve at that temperature, no matter how much they stir.

(a) What does it mean when sugar settles at the bottom and does not dissolve further?
Answer:
It means the solution is saturated; no more sugar can dissolve at that temperature.

(b) What type of solution is formed before and after the sugar settles?
Answer:
Before settling, the solution is unsaturated; after settling, it is saturated.

(c) How can temperature changes affect how much sugar dissolves in water in this scenario?
Answer:
Increasing temperature usually increases solubility; thus, more sugar can dissolve in warmer water, while decreasing temperature may cause sugar to crystallise out.

Picture Based Questions

I. Look at the picture and answer the following questions:

(a) Identify the picture.
(a) Bamboo raft
(b) Wooden raft
(c) Kayaka
(d) None of these
Answer:
(a) Bamboo raft

(b) Why it floats on water ?
Answer:
It floats on water because it is lighter than water.

(c) Why Bamboo was used in it ? Write its uses also?
Answer:
Bamboo was used because it is light, hollow and floats easily on water. People tied bamboo poles together to make rafts and small boats for fishing, trading and crossing water bodies.

The Amazing World of Solutes, Solvents, and Solutions Class 8 MCQ

Multiple Choice Questions (Mcqs)

Question 1.
Which of the following is an example of solution?
(a) Sugar in water
(b) Sand in water
(c) Muddy water
(d) Oil and water
Answer:
(a) Sugar in water

Question 2.
Which of the following is a characteristic of solute ?
(a) It is present in the largest quantity.
(b) It dissolved in the solvent
(c) Both (c) and (d)
(d) It is present in smallest quantity
Answer:
(c) Both (c) and (d)

Question 3.
A solution that contains the maximum amount of solute that can be dissolved at a given temperature is called :
(a) Unsaturated Solution
(b) Saturated Solution
(c) Supersaturated Solution
(d) Dilute Solution
Answer:
(b) Saturated Solution

Question 4.
If a solution has a high concentration of solute, it is considered :
(a) Dilute
(b) Concentrated
(c) Saturated
(d) Unsaturated
Answer:
(b) Concentrated

Question 5.
The SI unit of density is: ………….
(a) gram/cm³
(b) kg/m³
(c) cubic metre
(d) g/L
Answer:
(b) kg/m³

Assertion and Reasoning

These questions consist of two statements, each printed as Assertion (A) and Reason (R). While answering these questions, you are required to choose any one of the following four responses.
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.
(b) Assertion (A) and Reason (R) both are correct and reason is not correct explanation for assertion.
(c) Assertion (A) is correct but the Reason (R) is wrong.
(d) Assertion (A) is wrong but the Reason (R) is correct.

1. Assertion (A): Mass is the amount of matter contained in body.
Reason (R): Mass is measured in newton (N) unit.
Answer:
(c) Assertion (A) is correct but the Reason (R) is wrong.

2. Assertion (A): SI unit of density is kg/m
Reason (R): Relative density has no unit.
Answer:
(b) Assertion (A) and Reason (R) both are correct and reason is not correct explanation for assertion.

Fill in the blanks

1. The concentration of a solution is the amount of ………… present per unit volume or per unit mass of the solution/solvent.
Answer:
solute

2. A homogeneous mixture of two or more substances is called ………….
Answer:
solution

3. ………… is the maximum amount of the solute that can be dissolved in a given solution at a given temperature.
Answer:
solubility

4. SI unit of density is ………….
Answer:
kg/m³

5. Relative density is the ratio of the density of the substance to the density of ………….
Answer:
water.

True or False

1. The maximum amount of solute that dissolves in a fixed quantity of the solvent is called its solubility.
Answer:
True

2. Relative density do not have any units.
Answer:
True

3. Volume of liquids cannot be measured by a measuring cylinder.
Answer:
False

4. Volume of a solid object with regular shapes are calculated with the help of formulas.
Answer:
True

5. The unit of density is cubic metre.
Answer:
True

Students can refer to BSE Odisha Class 8 Math Solution and Ganita Prakash Chapter 7 Proportional Reasoning 1 Class 8 Question Answer to understand textbook questions step by step.

Class 8 Maths Chapter 7 Proportional Reasoning 1 Solutions

Ganita Prakash Class 8 Chapter 7 Solutions

Class 8 Maths Ganita Prakash Chapter 7 Solutions Proportional Reasoning 1

1. PROBLEM SOLVING WITH PROPORTIONAL REASONING
Figure it Out (Page 165 – 167) :

Question 1.
Circle the following statements of proportion that are true.
(i) 4 : 7 :: 12 : 21
(ii) 8 : 3 :: 24 : 6
(iii) 7 : 12 :: 12 : 7
(iv) 21 : 6 :: 35 : 10
(v) 12 : 18 – 28 : 12
(vi) 24 : 8 :: 9 : 3
Answer:
(i) Given statement is 4 : 7 :: 12 : 21.
This is true if \(\frac{4}{7}\) = \(\frac{12}{21}\) or if \(\frac{4}{7}\) = \(\frac{4}{7}\), which is true.
∴ The given statement is true.

(ii) Given statement is 8 : 3 :: 24 : 6.
This is true if \(\frac{8}{3}\) = \(\frac{24}{6}\) or if \(\frac{8}{3}\) = 4, which is false.
∴ The given statement is not true.

(iii) Given statement is 7 : 12 :: 12 : 7.
This is true if \(\frac{7}{12}\) = \(\frac{12}{7}\) which is false.
∴ The given statement is not true.

(iv) Given statement is 21 : 6 :: 35 : 10.
This is true if \(\frac{21}{6}\) = \(\frac{35}{10}\) or if \(\frac{7}{2}\) = \(\frac{7}{2}\), which is true.
∴ The given statement is true.

(v) Given statement is 12 : 18 :: 28 : 12.

This is true if \(\frac{12}{18}\) = \(\frac{28}{12}\) or if \(\frac{2}{3}\) = \(\frac{7}{3}\) or 2 = 7, which is false.
∴ The given statement is not true.

(vi) Given statement is 24 : 8 :: 9 : 3.
This is true if \(\frac{24}{8}\) = \(\frac{9}{3}\) or if 3 = 3, which is true.
∴ The given statement is true.

Question 2.
Give 3 ratios that are proportional to 4 : 9.
__________ : ____________ __________ : ____________ __________ : ____________
Answer:
To find ratios proportional to 4 : 9, we multiply both terms by the same number:
4 × 2 : 9 × 2 = 8 : 18.
4 × 3 : 9 × 3 = 12 : 27.
4 × 5 : 9 × 5 = 20 : 45.
So, three ratios proportional to 4 : 9 are 8 : 18; 12 : 27 and 20 : 45.

Question 3.
Fill in the missing numbers for these ratios that are proportional to 18 : 24.
3 : ________, 12 : ________ ,20 : ________ , 27 : ________
Answer:
(i) Given ratio is 18 : 24.
Let 18 : 24 : : 3 : x
∴ \(\frac{18}{24}\) = \(\frac{x}{3}\) or \(\frac{3}{4}\) = \(\frac{3}{x}\) or x = 4
∴ 24 X or 4 = X or x = 4
∴ Missing number in the ratio : 3 _________ is 4.

(ii) Let 18 : 24 : : 12 : x.
∴ \(\frac{18}{24}\) = \(\frac{12}{x}\) or \(\frac{3}{4}\) = \(\frac{12}{x}\)
or 3x = 48 or x = \(\frac{48}{3}\) = 16
∴ Missing number in the ratio 12 : _________ is 16.

(iii) Let 18 : 24 : : 20 : x.
∴ \(\frac{18}{24}\) = \(\frac{20}{x}\) or \(\frac{3}{4}\) = \(\frac{20}{x}\)
or 3x = 80 or x = \(\frac{80}{3}\)
Missing number in the ratio 20 : ___________ is \(\frac{80}{3}\)

(iv) Let 18 : 24 : : 27 : x.
∴ \(\frac{18}{24}\) = \(\frac{27}{x}\) or \(\frac{3}{4}\) = \(\frac{27}{x}\)
or 3x = 108 or x = \(\frac{108}{3}\) = 36
∴ Missing number in the ratio 27 : ____________ is 36.

Question 4.
Look at the following rectangles. Which rectangles are similar to each other? You can verify this by measuring the width and height using a scale and comparing their ratios.

Answer:

Rectangle	Width	Height	Ratio
A	0.5 cm	1.5 cm	0.5 : 1.5 = 1 : 3
B	1.5 cm	1 cm	1.5 : 1 = 3 : 2
C	4.5 cm	2 cm	4.5 : 2 = 9 : 4
D	3.5 cm	1 cm	3.5 : 1 = 7 : 2
E	0.5 cm	1.5 cm	0.5 : 1.5 = 1 : 3

Since rectangles A and E have the same simplified ratio 1 : 3. So, they are similar to each other.

Question 5.
Look at the following rectangle. Can you draw a smaller rectangle and a bigger rectangle with the same width to height ratio in your notebooks? Compare your rectangles with your classmates’ drawings.
Are all of them the same? If they are different from yours, can you think why? Are they wrong?

Answer:
For the given rectangle;
Width = 32 mm and height = 18 mm
∴ Ratio is 32 : 18.
We shall draw smaller and bigger rectangles and similar to the given rectangle by considering different ‘factors of change’.
Let the factor of change be \(\frac{1}{2}\).
∴ New width = \(\frac{1}{2}\) × 32 = 16 mm
and New height = \(\frac{1}{2}\) × 18 = 9 mm
A new, similar rectangle is shown in the figure.

Let ‘factor of change’ be 2.
∴ New width = 2 × 32 = 64 mm and new height = 2 × 18 = 36 mm

A new, similar rectangle is shown in the figure. The rectangles drawn by other classmates are all different, but they are all similar to the given rectangle.

Question 6.
The following figure shows a small portion of a long brick wall with patterns made using coloured bricks. Each wall continues this pattern throughout the wall. What is the ratio of grey bricks to coloured bricks? Try to give the ratios in their simplest form.

Answer:
(a) We consider one set of patterns in the given wall.

Number of grey bricks in one set of pattern = 2 + 3 + 4 = 9
Number of coloured bricks in one set of pattern = 3 + 2 + 1 = 6
∴ Ratio of grey bricks to coloured bricks = 9 : 6
We have 9 : 6 = 3 : 2
∴ Ratio in the simplest form = 3 : 2

(b) We use one set of patterns on the given wall

One set of pattern
Number of grey bricks in one set of pattern
= (\(\frac{1}{2}\) + 1 + 1 + \(\frac{1}{2}\)) + (1 + 1) + (\(\frac{1}{2}\) + 1 + \(\frac{1}{2}\)) + (1 + 1) + (\(\frac{1}{2}\) + 1 + \(\frac{1}{2}\)) + (1 + 1) + (\(\frac{1}{2}\) + 1 + 1 + \(\frac{1}{2}\))
= 3 + 2 + 2 + 2 + 2 + 2 + 3 = 16
Number of coloured bricks in one set of pattern
= 1 + (1 + 1) + (1 + 1) + (1 + 1) + (1 + 1) + (1 + 1) + 1
= 1 + 2 + 2 + 2 + 2 + 2 + 1 = 12
∴ Ratio of grey bricks to coloured bricks = 16 : 12
We have 16 : 12 = 4 : 3
∴ Ratio in the simplest form = 4 : 3.

Question 7.
Let us draw some human figures. Measure your friend’s body-the lengths of their head, torso, arms, and legs. Write the ratios as mentioned below-

Answer:
My friend’s body measurements :
(i) Head = 22 cm

(ii) Torso (neck to hip) = 50 cm

(iii) Arms (shoulder to fingertip) = 60 cm

(iv) Legs (hip to foot) = 80 cm

1. Head : Torso = 22 : 50
Simplify by dividing both by 2 → 11 : 25.

2. Torso : Arms = 50 : 60
Simplify by dividing both by 10 → 5 : 6.

3. Torso : Legs = 50 : 80
Simplify by dividing both by 10 → 5 : 8.
So the ratios are:

• Head : Torso = 11 : 25
• Torso : Arms = 5 : 6
• Torso : Legs = 5 : 8

Figure it Out (Page 170 – 171) :

Question 1.
The Earth travels approximately 940 million kilometres around the Sun in a year. How many kilometres will it travel in a week?
Answer:
We know that:
1 million = 10 lakh = 10,00,000
and 1 year = \(\frac{365}{7}\) weeks.
940 million kilometres, i.e., 940 × 10,00,000 kilometres, are travelled by the Earth in 1 year, i.e., in \(\frac{365}{7}\) weeks.
Let the fiarth travel x kilometres in 1 week
∴ The ratios 940 × 10,00,000 : \(\frac{365}{7}\) and x : 1 are
in proportion.
⇒ \(\frac{940 \times 10,00,000}{\frac{365}{7}}\) = \(\frac{x}{1}\)
⇒ x = \(\frac{940 \times 10,00,000 \times 7}{365}\)
⇒ x = \(\frac{188 \times 70,00,000}{73}\)
⇒ x = 1,80,27,397 (nearly)
∴ In 1 week, Earth travels nearly 1,80,27,397 kilometres around the Sun.

Question 2.
A mason is building a house in the shape shown in the diagram. He needs to construct both the outer walls and the inner wall that separates two rooms. To build a wall of 10-feet, he requires approximately 1450 bricks. How many bricks would he need to build the house? Assume all walls are of the same height and thickness.

Answer:
Number of bricks required for a 10ft wall =1450
∴ Ratio of length of wall to number of bricks = 10 : 1450

Total length of walls = AI + CH + DE + FG + IG + AF + CD
= 12 + (9 + 12) + 9 + 12 + (9 + 15) + (9 + 15) + 6 = 108 ft
Let x bricks be required for a 108 ft long wall.
∴ Ratio of length of wall to number of bricks = 108 : x
These ratios are in proportion.
∴ 10 : 1450 :: 108 : x
⇒ \(\frac{10}{1450}\) = \(\frac{108}{x}\)
⇒ \(\frac{1}{145}\) = \(\frac{108}{x}\)
⇒ x = 145 × 108 = 15,660
∴ Number of required bricks = 15,660.

Figure it Out (Page 175) :

Question 1.
Divide ₹4,500 into two parts in the ratio 2 : 3.
Answer:
Given ratio = 2 : 3
Amount to be divided = ₹ 4,500
∴ First part = \(\frac{2}{2 + 3}\) × 4,500
= \(\frac{2}{5}\) × 4,500 = 2 × 900 = ₹ 1,800
∴ Second part= \(\frac{3}{2 + 3}\) × 4,500 = \(\frac{3}{5}\) × 4,500
= 3 × 900 = ₹ 2,700
∴ Two parts are ₹ 1,800 and ₹ 2,700.
Verification:
1,800 : 2,700 = \(\frac{1,800}{2,700}\)
\(\frac{18}{27}\) = \(\frac{2}{3}\) = 2 : 3 and 1,800 + 2,700 = 4,500.

Question 2.
In a science lab, acid and water are mixed in the ratio of 1 : 5 to make a solution. In a bottle that has 240 mL of the solution, how much acid and water does the solution contain?
Answer:
Ratio of acid and water = 1 : 5
Quantity of solution = 240 mL
∴ Quantity of acid = \(\frac{1}{1 + 5}\) × 240
= \(\frac{1}{6}\) × 240 = 40 mL
∴ Quantity of water = \(\frac{1}{1 + 5}\) × 240
= \(\frac{5}{6}\) × 240 = 200 mL
∴ Quantities of acid and water in the solution are 40 mL and 200 mL.
Verification: 4Q
40 : 200 = \(\frac{40}{200}\)
\(\frac{1}{5}\) = 1 : 5 and 40 + 200 = 240.

Question 3.
Blue and yellow paints are mixed in the ratio of 3 : 5 to produce green paint. To produce 40 mL of green paint, how much of these two colours are needed? To make the paint a lighter shade of green, I added
20 mL of yellow to the mixture. What is the new ratio of blue and yellow in the paint?
Answer:
Ratio of blue and yellow paints = 3 : 5
Quantity of green paint = 40 mL
∴ Quantity of blue paint = \(\frac{3}{3 + 5}\) × 40
= \(\frac{3}{8}\) × 40 = 15 mL
∴ Quantity of yellow paint = \(\frac{5}{3 + 5}\) × 40
= \(\frac{5}{8}\) × 40 = 25 mL
Addition of yellow paint to the mixture = 20 mL
∴ New quantity of blue paint =15 mL
∴ New quantity of yellow paint = 25 mL + 20 mL = 45 mL
∴ New ratio of blue and yellow paints
= 15 : 45 = 1 : 3.

Question 4.
To make soft idlis, you need to mix rice and urad dal in the ratio of 2 : 1. If you need 6 cups of this mixture to make idlis tomorrow morning, how many cups of rice and urad dal will you need?
Answer:
Ratio of rice and urad dal = 2 : 1
Total number of cups of mixture = 6
∴ Number of cups of rice = \(\frac{2}{2 + 1}\) × 6
= \(\frac{2}{3}\) × 6 = 4
∴ Number of cups of urad dal = \(\frac{1}{2 + 1}\) × 6
= \(\frac{1}{3}\) × 6 = 2
∴ 4 cups of rice and 2 cups of urad dal are to be mixed.

Question 5.
I have one bucket of orange paint that I made by mixing red and yellow paints in the ratio of 3 : 5. I added another bucket of yellow paint to this mixture. What is the ratio of red paint to yellow paint in the new mixture?
Answer:
Let the capacity of one bucket be x L.
Ratio of red paint and yellow paint = 3 : 5
∴ Quantity of red paint in the bucket = \(\frac{3}{3 + 5}\) × x = \(\frac{3 x}{8}\)
∴ Quantity of yellow paint in the bucket = \(\frac{5}{3 + 5}\) × x = \(\frac{5 x}{8}\)
One bucket of yellow paint is added to the mixture.
∴ New quantity of red paint in the mixture = \(\frac{3 x}{8}\)
∴ New quantity of yellow paint in the mixture = \(\frac{5 x}{8}\)
+ x = \(\frac{13 x}{8}\)
∴ New ratio of red paint and yellow paint in the mixture = \(\frac{3 x}{8}\) : \(\frac{13 x}{8}\) = 3 : 13

Figure it Out (Page 176 – 177) :

Question 1.
Anagh mixes 600 mL of orange juice with 900 mL of apple juice to make a fruit drink. Write the ratio of orange juice to apple juice in its simplest form.
Answer:
Quantity of orange juice = 600 mL
Quantity of apple juice = 900 mL
∴ Ratio of orange juice to apple juice = 600 : 900
Ratio in the simplest form = 600 : 900 = 2:3

Question 2.
Last year, we hired 3 buses for the school trip. We had a total of 162 students and teachers who went on that trip and all the buses were full. This year we have 204 students. How many buses will we need? Will all the buses be full?
Answer:
Number of buses for 162 students and teachers = 3
Since the buses were full, the capacity of 1 bus = \(\frac{162}{3}\) = 54
∴ Ratio of number of seats to the number of buses is 54 : 1.
We have
54 : 1 = 2(54) : 2(1) = 108 : 2
54 : 1 = 3(54) : 3(1) = 162 : 3
54 : 1 = 4(54): 4(1) = 216 : 4
∴ Capacity of 4 buses = 216
∴ For 204 students, we shall need 4 buses.
Since 216 – 204 = 12, we have 12 vacant seats in the buses.

Question 3.
The area of Delhi is 1,484 sq. km and the area of Mumbai is 550 sq. km. The population of Delhi is approximately 30 million and that of Mumbai is 20 million people. Which city is more crowded? Why do you say so?
Answer:
Area of Delhi = 1,484 sq.km
Population of Delhi = 30 million
Area of Mumbai = 550 sq. km
Population of Mumbai = 20 million
∴ Ratio of area to population for Delhi = 1484 : 30
∴ Ratio of area to population for Mumbai = 550 : 20
Factor of change of area = \(\frac{550}{1484}\) = 0.371 (nearly)
Factor of change of population = \(\frac{20}{30}\) = 0.667 (nearly)
Since 0.667 > 0.371, Mumbai is more crowded than Delhi.
Alternative Method:
Ratio of area to population for Delhi = 1484 : 30
Let the density of Delhi and Mumbai be the same, and there be x people in Mumbai.
∴ The ratios 1,484 : 30 and 550 : x are in proportion.
∴ \(\frac{1,484}{30}\) = \(\frac{550}{x}\)
⇒ 1484x = 30 × 550 = 16,500
⇒ x = \(\frac{16500}{1484}\) = 11.118
There should be 11.118 million people in Mumbai. But the population of Mumbai is 20 million.
∴ Mumbai is more crowded than Delhi.

Question 4.
A crane of height 155 cm has its neck and the rest of its body in the ratio 4 : 6. For your height, if your neck and the rest of the body also had this ratio, how tall would your neck be?

Answer:
The ratio of the height of the neck and the height of the rest of the body of a crane is 4 : 6. My height is 65 inches, i.e., 165 cm.
Let the ratio of the height of my neck and the height of the rest of my body also be 4 : 6.
∴ Height of my neck = (\(\frac{4}{4 + 6}\) × 165)cm = 66 cm

Question 5.
Let us try an ancient problem from Lilavati. At that time weights were measured in a unit named palas and niskas was a unit of money. “If 2\(\frac{1}{2}\) palas of saffron costs \(\frac{3}{7}\) niskas, O expert businessman! tell me quickly what quantity of saffron can be bought for 9 niskas?”
Answer:
A proportional relationship between the quantity of saffron and its cost is described. The cost of a known quantity of saffron is provided, and the quantity of saffron that can be purchased for a different amount of money is to be determined.

Step 1: Convert Mixed Numbers to Improper Fractions
= The given quantity of saffron, 2\(\frac{1}{2}\) palas, converted to an improper fraction:
→ 2\(\frac{1}{2}\) = \(\frac{2 \times 2+1}{2}\) = \(\frac{5}{2}\)
= The given cost, \(\frac{3}{7}\) niskas.

Step 2 : Set Up the Proportion
A proportion is established relating the quantity of saffron to its cost. Let x be the unknown quantity of saffron.

Question 6.
Harmain is a 1-year-old girl. Her elder brother is 5 years old. What will be Harmain’s age when the ratio of her age to her brother’s age is 1 : 2?
Answer:
The current ages are given as Harmain being 1 year old and her brother being 5 years old.
→ The age difference = 5 – 1 = 4 years.
= The difference will remain constant over time.
= Let x be the number of years that pass untill the desired ratio is achieved.
= After x years, Harmain age will be 1 + x years
= After x years, her brother’s age will be 5 + x years
= Ratio is given as 1 : 2.
Can be expressed as \(\frac{1 + x}{5 + x}\) = \(\frac{1}{2}\)
→ 2(1 + x) = 1(5 + x)
→ 2 + 2x = 5 + x
→ 2x – x = 5 – 2
→ x = 3
= Harmain’s age when the ratio is 1 : 2 is found by adding * to her current age.
→ 1 + 3 = 4 years.
Harmain’s age will be 4 years when the ratio of her age to her brother’s age is 1 : 2

Question 7.
The mass of equal volumes of gold and water are in the ratio 37 : 2. If 1 litre of water is 1 kg in mass, what is the mass of 1 litre of gold?
Answer:
The given ratio of the mass of the gold to the mass of water for equal Volumes is 37 : 2
This Can be expressed as \(\frac{\text { Mass of gold }}{\text { Mass of water }}\) = \(\frac{37}{2}\)
= It is stated that 1 litre of water has a mass of 1kg.
Mass of 1 litre of gold = 1kg × \(\frac{37}{2}\)
= \(\frac{37}{2}\) kg = 18.5 kg
= The mass of 1 litre of gold is 18.5 kg.
= Mass of 1L of water is given as 1kg.
= Ratio to find the mass of 1L of gold
= Ratio of mass of equal volumes of gold to water is 37 : 2
\(\frac{\text { Mass of gold }}{\text { Mass of water }}\) = \(\frac{37}{2}\)
= Mass of Gold = Mass of water × \(\frac{37}{2}\)
= Mass of Gold = \(\frac{37}{2}\)kg = 18.5 kg.
So… Mass of 1 litre of gold is 18.5 kg.

Question 8.
It is good farming practice to apply 10 tonnes of cow manure for 1 acre of land. A farmer is planning to grow tomatoes in a plot of size 200 ft by 500 ft. How much manure should he buy? (Please refer to the section on Unit Conversions earlier in this chapter).
Answer:
Let’s Calculate the area of the plot in square feet.
Area = Length × width
Area = 500ft × 200ft = 100,000ft²
Convert the area from Square feet to acres we knew 1 acre = 43560ft²
Area in acres = \(\frac{100000 f^2}{43560 f^2}\) = 2.2956 acres
Calculate the total amount of manure required.
Manure required = 2.2956 acres × 10 tonnes/acres
= 22.956 tonnes.

Question 9.
A tap takes 15 seconds to fill a mug of water. The volume of the mug is 500 mL. How much time does the same tap take to fill a bucket of water if the bucket has a 10-litre capacity?
Answer:
Time taken by the tap for 500 mL of water = 15 seconds
∴ Ratio of volume to time = 500 : 15
We know 1 litre = 1,000 mL
10 litre = 10 × 1,000 = 10,000 mL
Let the time taken to fill a bucket of 10,000 mL be x seconds.
∴ Ratio of volume to time = 10,000 : x
These ratios are proportional.
∴ 500 : 15 :: 10,000 : x
⇒ \(\frac{500}{15}\) = \(\frac{10,000}{x}\)
⇒ 500x = 1,50,000
⇒ x = 300
∴ Time to fill bucket = 300 seconds \(\frac{300}{60}\) = minutes = 5 minutes.

Question 10.
One acre of land costs ₹15,00,000. What is the cost of 2,400 square feet of the same land?
Answer:
We know that 1 acre = 43,560 square feet.
∴ Cost of 43,560 sq. ft. land = ₹15,00,000
∴ Ratio of area of land to cost = 43,560 : 15,00,000
Let the cost of 2,400 sq. ft. of land be ₹x.
∴ Ratio of area of land to cost = 2,400 : x
These ratios are proportional.
∴ 43,560 : 15,00,000 :: 2,400 : x
⇒ \(\frac{43,560}{15,00,000}\) = \(\frac{2,400}{x}\)
⇒ 43,560x = 2,400 × 15,00,000
⇒ x = \(\frac{2,400 \times 15,00,000}{43,560}\)
⇒ x = 82,664.63
∴ Cost of land = ₹82,664.63.

Question 11.
A tractor can plough the same area of a field 4 times faster than a pair of oxen. A farmer wants to plough his 20-acre field. A pair of oxen takes 6 hours to plough an acre of land. How much time would it take if the farmer used a pair of oxen to plough the field? How much time would it take him if he decides to use a tractor instead?
Answer:
Ratio of efficiency of a tractor to a pair of oxen = 4 : 1
Time taken by a pair of oxen to plough 1 acre of field = 6 hours
∴ Time taken by a tractor to plough 1 acre field = \(\frac{6}{4}\) = 1.5 hours
∴ Time taken by a pair of oxen to plough 20 20- acre field = 20 × 6 = 120 hours
∴ Time taken by a tractor to plough a 20-acre field = 20 × 1.5 = 30 hours

Question 12.
The ₹10 coin is an alloy of copper and nickel called ‘cupro-nickel’. Copper and nickel are mixed in a 3 : 1 ratio to get this alloy. The mass of the coin is 7.74 grams. If the cost of copper is ₹906 per kg and the cost of nickel is ₹1,341 per kg, what is the cost of these metals in a ₹10 coin?
Answer:
Ratio of copper and nickel in ₹10 coin = 3 : 1
Mass of one ₹10 coin = 7.74 grams
∴ Mass of copper in one ₹10 coin = \(\frac{3}{3 + 1}\) × 7.74
\(\frac{3}{4}\) × 7.74 = 5.805 grams
Maas of nickel in one ₹10 coin = \(\frac{1}{3 + 1}\) × 7.74
= \(\frac{1}{4}\) × 7.74 = 1.935 grams
Cost of 1 kg copper = ₹ 906
∴ Cost of 1000 grams of copper = ₹ 906
∴ Cost of 5.805 grams copper = \(\frac{906}{1000}\) × 5.805
= ₹5.26
Cost of 1 kg nickel = ₹1341
∴ Cost of 1000 grams of nickel = ₹1341
∴ Cost of 1.935 grams nickel = \(\frac{1341}{1000}\) × 1.935
= ₹2.59
∴ In one ₹10 coin, the cost of copper and the cost of nickel are respectively ₹5.26 and ₹2.59.

Proportional Reasoning 1 Class 8 Extra Questions

Multiple Choice Questions

Question 1.
The ratio 72 : 96 in its simplest form is:
(a) 2 : 3
(b) 3 : 4
(c) 2 : 5
(d) 1 : 2
Solution:
HCF of 72 and 96 = 24
Now, \(\frac{72}{96}\) = \(\frac{72 \div 24}{95 \div 24}\) = \(\frac{3}{4}\)
(b) 3 : 4

Question 2.
The equivalant ratio, for the ratio 2 : 3 in the simplest form, is :
(a) 24 : 48
(b) 13 : 39
(c) 50 : 75
(d) 36 : 90
Solution:
HCF of 50 and 75 = 25
∴, \(\frac{50}{75}\) = \(\frac{50 \div 25}{75 \div 25}\) = \(\frac{2}{3}\)
∴, equivalant ratio of 2 : 3 is 50 : 75

Question 3.
If 14 : 21 :: 2 : x, then the value of x is :
(a) 1
(b) 2
(c) 14
(d) 3
Solution:
Since, 14 : 21 in the simplest form is 2 : 3.
Hence, x = 3
(d) 3

Question 4.
If 24 : x :: 48 : 72, then the value of x is:
(a) 36
(b) 30
(c) 48
(d) 32
Solution:
For 24 : x : : 48 : 72, we write
\(\frac{24}{x}\) = \(\frac{48}{72}\) ⇒ \(\frac{24}{x}\) = \(\frac{2}{3}\) ⇒ 2x = 2 × 3
⇒ x = \(\frac{24 \times 3}{2}\) ⇒ x = 36
(a) 36

Question 5.
If 15 : 35 = x : y, then x : y is :
(a) 5 : 7
(b) 3 : 7
(c) 1 : 3
(d) 3 : 4
Solution:
Hence, 15 : 35 = \(\frac{15}{35}\) = \(\frac{3}{7}\) = 3 : 7
(b) 3 : 7

Assertion and Reasoning

(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).
(b) Assertion (A) and Reason (R) both are true but Reason (R) is not the correct explanation for the Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.

Question 1.
Assertion (A) : The ratio 60 : 40 can be written as 3 : 2 in simplest form.
Reason (R) : To get the ratio in simplest form we divide both numerator and denominator by the HCF of them.
Solution:
\(\frac{60}{40}\) = \(\frac{60 \div 20}{40 \div 20}\) = \(\frac{3}{2}\) (HCF (60, 40) = 20)
(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).

Question 2.
Assertion (A) : 15 : 20 : : 12 : 16
Reason (R) : a : b :: c : d ⇔ \(\frac{a}{b}\) = \(\frac{c}{d}\)
Solution:
If \(\frac{x}{y}\) = \(\frac{z}{u}\), then x : y and z : u are in proportion.
Hence, x : y : : z : u
(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).

Case Based Questions

Question 1.
For the mid-day meal in a school with 600 students, the cook usually makes 75 kg of rice. On a certain day, only 120 students came to school.

(i) How much rice in grams is cooked for each student?
(ii) How much rice will be cooked if 120 students came to school?
(iii) What is the factor of change in the first term of 600 : 75 : 120:?
(iv) If on a certain day 180 students came to school, then how much rice will be cooked on that day?
Answer:
(i) Since, 75 kg of rice is cooked for 600 students Hence, for 1 student the amount of rice cooked

(ii) If 120 students come to school, then the amount of rice to be cooked 15.

(iii) Factor of change in the first term is : \(\frac{120}{600}\) = \(\frac{1}{5}\)

(iv) If 180 students come to school, then the amount of rice to be cooked on that day = \(\frac{1}{8}\) × 180 kg = 22.5 kg

Go through BSE Odisha Class 8 Science Solutions Chapter 8 Nature of Matter: Elements, Compounds, and Mixtures Question Answer to understand textbook questions more clearly.

Class 8 Science Curiosity Chapter 8 Question Answer

Class 8 Science Ch 8 Nature of Matter: Elements, Compounds, and Mixtures Question Answer

Class 8 Science Chapter 8 Nature of Matter: Elements, Compounds, and Mixtures Question Answer

Probe and Ponder Questions

Question 1.
Which of the entities in the picture consist of matter and which of them do not?

Answer:

• Entities consisting of matter include physical objects like staircases, air, water, food, clothes, shoes, books, trees, balls and sticks as these have mass and occupy space.
• They are made of tiny particles. Entities that do not consist of matter include light, heat, electricity, thoughts and emotions as they lack mass and do not occupy space.

Question 2.
How can elements be combined to form a compound?
Answer:

• Elements combine chemically in fixed ratios to form compounds.
• For example, hydrogen and oxygen combine in a 2:1 ratio to form water, where the atoms bond tightly, creating a new substance with properties different from the original elements.
• This requires a chemical reaction, not just physical mixing.

Question 3.
How could the discovery of a compound that absorb carbon dioxide from the air contribute to solving environmental challenges?
Answer:

• Such a compound could reduce atmospheric carbon dioxide levels, mitigating global warming and climate change.
• For instance, it might be used in technologies to capture emissions from industries or vehicles, similar to how calcium hydroxide reacts with carbon dioxide to form calcium carbonate.
• This could help address air pollution and support environmental cleanup efforts.

Question 4.
Share your questions …………………….
Answer:
Based on the chapter, questions could include:

• What happens when elements like iron and sulfur are heated together?
• Why does water extinguish fire while its components (hydrogen and oxygen) support combustion?
• How do alloys like stainless steel improve everyday material?

InText Questions

Question 1.
According to science, how would you classify milk, packed fruit juice, baking soda, sugar, and soil as mixtures or pure substances? (Page 121)
Answer:

• Milk: Mixture (contains water, fats, proteins, etc.)
• Packaged fruit juice: Mixture (water sugars, flavors, vitamins, etc.)
• Baking soda: Pure substance (if chemically pure-only sodium bicarbonate)
• Sugar: Pure substance (if only sucrose)
• Soil: Mixture (sand, clay, minerals, organic matter, water, air)

In science, “pure” means that the substance consists of the same kind of particle everywhere in the sample.

Question 2.
Electrolysis of water produces two different gases. Can these collected gases be water vapour ? (Page 122)
Answer:
These gases are Hydrogen and oxygen not water vapour otherwise they would have condensed back to form water.

Question 3.
When electric current is passed through water, it breaks down into hydrogen and oxygen. Is this a chemical change or a physical change? (Page 123)
Answer:
This is a chemical change because the properties of hydrogen and oxygen are different from original substance water and it is irreversible by simple physical method.

Question 4.
After heating sugar in a boiling tube what is left behind? Also, we observe a small droplets of water inside the boiling tube. Where did this water come from?
Answer:
Charcoal (carbon) is left behind in the boiling tube. Water must have come from the dry sugar and not from the air.

Nature of Matter: Elements, Compounds, and Mixtures Class 8 Questions and Answers

Keep the Curiosity Alive (Pages 131-132)

Question 1.
Consider the following reaction where two substances, A and B, combine to form a product C :
A+B →C
Assume that A and B cannot be broken down into simpler substances by chemical reactions. Based on this information, which of the following statements is correct?
(i) A, B, and C are all compounds, and only C has a fixed composition.
(ii) C is a compound, and A and B have a fixed composition.
(iii) A and B are compounds, and C has a fixed composition.
(iv) A and B are elements, C is a compound, and has a fixed composition.
Answer:
(iv) A and B are elements, C is a compound, and has a fixed composition.
A and B are elements, because elements are pure substances made of only one kind of atom and cannot be broken down chemically. When A and B combine chemically to form C, the result is a compound. A compound is formed when two or more elements combine in a fixed ratio through a chemical reaction. Therefore, A and B are elements, and C is a compound with a fixed composition.

Question 2.
Assertion: Air is a mixture.
Reason: A mixture is formed when two or more substances are mixed, without undergoing any chemical change.
(i) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(iii) Assertion is true, but Reason is false.
(iv) Assertion is false, but Reason is true.
Answer:
(i) Both Assertion and Reason are true and Reason is the correct explanation for assertion.
Air is indeed a mixture because it is composed of various gases like nitrogen. oxygen, and carbon dioxide, which are mixed without any chemical reaction between them. The properties of these individual gases are retained within the air.

Question 3.
Water, a compound, has different properties compared to those of the elements oxygen and hydrogen from which it is formed. Justify this statement.
Answer:
Water has properties which is completely different from hydrogen and oxygen. Like water is liquid in form, whereas hydrogen (H) and oxygen (O) are gases. This is because a compound’s properties depends on its molecular structure.

Question 4.
In which of the following cases are all the examples correctly matched? Give reasons in support of your answers.
(i) Elements – water, nitrogen, iron, air.
(ii) Uniform mixtures – minerals, seawater. bronze, air.
(iii) Pure substances – carbon dioxide, iron, oxygen, sugar.
(iv) Non-uniform mixtures – air, sand, brass, muddy water.
Answer:
(iii) Pure substances – Carbon dioxide, iron, oxygen, and sugar are all pure substances. (Correctly matched).
A pure substance is composed of only one type of particle. Carbon dioxide, iron, and oxygen are all elements, meaning they are made up of only one type of atom. Sugar is a compound. but it is still considered a pure substance because it consists of only one type of molecule.

Question 5.
Iron reacts with moist air to form iron oxide, and magnesium burns in oxygen to form magnesium oxide. Classify all the substances involved in the above reactions as elements, compounds, or mixtures, with justification.
Answer:

• Iron: Element (pure metal, cannot be broken down).
• Moist air: Mixture (air gases plus water vapor, components retain properties).
• Iron oxide: Compound (iron and oxygen combined chemically).
• Magnesium: Element (pure metal).
• Oxygen: Element (pure gas).
• Magnesium oxide: Compound (magnesium and oxygen in fixed ratio).
• Justification: Elements are simplest substance; compounds form from elements via chemical reactions with new properties; mixtures do not involve chemical bonding.

Question 6.
Classify the following as elements, compounds, or mixtures in the Table.
Carbon dioxide, sand, seawater, magnesium oxide, muddy water, aluminum, gold, oxygen, rust, iron sulfide, glucose, air, water, fruit juice, nitrogen, sodium chloride, sulfur, hydrogen, and baking soda.

Elements	Compounds	Mixtures

Identify pure substances amongst these and list them below.

pure substances

Answer:
Pure Substances: Aluminium, gold, oxygen, nitrogen, sulfur, hydrogen, carbon dioxide, magnesium oxide, iron sulfide, glucose, water, sodium chloride, baking soda.

Elements	Compounds	Mixtures
Aluminium	Carbon Dioxide CO2	Sand
Gold	Magnesium Oxide (MgO)	Seawater
Oxygen	Rust (Fe2O3)	Muddy Water
Nitrogen	Iron Sulfide (FeS)	Air
Sulfur	Glucose (C6 H12O6 )	Fruit Juice
Hydrogen	Water (H2O)
	Sodium Chloride(NaCl)
	Baking Soda NaHCO3

Question 7.
What new substance is formed when a mixture of iron filings and sulfur powder is heated, and how is it different from the original mixture? Also, write the word equation for the reaction.
Answer:
When iron filings and sulfur powder are heated, they react to form a new substance called ferrous sulfide (FeS), also known as iron sulfide. This is a chemical change, and the resulting compound has different properties from the original iron and sulfur.
The word equation for the reaction is :
Iron + Sulfur → Ferrous Sulfide.

Question 8.
Is it possible for a substance to be classified as both an element and a compound? Explain why or why not.
Answer:
No, a substance cannot be classified as both an element and a compound. Elements are pure substances that cannot be broken down into simpler substances by chemical means, while compounds are formed when two or more different elements are chemically bonded together. The defining characteristic of a compound is that it is composed of multiple elements, whereas an element is a single type of atom. Therefore, a substance cannot be both a single type of atom and a combination of different types of atoms simultaneously.

Question 9.
How would our daily lives be changed if water were not a compound but a mixture of hydrogen and oxygen?
Answer:
Water’s role in life and nature depends on it being a compound with stable properties. If it were a mixture, it would be dangerous and unusable, making life as we know it impossible.

Impact on Daily Life

• No safe drinking water → Life would not be possible.
• No water for agriculture →Crops would not grow.
• No water for cleaning or cooking → Daily tasks would be unsafe.
• No aquatic life → Fish and underwater plants would die.
• Increased fire hazards → Hydrogen and oxygen together are explosive.

Question 10.
Analyse the figure. Identify Gas A. Also, write the word equation of the chemical reaction.

Answer:
By analysing the figure, it is found that there will be a chemical reaction inside the test tube between dilute HCl and Fe.
2HCl+Fe →FeCl₂+H₂ ↑
So the reaction forms Iron Chloride (FeCl₂) and the gas above will be Hydrogen (H₂).
Hydrochloric Acid + Iron filing → Iron Chloride + Hydrogen (g)
Thus, Gas A = Hydorgen

Question 11.
Write the names of any two compounds made only from non-metals, and also mention two uses of each of them.
Answer:
1. Carbon Dioxide (CO₂)
Made of: Carbon and Oxygen (both nonmetals)
Uses:

• Used in fire extinguishers to put out flames.
• Used by plants during photosynthesis to make food.

2. Sulfur Dioxide (SO₂)
Made of: Sulfur and Oxygen (both nonmetals)
Uses:

• Used as a preservative in dried fruits and wines.
• Used in the manufacture of sulfuric acid, an important industrial chemical.

Question 12.
How can gold be classified as both a mineral and a metal?
Answer:
A mineral is a naturally occurring substance with a definite chemical composition.
Gold is found in nature in its native form, often embedded in rocks or alluvial deposits. It is extracted through mining, making it a metallic mineral. Minerals like gold are formed by natural geological processes.

Gold as a Metal
After extraction, gold is refined and used as a metal. It is a pure element (symbol: Au ) with typical metallic properties :

• Lustrous (shiny)
• Malleable (can be beaten into sheet)
• Ductile (can be drawn into wires)
• Good conductor of electricity
• Used in jewelry, electronics, and currency.

Class 8 Science Chapter 8 Question Answer

Activity 1.

Let us experiment

Aim: To demonstrate the presence of carbon dioxide in the air.
Materials Required: Calcium oxide (Quick lime), a petri dish a glass tumbler, a glass rod.

Procedure:

• For this activity we need a glass tumbler which is half filled with water.
• Now, add a small amount of calcium oxide (quick lime) slowly to it.
• Note down your observation.
• Calcium oxide reacts vigorously with water to form calcium hydroxide and releases heat.
• Now, stir the mixture with a glass rod to make a solution of calcium hydroxide. This solution is called lime water.
• Filter it using a filter paper and observe its colour.
• Leave this colourless solution in a petri dish for a few hours [Figure (a)].
• We should stirr the solution at regular intervals.
• Note down your observation. [Figure (b)]

Observations:

• Glass tumbler becomes hot when calcium oxide is added to it.
• After sometime, the clear solution of lime water turns milky.

Inferences:

• Lime water turns milky because carbon dioxide in the air reacts with calcium hydroxide to produce insoluble calcium carbonate (which looks milky).
  Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water
• It shows the presence of carbon dioxide in the air.

Activity 2.

Let us explore

Aim: To show that air contains dust particles.
Materials Required: A black sheet of paper, a magnifying glass.

Procedure:

• Take a black sheet of paper. Remember that it should be free from any visible dust particles.
• Fix the black sheet of paper on an open window [Figure (a)], or in the near by garden, for a few hours.
• Note down your observation.
• You may use a magnifying glass to see the particles.

Observations: We observed that tiny particles settled on its surface.

Inference:

• This shows that dust particles are suspended in the air.
• Since they are not an integral part of the air therefore are considered as pollutants. The nature and the amount of dust particles in the air may vary from time to time and from place to place.

Activity 3.

Let us experiment (Demonstration activity)
Aim: To demonstrate that water is composed of two different constituents by passing electricity through it.
Materials Required: 9 V battery, a beaker or a glass tumbler, dilute sulphuric acid.

Procedure:

• First of all we will take two small test tubes, a beaker or a glass tumbler, and a 9 V battery.
• Now, fill about 2/3rd of the beaker with water and add a few drops of dilute sulfuric acid to it.
  
  
• Fill both the small test tubes completely with water taken from the beaker [Figure (a)].
• Now, keep a 9V battery inside the beaker [Figure (b)].
• We should take precaution that water cannot be spilled out.
• Now, carefully place the water-filled test tubes on each of the terminals of the battery [Fig (c)].
• Now, we will wait for a few minutes.
• Are you observing the formation of any gas bubbles at both the terminals inside the test tubes?
• Now, you will continue it for 10-15 minutes.
• Observe the volume of gas collected in each test tube [Figure (d)].
• Is the volume of the gas collected the same in both the test tubes?
• Remove these test tubes one-by-one carefully.
• Test these gases one-by-one by bringing a burning candle close to the mouth of the test tubes.

Observations:

• Bubbles at both the terminals inside the test tube are formed.
• Volume of gas collected in each test tube in 2:1.
• Electrolysis of water produces two different gases (not water vapor): one that makes a “pop” sound with a flame (hydrogen), the other that makes a flame glow brighter (oxygen).

Inference:

• Water breaks down (chemically) into hydrogen and oxygen, proving it is a compound made of two elements.
• Water is composed of two different constituents-hydrogen and oxygen.

Precaution:

• This activity must be performed under the supervision of the teacher.
• Be careful while handling sulfuric acid. Do not use a lithium-ion battery.
• Perform gas testing with care. Maintain a safe distance from the set-up.

Activity 4.

Let us experiment

Aim: To show that sugar is a chemical compound and after heating it gives carbon (charcoal) and water.
Materials Required: A test tube, a test tube holder, a teaspoon of sugar.

Procedure:

• Take a teaspoon of sugar and put in a boiling tube.
• Now, start heating gently see figure (a).
• Note your observation.

Observations:

• After heating the sugar, it turns into brown [see figure (b)]. After sometime, it begins to char, i.e., it turns blackish [see figure (c)].
• We can observe a small droplets of water inside the boiling tube near its open end.

Inference:

• Since we are heating the tube, the water must have come from the dry sugar and not from the air.
• Charcoal (carbon) is left behind in the boiling tube. We can scoop it out in a watch glass [see figure (c)] and explore if it burns like coal.
• Sugar decomposes on heating and gives carbon and water.
• We may conclude that sugar is a chemical compound consisting of the elements carbon, hydrogen, and oxygen.

Precaution: This activity must be performed in the presence of a teacher.

Activity 5.

Let us experiment (Demonstration activity)
Aim: To differentiate between mixture and compound.

Materials Required: A tripod stand, wire gauze, iron filings, sulfur powder.

Procedure:

• In this experiment, we explore how iron and sulfur behave as a mixture before heating and form a compound after heating.
• This highlights key differences between mixtures (where substances stay separate) and compounds (where they combine chemically into something new). Let’s break it down step by step, starting with the initial mixture.

Demonstration:
Before Heating: Forming Sample A (The Mixture)
To begin, mix iron filings and sulfur powder together to create Sample A. This is a classic example of a mixture, where two substances are simply combined without any chemical change.

Observations:
Step 1 Appearance

• You can clearly see both components as separate substances-dark grey (or greyish-black) iron filings and yellow sulfur powder-making it look non-uniform.

Step 2 Magnet test

• When you bring a magnet near Sample A, it attracts only the iron filings, leaving the sulfur behind. This shows the components retain their individual properties.

Step 3 Acid test

• Add dilute hydrochloric acid to Sample A. The iron reacts to produce hydrogen gas (which makes a pop sound when ignited); but the sulfur does not react and remains as a yellow solid.

Inference:

• The above three tests confirm that in a mixture, substances can be separated easily and keep their original traits or properties.
• The reaction can be represented as – Iron + Dilute Hydrochloric acid → Iron chloride + Hydrogen gas

After Heating: Forming Sample B (The Compound)

• Take half of Sample ‘A’ in a China dish and heat it gently with continuous stirring. This causes a chemical reaction, resulting in a new black mass called iron sulfide (Sample ‘B’). The transformation shows how elements combine to form a compound with entirely new properties.
• Let the content of the China dish cool.
• Place this black mass in a mortar and grind it with the help of a pestle.
• Observe the appearance, result of magnetic test and acid test.

Observations:

• Step 1 – Appearance
  The black mass looks uniform throughout. No separate iron or sulfur visible anywhere.
• Step 2 – Magnet test
  Unlike Sample A, a magnet has no effect on Sample B. The iron is now chemically bounded and doesn’t lost their magnetic property.
• Step 3 Acid test
  Add dilute hydrochloric acid to Sample B. It produces hydrogen sulfide gas, which has a distinct rotten egg smell-completely different from the odourless hydrogen.

Inference:

• The above three tests confirms that Sample B is a compound, the properties of its constituents do not retain.
• At this point, iron and sulfur can no longer be separated by physical methods like magnets or simple filtering. A compound has formed, with fixed ratios and unique characteristics that differ from the original elements.
  The reaction can be represented as –

Iron sulfide Dilute Hydrochloric acid → Iron chloride + Hydrogen sulfide

Answer:

Precautions:

• Be careful while handling hydrochloric acid.
• Never smell anything directly.
• This activity may be demonstrated under the supervision of the teacher. It may be performed in a fume hood or a well-ventilated area. Do not inhale the gases.

Nature of Matter: Elements, Compounds, and Mixtures Class 8 Extra Questions and Answers

Short Answer Type Questions

Question 1.
What are metals and non-metals?
Answer:
Metals are elements that are shiny, good conductors of heat and electricity. Whereas. non-metals are dull in appearance and poor conductors of heat and electricity.

Question 2.
What is the main difference between physical and chemical change ?
Answer:
When physical change happens, no new substance forms and it is reversible whereas in a chemical change, new substance is formed and it is irreversible.

Question 3.
What do you mean by metalloids ?
Answer:
Metalloids are elements that have properties of both metals and non-metals. They are known as semimetals. For example; Silicon, Germ.anium.

Question 4.
What do you mean by adulteration ?
Answer:
Adulteration is the illegal act of mixing lower-quality or harmful substances into foods or products that increase quantity or cut costs, but this reduces quality and can be dangerous to health.

Question 5.
How do metals form compounds ?
Answer:
Metals form compounds by donating electrons to non-metals during chemical reactions. This results in ionic bonding, creating substances like metal oxides and metal chlorides.

Question 6.
What is a mineral and how is it different from rock?
Answer:
A mineral is a naturally occurring, inorganic substance with a definite chemical composition and a crystalline structure. In contrast, a rock is a solid material made up of one or more minerals.

Question 7.
What are pure substances ?
Answer:
A pure substance is a type of matter that has a uniform and definite composition. It contains only one kind of particle, either a single element (like oxygen or gold) or a single compound (water or salt) and cannot be separated into other substances by physical means.

Long Answer Type Questions

Question 1.
Discuss the importance and applications of elements, compounds and mixtures in our daily lives.
Answer:
Elements, compounds and mixtures are the basic building blocks of all matter. They play key roles in daily life and various industries.

Elements Importance: Elements are the simplest form of matter and cannot be further broken down, making them the foundation of all other substances.

Applications:

• Metals like iron, copper and aluminium are used in construction, wiring and packaging due to their strength, conductivity and malleability.
• Non-metals like oxygen are essential for respiration and combustion.
• Silicon is vital for electronics and computer technology.
• Gold and silver are valued for jewelry and in some electronic components.

Compounds
Importance: Compounds are formed by the chemical combination of elements, resulting in substances with unique properties necessary for life and various technological advancements.

Applications:

• Water (H₂O) is indispensable for life, used for drinking, cooking and industrial processes.
• Table salt (NaCl) is a fundamental seasoning and food preservative.
• Sugar (C₁₂H₂₂ O₁₁) is a crucial energy source.
• Many medicines and pharmaceuticals are pure compounds designed to interact with the body in specific ways.
  Carbon dioxide (CO₂) is involved in respiration and photosynthesis.

Mixtures
Importance: Most of the matter encountered daily exists as mixtures. Understanding mixtures is essential for various applications.

Applications:

• Air, a mixture of gases like Nitrogen and Oxygen, is vital for breathing and weather phenomena.
• Alloys, like steel and bronze, are mixtures of metals that possess enhanced properties like strength or corrosion resistance, used in construction and various manufactured goods.
• Food products like milk, juice and granola are mixtures, combining different components for taste, nutrition or texture.
• Soil is a complex mixture of minerals, organic matter and living organisms, crucial for plant growth.
• Many everyday products like paints, cleaning solutions and cosmetics are mixtures designed for specific purposes.

Question 2.
Why are compounds considered pure substances, while mixtures are not?
Answer:
Compounds are pure substances because they are made up of only one type of molecule and have a uniform and definite composition throughout. For example, every molecule of water (H₂O) is identical, consisting of two hydrogen atoms and one oxygen atom chemically bonded together. This fixed composition results in consistent physical and chemical properties, like a specific boiling point and density.

Mixtures are not pure substances because they consist of two or more substances that are physically blended, not chemically bonded. The components of a mixture retain their individual properties and can be present in varying proportions. For example, air is a mixture of Nitrogen, Oxygen and other gases and amount of each gas can vary.

Case-Study Based Questions

Question 1.
Read the following passage carefully and answer the questions that follow: A mixture contains more than one susbtance (element and/or compound) mixed in any proportion. Mixtures can be separated into pure substances using appropriate separation techniques. Pure substances can be elements or compounds.

An elements is a form of matter that cannot be broken down by chemical reactions into simpler substances. A compound is a substance composed of two or more different types of elements, chemically combined in a fixed proportion. Properties of a compound are different from its constituent elements where as a mixture shows the properties of its constituting elements or compounds.

(i) Which of the following are homogeneous in nature ?
A. Ice
B. Wood
C. Soil
D. Air
(a) A and C
(b) B and D
(c) A and D
(d) C and D
Answer:
(c) A and D

(ii) Two chemical species X and Y combine together to form a product P which contains both X and Y.
X+Y → P
X and Y cannot be broken down into simpler substances by simple chemical reactions. Which of the following concerning the species X, Y and P are correct?
A. P is a compound
B. X and Y are compounds
C. X and Y are elements
D. P has a fixed composition
(a) A, B and C
(b) A, B and D
(c) B, C and D
(d) A, C and D
Answer:
(d) A, C and D

(iii) Give two points of differences between an element and a compound.
Answer:

Element	Compound
1. An element is made up of same kind of atoms.	1. A compound is obtained from different kinds of atoms.
2. An element cannot be split by physical or chemical methods.	2. A compound can be split into new substances by chemical methods.

(iv) Which of the following are not compounds?
(a) Chlorine gas
(b) Potassium chloride
(c) Iron
(d) Iron sulphide
(e) Aluminium
(f) Iodine
(g) Carbon
(h) Carbon monoxide
(i) Sulphur powder
Answer:
Chlorine gas, iron, aluminium, iodine, carbon, sulphur powder.

Picture Based Questions

I. Look at the pictures and answer the following questions :
(a) Identify the pictures (i) and (ii).

Answer:
(i) Graphene aerogel
(ii) Dhokra art

(b) Write the use of figure (i).
Answer:
(i) It is used as an environment cleaner
(ii) It is useful in fabricating energy saving devices and special coating for buildings.

(c) In which states this craft is popular [see figure (ii)] ?
(a) Bihar
(b) Odisha
(c) both (a) and (b)
(d) None of these
Answer:
(c) both (a) and (b)

Nature of Matter: Elements, Compounds, and Mixtures Class 8 MCQ

Multiple Choice Questions (Mcqs)

Question 1.
What is a mixture?
(a) A substance formed by chemical reaction
(b) A single pure substance
(c) A combination of substances without chemical reaction
(d) A new element
Answer:
(c) A combination of substances without chemical reaction

Question 2.
What is a component in a mixture ?
(a) A new element
(b) An individual substance in the mixture
(c) A type of atom
(d) A compound
Answer:
(b) An individual substance in the mixture

Question 3.
Which of the following is a compound ?
(a) Brass
(b) Salt (NaCl)
(c) Air
(d) Lemonade
Answer:
(b) Salt (NaCl)

Question 4.
What is a pure substance ?
(a) Any liquid
(b) Substance with only one type of particle
(c) Mixture of water and sugar
(d) A combination of many substances
Answer:
(b) Substance with only one type of particle

Question 5.
Which of the following statements are true for pure substances?
(i) Pure substances contain only one kind of particles.
(ii) Pure substances may be compounds or mixtures.
(iii) Pure substances have the same composition throughout.
(iv) Pure substances can be exemplified by all elements other than nickel.
(a) (i) and (ii)
(b) (i) and (iii)
(c) (iii) and (iv)
(d) (ii) and (iii)
Answer:
(b) (i) and (iii)

Assertion and Reasoning

These questions consist of two statements, each printed as Assertion (A) and Reason (R). While answering these questions, you are required to choose any one of the following four responses.
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.
(b) Assertion (A) and Reason (R) both are correct and reason is not correct explanation for assertion.
(c) Assertion (A) is correct but the Reason (R) is wrong.
(d) Assertion (A) is wrong but the Reason (R) is correct.

1. Assertion (A): Copper is called an element.
Reason (R): Copper cannot be broken down to simpler substances by chemical reactions.
Answer:
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.

2. Assertion (A): Elements combine and react to form a compound.
Reason (R): The constituents of a compound can be separated easily by physical methods.
Answer:
(c) Assertion (A) is correct but the Reason (R) is wrong.

Fill in the blanks

1. The atoms of most of the elements cannot exist ………….
Answer:
independently

2. Two or more atoms combine and form a stable particle of that element called a ………….
Answer:
molecule

3. Baking powder is a mixture of baking soda and …………acid.
Answer:
tartaric

4. In ancient Indian Texts, Bronze is also known as ………….
Answer:
Kamsya

5. Bronze is an alloy made of copper and ………….
Answer:
tin

True or False

1. The properties of a mixture depend upon the properties of its components and no new substances is formed.
Answer:
True

2. The composition of a compound is always fixed.
Answer:
True

3. Brass is a compound of copper and zinc.
Answer:
False

4. Air is not a mixture.
Answer:
False

5. Lime water turns milky in the presence of carbon dioxide.
Answer:
True

Students can refer to BSE Odisha Class 8 Math Solution and Ganita Prakash Chapter 6 We Distribute Yet Things Multiply Class 8 Question Answer to understand textbook questions step by step.

Class 8 Maths Chapter 6 We Distribute Yet Things Multiply Solutions

Ganita Prakash Class 8 Chapter 6 Solutions

Class 8 Maths Ganita Prakash Chapter 6 Solutions We Distribute Yet Things Multiply

IS THIS A MULTIPLF OF?
Figure it Out (Page 142 – 143) :

Question 1.
Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 x 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid.

Answer:

3 × 5	3 × 6	3 × 7
4 × 5	4 × 6	4 × 7
5 × 5	5 × 6	5 × 7

(p – 1)(q – 1)	(p – 1)q	(p – 1) (q + 1)
P(q – 1)	pq	P(q + 1)
(p + 1) (q – 1)	(p + 1)q	(p + 1) (q + 1)

Question 2.
Expand the following products.
(i) (3 + u) (v – 3)
(ii) \(\frac{2}{3}\)(15 + 6a)
(iii) (10a + b) (10c + d)
(iv) (3 – x) (x – 6)
(v) (-5a + b) (c + d)
(vi) (5 + z) (y + 9)
Answer:
(i) (3 + u) (v – 3) = 3(v – 3) + u(v – 3)
= 3v – 9 + uv – 3u = 3v – 3u + uv – 9

(ii) \(\frac{2}{3}\) (15 + 6a) = \(\frac{2}{3}\) × 15 + \(\frac{2}{3}\) × 6a = 10 + 4a

(iii) (10a + b) (10c + d)
= 10a × 10c + 10a × d + b × 10c + b × d
= 100ac + 10ad + 10bc + bd

(iv) (3 – x) (x – 6) = 3(x – 6) – x(x – 6)
= 3x – x² – 18 + 6x = – x² + 9x – 18.

(v) (- 5a + b) (c + d)
= (- 5a + b)c + (- 5a + b)d
= – 5ac + bc – 5ad + bd
= – 5ac – 5ad + bc + bd.

(vi) (5 + z) (y + 9)
= (5 + z)y + (5 + z)9
= 5y + zy + 45 + 9z
= 5y + 9z + zy + 45.

Question 3.
Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4.
Answer:
Let the two numbers be x and y, then:
x × y = (x + 2) × (y – 4)
xy = (x + 2)y – (x + 2)4
xy = xy + 2y – (4x + 8)
xy = xy + 2y – 4x – 8
xy – xy = 2y – 4x – 8
0 = 2y – 4x – 8
4x + 8 = 2y
2(2x + 4) = 2y
y = 2x + 4.
Examples :
(i) x = 1, y = 6 → Product = 1 × 6 = 6
Check : (1 + 2) × (6 – 4) = 3 × 2 = 6.

(ii) x = 2, y = 8 → Product = 16
Check : (2 + 2) × (8 – 4) = 4 × 4 =16.

(iii) x = 5, y = 14 → Product = 5 × 14 = 70
Check : (5 + 2) × (14 – 4) = 7 × 10 = 70.
Therefore, (1, 6), (2, 8), and (5, 14) are three examples for the given situation.

Question 4.
Expand
(i) (a + ab – 3b²) (4 + b), and
(ii) (4y + 7)(y + 11z – 3).
Answer:
(i) (a + ab – 3b²) (4 + b)
= (a + ab – 3b²)4 + (a + ab – 3b²)b
= 4a + 4ab – 12b² + ab + ab² – 3b³
= – 3b³ – 12b² + ab² + 4ab + ab + 4a
= – 3b³ – 12b² + ab² + 5ab + 4a.

(ii) (4y + 7) (y + 11z – 3)
= (4y + 7)y + (4y + 7)11z – (4y + 7)3
= 4y² + 7y + 44yz + 77z – (12y + 21)
= 4y² + 7y + 44yz + 77z – 12y – 21
= 4y² + 7y – 12y + 44yz + 77z – 21
= 4y² – 5y + 44yz + 77z – 21.

Question 5.
Expand (i) (a – b) (a + b),
(ii) (a – b) (a² + ab + b²) and
(iii) (a – b)(a³ + a²b + ab² + b³), Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding?
Answer:
(i) (a – b)(a + b) = (a – b)a + (a – b)b
= a² – ab + ab – b² = a² – b².

(ii) (a – b) (a² + ab + b²)
= (a – b)a² + (a – b)ab + (a – b)b²
= a³ – a²b + a²b – ab² + ab² – b³ = a³ – b³.

(iii) (a – b)(a³ + a²b + ab² + b³)
= (a – b)a³ + (a – b)a²b + (a – b)ab² + (a – b)b³
= a⁴ – a³b + a³b – a²b² + a²b² – ab³ + ab³ – b⁴
= a⁴ – b⁴.
The next identity would be : (a – b)(a⁴ + a³b + a²b² + ab³ + b⁴) = a⁵ – b⁵.
By expanding we can check it as :
(a – b) (a⁴ + a³b + a²b² + ab³ + b⁴)
= a(a⁴ + a³b + a²b² + ab³ + b⁴) – b(a⁴ + a³b + a²b² + ab³ + b⁴)
= a⁵ + a⁴b + a³b² + a²b³ + ab⁴ – a⁴b – a³b² – ab⁴ – b⁵ = a⁵ – b⁵

2. SPECIAL CASES OFTHE DISTRIBUTIVE PROPERTY
Figure it Out (Page 149) :

Question 1.
Which is greater: (a – b)² or (b – a)²? Justify your answer.
Answer:
Here, (a – b)² = a² + b² – 2ab ……….. (1)
and (b – a)² = b² + a² – 2ba
b² + a² = a² + b² and ba = ab
(b – a)² = a² + b² – 2ab ……… (2)
Comparing (1) and (2), we get
(a – b)² = (b – a)²

Question 2.
Express 100 as the difference of two squares.
Answer:
a² – b² = 100
(a + b) (a – b) = 100
[100 = 1 × 100, 2 × 50, 4 × 25, 5 × 20, 10 × 10]
We can take anyone
Let us take 50 × 2 = 100
Hence, (a + b) (a – b)= 50 × 2
a + b = 50 ……… (1)
a – b = 2 …….. (2)
Adding (1) and (2)
2a = 52
⇒ a = 26
Substituting a = 26 in (1)
26 + b = 50
⇒ b = 50 – 26 = 24
Let us check 26² – 24² = 676 – 576 = 100
Hence 26² – 24² = 100

Question 3.
Find 406², 72², 145², 1097² and 124² using the identities you have learnt so far.
Answer:
(i) 406² = (400 + 6)²
= 400² + 2 × 400 × 6 + 6²
= 160000 + 4800 + 36 = 164836

(ii) 72² = (50 + 22)²
= 50² + 2 × 50 × 22 + 22²
= 2500 + 2200 + 484 = 5184

(iii) 145² = (150 – 5)²
= 150² – 2 × 150 × 5 + 5²
= 22500 – 1500 + 25
= 21025

(iv) 1097² = (1100 – 3)²
= 1100² – 2 × 1100 × 3 + 3²
= 1210000 – 6600 + 9
= 1203409

(v) 124² = (100 + 24)²
= 100² + 2 × 100 × 24 + 24²
= 10000 + 4800 + 576
= 15376

Question 4.
Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer.
Answer:
Pattern 1
2(a² + b²) – (a + b)² + (a – b)²
Case-I
Let a = 4, b = 2
LHS = 2(4² + 2²)
= 2 × (16 + 4) = 40
RHS = (4 + 2)² + (4 – 2)²
= 36 + 4 = 40
∴ Pattern 1 holds for counting numbers.

Case-II
Let a = -4, b = -2
LHS = 2((-4)² + (-2)²)
= 2 × (16 + 4) = 2 × 20 = 40
RHS = (-4 + (-2))² + (-4 – (-2))²
= (- 4 – 2)² + (- 4 + 2)²
= (-6)² + (-2)² = 36 + 4 = 40
LHS = RHS
∴ Pattern 1 holds for negative integers also.

Case-III

The pattern holds for fractions also.

Pattern 2
a² – b² = (a + b) (a – b)
Case – I
Let a = 5, b = 3
LHS = 5² – 3² = 25 – 9 = 16
RHS = (5 + 3) (5 – 3) = 8 × 2 = 16
∴ LHS = RHS
∴ Pattern 2 holds for counting numbers.

Case-II
Let a = -5, b = -3
Now, LHS = (-5)² – (-3)² = 25 – 9 = 16
and RHS = [(-5) + (-3)] [(-5) – (-3)]
= (- 5 – 3) (- 5 + 3)
= (-8)(-2) = 16
∴ LHS = RHS
∴ Pattern 2 holds for negative integers also.

Case-III

and RHS = (\(\frac{1}{2}\) + \(\frac{1}{3}\)) (\(\frac{1}{2}\) – \(\frac{1}{3}\))
= (\(\frac{3 + 2}{6}\)) (\(\frac{3 – 2}{6}\))
= \(\frac{5}{6}\) \(\frac{1}{6}\) = \(\frac{5}{36}\)
∴ LHS = RHS
∴ Pattern 2 holds for fractions also.

3. THIS WAY OR THAT WAY, ALL WAYS LEAD TO THE BAY
Figure it Out (Page 154 – 156) :

Question 1.
Compute these products using the suggested identity.
(i) 46² using Identity 1A for (a + b)²
(ii) 397 × 403 using Identity 1C for (a + b) (a – b)
(iii) 91² using Identity 1B for (a – b)²
(iv) 43 × 45 using Identity 1C for (a + b) (a – b)
Answer:
(i) 46² = (40 + 6)² = 40² + 2 × 40 × 6 + 6²
[∵ (a + b)² = a² + 2ab + b²]
= 1600 + 480 + 36 = 2116

(ii) 397 × 403 = (400 – 3) (400 + 3)
[∵ (a + b) × (a – b) = a² – b²] = 400² – 3² = 160000 – 9 = 159991

(iii) 91² = (100 – 9)² = 100² – 2 × 100 × 9 + 9²
[∵ (a – b)² = a² + b² – 2ab] = 10000 – 1800 + 81 = 8281

(iv) 43 × 45 = (44 – 1) (44 + 1)
[∵ a² – b² = (a + b) × (a – b)]
= 44² – 1² = 1936 – 1 = 1935

Question 2.
Use either a suitable identity or the distributive property to find each of the following products.
(i) (p – 1) (p + 11)
(ii) (3a – 9b) (3a + 9b)
(iii) -(2y + 5) (3y + 4)
(iv) (6x + 5y)2
(v) (2x – \(\frac{1}{2}\))²
(vi) (7p) × (3r) × (p + 2)
Answer:
(i) (p – 1) (p + 11) = p(p + 11) – 1(p + 11)
= p² + 11p – p – 11 = p² + 10p – 11

(ii) (3a – 9b) (3a + 9b) = (3a)² – (9b)² = 9a² – 81b²

(iii) – (2y + 5)(3y + 4) = (- 2y – 5) (3y + 4)
= – 2y(3y + 4) – 5(3y + 4)
= – 6y² – 8y – 15y – 20 = – 6y² – 23y – 20

(iv) (6x + 5y)² = (6x)² + 2(6x) (5y) + (5y)²
= 36x² + 60xy + 25y²

(v) (2x – \(\frac{1}{2}\))² = (2x)² – 2 × 2x × \(\frac{1}{2}\) + (\(\frac{1}{2}\))²
= 4x² – 2x + \(\frac{1}{4}\)

(vi) (7p) × (3r) × (p + 2) = 7p × 3r × (p + 2)
= 21pr(p + 2) = 21pr × p + 21pr × 2
= 21p²r + 42pr

Question 3.
For each statement identify the appropriate algebraic expression(s).
(i) Two more than a square number.
2 + s (s + 2)² s² + 2 s² + 4 2s² 2²s

(ii) The sum of the squares of two consecutive numbers
m² + n² (m + n)²
m² + 1 m² + (m + 1)²
m² + (m – 1)²
(m + (m + 1))² (2m)² + (2m + 1)²
Answer:
(i) For “Two more than a square number”: The correct expression is s² + 2.

(ii) For “The sum of the squares of two consecutive numbers”: The correct expression is m² + (m + 1)².

Question 4.
Consider any 2 by 2 square of numbers in a calendar, as shown in the figure.

Find products of numbers lying along each diagonal – 4 × 12 = 48, 5 × 11 = 55. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens.
Hint: Label the numbers in each 2 by 2 square as

a	(a + 1)
a + 7	(a + 8)

Answer:
Case – I

6	7
13	14

Here, 6 × 14 = 84
13 × 7 = 91
Difference = 91 – 84 = 7

Case – II

9	10
16	17

Here, 9 × 17 = 153
16 × 10= 160
Difference = 160 – 153 = 7

We observe that the difference of the diagonal products in both cases is always 7.

Question 5.
Verify which of the following statements are true.
(i) (k + 1) (k + 2) – (k + 3) is always 2.

(ii) (2q + 1) (2q – 3) is a multiple of 4.

(iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.

(iv) (6n + 2)² – (4n + 3)² is 5 less than a square number.
Answer:
(i) (k + 1)(k + 2) – (k + 3) is a multiple of 2
Let k = 5, Then (5 + 1) (5 + 2) – (5 + 3)
= 6 × 7 – 8 = 42- 8 = 34
34 is a multiple of 2.
∴ The statement is true.

(ii) (2q + 1) (2q – 3) is a multiple of 4.
Let q = 3, Then (6 + 1) (6 – 3)
= 7 × 3 = 21
21 is not a multiple of 4
∴ The statement is false.

(iii) The square of an even number is a multiple of 4.
2² – 4 = 4 × 1
4² = 16 = 4 × 4
6² = 36 = 4 × 9
∴ The statement is true.
The square of an odd number is 1 more than a multiple of 8.
3² = 9 = 8 × 1 + 1
5² = 25 = 8 × 3 + 1
7² = 49 = 8 × 6 + 1
∴ The statement is true.

(iv) (6n + 2)² – (4n + 3)² is 5 less than a square number.
Let n = 2, (6 × 2 + 2)² – (4 × 2 + 3)²
= 14² – 11² = 196 – 121 = 75 = 80 – 5
But 80 is not a square number.
∴ The statement is false.

Question 6.
A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7?
Answer:
Let the numbers be x and y.
x = 7a + 3, y = 7b + 5
Sum = x + y
= 7a + 3 + 7b + 5 = 7(a + b) + 8
= 7(a + b) + 7 + 1 = 7(a + b + 1) + 1
∴ The remainder on division by 7 is 1.
Difference = x – y
= (7a + 3) – (7b + 5)
= 7a + 3 – 7b – 5 = 7(a – b) – 2
= 7(a – b) – 1 + 5 (∵ -2 = – 7 + 5)
= 7(a – b – 1) + 5
∴ The remainder on division by 7 is 5.
Product = xy
= (7a + 3) (7b + 5)
= 49ab + 35a + 21b + 15
= (49ab + 35a + 21b + 14) + 1
= 7(7ab + 5a + 3b + 2) + 1
∴ The remainder on division by 7 is 1.

Question 7.
Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.
Answer:
Let us take the numbers 7, 8, 9
Now, 8² – 7 × 9 = 64 – 63 = 1
Let us take the numbers 10, 11, 12
Then 11² – 10 × 12 = 121 – 120 = 1
Generalizing:
Let the numbers be a – 1, a, a + 1
Then a² – (a + 1) (a – 1) = 1
LHS = a² – (a + 1)(a – 1)
= a² – (a² – 1)
= a² – a² + 1 = 1
LHS = RHS
∴ Hence, the identity is correct.

Question 8.
What is the algebraic expression describing the following steps – add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers.
Answer:
Let the two numbers be a and b.
Step 1: a + b
Step 2: (a + b) × \(\frac{1}{2}\) (a + b)
∴ (a + b) × \(\frac{1}{2}\) (a + b) = \(\frac{1}{2}\) (a + b)²

Question 9.
Which is larger? Find out without fully computing the product.
(i) 14 × 26 or 16 × 24
(ii) 25 × 75 or 26 × 74
Answer:
(i) Let p = 14 × 26
p’ = 16 × 24
= (14 + 2) (26 – 2)
= 14 × 26 + 2 × 26 – 14 × 2 – 2 × 2
= 14 × 26 + 2(26 – 14 – 2)
= 14 × 26 + 2 × 10
p’ = p + 2 × 10
∴ p’ > p or 16 × 24 > 14 × 26

(ii) Let p = 25 × 75
p’= 26 × 74
=(25 + 1) (75 – 1)
=25 × 75 + 75 × 1 – 25 × 1 – 1 × 1
= p + (75 – 25 – 1) = p + 49
∴ p’ > p or 26 × 74 > 25 × 75

Question 10.
A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area g² sq. ft., will have a green cover. All the remaining area is a walking path w ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled.

Answer:
Length = w + g + 2w + g + w = 4w + 2g
Breadth = w + g + w = 2w + g
Area of park = (4w + 2g) (2w + g)
= 8w² + 4wg + 4wg + 2g²
= 8w² + 8wg + 2g²
Area of path = Area of park – Area of green cover
= 8w² + 8wg + 2g² – 2g²
= 8w² + 8wg
∴ (8w² + 8wg) sq. feet area needs to be tiled.

Question 11.
For each pattern shown below,
(i) Draw the next figure in the sequence.
(ii) How many basic units are there in Step 10?
(iii) Write an expression to describe the number of basic units in Step y.

Answer:

Step 1: 2 vertical strips of 3 units each + 1 vertical strip of 3 units
= 3 strips of 3 units each
= 9 units squares = (1 + 2)² unit squares

Step 2: 4 strips of 4 units each
= 16 units squares = (2 + 2)² unit squares

Step 3: 5 strips of 5 units each
= 25 units squares = (3 + 2)² unit squares

Step 4: (i) 6 strips of 6 units each = 2 are vertical and 4 are horizontal
(ii) Number of unit squares in step 10
= (10 + 2)² = 144

(iii) Number of unit squares in step y = (y + 2)²
(b) (i)
Number of unit squares in step 1 = 5 = 2² + 1
Number of unit squares in step 2 = 11 = 3² + 2
Number of unit squares in step 3 = 19 = 4² + 3

(ii) Step 1 has (1 + 1)² + 1 or 5 squares
Step 2 has (2 + 1 )² + 2 or 11 squares
Step 3 has (3 + 1)² + 3 or 19 squares
Hence step 10 has (10 + 1)² + 10 or 131 squares

(iii) Step y has [(y + 1)² + y] squares

We Distribute Yet Things Multiply Class 8 Extra Questions

Multiple Choice Questions

Question 1.
The product of 28 and 17 increase by which number if the value of 28 is increased by 1?
(a) 17
(b) 28
(c) 45
(d) 11
Solution:
Here, (28 + 1) × 17 = (28 × 17) + 17, which is 17 more than the product 28 × 17.
(a) 17

Question 2.
The product of 28 × 17 increases by what value if 17 is increased by 1?
(a) 17
(b) 28
(c) 45
(d) 11
Solution:
Here, 28 × (17 + 1) = 28 × 17 + 28, which is 28 more than the product 28 × 17.
(b) 28

Question 3.
The product 12 × 15 will increase by what value if both the numbers are increased by 1?
(a) 12
(b) 15
(c) 27
(d) 28
Solution:
Here, (12 + 1) (15 + 1)
= (12 × 15) + 12 × 1 + 1 × 15 + 1 × 1
= (12 × 15) + 12 + 15 + 1
= (12 × 15) + 28
(d) 28

Question 4.
Let a, b, and c be three numbers.
a × (b + c) = a × b + a × c
The propery by which the above happens is :
(a) Commutative
(b) Associative
(c) Distributive
(d) Closure
Solution:
a × (b + c) = a × b + a × c is by distributive property.
(c) Distributive

Question 5.
Expanded form of a (b + c – 1) is :
(a) ab + bc – 1
(b) ab + ab – 1
(c) ab + bc – a
(d) ab + ac – a
Solution:
Here, a (b + c – 1) = ab + ac – a
Answer:
(d) ab + ac – a

Assertion and Reasoning

(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).
(b) Assertion (A) and Reason (R) both are true but Reason (R) is not the correct explanation for the Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.

Question 1.
Assertion (A) : (a + 4) (b + 4) = ab + 4
Reason (R) : (x + y) + (u) = x + y + u
Solution:
∵ (a + 4) (b + 4) = ab + 4a + 4b + 16, So,
Assertion (A) is false.
Answer:
(d) Assertion (A) is false but Reason (R) is true.

Question 2.
Assertion (A) : A number divisible by 4 and another number divisible by 3, have sum which is always divisible by 12.
Reason (R) : A number divisible by 12 can be algebraically represented by ‘12k’, where ‘k’ is an integer.
Solution:
Let the number represented by 4 be represented by 4 m and the number represented by 3 be represented by 3n, where m and n be integers.
Now, 4m + 3n may not be divisible by 12 for all values of m and n.
Answer:
(d) Assertion (A) is false but Reason (R) is true.

Case Based Questions

Question 1.
Consider any 2 × 2 square numbers (grid) in a calender, as shown in the figure.

Answer the following questions based on the above assumptions:
(i) Write the numbers in the highlighted box as they appear.
(ii) Find the products of numbers lying along each diagonal. Find their difference.
(iii) Take any other 2 × 2 square of numbers and write it as the numbers appear- in it.
(iv) Find the product of numbers lying along each diagonal. Find their difference. Are the difference obtained in (ii) and now same?
Answer:
(i)

6	7
13	14

(ii) 6 × 14 = 84, 13 × 7 = 91
Difference = 91 – 84 = 7

(iii) Let us consider

2	3
9	10

(iv) 2 × 10 = 20, 9 × 3 = 27
Difference = 27 – 20 = 7

Go through BSE Odisha Class 8 Science Solutions Chapter 7 Particulate Nature of Matter Question Answer to understand textbook questions more clearly.

Class 8 Science Curiosity Chapter 7 Question Answer

Class 8 Science Ch 7 Particulate Nature of Matter Question Answer

Class 8 Science Chapter 7 Particulate Nature of Matter Question Answer

Probe and Ponder Questions

Question 1.
Why is it possible to pile up stones or sand, but not a liquid like water?
Answer:
Stones and sand are solids. In solids, particles are tightly packed and held together by relatively strong interparticle attractions. This fixed arrangement gives solids a definite shape and allows them to rest on one another, so they can be piled up. Water is a liquid; its partcles have weaker attractions and can move past one another, so liquid flows and cannot keep a free-standing pile of its own shape.

Question 2.
Why does water take the shape of folded hands but lose that shape when released?
Answer:
Water is a liquid. Its particles can move around and rearrange themselves to fit the shape of the container (in this case, folded hands). When the hands are opened, gravity and the ability of the particles to move cause the water to flow and change shape again, because liquids have a definite volume but no fixed shape.

Question 3.
We cannot see air, so how does it add weight to an inflated balloon?
Answer:
Air is a mixture of gases made of tiny partcles (molecules) that cannot be seen individually. When a balloon is inflated, these gas particles occupy space inside the balloon and add mass to it. Because mass is present. the balloon becomes heavier – that is, air inside the balloon contributes to its weight.

Question 4.
Is the air we breathe today the same that existed thousands of years ago?
Answer:
Yes. Matter, including air, is continuously recycled in nature through processes such as respiration, photosynthesis and weather cycles. The atoms and molecules in the air today have existed for a very long time and keep circulating through the envvironment.

Question 5.
Share your questions?
Answer:

• How small are the tiniest particles of matter and can we ever see them?
• Why do some solids melt easily while others need very high temperatures?
• If gases have no fixed volume, how do they stay contained in the atmosphere?
• What happens to the partcles when a substance changes from solid to liquid?
• Why don’t all solids dissolve in water like sugar does?

InText Questions

Question 1.
Is every speck of this fine chalk powder still composed of the same substance, or has it changed into something else on breaking or grinding? (Page 99)
Answer:
Yes, even after breaking or grinding, each speck of chalk powder (fine-grinded) is the same as the previous state because this change is a physical change in which only the size of chalk changes, not any chemical change occurred.

Question 2.
Are the units of chalk obtained in this manner considered the smallest units of chalk? (Page 100)
Answer:
No, the obtained units of chalk in the process of grinding are not the smallest unit. Every unit of chalk is even consists of constituent particles, which are the basic units of chalk.

Question 3.
Chalk and sugar can both be broken down into their constituent particles. But how are the constituent particles held together to form the solid pieces we see? (Pages 101)
Answer:
The constituent particles in solids are held together by interparticles forces of attraction. These forces keep the particles closely packed and fixed in position, giving solids a definite shape and volume.

Question 4.
In the solid state, is there any way to move these particle apart? (Page 102)
Answer:
In a solid, particles can only vibrate about fixed positions because they are very close together and strongly attracted to one another. To move them further apart, we must supply energy (for example by heating) so that the solid melts into a liquid, where particles can move more freely.

Question 5.
Solids have a definite volume; what about liquids and gases? (Page 103)
Answer:
Liquids: Liquids have a definite volume but no definite shape; they take the shape of the container that holds them.
Gases: Gases have neither definite shape nor definite volume; they expand to fill the entire contain or space available to them.

Question 6.
Do gases also have a fixed volume? (Page 105)
Answer:
No, gases don’t have a fixed shape or volume. The volume of gas changes with the amount of closeness of particles or the interparticle attraction between particles.

Question 7.
Sugar and sand are both solids. Why does sugar dissolve in water, but sand does not? (Pages 108)
Answer:
Sugar particles are solid, but they dissolve in water and occupy some space between the water molecules. Because water can break down sugar particles, which reduces the total volume of the mixture. Whereas sand particles have a rigid crystal structure, which cannot be broken down by water molecules, and hence settle down in water and increasing the total volume.

Question 8.
How can we demonstrate the movement of gas particles that cannot be seen with the naked eye? (Page 110)
Answer:
We can use visible tracers such as smoke or coloured vapours to show gas motion. For example. Smoke from incense spreads through the air and its movement shows that gases particles move randomly and can carry other particles with them.

Particulate Nature of Matter Class 8 Questions and Answers

Keep the Curiosity Alive (Pages 113-114)

Question 1.
The primary difference between solids and liquids is that the constituent particles are :
(i) closely packed in solids, while they are stationary in liquids.
(ii) far apart in solids and have fixed positions in liquids.
(iii) always moving in solids and have a fixed position in liquids.
(iv) closely packed in solids and move past each other in liquids.
Answer:
(iv) closely packed in solids and move past each other in liquids.
Explanation: In solids, particles are closely packed and fixed in position due to strong interparticle attractions. In liquids, particles are still close but can move or slide past each other, allowing the liquid to flow and take the shape of its container.

Question 2.
Which of the following statements are true? Correct the false statements.
(i) Melting ice into water is an example of the transformation of a solid into a liquid.
Answer:
True: Melting ice into water is an example of the transformation of a solid into a liquid.

(ii) The melting process involves a decrease in interparticle attractions during the transformation.
Answer:
True: The Melting process involves a decrease in interparticle attraction during the transformation.

(iii) Solids have a fixed shape and a fixed volume.
Answer:
True: Solids have a fixed shape and a fixed volume.

(iv) The interparticle interactions in solids are very strong, and the interparticle spaces are very small.
Answer:
True: The interparticle interactions in solids are very strong, and the interparticle spaces are very small.

(v) When we heat camphor in one corner of a room, the fragrance reaches all corners of the room.
Answer:
True: When we heat camphor in one corner of a room, the fragrance reaches all corners of the room.

(vi) On heating, we are adding energy to the camphor, and the energy is released as a smell.
Answer:
False: The correct statement is: On heating, energy is added to camphor, causing it to undergo sublimation. The camphor directly converts into gas, and the vapour carries its characteristic smell.

Question 3.
Choose the correct answer with justification. If we could remove all the constituent particles from a chair, what would happen?
(i) Nothing will change.
(ii) The chair will weigh less due to lost particles.
(iii) Nothing of the chair will remain.
Answer:
Correct option is (iii) Nothing of the chair will remain.
Justification: A chair is made up of constituent particles (atoms and molecules). If you remove all the particles from the chair, there is nothing left to form the structure, shape, weight, or existence of the chair.

Question 4.
Why do gases mix easily, while solids do not?
Answer:
Gas particles are far apart from each other, and that’s why they move very fast in all directions. Gases have weak intermolecular forces, so they don’t attach. Due to this reason, gas particles spread easily around other particles.

Question 5.
When spilled on the table, milk in a glass tumbler flows and spreads out, but the glass tumbler stays in the same shape. Justify this statement.
Answer:
In this case, milk is spilled on the table, and it spreads around the table because its state is liquid. Liquids can take the shape of their surrounding because their molecules are free to move. This is the reason the milk flows around the table. Whereas the glass tumbler’s shape does not change because it is a solid. In solids, the molecules are closely packed.

Question 6.
Represent diagrammatically the changes in the arrangement of particles as ice melts and transforms into water vapour.
Answer:
As ice melts into water and then vaporizes into steam, the arrangement of water particles changes significantly. Initially, in ice (a solid), water molecules are tightly packed in a fixed, crystalline structure with limited movement (vibrations). As ice melts, the particles gain kinetic energy, breaking free from their fixed positions and becoming able to slide past each other, forming liquid water.

Further heating increases the kinetic energy, causing the particles to move more rapidly and spread out, eventually breaking free from the liquid and becoming water vapour, a gas with particles moving randomly and freely.

Ice (Solid)

• Arrangement: Water molecules are tightly packed in a regular, crystalline structure.
• Movement: Molecules vibrate in fixed positions.

Liquid Water

• Arrangement: Molecules are closer together than in a gas, but not in a regular structure. They can move around and slide past each other.
• Movement: Molecules can move around and slide past each other.

Water Vapor (Gas)

• Arrangement: Molecules are far apart and move randomly and freely in all directions.
• Movement: Molecules move rapidly and randomly, colliding with each other and the container walls.

Question 7.
Draw a picture representing particles present in the following :
(i) Aluminium foil
(ii) Glycerin
(iii) Methane gas
Answer:
Pictorial representation of particles of Aluminium foil, Glycerin, and Methane gas.

Question 8.
Observe figure (a), which shows the image of a candle that was just extinguished after burning for some time. Identify the different states of wax in the figure and match them with figure (b), showing the arrangement of particles.
Answer:
Different states of wax

Question 9.
Why does the water in the ocean taste salty, even though the salt is not visible? Explain.
Answer:
Ocean water tastes salty because it contains a high concentration of dissolved salts, primarily sodium chloride (common table salt). These salts are not visible because they are dissolved at a molecular level, meaning the individual salt molecules are dispersed throughout the water, making it appear clear.

Question 10.
Grains of rice and rice flour take the shape of the container when placed in different jars. Are they solids or liquids? Explain.
Answer:
Grains of rice and rice flour are considered solids, despite appearing to take the shape of their container. This is because each grain retains its shape and volume, even when mixed. The “flowing” behavior is due to the ability of these small, irregularly shaped particles to move past each other with minimal friction.

Class 8 Science Chapter 7 Question Answer

Activity 1

Aim: To show that matter is made of tiny units which is called constituent particles or building blocks.
Materials Required: A stick of chalk, magnifying glass.

Procedure:

• Break a piece of chalk into two pieces see figure (a) and figure (b).
• Then break the chalk till it becomes difficult to break it further by hand.
• Finally grind the small pieces of chalk thus obtained [Fig. (c) using a mortar and pestle.
• Look at the fine powder of chalk with a magnifying glass and note down your observations. [Fig. (d)

Observations:

• Even finest chalk powder you can make still looks like chalk under a magnifying glass.
• By breaking or grinding the chalk no new substance is formed. Grinding is a physical change in which only the size of each speck of chalk has reduced further.

Inferences:

• If you could keep breaking the chalk smaller and smaller you would eventually reach the constituent particles that can’t be broken down further by normal means.
• The constituent particle is the basic unit that makes up a substance.What happen to sugar when dissolve in water?

Activity 2.

Let us perform
Aim: To show that particles have a lots of space between them and each tiny particle is made up of millions of constituent particles.
Materials Required: A glass tumbler, sugar.

Procedure:

• First of all we will take a glass tumbler and fill it with drinking water.
• Now, add two or three teaspoons of sugar into it.
• Do not stir the water. Taste a small spoonful of water from the top layer of the glass.
• When you taste without stirring, the top layer does not taste sweet.
• Now, dissolve the sugar with the help of a spoon by stiring the Fig.: Dissolving sugar in water solution.
• Again taste a spoonful of water from the top layer.

Observations:

• After stirring the whole solution tastes sweet.
• Sugar particles dissolves completely and no longer can be seen.

Inference:

• The sugar has separated into its constituent particles, which spread out among the water particles and occupy the space called interparticle spaces.
• Constituent particles or basic particles are so small that they are invisible to the naked eye or even ordinary microscope.

Conclusion: Both chalk and sugar can be broken down to pieces made of their basic particles, and these are so small they are invisible to the naked eye or even ordinary microscopes.

Activity 3.

Let us find out
Aim: To show that solids are hard and keep their shape.
Materials Required: Some solid objects like stone, iron nail, etc. hammer.

Procedure:

• First of all we will collect a few solid objects, like a piece of iron or an iron nail, a piece of rock salt, a stone, a piece of wood, a key, and a piece of aluminium (See figure).
• Now look at their shapes and sizes.
• Take one by one and try hammering them.
• Tabulate your observation and note down that which of the above six objects particles are strongly held together?

Observations:

Objects	After hammering	Particles are strongly held together (Yes/No)
1. Iron nail 2. Rock salt 3. Stone 4. Wooden block 5. Key 6. Piece of aluminium	Shape can change May break May break No change Shape can change May convert into sheet	Yes Yes Yes Yes Yes Yes

Inference:

• They have definite shape and volume.
• They are tightly packed.
• This is due to strong interparticle attraction.
• The particles can only move to and fro about their positions (vibrate or oscillate) but cannot move past each other.

Activity 4.

Let us try and find out
Aim: To show that liquid have no fixed shape but have a fixed volume.
Materials Required: Containers of different shapes, marker, strip of paper.

Procedure:

• First of all we will take three clean and dry containers of different shapes.
• Label them as X, Y and Z as in figure.
• Now, mark the 250 mL level in each container with the help of a marker or by pasting a thin strip of paper.
• Then, fill the water in X up to the marked level.
• Be careful when you are transfering water from container ‘X’ to container ‘Y’. Water should not be spill out.
• Observe the shapes and level of the water.
• Similarly, transfer the same water from Container Y to Container Z, carefully, and once again observe the shape and level of the water.

Observations:

• The volume stays the same.
• Liquids have no fixed shape it takes the shape of the container into which it is poured.

Inference:

• Liquids have no fixed shape but have a fixed volume.
• This happes because the particles of liquids are free to move.

Activity 5.

Let us investigate
Aim: To show that gases do not have fixed shape or volume and particles of gases move freely in all directions.
Materials Required: Two transparent gas jar or glass tumblers, incense stick.

Procedure:

• First of all we will take two transparent gas jars or glass tumblers and mark them A and
• Now, burn an incense stick to create some smoke.
• Collect the smoke by holding the Gas Jar A upside down. [See figure (a)]
• We can see that the gas jar is filled with smoke.
• Now, turn it over and cover it with a glass plate. [See figure (b)]
• Then, take another Gas Jar B and turn it upside down and gently place it over the glass plate covering the Gas Jar A.
• Now, we will remove the glass plate slowly.
• We should take precaution that both gas jars are close and there is no gap for smoke to escape out. [See figure (c)]
• This experiment can be demonstrated by using an Iodine Vapour also.
• Note down your observations.

Observations:

• The smoke fills the entire space in the Gas Jar B, see figure (d).
• Particles in gases move freely in all directions.

Inference:

• Gases do not have a fixed shape or fixed volume.
• They acquire the shape of the vessel in which they are kept.
• Particles of gases are always in rapid, random motion.

Activity 6.

Let us experiment
Aim: To understand the compressibility of fluid (Gas and liquid) using a syringe.
Materials Required: A syringe without needle.

Procedure:

• First of all we will take a syringe without a needle.
• Pull the plunger of the syringe in the outwards direction in a fully extended position (See figure (a)].
• Now, place our thumb on the open end of the syringe so that the air present inside the syringe may not escape.
• Try to push the plunger slowly and steadily inward [See figure (c)].
• Note down your observation.
• Repeat the activity using water and once again note down your observations.

Observations :

• Volume of air inside the syringe decreases because after compressing the air by pushing the plunger, the particles are forced to come closer.
• The plunger cannot move much when we do the same experiment with water.

Inference:

• Gas particles have large gaps between them, so gases are easily compressible.
• Liquids have much smaller gaps between particles, so they are almost incompressible.

Activity 7.

Let us observe
Aim: To show that liquids have enough interparticle spaces.
Materials Required: A glass vessel, a marker, glass rod.

Procedure:

• First of all we have to take a glass vessel, fill it about half with water and mark the level of water as A with the help of a marker. [See figure (a)]
• Now, add two teaspoons of sugar into it.
• Obviously the water level will rise. So, mark the new water level on the glass vessel as B. (See figure (b)]
• Take a glass rod and stir the water so that the sugar can be dissolved. [See figure (c)]
• Guess whether the water level will increase or decrease with respect to the mark B.
• Now, mark this water level again as C. [See figure (d)]
• Repeat the above activity with some other soluble solids, such as common salt or glucose, and insoluble solids, like sand and stone pieces.

Observations :

• Water level first rises (as sugar is added).
• After stirring, sugar dissolves, the final liquid level (C) is less than the expected sum of water plus sugar (i.e., level B).

Inference:

• There are some empty spaces between water particles (inter particle spaces). The particles of the dissolved substance occupy these spaces.
• In case of insoluble substances like sand water level stays high or rises because sand does not dissolve or fill the interparticle spaces.

Activity 8.

Let us experiment
Aim: To show that particles of matter are continuously moving.
Materials Required: A glass tumbler, a few grains of potassium permanganate.

Procedure:

• Take a few grains of potassium permanganate and dissolve it in a glass tumbler filled with water.
• What did you observed?

Observations:

• At first, you will see pink coloured streaks spreading out. [See figure (a)]
• Very soon, the entire glass of water will acquire a uniform pink colour. [See figure (b)]

Inference:

• Water particles are in motion constantly.
• First they pull out the particles of potassium permanganate from its grain and then hit these particles so that they get spread throughout the liquid.
• Key concept: Particles of liquids are always moving, pulling apart and mixing other particles this is why substances can dissolve and diffuse in water.

Activity 9.

Let us find out
Aim: Particles of air are moving constantly.
Materials Required: Incense stick match stick.

Procedure:

• Put an incense stick in a corner of your room. You have to go near it to get its smell
• Now light the stick with a match stick.

Observations: The fragrance spread immediately and can be felt even from a distance.
Inference: This shows that the particles of air are moving constantly. The air particles hit the particles of the fragrance i.e., got mixed and help them spread throughout the room.

Particulate Nature of Matter Class 8 Extra Questions and Answers

Short Answer Type Questions

Question 1.
Define ‘particulate nature of matter’.
Answer:
The particulate nature of matter means that all matter is made up of tiny particles. These particles are constantly moving and have space between them.

Question 2.
How does temperature affect the state of matter ?
Answer:
Increasing temperature give particles more energy. Which can change solids to liquids and liquids to gases.

Question 3.
What is the significance of interparticle spacing in matter ?
Answer:
Interparticle spacing affects properties like shape, volume and compressibility of a substance.

Question 4.
Give reasons :
(a) A gas fills completely the vessel in which it is kept.
(b) A wooden table should be called a solid.
Answer:
(a) A gas fills completely the vessel in which it is kept because the force of attraction between the particles of gas is very-very less and particles are free to move in all directions.
(b) A wooden table is called a solid because particles of the wood are tightly packed and it has definite shape and volume. It cannot be compressed easily.

Question 5.
A rubber band can change its shape on stretching. Will you classify it as solid or not? Justify your answer.
Answer:
Rubber band changes shape under force and regains the shape when the force is removed. So, it is classified as a solid.

Question 6.
How does particle of soap help in cleaning clothes?
Answer:
When we wash clothes stained with oil using soap, there are many soap particles which surround the oil particles on the fabric. One end of the soap particle attaches to the oil, and the other mixes with water, thus helping lift the oil off and wash it away (See figure).

Question 7.
How does particles behave when they are heated ?
Answer:
(a) Particles move more vigorously and separate from each other.
(b) This separation results in a decrease in interparticle forces of attraction allowing particles to escape the liquid and form vapour.
(c) The overall transformation : The liquid converts into its gaseous state (vapor), with boiling being rapid. At the boiling point, the formation of vapour is very fast and occurs not only at the surface but also within the liquid.

Long Answer Type Questions

Question 1.
Give reasons :
(a) A gas exerts pressure on the walls of the container.
(b) We can easily move our hand in air but to do the same through a solid block of wood we need a karate expert.
Answer:
(a) The molecules of a gas are free to inove randomly in all directions. During their motion, they collide with one another and also with the walls of the container. The constant bombardment of the molecules on the walls of the container exerts a steady force. The force acting per unit area on the walls of the container is called pressure. Thus, gases exert pressure.

(b) In air there is a lot of empty space between the molecules and the forces between the particles are almost negligible. Hence we can move our hand in air. Through a solid block of wood only a karate expert can do this because there are strong forces of attraction between particles in a solid block of wood and there is no empty space between them.

Question 2.
Give reasons for the following :
(a) A gas does not have a fixed shape.
(b) A gas does not have a fixed volume.
(c) A gas can be compressed easily.
Answer:
(a) A gas does not have a fixed shape because the positions of its particles (molecules) are not fixed and particles move freely.
(b) A gas does not have a fixed volume because the spaces between its particles (molecules) are not fixed. Since the particles (molecules) of a gas are free to move anywhere, it takes the shape and volume of its container.
(c) A gas can be compressed easily because its molecules are far apart and there are large spaces between them which can be reduced by compression.

Case-Study Based Questions

1. Read the following passage carefully and answer the questions that follow : The force of attraction between the particles are maximum in solids, intermediate in liquids and minimum in gasses. The space in between the constituent particles and kinetic energy of the particles are minimum in the case of solids, intermedicate in liquids and maximum in gases.

(i) Which one of the following represents a correct arrangement of increasing order of forces of attraction between their particles?
(a) Water, air, wind
(b) Air, sugar, oil
(c) Oxygen, water, sugar
(d) Salt, juice, air
Answer:
(c) Oxygen, water, sugar

(ii) Which one of the following represents a correct arrangement of increasing order of forces of attraction between the particles?
(a) Water, common salt, carbondioxide
(b) Carbondioxide, water, common salt
(c) Carbondioxide, common salt, water
(d) Common salt, water, carbondioxide
Answer:
(b) Carbondioxide, water, common salt

(iii) The space in between the constituent particles is :
(a) Least of solids, intermediate in liquids and maximum in gases
(b) Least in liquids, intermediate in solids and maximum in gases
(c) Least in gases, intermediate in liquids and maximum in solids
(d) Least in solids, intermediate in gases and minimum in liquids
Answer:
(a) Least of solids, intermediate in liquids and maximum in gases

(iv) In which of the following conditions, the distance between the molecules of hydrogen gas could increase?
A. Increased pressure in hydrogen contained in a closed container
B. Some hydrogen gas leaking out of the container.
C. Increasing the volume of the container of hydrogen gases.
D. Adding more hydrogen gas to the container without increasing the volume of the container.
(a) A and C
(b) A and D
(c) B and C
(d) B and D
Answer:
(c) B and C

Picture Based Questions

I. Look at the picture and answer the following questions :
(a) Which phenomenon is displayed by figure (A) and figure (B).

Answer:
A → Evaporation
B → Boiling

(b) How did you identified it ?
Answer:
Evaporation is slower process of vapour formation and occur at all temperature. Also, bubbles do not forms. But, boiling is the fast process of vapour formation and bubbles are formed.

(c) Do boiling take place at all temperature?
Answer:
No, it occur at boiling point only.

Particulate Nature of Matter Class 8 MCQ

Multiple Choice Questions (Mcqs)

Question 1.
Which of the following is the basic unit of matter?
(a) Molecule
(b) Atom
(c) Element
(d) Compound
Answer:
(b) Atom

Question 2.
Which of these is not a form of matter?
(a) Solid
(b) Liquid
(c) Gas
(d) Energy
Answer:
(d) Energy

Question 3.
Which of the following is true about matter?
(a) It occupies space
(b) It has mass
(c) It is made up of particles
(d) All of the above
Answer:
(d) All of the above

Question 4.
The particles of matter are:
(a) stationary
(b) invisible and always moving
(c) not attracted to each other
(d) fixed in space
Answer:
(b) invisible and always moving

Question 5.
When sugar dissolves in water, it shows:
(a) sugar disappears
(b) particles are stationary
(c) matter is continuous
(d) matter is made up of particles
Answer:
(d) matter is made up of particles

Assertion and Reasoning

These questions consist of two statements, each printed as Assertion (A) and Reason (R). While answering these questions, you are required to choose any one of the following four responses.
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.
(b) Assertion (A) and Reason (R) both are correct and reason is not correct explanation for assertion.
(c) Assertion (A) is correct but the Reason (R) is wrong.
(d) Assertion (A) is wrong but the Reason (R) is correct.

1. Assertion (A): Oxygen is called a gas. Reason (R): Oxygen has neither fixed shape nor fixed volume.
Answer:
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.

2. Assertion (A): Solids are incompressible. Reason (R): The forces of attraction between the particles are maximum and spaces in between the constituent particles are least in the case of solids.
Answer:
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.

Fill in the blanks

1. In gaseous state interparticle spacing is ………….
Answer:
maximum

2. The ………… energy is used to overcome the attractive forces between particles.
Answer:
thermal

3. Movement of particles are ………… in solids.
Answer:
negligible

4. Evaporation is a ………… phenomenon.
Answer:
surface

5. Matter is made up of very tiny ………….
Answer:
particles

True or False

1. All matter is made up of tiny particles.
Answer:
True

2. Particles of matter are visible to the naked eye.
Answer:
False

3. The spaces between particles are the same in all states of matter.
Answer:
False

4. Matter is anything that has mass and occupies space.
Answer:
True

5. Water is not considered matter because it flows.
Answer:
False

Students can refer to BSE Odisha Class 8 Math Solution and Ganita Prakash Chapter 5 Number Play Class 8 Question Answer to understand textbook questions step by step.

Class 8 Maths Chapter 5 Number Play Solutions

Ganita Prakash Class 8 Chapter 5 Solutions

Class 8 Maths Ganita Prakash Chapter 5 Solutions Number Play

1. IS THIS A MULTIPLE OF?
Figure it Out (Page 122 – 123) :

Question 1.
The sum of four consecutive numbers is 34. What are these numbers?
Answer:
Let four consecutive numbers be x, (x + 1), (x + 2) and (x + 3) respectively.
x + x + 1 + x + 2 + x + 3 = 34
⇒ 4x + 6 = 34
⇒ 4x = 34 – 6
⇒ 4x = 28
x = \(\frac{28}{4}\) = 7.
So, (x + 1) = 7 + 1 = 8
(x + 2) = 7 + 2 = 9
(x + 3) = 7 + 3 = 10
Therefore, the given four consecutive numbers are 7, 8, 9, and 10.

Question 2.
Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.
Answer:
If p is the greatest office consecutive numbers, then the other four numbers in terms of p are (p – 1), (p – 2), (p – 3), and (p – 4).

Question 3.
For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.

(i) The sum of two even numbers is a multiple of 3.
Answer:
Let the two even numbers be 2a + 2b
Sum = 2a + 2b = 2(a + b)
For 2(a + b) to be a multiple of 3, (a + b) must be multiple of 3.
Example:
2 + 4 = 6 → divisible by 3
2 + 8 = 10 → not divisible by 3
Conclusion: Sometimes true.

(ii) If a number is not divisible by 18, then it is also not divisible by 9.
Answer:
If a number is divisible by 18, then it is also divisible by 9 because 9 is a factor of 18.
18 ÷ 9 = 2 → divisible by 9.
But if a number is divisible by 9, it is not always divisible by 18.
9 ÷ 18 = 0.5 → not divisible by 9.
Example: 9 is divisible by 9 but not divisible by 18.
27 is divisible by 9, but not 18.
Conclusion : Sometimes true.

(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
Answer:
Let the two numbers be a and b.
Not divisible by 6 means they do not satisfy
\(\frac{a}{6}\) or \(\frac{b}{6}\)
But their sum can still be divisible by 6.
Example :
• 8 and 10 are not divisible by 6.
The sum of two numbers = 8 + 10 = 18, is divisible by 6.
• 10 and 13 are not divisible by 6.
The sum of 10 and 13 = 10 + 13 = 23, which is not divisible by 6.

(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
Answer:
Let the multiple of 6 be 6a, the multiple of 9 be 9b.
Sum: 6a + 9b = 3(2a + 36) → clearly divisible by 3.
Example :
6 + 9 = 15 → divisible by 3.
12 + 18 = 30 → divisible by 3.
Conclusion : Always true.

(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Answer:
Let multiple of 6 be 6a, multiple of 3 be 3b.
Sum : 6a + 3b = 3(2a + b).
For it to be divisible by 9, 2a + b must be divisible by 3.
Example :
6 (6 × 1) + 3 (3 × 1) = 9 → divisible by 9
6 + 6 = 12 → not divisible by 9
Conclusion : Sometimes true.

Question 4.
Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.
Answer:
Here, Remainder = 2, Dividend = 3
∴ Number = (Quotient × Dividend) + Remainder
= (K × 3) + 2
where, K = 1, 2, 3,…..
Numbers = 1 × 3 + 2 = 3 + 2 = 5
Numbers = 2 × 3 + 2 = 6 + 2 = 8
Numbers = 3 × 3 + 2 = 9 + 2 = 11
Thus, 5, 8, and 11 are numbers that leave a remainder of 2 when divided by 3.
Algebraic expression = 3K + 2
Here, Remainder = 2, dividend = 4
Number = 4K + 2, where K = 1, 2, 3, 4,…
Numbers = 4 × 1 + 2 = 4 + 2 = 6
Numbers = 4 × 2 + 2 = 8 + 2 = 10
Numbers = 4 × 3 + 2 = 12 + 2 = 14
Algebraic expression = 4K + 2
Thus, 6, 10, and 14 are numbers that leave a remainder of 2 when divided by 4.

Question 5.
“I hold some pebbles, not too many, When I group them in 3’s, one stays with me. Try pairing them up – it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?”

Answer:
The LCM of 3, 5, and 7
= 3 × 5 × 7 = 105 [∵ 3, 5, and 7 are prime numbers]
No. of pebbles = 105 + 1 = 106

Question 6.
Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?
Answer:
The expression has been written by Tathagat = 6k + 2
where, k = 1, 2, 3, 4, 5, 6,…
6 × 1 + 2 = 8
6 × 2 + 2 = 14
6 × 3 + 2 = 20
6 × 4 + 2 = 26
The sum of three numbers
8 + 14 + 20 = 42, it is a multiple of 6.
14 + 20 + 26 = 60, it is a multiple of 6.
Yes, Tathagat’s claim is true.

Question 7.
When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.
(i) 4779 + 661
(ii) 4779 – 661
Answer:
(i) 4779 + 661
= Remainder 5 + Remainder 3
= Remainder 8
8 divided by 7 → remainder 1.
Visualization Method:
4779 + 661
= (682 × 7) + 5 + (94 × 7) + 3
= 7 × (682 + 94) + 5 + 3
= 7 × 776 + 8
= Divisible by 7 + 87
= 1, Remainder

(ii) 4779 – 661
= Remainder 5 → Remainder 3
= Remainder 2
Visualization Method:
4779 – 661
= (682 × 7) + 5 – (94 × 7) – 3
= 7 × (682 – 94) + 5 – 3
= 7 × 588 + 2
= Divisible by 7 + 2
= 2, Remainder

Question 8.
Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?
Answer:
A number that leaves a remainder of 2 when divided by 3 is = 3x + 2
A number that leaves a remainder of 3 when divided by 4 is = 4x + 3
A number that leaves a remainder of 4 when divided by 5 is = 5x + 4
L.C.M of 3, 4, and 5 = 60
All the numbers are the same,
so 4x + 3 = 3x + 2
4x – 3x = 2 – 3
x = -1
Each remainder is 1 less than the divisor.
Hence, the number is 1 less than the L.C.M = (60 – 1) = 59.
So, 59 is the smallest number that satisfies all the given conditions.

2. CHECKING DIVISIBILITY QUICKLY
Figure it Out (Page 126) :

Question 1.
Find, without dividing, whether the following numbers are divisible by 9.
(i) 123
(ii) 405
(iii) 8888
(iv) 93547
(v) 358095
Answer:
If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
(i) Sum of the digits = 1 + 2 + 3 = 6, is not divisible by 9.
Thus, 123 is not divisible by 9.

(ii) Sum of the digits = 4 + 0 + 5 = 9, is divisible by 9.
Thus, 405 is divisible by 9.

(iii) Sum of the digits = 8 + 8 + 8 + 8 = 32, is not divisible by 9.
Thus, 8888 is not divisible by 9.

(iv) Sum of the digits = 9 + 3 + 5 + 4 + 7 = 28, is not divisible by 9.
Thus, 93547 is not divisible by 9.

(v) Sum of the digits = 3 + 5 + 8 + 0 + 9 + 5 = 30, is not divisible by 9.
Hence, 358095 is not divisible by 9.

Question 2.
Find the smallest multiple of 9 with no odd digits.
Answer:
Multiples of 9 = 9, 18, 27, 36, …, 288, ……….
The smallest multiple of 9 with an odd digit is 9.
The smallest multiple of 9 that can be formed by summing even digits is 18 (since 9 is odd).
Thus, the smallest multiple of 9 with no odd digits is 288.

Question 3.
Find the multiple of 9 that is closest to the number 6000.
Answer:
Given, 6000
Sum of the digits = 6 + 0 + 0 + 0 = 6
We know that, if the number is divisible by 9, then the sum of the digits is divisible by 9.
If we add 3 to the number 6000.
6000 + 3 = 6003, it is divisible by 3.
Thus, the multiple of 9 that is closest to the number is 6003.

Question 4.
How many multiples of 9 are there between the numbers 4300 and 4400?
Answer:
The multiples of 9 are there between the numbers 4300 and 4400 are 4302, 4311, 4320, ………… , 4392
The number of multiples of 9
= \(\frac{\text { Last term }- \text { First term }}{\text { Difference }}\) + 1
= \(\frac{4392 – 4302}{9}\) + 1
= \(\frac{90}{9}\) + 1 = 10 + 1 = 11
Thus, the multiples of 9 are 11.

Figure it Out (Page 131) :

Question 1.
The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?
Answer:
Consider the 8-digit number 80000006.
The digital root of 80000006 = 8 + 0 + 0 + 0 + 0 + 0 + 0 + 6 = 14
= 1 + 4 = 5
10 more than 80000006 = 80000006 + 10 = 80000016
The digital root of 80000016 = 8 + 0 + 0 + 0 + 0 + 0 + 1 + 6
= 15 = 1 + 5 = 6
Thus, the digital root of 10 more than 80000006 is 6.

Question 2.
Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.
Answer:
Consider the number = 40
The sequence of numbers by repeatedly adding 11 are 40, 51(40 + 11), 62(51 + 11), 73(62 + 11), 84(73 + 11), 95(84 + 11), 106(95 + 11), 117(106 + 11), 128(117 + 11), 139(128 + 11), etc.
The digital roots of this sequence of numbers are:
40 = 4 + 0 = 4;
51 = 5 + 1 = 6;
62 = 6 + 2 = 8;
73 = 7 + 3 = 10 = 1 + 0 = 1;
84 = 8 + 4 = 12 = 1 + 2 = 3;
95 = 9 + 5 = 14 = 1 + 4 = 5;
106 = 1 + 0 + 6 = 7;
117 = 1 + 1 + 7 = 9;
128 = 1 + 2 + 8 = 11 = 1 + 1 = 2;
139 = 1 + 3 + 9 = 13 = 1 + 3 = 4, …. etc.
Thus, the digital roots of this sequence of numbers are 4, 6, 8, 1, 3, 5, 7, 9, 2, 4, ………..
Observations:
The digital roots are 4. 6, 8, 1, 3, 5, 7, 9, 2, 4, ………..
This sequence starts repeating after 9 steps.
So the digital roots form a cycle: 4, 6, 8, 1, 3, 5, 7, 9, 2, 4, ………..

Question 3.
What will be the digital root of the number 9a + 36b + 13?
Answer:
First Method:
The digital root of the number 9a + 36b + 13
= 9a + 366 + 9 + 4
= 9(a + 4b + 1)+ 4 = 9 + 4 = 13 [∵ The digital root of multiples of 9 is always 9.]
= 1 + 3 = 4
Thus, the digital root of the number 9a + 36b + 13 will be 4.
Second Method:
We have 9a + 36b + 13
Here, a and 6 are integers
Put a = 1, 6 = 1,
9a + 36b + 13 = 9 × 1 + 36 × 1 + 13 = 9 + 36 + 13 = 58
The digital root of 58 = 5 + 8 = 13 = 1 + 3 = 4
Put a = 2, 6 = 3,
9a + 36b + 13 = 9 × 2 + 36 × 3 + 13 = 18 + 108 + 13 = 139
The digital root of 139 = 1 + 3 + 9 = 13 = 1 + 3 = 4
Thus, the expression 9a + 36b + 13 always has a digital root of 4.

Question 4.
Make conjectures by examining if there are any patterns or relations between
(i) the parity of a number and its digital root.
(ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.
Answer:
Consider the pattern: 8, 16, 24, 32, 40, ……….
(i) 8 = 8 = digital root, parity → even
16 = 1 + 6 = 7 = digital root, parity → odd
24 = 2 + 4 = 6 = digital root, parity → even
32 = 3 + 2 = 5 = digital root, parity → odd
40 = 4 + 0 = 4 = digital root, parity → even

(ii) Divided by 3
8 ÷ 3 ⇒ 2, Remainder
24 ÷ 3 ⇒ 0, Remainder
32 ÷ 3 ⇒ 2, Remainder
40 ÷ 3 ⇒ 1, Remainder

Divided by 9
8 ÷ 9 ⇒ 8, Remainder
24 ÷ 9 ⇒ 6, Remainder
32 ÷ 9 ⇒ 5, Remainder
40 ÷ 9 ⇒ 4, Remainder

3. DIGITS IN DISGUISE
Figure it Out (Page 132 – 134) :

Question 1.
If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem.
Answer:
Here, 31z5
Sum of the digits = 3 + 1 + z + 5 = 9 + z (9 + z)
should be divisible by 9.
z = 0, 3105 is divisible by 9.
z = 9, 3195 is also divisible by 9.
∴ z = 0 or 9
There are two answers to this problem because, excluding z, the sum of the digits is divisible by 9.

Question 2.
“I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8”, claims Snehal. Examine his claim and justify your conclusion.
Answer:
A number that leaves a remainder of 8. when divided by 12 : 12k + 8, where k ≥ 1.
Also, another number 4 short of a multiple of 12: 12k – 4

Question 3.
When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.
Answer:
Multiples of 3 are: 3, 6, 9, 12, 15, 18, ………….
3 + 6 = 9, not a multiple of 6.
6 + 9 = 15, not a multiple of 6.
3 + 9 = 12, multiple of 6.
6 + 12 = 18, multiple of 6.
There are two possible cases.
• If both numbers are odd, then the sum is a multiple of 6.
• If both numbers are even, then the sum is a multiple of 6.

Question 4.
Sreelatha says, “I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9”.
(i) Examine if her conjecture is true for any multiple of 9.
(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?
Answer:
Consider a number that is divisible by 9 = 72
We know that,
If the sum of the digits is divisible by 9, then the number is divisible by 9.
If its digits are reversed
27 = 2 + 7 = 9, it is also divisible by 9.
(i) True
(ii) Yes, any other digit shuffle is possible that the number is still a multiple of 9.

Question 5.
If 48a23b is a multiple of 18, list all possible pairs of values for a and b.
Answer:
Given by question,
48a23b is a multiple of 18.
As we know that,
If the number is a multiple of 18, then it is also a multiple of 2 and 9.
∴ 48a23b
Sum of the digits = 4 + 8 + a + 2 + 3 + 5 = 17 + a + 5

Case 1: Put a = 1 and 5 = 0
481230, it is possible values of a and b.
Sum = 18, it is divisible by 9.

Case 2: Put a = 4 and 5 = 6
484236
Sum = 17 + 10 = 27, it is divisible by 9.
Thus, the possible values of a and 6 are a = 1 and b = 0, a = 4 and b = 6; there are two possible cases.

Question 6.
If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
Answer:
Given by question, 3p7q8 is divisible by 44.
As we know, if a number is divisible by 44, then it is also divisible by 4 and 11.
∴ 3p7q8

Case 1: Put p = 1 and q = 0 v
37708 is divisible by 4 and 11, then it is also divisible by 44.

Case 2: Put p = 5 and q = 2
35728 is divisible by 4 and 11, then it is also divisible by 44.

Case 3: Put p = 3 and q = 4
33748 is divisible by 4 and 11, then it is also divisible by 44.

Case 4: Put p = 1 and q = 6
31768 is divisible by 4 and 11, then it is also divisible by 11.
Thus, (p = 7, q = 0), (p = 5, q = 2), (p = 3, q = 4), and (p = 1 and q = 6) are the possible pairs of values for p and q.

Question 7.
Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?
Answer:
Let x, x + 1 and (x + 2) be the three numbers
Put x = 2, ⇒ 2, 3, 4
Put x = 14, ⇒ 14, 15, 6
Put x = 26, ⇒ 26, 27, 28
Put x = 38, ⇒ 38, 39, 40
Thus, the three consecutive numbers are (14, 15, 16),
Put x = 26, ⇒ 26, 27, 28
(26, 27, 28) and (38, 39, 40)
There are infinite numbers, spaced apart by 12.

Question 8.
Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.
Answer:
We know that if a number is a multiple of 36, then it is also a multiple of 4 and 9.
45000
Last two digits = 00, it is divisible by 4.
Sum of the digits = 4 + 5 + 0 + 0 + 0 = 9, it is also divisible by 9.
Thus, 45000 is completely divisible by 36.
The five multiples of 36 between 45,000 and 47,000.
(45,000 + 36), (45,000 + 2 × 36), (45,000 + 3 × 36), (45,000 + 4 × 36) and (45,000 + 5 × 36)
i.e., 45,036, 45,072, 45,108, 45,144, and 45,180.

Question 9.
The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.
Answer:
Given the middle number in the sequence of 5 consecutive even numbers 5p.
The other four numbers in the sequence in terms of p are 5p – 4, 5p – 2, 5p + 2, 5p + 4
Hence, the other four numbers in sequence are p, 3p, 7p and 9p.

Question 10.
Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.
Answer:
We know that if the number is divisible by 3 and 5, then it is also divisible by 15.
Consider the number 643215.
Sum of the digits = 6 + 4 + 3 + 2 + 1 + 5 = 21,
which is divisible by 3.
Thus, 643215 is divisible by 3.
One’s place = 5, it is also divisible by 5.
Hence, 643215 is divisible by 15.
One’s place is not 0, because the digits are reversed, it becomes a 5-digit number.
Lakhs place is always taken as an even number.
Reversed the digits: 512346
One’s place = 6, 512346 is divisible by 2.
Sum of the digits = 5 + 1 + 2 + 3 + 4 + 6 = 21.
It is also divisible by 3.
Hence, 512346 is divisible by 6.

Question 11.
Deepak claims, “There are some b multiples of 11 which, when doubled, are still multiples of 11. But other multiples e of 11 don’t remain multiples of 11 when doubled”. Examine if his conjecture is true; explain your conclusion.
Answer:
The multiples of 11 are: 11, 22, 33, 44, 55, … When doubled, 22, 44, 66, 88, 110, …….
i.e. (11) × 2, 11 × 4, 11 × 6, 11 × 8, 11 × 10, …. are also multiples of 11.
False, if multiples of 11 are doubled, then the multiples of 11 are these numbers.

Question 12.
Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’. Explain your reasoning.
(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9.
(ii) The sum of three consecutive even numbers will be divisible by 6.
(iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6.
(iv) 8(7b – 3) – 4 (11b + 1) is a multiple of 12.
Answer:
(i) Always True,
The multiple of 6 can be written as 6a, where a is an integer.
The multiple of 3 can be written as 36, where 6 s is an integer.
∴ Product = (6a) × (36) = 18(ab) is a multiple of 9. e

(ii) Always True,
The sum of three consecutive even numbers will be divisible by 6.
For example 2 + 4 + 6 = 12, 4 + 6 + 8 = 18, 6 + 8 + 10 = 24, 8 + 10 + 12 = 30,…
These numbers are divisible by 6.

(iii) Always True, because one’s place does not change.

(iv) Sometimes true,
Conclusion:
8(7 × 1 – 3) – 4(11 × 1 + 1) = -16, not divisible by 12. 8(7 × 10 – 3) – 4(4 × 10 + 1) = 536 – 164 = 372, divisible by 12.

Question 13.
Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.
Answer:
Let the three numbers be n₁, n₂, and n₃.
Let their remainders when divided by 3 be r₁, r₂, and r₃.

The sum n₁ + n₂ + n₃ is divisible by 3 if and only if r₁ + r₂ + r₃ is divisible by 3.

Case 1: All remainders are 0.
r₁ = 0, r₂ = 0, r₃ = 0
Sum of remainders = 0 + 0 + 0 = 0, which is divisible by 3.

Case 2: All remainders are 1.
r₁ = 1, r₂ = 1, r₃ = 1
Sum of remainders = 1 + 1 + 1 = 3, which is divisible by 3.

Case 3: All remainders are 2.
r₁ = 2, r₂ = 2, r₃ = 2
Sum of remainders = 2 + 2 + 2 = 6, which is divisible by 3.

Case 4: One remainder is 0, one is 1, and one is 2.
r₁ = 0, r₂ = 1, r₃ = 2 (in any order).
Sum of remainders = 0 + 1 + 2 = 3, which is divisible by 3.
The sum of three numbers is divisible by 3 if and only if all three numbers have the same remainder when divided by 3, or if they all have different remainders when divided by 3.

Question 14.
Is the product of two consecutive integers always multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?
Answer:
Yes, the product of two consecutive integers is always a multiple of 2.
1 × 2 = 2, 2 × 3 = 6, 5 × 6 = 30, 10 × 11 = 110, and so on.
Since we know that multiplying by an odd number and an even number is always an even number.
No, it is not always a multiple of 6.
1 × 2 = 2, 4 × 5 = 20, 7 × 8 = 56
Since it is not divisible by 6.
The product of 4 consecutive integers
2 × 3 × 4 × 5 = 120,
4 × 5 × 6 × 7 = 840,
5 × 6 × 7 × 8 = 1680
We can say that the product of 4 consecutive integers, divisible by 12.

The product of five consecutive integers is:
1 × 2 × 3 × 4 × 5 = 120,
2 × 3 × 4 × 5 × 6 = 720,
3 × 4 × 5 × 6 × 7 = 2520
Hence, we can say that the product of five consecutive integers is always divisible by 24.

Question 15.
Solve the cryptarithms –
(i) EF × E = GGG
(ii) WOW × 5 = MEOW
Answer:
(i) This means a 2-digit number multiplied by 5 gives a 3-digit number.
2-digit number = 20, 21,…., 99
37 × 3 = 111, all conditions are satisfied,

(ii) This means a 3-digit number multiplied by 5 gives 4-digit numbers.
Pick 3-digit number = 200, 201,…., 999
525 × 5 = 2625

Question 16.
Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?

Answer:
(iv) Multiples of 4 are: 4, 8, 12,16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64,…
Multiples of 8 are: 8, 16, 24, 32, 40, 48, 56, 64,….
Multiples of 32 are: 32, 64, 96, 128,…
The Venn diagram captures the relationship between the multiples of 4, 8, and 32 :

Number Play Class 8 Extra Questions

Multiple Choice Questions

Question 1.
Which of the following arithmetic expressions is even?
(a) 119 × 303
(b) (513)³
(c) 708 – 477
(d) 4 × 347 × 3
Solution:
Here, 119 × 303 = odd, as 119 and 303 both have odd parity.
(513)³ = odd, as cube of odd number has odd parity.
708 – 477 = 231, which is odd.
4 × 347 × 3 = even, as 4 has even parity, 347 and 3 have odd parity and thus the product will have even parity.

(d) 4 × 347 × 3

Question 2.
Which of the following arithmetic expression is odd?
(a) 2 × 1037
(b) 24 × 7
(c) 365 × 7
(d) 365 × 24 × 7
Solution:
Here, 2 × 1037 = even, as 2 has even parity and 1037 has odd parity. So, the product will have an odd parity.
24 × 7 = even, as 24 has an even parity and hence its product with 7 having an odd parity will be having an even parity.
In 365 × 24 × 7, 24 has even parity, so the product will have the even parity.
Finally, in 365 × 7, both have odd parity so the product will have the odd parity.
(c) 365 × 7

Question 3.
Which of the following algebraic expressions gives an even number for any integer values for the letter-numbers?
(a) 4a + 3b
(b) 2x – 5y
(c) x² + 2
(d) 2u – 4υ
Solution:
2u – 4υ = 2(u – 2υ), which has even parity.
∴, it will give an even number for any integer value.
(d) 2u – 4υ

Question 4.
Which of the following algebraic expressions give an odd number for any integer values for the letter-numbers?
(a) 2x + 1
(b) 2x + 2
(c) 2x
(d) 2x – 2
Solution:
2x + 1 has odd parity as 2x has even parity while 1 has an odd parity.
If we add an even number with an odd number we get a number whose parity is odd.
(a) 2x + 1

Question 5.
Three consecutive numbers have sum 96. Smallest number amongst them is :
(a) 29
(b) 30
(c) 31
(d) 33
Solution:
Let the three consecutive numbers be a, a + 1 and a+ 2.
Now sum of the numbers
= a + (a + 1) + (a + 2)
= 3a + 3 = 96 (given)
∴, 3a = 96 – 3 = 93
⇒ a = \(\frac{93}{3}\) = 31
∴, smallest number = a = 31
(c) 31

Assertion and Reasoning

(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).
(b) Assertion (A) and Reason (R) both are true but Reason (R) is not the correct explanation for the Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.

Question 1.
Assertion (A) : Algebraic expression 2u – 6v will always give an even number for any integer value for the letter number.
Reason (R) : Difference of two expressions or terms having even parity has an even parity.
Solution:
Here, 2u – 6v = 2(u – 3v), which has even parity. So Assertion (A) is true.
Also, the Reason (R) is true and it explains the truthness of the Assertion (A).
Answer:
(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).

Question 2.
Assertion (A) : 252525 is divisible by 3.
Reason (R) : Any number having 5 at units place digit is divisible by 5.
Solution:
Here, 252525 is divisible by 3 as sum of digits of 252525 = 2 + 5 + 2 + 5 + 2 + 5 = 21, is divisible by 3.
Reason (R) is also true as any number having 5 at its units place is divisible by 5.
But Reason does not explain the divisibility of the number 252525 by 3.
Answer:
(b) Assertion (A) and Reason (R) both are true but Reason (R) is not the correct explanation for the Assertion (A).

Case Based Questions

Question 1.
Aadya was trying to solve some cryptarithms which are as follows :

Based on the above, answer the following:
(a) What is the value of N? Are they more than one?
(b) What is the value of R? How many such values are there?
(c) What is the value of P in the first cryptarithm?
(d) What is the value of P in the second cryptarithm? Is it same as that for the first cryptarithm?
(e) What are the values of 0 and Q?
Answer:
To answer these questions, we first solve the two cryptarithms:

Here we have to consider a two digit number when added thrice to itself gives a two digit number having PO units digit same as the tens digit of the number
The two digit number must be less than 33. We can consider the numbers 17, 24 and 31.
Here,

For the cryptarithm,

We should consider the two digit numbers greater than 33 as we need the sum a three-digit number.
Here, R can be 0 or 5 as in only these two cases the sum will be 0 or 5, when added thrice. If we consider 85, we get

which is the answer for the given cryptarithm.
Now we can answer any question on the given cryptarithms.
(a) N = 7, 4 or 1
They are more than one in number.

(b) R = 5
They are more than one in number.

(c) The value of P in the first cryptarithm is 5, 7 or 9.

(d) The value of P in the second cryptarithm is 2. No. The values of P are different in the two cryptarithms.
(e) O = 1, 2 or 3
Q = 8

Go through BSE Odisha Class 8 Science Solutions Chapter 6 Pressure, Winds, Storms, and Cyclones Question Answer to understand textbook questions more clearly.

Class 8 Science Curiosity Chapter 6 Question Answer

Class 8 Science Ch 6 Pressure, Winds, Storms, and Cyclones Question Answer

Class 8 Science Chapter 6 Pressure, Winds, Storms, and Cyclones Question Answer

Probe and Ponder Questions

Question 1.
Why are winds stronger on some days than on others?
Answer:
Winds are stronger on some days due to greater differences in air pressure (pressure gradients) between different locations. When a weather system, such as storm or a cyclone, creates steep pressure differences, air moves rapidly from high-pressure areas to low-pressure areas, resulting in strong winds. Temperature contrasts (such as between land and sea, or during weather fronts), and local topography can intensify these differences and wind speeds.

Question 2.
Why are water tanks usually placed at a height?
Answer:
Water tanks are placed at a height so that water can flow down easily with pressure due to gravity, ensuring a steady supply to all parts of a building or area without needing extra pumping.

Question 3.
Can air pressure really crush us?
Answer:
No, air pressure cannot crush us because the pressure inside our bodies is equal to the air pressure outside. The atmosphere presses on us with great force, but our body fluids and gases push outward with the same amount of pressure, keeping everything balanced. That’s why we don’t feel the weight of the air or get crushed by it under normal conditions on Earth.

Question 4.
What causes storms and cyclones? If the Earth stopped rotating, would cyclones still form?
Answer:
Storms and cylones are formed when warm, moist air from the sea rises and cools, causing strong winds and heavy rain. The spinning of the Earth makes these winds rotate, forming a cyclone. If the Earth stopped rotating, there would be no spinning effect, so cyclones would not form, though normal storms could still happen.

Question 5.
Share your questions ………………
Answer:
Possible questions you might ask after exploring these ideas:

• Why don’t we feel atmospheric pressure even though it is so high?
• How do buildings and bridges withstand strong winds during storms?
• Do animals sense changes in air pressure before storms?
• What scientific instruments are used to measure wind speed and pressure?
• How do disaster warning systems work for cyclones?

InText Questions

Question 1.
Why do fall leaves rise in the air or trees bend when a strong wind blows? (Page 81)
Answer:
The force exerted by the wind creates wind pressure, which causes fallen leaves to rise in the air and the bending or swinging of trees when a strong wind blows.

Question 2.
Can the shape or size of the straps make a difference, provided both bags are equally heavy? (Page 81)
Answer:
When we carry a bag, we feel its weight because of the force of gravity acting on our shoulders. The weight of the bag with narrow straps acts on a smaller area of our shoulders, whereas the weight of the bag with broad straps is spread over a larger area of our shoulders. Although both bags have equal weight, we feel more comfortable carrying a bag with broad straps. Broad straps reduce the pressure exerted by the bag on the shoulders.

Question 3.
Do liquids also exert pressure? (Page 83)
Answer:
Yes, liquids also exert pressure. The pressure from a liquid acts in all directions and increases with depth – the deeper you go, the greater the pressure.

Question 4.
What will happen to the bulge of the balloon if we increase the height of the water column? (Page 84)
Answer:
If we increase the height of the water column, the bulge of the balloon will become bigger because the pressure of the water increases with height, pushing more strongly on the balloon.

Question 5.
(a) Why are overhead tanks kept at a height?
(b) Suppose you are living on the second floor of a three-story building and an overhead tank is placed on the top floor. Will you or your friend on the first floor receive a more powerful stream of tap water? Give reasons. (Page 84)
Answer:
(a) Overhead tanks are placed at a height so that the pressure in the taps is increased, resulting in a good stream of water from the taps.

(b) Pressure exerted by water increases with the increase of height of the liquid column. Suppose I live on the second floor of the building that has three floors, and the overhead tank is installed on the top floor. The pressure of water and hence the power of water stream from the taps will depend on the height of the overhead tank above the taps of the second floor. My friend living on the first floor will have a longer distance from the taps to the overhead tank, which means the height of the water column above his taps will be more compared to my water taps on the second floor. Therefore, my friend will receive more powerful stream of tap water.

Question 6.
Why does water spurt out like a fountain from leaking joints or holes in water pipes? (Page 85)
Answer:
Water in water pipes has long water columns, as the water tanks to which the water pipes are connected are placed at heights. The pressure inside the pipes is high. The water exerts pressure in all sides of the container including the bottom and walls of the pipe. When this water exerting pressure on all sides, finds a narrow opening like a hole or a leaking joint, it spurts like a fountain.

Question 7.
What happens when an inflated balloon is kept without closing its mouth? (Page 86)
Answer:
The inflated balloon has air inside, which exerts pressure on the walls of the balloon, resulting in expansion of the balloon on all sides. The elastic walls of the balloon exert equal but opposite pressure on the air. When we keep the inflated balloon without closing its mouth, the air from within the balloon escapes through its mouth from high pressure inside the balloon to low pressure outside the balloon.

Question 8.
Does the difference in air pressure have anything to do with the formation of winds? (Page 88)
Answer:
Air moves from a high-pressure region to a low-pressure region. Moving air is called wind. Thus, it is the difference in air pressure that results in the formation of wind.

Pressure, Winds, Storms, and Cyclones Class 8 Questions and Answers

Keep the Curiosity Alive (Pages 94-96)

Question 1.
Choose the correct statement.
(i) Look at figure carefully. Vessel R is filled with water. When pouring of water is stopped, the level of water will be ………………

(a) the highest in vessel P
(b) the highest in vessel Q
(c) the highest in vessel R
(d) equal in all three vessels
Answer:
(d) equal in all three vessels
Reason: The level of water in connected vessels (communicating vessels) will be the same regardless of shape, as liquid pressure depends on the height of the column, not the vessel’s shape or width.

(ii) A rubber sucker (M) is pressed on a flat smooth surface and an identical sucker (N) is pressed on a rough surface:
(a) Both M and N will stick to their surfaces.
(b) Both M and N will not stick to their surfaces.
(c) M will stick but N will not stick.
(d) M will not stick but N will stick.
Answer:
(c) M will stick but N will not stick.
Reason: The rubber sucker sticks due to atmospheric pressure creating a vacuum on a smooth surface. On a rough surface, air leaks in, preventing the vacuum and thus the sticking.

(iii) A water tank is placed on the roof of a building at a height ‘H’. To get water with more pressure on the ground floor, one has to
(a) increase the height ‘H’ at which the tank is placed.
(b) decrease the height ‘H’ at which the tank is placed.
(c) replace the tank with another tank of the same height that can hold more water.
(d) replace the tank with another tank of the same height that can hold less water.
Answer:
(a) increase the height ‘ H ‘ at which the tank is placed.
Reason: Liquid pressure increases with the height of the water column. Raising the tank increases the pressure, resulting in a stronger stream of water.

(iv) Two vessels, A and B contain water up to the same level as shown in figure. P_A and P_B is the pressure at the bottom of the vessels. FA and FB is the force exerted by the water at the bottom of the vessels A and B.

(a) P_A=P_B, F_A=F_B
(b) P_A=P_B, F_A<F_B
(c) P_A<P_B, F_A=F_B
(d) P_A>P_B, F_A>F_B
Answer:
(b) P_A=P_B, F_A<F_B
Reason: Pressure at the bottom depends on the height of the water, which is the same in both vessels, so PA=PB. Force equals pressure times area; since vessel A is narrower (smaller area), FA<FB.

Question 2.
State whether the following statements are True [T] or False [F].
(i) Air flows from a region of higher pressure to a region of lower pressure.
Answer:
True.

(ii) Liquids exert pressure only at the bottom of a container.
Answer:
False (liquids exert pressure in all directions, including on the walls of the container.)

(iii) Weather is stormy at the eye of a cyclone.
Answer:
False (it is calm at the eye of a cyclone)

(iv) During a thunderstorm, it is safer to be in a car.
Answer:
True.

Question 3.
Figure (a) shows a boy lying horizontally, and Figure (b) shows the boy standing vertically on a loose sand bed. In which case does the boy sink more in sand? Give reasons.

Answer:
The boy in Figure (b) will sink more. The weight of the boy is the same, but in Fig. (a), the force (of the weight) is acting on a large area. The pressure in this case is less. In Fig. (b), the same force (of the weight) is acting on a small area. The pressure, therefore, is more. The boy will sink more into the sand in this case.

Question 4.
An elephant stands on four feet. If the area covered by one foot is 0.25 m², calculate the pressure exerted by the elephant on the ground if its weight is 20000 N.
Answer:
Force (the weight) of the elephant acting on the ground =20000 N
Area on the ground covered by the four feet of the elephant =4 × 0.25 m² = 1 m²
Pressure exerted by the elephant on the ground
= \(\frac{\text { Force }}{\text { Area }}=\frac{200000 \mathrm{~N}}{1 \mathrm{~m}^2}\)=20000 Pa

Question 5.
There are two boats, A and B. Boat A has a base area of 7 m², and 5 persons are seated in it. Boat B has a base area of 3.5 m², and 3 persons are seating in it. If each person has a weight of 700 N, find out which boat will experience more pressure on its base and by how much?
Answer:
Force of the weight of 5 persons acting on the base of boat A=5 × 700 N=3500 N
Base area of boat A=7 m²
Pressure exerted on the base of boat A
\(= \frac{\text { Force }}{\text { Area }}=\frac{3500 \mathrm{~N}}{7 \mathrm{~m}^2}\) = 500 Pa
Force of the weight of 3 persons acting on the base of boat B = 3 ×700 N=2100 N
Base area of boat B = 3.0 m²
Pressure exerted on the base of boat B
= \(\frac{\text { Force }}{\text { Area }}=\frac{2100 \mathrm{~N}}{3.5 \mathrm{~m}^2}\) =600Pa
Therefore, boat B will experience more pressure on its base by 100 Pa.

Question 6.
Would lightning occur if air and clouds were good conductors of electricity? Give reasons for your answer.
Answer:
In case the air and the clouds were good conductors of electricity, the charges could not accumulate in the clouds (because they would flow into the air), there would be no charge buildup, which is necessary for lightning to occur. Therefore, if air and clouds were good conductors of electricity, lightning would not occur.

Question 7.
What will happen to the two identical balloons A and B as shown in figure when water is filled into the bottle up to a certain height. Will both the balloons bulge? If yes, will they bulge equally? Explain your answer.

Answer:
When water is filled into the bottle up to a certain height (sufficiently above the level of the entry points of water from the bottle) both the balloons will bulge. The entry points of water from the bottle to the balloons are at the same height. The balloons, being elastic, exert some force on the water. Assuming that the balloons are equally elastic, both balloons will bulge equally.

Question 8.
Explain how a storm becomes a cyclone.
Answer:
Cyclones are large storms that form over warm ocean waters.

• As the ocean water gets heated, the air above it becomes moist and warm and rises to a height where water vapor condenses to form raindrops.
• Condensing water vapor releases heat back into the atmosphere.
• This further warms the ascending air, leading to its further rise, creating an even lower pressure.
• Air from the surrounding regions rushes in, and it also starts rising.
• The moving air starts to spin under the influence of the Earth’s rotation.
• This cycle is repeated, resulting in the creation of a very low-pressure area with high-speed winds revolving around it.
• This spinning system of clouds, winds, and rain is called a Cyclone.

Question 9.
The figure shows trees along the sea coast in a summer afternoon. Identify which side is land A or B. Explain your answer.

Answer:
During the daytime on a summer afternoon, there is a sea breeze that blows from sea to land. This happens because the land gets heated faster and the air above land rises, resulting in a low-pressure region over land. Cooler air over sea moves from the high-pressure region towards the land. The bending of trees due to wind, as shown in the figure, suggests that wind is blowing in the direction from B to A. This suggests ‘A’ side is land.

Question 10.
Describe an activity to show that air flows from a region of high pressure to a region of low pressure.
Answer:
Activity to show that air flows from a region of high pressure to a region of low pressure.
Materials required: Two similar balloons made of thin rubber, a drinking straw, and some thread.

Procedure:

• Insert one end of the straw into one balloon and secure it with the thread.
• Inflate the second balloon and insert the free end of the straw into the neck of the inflated balloon, and secure it with the thread. (Ensure that the air from the inflated balloon does not leak.)

Observations: Some air moves from the inflated balloon to the uninflated balloon, and the sizes of both balloons change. After some time, both the balloons attain almost the same size, and the flow of air stops.

Conclusion: The inflated balloon has higher pressure inside it, and the uninflated balloon has low pressure inside. When the two balloons are connected through a straw, air moves from the high-pressure area (inside the inflated balloon) to the low-pressure area (inside the uninflated balloon).

Question 11.
What is a thunderstorm? Explain the process of its formation.
Answer:
(a) A Thunderstorm is a storm that produces thunder, lightning, heavy rain, and strong winds. It usually occurs during hot weather when the atmosphere becomes unstable.
(b) During a thunderstorm formation, under certain conditions, warm air rises to great heights, and the low temperature there changes water droplets into ice particles. Strong winds blowing upwards and downwards result in rubbing between water droplets and ice particles. This generates electric charges within clouds. Ice particles are positively charged, and they move upwards in the upper part of the clouds. Water droplets are negatively charged and occupy the lower part of the clouds.

When negatively charged water droplets in the lower part of the cloud move closer to the ground, trees, buildings, and the ground become positively charged. Normally, air acts as an electrical insulator and does not let opposite charges meet. This insulating property of the air breaks down when the build-up charges becomes very large.

A sudden flow of charges takes place, resulting in a bright flash of light called lightning. Lightning can occur as opposite charges collide within a cloud, between clouds, or between clouds and the ground. Lightning rapidly heats the air around it. This results in expansion of air to produce a loud sound called thunder. A storm accompanied by lightning and thunder is called a thunderstorm.

Question 12.
Explain the process that causes lightning.
Answer:
During a thunderstorm formation, under certain conditions, warm air rises to great heights, and the low temperature there changes water droplets into ice particles. Strong winds blowing upwards and downwards result in rubbing between water droplets and ice particles. This generates electric charges within clouds. Ice particles are positively charged, and they move upwards in the upper part of the clouds. Water droplets are negatively charged and occupy the lower part of the clouds. When negatively charged water droplets in the lower part of the cloud move closer to the ground, the trees, buildings, and the ground become positively charged.

Normally, air acts as an electrical insulator and does not let opposite charges meet. This insulating property of the air breaks down when the build-up charges becomes very large. A sudden flow of charges takes place, resulting in a bright flash of light called lightning. Lightning can occur as opposite charges collide within a cloud, between clouds, or between clouds and the ground. Lightning rapidly heats the air around it. This results in expansion of air to produce a loud sound called thunder.

Question 13.
Explain why holes are made in banners and hoardings.
Answer:
Holes are made in banners and hoardings to save them from blowing off with high-speed winds. High-speed winds are accompanied by a reduced pressure. If there are no holes in the banners or hoardings, they block the wind, but the wind blows on the sides of these banners and hoardings. This forms low-pressure area on sides and opposite side of the banners/hoardings.

There remains a high-pressure area on the side of the banner or hoarding that faces the direction from which wind is blowing. When the pressure difference is large the banners or the hoardings are blown away. With the holes in the banners or hoardings, the wind blows through these holes, and the pressure difference is minimised, keeping them intact.

Class 8 Science Chapter 6 Question Answer

Activity 1.

Aim: To show that the pressure exerted by water at the bottom of the container depends on the height of its column.
Materials Required: Two rubber balloons, glass or plastic pipes of different diameters.

Procedure:

Case I: When two pipes are of different diameters.

• First of all we will take two transparent glass or plastic pipes of the same length (about 25 cm), these pipes must be of different diameters, as shown in fig.
• Now, we will take two good-quality rubber balloons and attach them to one end of each pipe.
• Clamp the pipes on a stand as shown in fig. (1)
• Now, fill both the pipes with water up to the same level about halfway.

Case II: When two pipes are of same diameters.

• Take two pipes of same length and same diameter.
• Now, pour water in both the pipes.
• Water should be at different leves in the pipes. See fig. (2) Note down your observations.

Observations:

• When the water levels in both pipes are same. Two balloons bulge to the same extent.
• Higher heights of water column produce bigger bulge of the balloon.

Inferences:

• More height of water column produce bigger buldge (i.e., bulge of balloon increases as the height of water column increases).
• Bulging of balloons do not depends on the diameters of the pipes. It depends on height of water column.
• Hence, the pressure due to liquid increases with height of water column.

Activity 2.

Let us find out
Aim: To show that a liquid exert pressure on the walls of the container and also equal pressure at the same depth.
Materials Required: An empty plastic bottle, water, a needle or a nail.

Procedure:

• Take an empty plastic bottle.
• Drill four holes near the bottom of the bottle using a needle or a nail.
• Now, seal the holes with a tape and fill the bottle with water.
• Make sure that the holes are at the same height from the bottom.
•  Remove the tape from all holes at the same time.

Observations:

• Water flowing out from the holes on the side of the bottle.
• Different streams of water coming out of the holes falls at the same distance from the bottle.

Inference:

• Therefore, we can conclude that liquids exert pressure not only at the bottom of the container, but also on its sides. In fact liquids exert pressure in all directions.
• Liquid exert equal pressure at same depth.

Activity 3.

Let us explore
Aim: To show that air exert pressure.
Materials Required: A paper plate, two identical chart paper of size about 70 cm × 56 cm

Procedure:

• First of all we will take a paper plate and invert it and attach a stick to it see fig. (a).
• Now place it on a plain wooden surface.
• Then, take two sheets of chart paper of same size 70cm × 56cm each. Fold one sheet twice and make a hole in the centre of the folded chart paper sheet – big enough for the stick to come out. Place the folded sheet on top of the inverted paper plate as see fig. (b).
• Now, try to lift the paper plate covered with a folded sheet using the stick.
• What did you observed ? How much effort is needed to lift it.
• Now, put the second unfolded chart paper sheet in place of the folded sheet. Make a hole at the centre of this chart paper for the stick to pass through. Cover the paper plate with the unfolded chart paper as shown in fig. (c).
• Try to lift the paper plate once again and feel the effort needed to do so.
• Which of the above case, with the folded or the unfolded is easy to lift, either chart paper covering the paper plate.

Observations:

• We feel more effort to lift the paper plate when it is covered with the unfolded chart paper, than with the folded chart paper.
• The larger the surface area covered, the greater the force needed to life the plate.

Inference:

• Air exert force on the covering sheet.
• Force increases with increase in the area of covering sheets.
• It is clear that air is exerting a force on the paper plate, which increases as the area of the sheet covering it increases.

Activity 4.

Let us perform
Aim: To show the effect of pressure due to air.
Materials Required: A good quality rubber sucker.

Procedure:

• Take a good quality rubber sucker. It looks like a small rubber cup.
• Press it hard on a smooth, plane surface.
• Does it stick to the surface ? Yes, it sticks to the surface very hard.
• Now, try to pull it off the surface. Do you find it difficult to pull it off ?

Observations: It is very difficult to pull off a rubber sucker.

Inference:

• All gases exert pressure.
• The sucker sticks to the surface because the pressure of air surrounding the sucker is higher than the pressure exerted by the air inside the sucker.

Activity 5.

Let us observe
Aim: To show that air (winds) blows untill there is a pressure difference. (i.e., from a high pressure region to low pressure region)
Materials Required: Two balloons of same size, a drinking straw, thread.

Procedure:

• First of all we will take two balloons made of thin rubber of same size and a drinking straw.
• Now, we will insert one end of the straw into one balloon and tie its mouth properly with a rubber band or thread.
• Now, we will inflate the second balloon and hold its mouth with our fingers, in such away that air does not escape.
• Now, insert the free end of the straw into the neck of the inflated balloon and tie it with a rubber band or thread.
• We should take precaution at the time of insertion of straw that there is no leakage of air from the balloons.
• Now, we can see in fig. that one end of the straw is inside the inflated balloon and the other end inside the uninflated balloon.

Observations:

• Size of uninflated balloon increases and that of inflated balloon decreases.
• After sometime both the balloons attain almost the same size and flow of air stops.

Inference:

• We can conclude that the air pressure in the inflated balloon is higher than that in the uninflated balloon. So, air moves from the inflated balloon to the uninflated balloon, resulting in changes in the size of these two balloons.
• The air flow stops when the pressure in both balloons becomes equal.
• Hence, it shows that air moves or blows untill there is a pressure difference.

Activity 6.

Let us observe

Aim: To show that high speed winds are due to a reduced air pressure.
Materials Required: Two balloons of same size, string or thread, a stick.

Procedure:

• Take two balloons of the same size.
• Now, inflate both balloons and tie strings to them.
• Take a stick and hang the two balloons from it.
• We must leave a gap of 6-10 cm (See fig.).
• Now, blow air with your mouth into the narrow space between the balloons.
• What happens to the balloons? Note down your observations.
• Now blow harder and observe.

Observations:

• Both balloons move towards each other.
• When we blow harder the balloons approaches towards each other with more speed.

Inference:

• When you blow air between the balloons, a low pressure area is created between them.
• We conclude that high speed winds are accompanied by a reduced air pressure.

Pressure, Winds, Storms, and Cyclones Class 8 Extra Questions and Answers

Short Answer Type Questions

Question 1.
Why school bags have wide straps ?
Answer:
A school bag has wide strap so that the weight of a bag may fall over a large area of the shoulder of the child producing less force on the shoulder. Due to less pressure, it is more comfortable to carry the heavy school bag.

Question 2.
Why is the bottom part of the foundation of a building made wider ?
Answer:
Foundation of a building is made wider so that it may not sink under the extremely high pressure of building. The wider foundation distributes the weight of the building over a large area on the ground.

Question 3.
Why cutting instruments are sharpened?
Answer:
All cutting instruments such as blades, axes etc., are sharpened so that the area of cross-section decreases and hence pressure exerted by them increases. Thus they can easily cut and penetrate a given surface.

Question 4.
Calculate the pressure if a force of 8 N is applied on an area of 2 cm²
Answer:
Force =8 N
Area =2cm²
[ ∵ 1cm²= \(\frac{1}{10,000}\)=\(\frac{2 \mathrm{~m}^2}{10,000}\)
Pressure = \(\frac{8}{2 / 10,000}=40,000 \mathrm{~N} / \mathrm{m}^2 \)

Question 5.
Why a sharp knife cuts better than a blunt knife?
Answer:
A sharp knife cuts objects (like vegetables) better because due to its very thin edge, the force of our hand falls over a very small area of the object producing a large pressure. This large pressure cuts the object easily. On the other hand, a blunt knife has a thicker edge. A blunt knife does not cut an object easily because due to its thicker edge, the force of our hand falls over a larger area of the object and produces lesser pressure. This lesser pressure cuts the object with difficulty.

Question 6.
Why does water spurt out like a fountain from leaking joints or holes in water pipes?
Answer:
Water spurting out like a fountain from holes because:

• Water pipes are connected to tank at certain height by a long pipes and create a long water column and the pressure inside the pipes is high.
• Liquid exert pressure not only at the bottom of the container, but also on its side, i.e., liquid exert pressure in all directions. Due to the pressure exerted by water on the walls of the pipes we have seen water spurting out like a fountain from leaking joints or holes in water pipes.

Question 7.
How does rubber sucker stick to the surface like wall ?
Answer:
The sucker sticks to the surface because the pressure of air surrounding the sucker is higher than the pressure exerted by the air inside the sucker. To pull the sucker off the surface, the applied force should be strong enough to overcome the pressure difference between outside the sucker and inside the sucker.

Question 8.
Why do mountainers suffer from nose-bleeding at higher altitudes?
Answer:
This is because, at higher altitude, the atmospheric pressure suddenly drops. This leads to the rapture of blood vessels in the body causing bleeding from the nose.

Question 9.
How do thunderstorms produce lightning through charge separation?
Answer:
Strong updrafts and downdrafts make water droplets and ice rub and become charged, with opposite charges separating within the cloud. When the charge difference becomes very large, a sudden discharge occurs as lightning.

Long Answer Type Questions

Question 1.
Why it is easier to walk on soft sand if we have flat shoes rather than shoes with sharp heels (or pencil heels) ?
Answer:
This is because a flat shoe has a greater area in contact with the soft sand due to which there is less pressure on the soft ground. Due to this the ‘flat’ shoes do not sink much in soft sand and it is easy to walk on it.

On the other hand, a sharp heel has a small area in contact with the soft sand and so exerts a greater pressure on the soft sand. Due to this greater pressure, the sharp heels tend to sink deep into soft sand making it difficult for the wearer to walk on soft sand.

Question 2.
It is difficult to cut cloth using a pair of scissors with blunt blades. Explain.
Answer:
We know that more area of contact will produce less pressure and vice-versa. Blunt blades have larger area as compared to the sharp-edged blades. Thus, the applied force produces a lower pressure in case of blunt blades, which makes it difficult to cut the cloth.

Question 3.
Two rods of the same weight and equal length have different thickness. They are held vertically on the surface of sand as shown in figure. Which one of them will sink more? Why?
Answer:
Since rod B is thinner than ‘A’. Therefore, rod B will go more deeper as it has a smaller area of contact, therefore the same force (weight of the rod) will produce more pressure. But in case of rod A the same force produces less pressure.

Question 4.
Two women are of the same weight. One wears sandals with pointed heels while the other wears sandals with flat soles. Which one would feel more comfortable while walking on a sandy beach? Give reasons for your answer.
Answer:
Two women are of the same weight but the woman wearing sandals with flat soles will feel more comfortable while walking on the sandy beach. Because, the flat soles have larger area as compared to the sandals with pointed heels. Here, the two women are of the same weight, they will apply same force on the ground.

Hence, the pressure exerted by the pointed heels will be more compared to that with sandals having flat soles. So, the pointed heel sandals will sink more in the sand than the flat sole sandals. Hence, walking with flat sole sandals will be more comfortable.

Question 5.
It is much easier to burst an inflated balloon with a needle than by a finger. Explain.
Answer:
Smaller area produces large pressure. When we prick the surface of an inflated balloon with a needle it exerts a large pressure because it has a smaller area of contact compared to the finger. So, the large pressure pierces the surface of the balloon easily.

Case-Study Based Questions

1. Read the following passage carefully and answer the questions that follow:

Our earth is surrounded by a large amount of air consisting of Nitrogen, Oxygen, Argon, Carbon dioxide and traces of other gases. Layers of air surrounding the earth are called the atomsphere.

(a) What is atmosphere?
Answer:
The layers of air surrounding the earth are called the atmosphere.

(b) What is atmospheric pressure?
Answer:
The pressure exerted by air on the surface of the earth is known as the atmospheric pressure.

(c) Is the pressure inside our bodies equal to the atmospheric pressure? Why are we not crushed?
Answer:
Yes, the pressure inside our bodies balances the outside pressure of the atmosphere. That is why we are not crushed under the atmospheric pressure.

Picture Based Questions

I. Observe the picture and answer the questions.

(i) Do the different streams of water coming out of the holes fall over it the same distance from the bottle?
(a) Different distance
(b) Same distance
(c) Two holes same distance
(d) None of these
Answer:
(b) Same distance

(ii) Does liquid exert equal pressure on the same depth?
(a) No
(b) Some times
(c) Yes
(d) 50-50
Answer:
(c) Yes

(iii) What does this indicate?
Answer:
This indicates that liquids exert equal pressure at the same depth.

Pressure, Winds, Storms, and Cyclones Class 8 MCQ

Multiple Choice Questions (Mcqs)

Question 1.
What is the SI unit of pressure ?
(a) N/m²
(b) Newton (N)
(c) N/m
(d) m²
Answer:
(a) N/m²

Question 2.
Bulging of balloons depends on :
(a) height of water or liquid column
(b) diameter of pipes
(c) weight of liquid or water
(d) none of these
Answer:
(a) height of water or liquid column

Question 3.
The sucker sticks to the surface because :
(a) the pressure of air surrounding the sucker is lower than the pressure exerted by air inside the sucker.
(b) the pressure of air is equal to the pressure inside the sucker.
(c) the pressure of air outside is higher than the air pressure inside the sucker
(d) none of these
Answer:
(c) the pressure of air outside is higher than the air pressure inside the sucker

Question 4.
1 hectopascal (hpa) is equal to :
(a) 10 Pa
(b) 50 Pa
(c) 100 Pa
(d) 1000 Pa
Answer:
(c) 100 Pa

Question 5.
Figure shows a container filled with water. Which of the following statements is correct about pressure of water?
(a) Pressure at A> Pressure at B> Pressure at C
(b) Pressure at A= Pressure at B= Pressure at C
(c) Pressure at A<Pressure at B>Pressure at C
(d) Pressure at A<Pressure at B<Pressure at C
Answer:
(d) Pressure at A<Pressure at B<Pressure at C

Assertion and Reasoning

These questions consist of two statements, each printed as Assertion (A) and Reason (R). While answering these questions, you are required to choose any one of the following four responses.
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.
(b) Assertion (A) and Reason (R) both are correct and reason is not correct explanation for assertion.
(c) Assertion (A) is correct but the Reason (R) is wrong.
(d) Assertion (A) is wrong but the Reason (R) is correct.

1. Assertion (A): We feel more comfortable carrying a bag with broad straps as compared to bag of the same weight with narrow straps.
Reason (R): Broad straps reduce the pressure exerted by the bag on our shoulders as compared to narrow straps.
Answer:
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.

2. Assertion (A): The base of the dam is kept narrower than the top to keep the dam strong to hold a huge amount of water.
Reason (R): The water stored in the dam exerts pressure horizontally on the side walls of the dam and vertically on the floor due to the height of the water level. The pressure that acts horizontally is very large near its bottom.
Answer:
(d) Assertion (A) is wrong but the Reason (R) is correct.

Fill in the blanks

1. Force acting on a area is called …………
Answer:
pressure

2. The pressure exerted by a liquid ………… with depth.
Answer:
increases

3. Overhead water tanks are placed high to increase water ………… in the pipes.
Answer:
pressure

4. 1 Pascal (Pa) equals 1 ………… per square metre.
Answer:
Newton

5. Liquid exert pressure at the bottom and also on the ………… of a container.
Answer:
walls

True or False

1. The pressure exerted by a liquid depends on the area of a base of its container.|
Answer:
False

2. A drinking straw works on the pressure exerted by the liquid filled in a soft drink bottle in which it is placed.
Answer:
False

3. Atmospheric pressure decreases with altitude.
Answer:
True

4. A liquid exerts presure either in downward direction or sideways, but not in upward direction.
Answer:
False

5. When the height of the liquid column in a tall jar is doubled, the pressure at the bottom also gets doubled.
Answer:
True

Students can refer to BSE Odisha Class 8 Math Solution and Ganita Prakash Chapter 4 Quadrilaterals Class 8 Question Answer to understand textbook questions step by step.

Class 8 Maths Chapter 4 Quadrilaterals Solutions

Ganita Prakash Class 8 Chapter 4 Solutions

Class 8 Maths Ganita Prakash Chapter 4 Solutions Quadrilaterals

1. RECTANGLES AND SQUARES
Figure it Out : Page : 94

Question 1.
Find all the other angles inside the following rectangles.

Solution:

∠1 + ∠9 = 90° …. (All corner angles of a rectangle are 90°)
∠1 + 30° = 90°
∠1 = 90° – 30°
∠1 = 60°
∠1 = ∠5 = 60° … (Alternate interior angles)
∠9 = ∠4 = 30° … (Alternate interior angles)
In ΔAOB, OA = OB, then the angles opposite them are equal
∴ ∠9 = ∠7 = 30°
∠7 = ∠3 = 30° … (Alternate interior angles)
In ΔAOD, OA = OD, then the angles opposite them are equal
∴ ∠2 = ∠1 = 60°
∠2 = ∠6 = 60° … (Alternate interior angles)
In ΔAOB
∠9 + ∠7 + ∠AOB = 180° … (Sum of angles of a triangle)
30° + 30° + ∠AOB = 180°
60° + ∠AOB = 180°
∠AOB = 180° – 60°
∠AOB = 120°
∠AOB = ∠COD = 120° … (Vertically opposite angles)
∠AOB + ∠AOD = 180° … (Linear pair)
120° + ∠AOD = 180°
∠AOD = 180° – 120°
∠AOD = 60°
∠AOD = ∠BOC = 60° … (Vertically opposite angles)
Thus, ∠1 = ∠5 = ∠2 = ∠6= ∠AOD = ∠BOC = 60°.
∠AOB = ∠COD = 120°.
∠9= ∠4 = ∠7 = ∠3 = 30°.

(ii) The given rectangle is PSRQ.

∠POS = ∠ROQ = 110° … (Vertically opposite angles)
∠POS + ∠POQ = 180° … (Linear Pair)
110°+∠POQ = 180°
∠POQ = 180° – 110°
∠POQ = 70°
∠POQ = ∠SOR = 70° … (Vertically opposite angles)
In ΔPOS, OP = OS, then the angles opposite them are equal.
∴ ∠1 = ∠2 = a
In ΔPOS,
∠1 + ∠2 + ∠POS = 180° … (Sum of all angles of a triangle)
a + a + 110° = 180°
2a = 180°- 110°
2a = 70°
a = 35°
∠1 = ∠2 = a = 35°
∠1 = ∠5 = 35° …. (Alternate interior angles)
∠2 = ∠6 = 35° … (Alternate interior angles)
Since ABCD is a rectangle, ∠P = 90°
∠9 = ∠1 + ∠8
90° = 35° + ∠8
∠8 = 90° – 35°
∠8 = 55°
∠8 = ∠4 = 55° …. (Alternate interior angles)
In ΔPOQ, OP = OQ, then the angles opposite to them are equal
i. e. ∠7 = ∠8 = 55°
∠7 = ∠2 = 55° … (Alternate interior angles)
Thus, ∠POS = ∠ROQ = 110°.
∠POQ = ∠SOR = 70°.
∠1 = ∠2 = 5 = ∠6 = 35°.
∠3 = ∠4 = ∠7 = ∠8 = 55°.

Question 2.
Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of
(i) 30°
(ii) 40°
(iii) 90°
(iv) 140°
Solution:
(i) Draw a line AB equal to 8 cm.
Take point M on AB such that AM = BM = 4 cm.
Using a protractor, draw an angle of 30° at M on MB. On this line, take points C and D such that MC = MD = 4 cm.
Join AD, DB, BC, and CA.
ABCD is the required quadrilateral.

Since diagonals AB and CD are equal and are bisecting each other at M, ACBD is a rectangle.

(ii) Draw a line AB equal to 8 cm.
Take point M on AB such that AM = BM = 4 cm.
Using a protractor, draw an angle of 40° at M on MB.
On this line, take points C and D such that MC = MD = 4 cm.
Join AD, DB, BC, and CA.
ABCD is the required quadrilateral.

Since diagonals AB and CD are equal and are bisecting each other at M, ACBD is a rectangle.

(iii) Draw a line AB equal to 8 cm.
Take a point M on AB such that AM = BM = 4 cm.
Using a protractor, draw an angle of 90° at M on MB.
On this line, take points C and D such that MC = MD = 4 cm.
Join AD, DB, BC, and CA.
ACBD is the required square.

Since diagonals AB and CD are equal and are bisecting each other at M, and also the diagonals are perpendicular to each other, ACBD is a square.

(iv) Draw a line AB equal to 8 cm.
Take a point M on AB such that AM = BM = 4 cm.
Using a protractor, draw an angle of 140° at M on MB.
On this line, take points C and D such that MC = MD = 4 cm.
Join AD, DB, BC, and CA.
ACBD is the required quadrilateral.

Since diagonals AB and CD are equal and are bisecting each other at M, ACBD is a rectangle.

Question 3.
Consider a circle with centre O. Line segments PL and AM are two perpendicular diameters of the circle. What is the figure APML? Reason and/or experiment to figure this out.
Solution:
In the figure, PL and AM are two perpendicular diameters of the circle. Let r be the radius of the circle.

∴ PL = PO + OL
= r + r
= 2 r
and AM = AO + OM
= r + r
= 2 r
∴ PL = AM
∴ In the quadrilateral
APML, diagonals PL and
AM are equal and are perpendicular to each other.
Also, OP = OA = OL = OM = r
∴ Diameters PL and AM bisect each other at 0.
∴ Quadrilateral APML is a square.

Question 4.
We have seen how to get 90° using paper folding. Now, suppose we do not have any paper but two sticks of equal length, and a thread. How do we make an exact 90° using these?
Solution:

• Let AB and CD be two sticks of equal length, say 6 cm.
• Mark the midpoints of the sticks using a ruler.
• Fix a screw to the sticks at their midpoints.
• Using a thread, measure distances AD and BD.
• Keep on moving the sticks about the screw, so that the distances AD and BD are equal.
• In this position, fix the sticks by tightening the screw.
• The new positions of the sticks are shown in the figure.
• The pieces of thread along AD and BD.
  
  Consider ΔAMD and ΔBMD.
  We have AM = BM, AD = BD and MD is common
  ∴ By the SSS condition,
  ΔAMD and ABMD are congruent.
  ∴ ∠AMD = ∠BMD
  Also ∠AMD + ∠BMD = 180° (Linear angles)
  ∴ ∠AMD + ∠AMD = 180°
  ⇒ 2 ∠AMD = 180°
  ⇒ ∠AMD = 90°
  ∴ ∠AMD = ∠BMD = 90°
  ∴ Angle between the sticks is 90°.

Question 5.
We saw that one of the properties of a rectangle is that its opposite sides are parallel. Can this be chosen as a definition of a rectangle? In other words, is every quadrilateral that has opposite sides parallel and equal a rectangle?
Solution:
No, this can’t be the definition of a rectangle. A quadrilateral with opposite sides parallel and equal is a parallelogram, but not all parallelograms are rectangles. A rectangle needs all angles to be right angles.

2. QUADRILATERALS WITH EQUAL SIDELENGTHS
Figure it Out: Page : 102

Question 1.
Find the remaining angles in the following quadrilaterals.

Solution:
(i) Here PR || EA, and PE || RA
Therefore, PEAR is a parallelogram.
∠P = ∠A = 40° … (Opposite angles of a parallelogram are equal)
∠ P + ∠R = 180° … (The sum of the adjacent angles of a parallelogram is 180°)
40° + ∠R = 180°
∠R = 180° – 40°
∠R = 140°.
∠R = ∠E = 140°… (Oppositeangles of a parallelogram are equal)

(ii) Here PQ // SR, and PS // QR
∴ PQRS is a parallelogram.
∠P = ∠R = 110° … (Opposite angles of a parallelogram are equal)
∠P + ∠S = 180° … (The sum of the adjacent angles of a parallelogram is 180°)
110° + ∠S = 180°
∠S = 180° – 110°
∠S = 70°.
∠S = ∠Q = 70° … (Opposite angles of a parallelogram are equal)

(iii) Here, XWVU is a rhombus (all sides equal).
In ΔVUX, UV = UX, then the angles opposite them are equal.
∴ ∠UXV = ∠UVX = 30°
∠UXV = ∠WXV = 30° ………….. (The diagonals of a rhombus bisect its angles)
Also, ∠UVX = ∠WVX = 30° ………….. (The diagonals of a rhombus bisect its angles)
∠E = 2 × ∠UVX = 2 × 30° = 60°
∠V = ∠X = 60° ………….. (Opposite angles of a rhombus are equal)
∠V + ∠U = 180° ………….. (The sum of adjacent angles of a rhombus is 180°)
60° + ∠U = 180°
∠U = 180° – 60°
∠U = 120°
∠U = ∠W = 120° ………….. (Opposite angles of a rhombus are equal)

(iv) Here, AEIO is a rhombus (all sides equal).
In ΔEAO, AE = AO, then the angles opposite them are equal.
∴ ∠AOE = ∠AEO = 20°
∠AEO = ∠IEO = 20° ………….. (The diagonals of a rhombus bisect its angles)
Also, ∠AOE = ∠IOE = 20° ………….. (The diagonals of a rhombus bisect its angles)
∠E = 2 × ∠AEO = 2 × 20° = 40°
∠E = ∠O = 40° ………….. (Opposite angles of a rhombus are equal)
∠E + ∠A = 180° ………….. (The sum of adjacent angles of a rhombus is 180°)
40° + ∠A = 180°
∠A = 180° – 40°
∠A = 140°
∠A = ∠I = 140° ………….. (Opposite angles of a rhombus are equal)

Question 2.
Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of 140°.
Solution:

Steps of construction:
(i) Draw a line segment AC of length 7 cm and mark its midpoint as 0.
(ii) At point 0, draw an angle of 140° with respect to diagonal AC.
(iii) At 0, along the 140° angle’s free arms in both directions, mark OD = 2.5 cm and OB = 2.5 cm using a compass.
(iv) Join D to A and C.
Join B to A and C.
So, ABCD is the required parallelogram.

Question 3.
Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm.
Solution:

Steps of construction:
(i) Draw a line segment AC of length 5 cm.
(ii) Draw the perpendicular bisector of AC, intersecting it at 0.
(iii) With 0 as centre and radius 2 cm, mark points B (below) and D (above) on the perpendicular bisector.
(iv) Join A with D, D with C, B with A and C with B.
∴ ABCD is the required rhombus.

3. KITE AND TRAPE∠IUM
Figure it Out : Page : 107 – 109

Question 1.
Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm.
Solution:

Since all sides of an equilateral triangle are equal.
Thus, the lengths of all sides of the given quadrilateral are equal.
∴ PQ = QR = RS = SP = 4 cm.
Also, the measure of all angles of an equilateral triangle is 60°.
∠P = ∠R = 60°
∠S = ∠PSQ + ∠RSQ = 60° + 60° = 120°.
∠Q = ∠PQS + ∠RQS = 60° + 60° = 120°.

Question 2.
Construct a kite whose diagonals are of lengths 6 cm and 8 cm.
Solution:

(i) Draw a line segment AC = 6 cm.

(ii) Construct the perpendicular bisector of AC; let it meet AC at 0 (so 0 is the midpoint).

(iii) With centre at 0 and radius 3 cm draw an arc to cut the bisector above AC; label that point D. With centre 0 and radius 5 cm draw an arc to cut the bisector below AC; label that point B.

(iv) Join A with B, B with C, C with D and D with A.
ABCD is the required kite.

Question 3.
Find the remaining angles in the following trapeziums-

Solution:

Since AB // DC, and AD is a tranversal, then ∠A + ∠D = 180° … (Sum of angles on the same side of the transversal)
135° + ∠D = 180°
∠D = 180° – 135°
∠D = 45°
Also, since AB // DC, and BC is a tranversal, then So, ∠B + ∠C = 180°
… (Sum of angles on the same side of the transversal)
105° + ∠C = 180°
∠C = 180° – 105°
∠C = 75°
In the second figure :

Since PQ // SR, and PS is a tranversal, then ∠P + ∠S = 180° … (Sum of angles on the same side of the transversal)
∠P + 100° = 180°
∠P = 180° – 100° = 80°.
∠S = ∠R = 100° … (In an isosceles trape∠ium base angles are equal)
Also, since PQ // SR, and QR is a tranversal,
So, ∠Q + ∠R = 180° … (Sum of angles on the same side of the transversal)
∠Q + 100° = 180°
∠Q = 180° – 100° = 80°.

Question 4.
Draw a Venn diagram showing the set of parallelograms, kites, rhombuses, rectangles, and squares. Then, answer the following questions –
(i) What is the quadrilateral that is both a kite and a parallelogram?
(ii) Can there be a quadrilateral that is both a kite and a rectangle?
(iii) Is every kite a rhombus? If not, what is the correct relationship between these two types of quadrilaterals?
Solution:

(i) A rhombus is a quadrilateral that is both a kite and a parallelogram.

(ii) A kite is not a rectangle and a rectangle is not a kite.
∴ There can be no quadrilateral that is both a kite and a rectangle.
Also, there is no common portion of the set of kites and the set of rectangle.

(iii) No, every kite is not a rhombus.
Correct relationship:
Every rhombus is a kite, but not every kite is a rhombus.

Question 5.
If PAIR and RODS are two rectangles, find ∠IOD.

Solution:
Since PAIR and RODS are two triangles.
∠RIO = 90° … (Corner angle of a rectangle)
In ΔRIO,
∠IRO + ∠IOR + ∠RIO = 180° … (Sum of angles of a triangle)
30° + ∠IOR + 90° = 180°
120° + ∠IOR = 180°
∠IOR = 180° – 120° = 60°.
∴ ∠IOD = 90° – ∠IOR
= 90° – 60° = 30°.

Question 6.
Construct a square with diagonal 6 cm without using a protractor.
Solution:

Steps of construction:
(i) Draw a line segment AC = 6 cm and mark its midpoint as O.

(ii) With O as centre and radius greater than half of AC, draw arcs above and below AC from points A and C.

(iii) Join the arcs intersections to get a line perpendicular to AC and passing through 0.

(iv) Again, with 0 as centre and radius equal to 3 cm, mark points B and D on the perpendicular line.

(v) Join (A, B), (C, B), (A, D) and (C, D). Hence, ABCD is the required square with a diagonals of 6 cm.

Question 7.
CASE is a square. The points U, V, W and X are the midpoints of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b).

Solution:
(a) U, V, W, and X are the midpoints of the sides of the square.
In ΔVCU and ΔUAX,
we have VC = UA,
∠VCU = ∠UAX = 90°,
and CU = AX.
∴ By the SAS condition, ΔVCU and ΔUAX are congruent.
∴ VU = UX
Similarly, VU = XW, VU = WV.
∴ Sides of the quadrilateral UVWX are equal.

In ΔVCU, VC = CU
⇒ ∠1 = ∠2
Also, ∠1 + ∠C + ∠2 = 180°
⇒ ∠1 + 90° + ∠1 = 180°
⇒ 2∠1 = 90°
⇒ ∠1 = 45°
∴ ∠2 is also 45°.
Similarly, ∠3 = ∠4 = 45°
We have ∠2 + ∠VUX + ∠3 = 180°
⇒ 45° + ∠VUX + 45° = 180°
⇒ ∠VUX = 180° – 90°
⇒ ∠VUX = 90°
Similarly, ∠VXW = 90°,
∠XWV = 90°
and ∠WVU = 90°.
∴ By definition, the quadrilateral UVWX is a square.

(b) Let ABCD be a square.
Take points P, Q, R, and S such that AS = BP = CQ = DR.

Since the sides of squares are equal,
we have DS = AP = BQ = CR.
In ΔPAS and ΔSDR, we have
PA = SD,
∠PAS = ∠SDR = 90°,
and AS = DR.
∴ By the SAS condition, ΔPAS and ΔSDR are congruent.
∴ PS = SR
Similarly, PS = RQ, PS = QP.
∴ Sides of the quadrilateral PQRS are equal.
In ΔPAS, ∠1 + ∠2 + 90° = 180°
⇒ ∠1 + ∠2 = 90°
⇒ ∠3 + ∠2 = 90° (∵ ∠1 = ∠3)
Also, ∠2 + ∠4 + ∠3 = 180°
⇒ 90° + ∠4 = 180°
⇒ ∠4 = 180° – 90°
⇒ ∠4 = 90°
∴ Similarly, ∠5 = 90°,
∠6 = 90°,
and ∠7 = 90°.
By definition, the quadrilateral PQRS is a square.

Question 8.
If a quadrilateral has four equal sides and one angle of 90°, will it be a square? Find the answer using geometric reasoning as well as by construction and measurement.
Solution:

Let ABCD be a quadrilateral such that AB = BC = CD = DA and ∠DAB = 90°.
Join BD.
In ΔADB and ΔCDB, we have AD = CD, AB = CB, and DB is a common side.
∴ ΔADB and ΔCDB are congruent.
∴ ∠C = ∠A = 90°
In ΔDAB, ∠1 = ∠2 (∵ AB = AD)
Also, ∠1 + 90° + ∠2 = 180°
⇒ ∠1 + ∠2 = 90°
⇒ ∠1 = 45°
and ∠2 = 45° (∵ ∠1 = ∠2)
In ΔCDB, ∠3 = ∠4 (∵ CD = CB)
Also, ∠3 + 90° + ∠4 = 180°
⇒ ∠3 + ∠4 = 90°
⇒ ∠3 = ∠4 = 45° (∵ ∠3 = ∠4)
∴ ∠ABC = ∠1 + ∠4 = 45° + 45° = 90°
and ∠ADC = ∠2 + ∠3 = 45° + 45° = 90°.
∴ Each angle of the quadrilateral ABCD is 90°.
∴ ABCD is a square.
Also, by measurement, we find
AB = BC = CD = DA
and ∠A = ∠B = ∠C = ∠D = 90°.

Question 9.
What type of quadrilateral is one in which the opposite sides are equal? Justify your answer.
Hint: Draw a diagonal and check for congruent triangles.
Solution:

Let ABCD be a quadrilateral in which opposite sides are equal. Join AC.
In ΔADC and ΔCDA,
AD = CB (given)
DC = BA (given)
AC = AC (common side)
By SSS condition, ΔADC ≅ ΔCBA.
∴ ∠1 = ∠3 and ∠2 = ∠4
AC is a transversal of lines AB and DC, and alternate angles ∠1 and ∠3 are equal.
∴ Lines AB and DC are parallel.
AC is a transversal of lines AD and BC, and alternate angles ∠2 and ∠4 are equal.
∴ Lines AD and BC are parallel.
∴ By definition, the quadrilateral ABCD is a parallelogram.

Question 10.
Will the sum of the angles in a quadrilateral such as the following one also be 360°? Find the answer using geometric reasoning as well as by constructing this figure and measuring.

Solution:
In the given quadrilateral, join BD.
In ΔABD, we have
∠A + ∠3 + ∠1 = 180°
In ΔCBD, we have ∠C + ∠4 + ∠2 = 180°

Adding, we get
(∠A + ∠3 + ∠1) + (∠C + ∠4 + ∠2) = 180° + 180°
⇒ ∠A + (∠3 + ∠4) + ∠C + (∠1 + ∠2) = 360°
⇒ ∠A + ∠B + ∠C + ∠D = 360°
∴ The sum of the angles of the quadrilateral ABCD is 360°.
Also, by using a protractor, we find that the sum of all angles is 360°.

Question 11.
State whether the following statements are true or false. Justify your answers.
(i) A quadrilateral whose diagonals are equal and bisect each other must be a square.
Solution:
False.
A quadrilateral whose diagonals are equal and bisect each other is a rectangle. A square is a special case of a rectangle where all sides are also equal.

(ii) A quadrilateral having three right angles must be a rectangle.
Solution:
True.
Three right angles force the fourth to be right angle as well and a quadrilateral with four right angles is a rectangle.

(iii) A quadrilateral whose diagonals bisect each other must be a parallelogram.
Solution:
True.
If the diagonals bisect each other, then the two triangles formed by a diagonal are congruent, which gives pairs of opposite sides parallel. Hence the figure is a parallelogram.

(iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus.
Solution:
False.
Squares, kites, and some other quadrilaterals also have perpendicular diagonals. Therefore, having perpendicular diagonals does not necessarily mean the quadrilateral is a rhombus.

(v) A quadrilateral in which the opposite angles are equal must be a parallelogram.
Solution:
True.
If both pairs of opposite angles are equal, then each pair of adjacent angles are supplementary, which implies opposite sides are parallel. Hence the quadrilateral is a parallelogram.

(vi) A quadrilateral in which all the angles are equal is a rectangle.
Solution:
True.
If all four angles are equal, each angle must be 360°/4 = 90°. A quadrilateral with four right angles is a rectangle.

(vii) Isosceles trapeziums are parallelograms.
Solution:
False.
An isosceles trapezium has exactly one pair of parallel sides and equal non-parallel sides. While a parallelogram must have two pairs of parallel sides. So an isosceles trapezium is not a parallelogram.

Quadrilaterals Class 8 Extra Questions

Multiple Choice Questions

Question 1.
The angles in a square are :
(a) 90° each
(b) 70° each
(c) 60° each
(d) 100° each
Solution:
Each angle of a square is 90°.
(a) 90° each

Question 2.
ABCD is a quadrilateral. Sum of angles ∠A + ∠B + ∠C + ∠D is :
(a) 180°
(b) 270°
(c) 360°
(d) 540°
Solution:
Sum of all angles of any quadrilateral is 360°.
(c) 360°

Question 3.
In a square ABCD, AC and BD are its two diagonals. Then which of the following is true?
(a) AC > BD
(b) BD > AC
(c) AC = BD
(d) AC + BD = AB
Solution:
Diagonals of a square are equal.
(c) AC = BD

Question 4.
In the following figure, ABCD is a rectangle. ∠AOB =

(a) 90°
(b) 60°
(c) 45°
(d) 35°
Solution:
Here, ABCD is a rectangle.
Hence, ∠DAB = 90°
⇒ ∠DAO + ∠BAO = 90°
⇒ 30° + ∠BAO = 90°
⇒ ∠BAO = 90° – 30° = 60°
Now, if we consider ΔOAB, then .
OA = OB (∵, diagonals are equal and they bisect each other)
∴, ΔOAB is an equilateral Δ as ∠BAO = 60°
So, ∠AOB = 60°

Question 5.
In the following figure, ABCD is a rectangle. ∠OAB =

(a) 60°
(b) 30°
(c) 45°
(d) 90°
Solution:
In ΔOAB, if ∠AOB = 60° then all of its angles are 60° each.
(a) 60°

Assertion and Reasoning

(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).
(b) Assertion (A) and Reason (R) both are true but Reason (R) is not the correct explanation for the Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.

Question 1.
Assertion (A) : In a quadrilateral ABCD, if ∠A = 40°, ∠B = 90° and ∠C = 110°, then ∠D = 120°.
Reason (R) : Sum of all angles of a quadrilateral is 380°.
Solution:
Reason (R) is false as sum of all angles of a quadrilateral is 360°.
Answer:
(c) Assertion (A) is true but Reason (R) is false.

Question 2.
Assertion (A) : If PQRS is a square, then ∠P = 90°.
Reason (R) : Each angle of a square is 90°.
Solution:
In a square PQRS, ∠P = ∠Q = ∠R = ∠S = 90°.
So, ∠P = 90°.
Answer:
(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).

Case Based Questions

Question 1.
A carpenter needs to put together two thin strips of laminates, as shown in the following figure, So that when a thread is passed through their end points, it forms a rectangle.
She already has one 12 cm long strip.

Based on the above, answer the following:
(a) What should be the length of the other strip?
(b) At what point they should be joined?
(c) If thread is passed through B, E, S, T, then what will be the angle between the arms BT and BE?
(d) Will the lengths of BT and ES same? Why?
Answer:
(a) The lengths of both the strips must be same. Hence, the length of the other strip must be 12 cm.

(b) The two strips must be joined at O.

(c) The resultant figure is a rectangle. In a rectangle, the opposite sides are equal. Hence, BT = ES.

Go through BSE Odisha Class 8 Science Solutions Chapter 5 Exploring Forces Question Answer to understand textbook questions more clearly.

Class 8 Science Curiosity Chapter 5 Question Answer

Class 8 Science Ch 5 Exploring Forces Question Answer

Class 8 Science Chapter 5 Exploring Forces Question Answer

Probe and Ponder Questions

Question 1.
Why does it feel harder to pedal a bicycle when going uphill than on flat ground?
Answer:
When we cycle uphill, we’re constantly fighting against the Earth’s gravitational pull, which tries to pull us back down. This force acts perpendicular to the ground on a flat surface, meaning it doesn’t directly oppose our forward movement. But on a slope, a portion of the gravity acts against our direction of motion, requiring us to exert more force to move forward and upward.

Question 2.
Why is it easier to slip on a wet surface?
Answer:
It is easier to slip on a wet surface due to reduced friction between the foot and the surface. Water acts as a lubricant, minimizing the grip and making it easier to slide.

Question 3.
Why do we feel ‘light’ or like we are ‘floating’ just after our swing reaches its highest point and begins to come down?
Answer:
When the swing reaches its highest point, it momentarily comes to rest before moving downward. A this point, gravity still acts downward, but the support (normal) force exerted by the swing on our body decreases. Because the normal force is reduced, we feel less push from the swing, which makes us feel “light” or as if we are “floating” for a brief moment.

Question 4.
Share your questions
Answer:

• Why does sliding work better on ice than on sand?
• What would happen if friction did not exist at all?
• Why do heavier objects sink more in water than lighter ones of the same size?
• Does air also provide friction to moving objects?
• Why can some birds fly easily for long periods without getting tired?

InText Questions

Question 1.
Does this mean that whenever there is a change in speed or direction, or change in shape, a force is acting on the object? (Page 65)
Answer:
Yes, none of these take place without the action of force.

Question 2.
Suppose an object is at rest. Does it mean that no force is acting on this object? (Page 65)
Answer:
No, the forces are acting on this object. But all these forces acting on them are balancing one another. That is why an object is at rest.

Question 3.
Does this mean that the force of friction will be greater if the surfaces are rough? (Page 68)
Answer:
Yes, more the roughness of a surface, larger is the number of irregularities on its surfaces and hence greater will be the friction.

Question 4.
Is it essential for an object applying force on another object to always be in contact with it? (Page 69)
Answer:
No. Forces can be applied either through direct contact (contact forces) or without direct contact (non-contact forces) such as gravitational or magnetic forces.

Question 5.
Does it mean that there are two kinds of electrical charges?(Page 71)
Answer:
Yes, there are two kinds of electrical charges ‘positive’ and ‘negative’.

Question 6.
Why do all the objects fall towards the Earth? (Page 72)
Answer:
All the objects fall towards the Earth because the Earth attracts (pull) them. This force is called gravitational force.

Question 7.
Is there any force which acts on them. ? What exerts this force ?(Page 72)
Answer:
Yes, there is a force acting on any object in the universe, and it is called gravity. Gravity is a force of attraction that exists between any two objects with mass. The Earth, due to its large mass, exerts a gravitational orce on all objects near it, pulling them towards its center.

Question 8.
Does the Earth pull every object with equal force ? (Page 72)
Answer:
No, the Earth does not pull every object with equal force. The force of gravity is stronger on objects with greater mass. While the Earth exerts a gravitational pull on all objects, the strength of that pull depends on the mass of the object being attracted.

Question 9.
What is the difference between weight and mass ? (Page 75)
Answer:
Mass is the amount of matter in an object and is measured in grams (g) or kilograms (kg). Its value remains the same at every place. Weight, on the other hand, is the gravitational force with which the Earth (or another planet) pulls an object.

Question 10.
If we place some objects on water, some of them float, while others fall to the bottom. The gravitational force of the Earth is acting on all objects, then why don’t all objects fall to the bottom? (Page 76)
Answer:
While the Earth’s gravitational force acts on all objects, whether they sink or float in water depends on the buoyant force and the density of the object relative to water. Objects with a density lower than water experience a stronger buoyant force, causing them to float, while those with a higher density experience a weaker buoyant force and sink.

Exploring Forces Class 8 Questions and Answers

Keep the Curiosity Alive (Pages 25-26)

Question 1.
Match items in Column A with the items in Column B.

Answer:

Column A (Type of force)	Column A (Type of force)
(i) Muscular force	(b) A child lifting a school bag
(ii) Magnetic force	(e) A compass needle pointing North
(iii) Frictional force	(a) A cricket ball stopping on its own just before touching the boundary line
(iv) Gravitational force	(c) A fruit falling from a tree
(v) Electrostatic force	(d) Balloon rubbed on woollen cloth attracting hair strands

Question 2.
State whether the following statements are True or False.
(i) A force is always required to change the speed of motion of an object.
(ii) Due to friction, the speed of the ball rolling on a flat ground increases.
(iii) There is no force between two charged objects placed at a small distance apart.
Answer:
(i) True: A force is indeed always required to change the speed of motion of an object. If there is no force acting on an object, it will maintain its current speed and direction (unless it’s already at rest, in which case it will stay at rest).
(ii) False: Friction opposes motion, so it will decrease the speed of a rolling ball.
(iii) False: There is a force between charged objects. This force can be attractive or repulsive depending on the charges. But it is always present when charges are close to each other.

Question 3.
Two balloons rubbed with a woollen cloth are brought near each other. What would happen and why?
Answer:
When two balloons are rubbed with a woollen cloth and brought near each other, they will repel each other. This happens because both balloons will acquire a negative charge when rubbed with wool, and like charges repel.

Question 4.
When you drop a coin in a glass of water, it sinks, but when you place a bigger wooden block in water, it floats. Explain.
Answer:
A coin sinks in water because its density (mass per unit volume) is greater than that of water. A wooden block floats because its density is less than that of water, causing it to be buoyed up by the water.

Question 5.
If a ball is thrown upwards, it slows down, stops momentarily, and then falls back to the ground. Name the forces acting on the ball and specify their directions.
(i) During its upward motion
(ii) During its downward motion
(iii) At its topmost position
Answer:
When a ball is thrown upwards, the only force acting on it throughout its entire motion is gravity, which pulls it downwards. However, depending on the motion of the ball, the direction of this force relative to the ball’s velocity changes.
(i) During its upward motion: The force of gravity is downwards, opposing the upward motion of the ball, causing it to slow down.
(ii) During its downward motion: The force of gravity is still downwards, but now it aligns with the ball’s direction of motion, accelerating it downwards.
(iii) At its topmost position: The ball has zero velocity, meaning it’s momentarily stationary. At this point, the force of gravity is still downwards, but since the ball is not moving upwards or downwards, it has no net effect on the ball’s motion.

Question 6.
A ball is released from the point P and moves along an inclined plane and then along a horizontal surface as shown in the fig. It comes to stop at the point A on the horizontal surface. Think of a way so that when the ball is released from the same point P, it stops
(i) before the point A
(ii) after crossing the point A.

Answer:
The ball’s motion is governed by the forces of gravity and friction. On the inclined plane, gravity provides the acceleration for the motion. On the horizontal surface, only friction acts on the ball.

• Stopping before A: Increasing friction on the horizontal surface will cause the ball to decelerate more rapidly, meaning it will come to a stop sooner, potentially before reaching point A.
• Stopping after A: Decreasing friction on the horizontal surface will reduce the deceleration, allowing the ball to travel further before losing all its kinetic energy and stopping.

Question 7.
Why do we sometimes slip on smooth surfaces like ice or polished floors? Explain.
Answer:
When we walk on surfaces like ice, we often slip, which means we lose our balance and fall. This happens because the force that helps us stay upright and move forward (friction) is not enough. These surfaces have fewer irregularities. Minimizing the contact area and the force of friction between the surface and our shoes makes it easier to slide instead of grip. A layer of water, even a thin one on ice, can further reduce friction by acting as a lubricant, making the surface even more slippery.

Question 8.
Is any force being applied to an object in a non-uniform motion?
Answer:
Yes, for an object to be in non-uniform motion, a force must be acting upon it. Nonuniform motion, also known as accelerated motion, means the object’s velocity is changing, either in speed or direction, or both. This change in velocity requires a force to be applied.

Question 9.
The weight of an object on the Moon becomes one-sixth of its weight on the Earth. What causes this change? Does the mass of the object also become one-sixth of its mass on the Earth?
Answer:
The change in an object’s weight on the moon compared to Earth is due to the difference in gravitational force. The moon’s gravity is significantly weaker than Earth’s, roughly one-sixth as strong. However, the mass of an object remains the same regardless of location; only weight changes with gravitational pull.

Question 10.
Three objects 1,2 and 3 of the same size and shape but made of different materials are placed in the water. They dip to different depths as shown in fig. If the weights of the three objects 1,2 and 3 are w₁, w₂, and w₃, respectively, then

(i) w₁=w₂=w₃
(ii) w₁>w₂>w₃
(iii) w₂>w₃>w₁
(iv) w₃>w₁>w₂
Answer:
(ii) The relationship between the weights of the objects is w₁>w₂>w₃.
Object 1 is the deepest meaning it displaces the most water.
Object 2 is less deep than object 1 but deeper than object 3.
Object 3 is the least deep, meaning it displaces the least amount of water.
Since the objects have the same size and shape, the greater the depth, the greater the weight of the object (assuming they are all made of the same material).
Hence, the object with the greatest weight will sink the deepest, and the object with the least weight will be the closest to the surface.

Class 8 Science Chapter 5 Question Answer

Activity 1.

Let us explore
Aim: To experience the push or pull.
Materials Required: A large cardboard box, a rope.

Procedure :

• Take a heavy and large cardboard box.
• Try to push or pull it by yourself.
• Try to move the box in as many different ways as you can do.
• Are you able to move the box in any other way than shown in fig?

Answer:
It is really very difficult to move.
Observations: In every method you are either pushing or pulling the box.

Inferences:

• No matter how you chose to move the box – by dragging, sliding, rolling, lifting etc. Some form of push or pull is always involved.
• Generally, the push or pull applied on an object is called force.

Activity 2.

Let us analyse
Aim: To analyse and study the effect of force.

Procedure:

• Think of different examples and situations where a force either push or pull is applied.
• List them in the table given below.
• Write the effect of the force in the table also.
• Some of them are already listed for you.

Observations :
Different actions and their effects

Answer:

Inferences:
We have noticed that the effect of the force are :

• A force can make an object move from rest.
• A force can change the speed of an object if it is moving. Either increase or decrease.
• A force can change the direction of motion of an object.
• A force can bring about a change in the shape of an object.
• A force can cause some or all of these effects.

Activity 3.

Let us investigate
Aim : To show that friction always opposes the relative motion between the two bodies, irrespective of the direction of motion.
Materials Required : A book.

Procedure :

• Gently push a book on a table.
• It stops after moving for some distance.
• Repeat this activity pushing the book from the opposite direction. [Fig. (b).
• Does the book stop this time, too? Yes, the book stop this time too.
• Can you think of an explanation ?
  The book slides for some time and then stops. The reason is that

Inference:
The force acting along the two surfaces in contact which opposes the motion of one body over the other, is called the force of friction.

Activity 4.

Let us explore
Aim: To study that force of friction depends upon the nature of the two surfaces in contact.
Materials Required: A wooden board, a pencil cell, bricks or books, a piece of cloth, sand.

Procedure:

• Make an inclined plane on a smooth floor, or on a table. (You may use a wooden board supported by bricks or books).
• Put a mark with a pen at any point ‘A’ on the inclined plane.
• Now let a pencil cell move down from this point.
• How far does it move on the plane before coming to rest ?
  It moves up to the end of the inclined plane and comes to rest on the table.
• Note down the distance.
• Now spread a piece of cloth over the table. Make sure that there are no wrinkles in the cloth.
  Try this activity again [Fig. (b).
• Repeat this activity by spreading a thin layer of sand over the table. Maintain the same slope throughout the activity.
  Inference: Friction depends upon the nature of surfaces in contact.

Activity 5.

Let us test
Aim : To show that a magnet can exert force on another magnet without being in contact with it.
Materials Required : Two ring magnets, a wooden stick.

Procedure :

• First of all we will take two ring magnets and a wooden stick.
• Now, hold the stick in a vertical position on a wooden table.
• Insert one ring magnet into the stick. (See fig.)
• Now, take another ring magnet and insert in such a way that like poles of two magnets face each other.

Observations:

• Second magnet stay floating above the first magnet. It means that they repel each other.
• The second magnet still remains floating after reversing the poles of both magnets.

Inference:

• A magnet can exert force on another magnet without being in contact with it.
• The force exerted by a magnet on another magnet or a magnetic material is called magnetic force. Since a magnet can exert a force from a distance without being in contact it is called a non-contact force.

Activity 6.

Let us experiment
Aim: To show electrostatic force.
Materials Required : A plastic straw, a plastic scale, a piece of polythene and small piece of paper.

Procedure :

• Take a plastic scale/straw and rub it vigorously with polythene.
• It is advised that never touch the rubbed part with your hand or any other metal object.
• Bring the rubbed straw or scale close to the small pieces of paper which is kept on the table.
• Do not touch the paper.
• Write down your observations.

Observations:
The paper pieces get pulled towards the plastic scale/straw and stick to it when it is brought close to paper pieces.

Inference:

• When two objects are rubbed, electrical charges are build up on their surfaces which is called static charges.
• A charged object attract an uncharged objects like small pieces of paper.
• This is called electrostatic force and comes into play even when the two bodies are not in contact.

Activity 7.

Let us experiment
Aim: To show that like charges repel and unlike charges attract each other.
Materials Required: Two balloons, a length of thread, and a woollen cloth.

Procedure:

• First of all we will take two balloons, a thread and a woollen cloth.
• Now, inflate the two balloons and hang them.
• We should be careful that the two balloons do not touch each other. (See fig. (a))
• Now, rub both balloons with the woollen cloth and release them. Remember that these two balloons should not be touched with your fingers. Note down your observation.
• Now, bring the woollen cloth used for rubbing the balloons close to one of the rubbed balloons. Note down your observation.

Observations :

• Case-I: The balloons move away from each other i.e., repelling each other.
• Case-II: They attract each other.

Inference:

• The force exerted by a charged body on another charged body or an uncharged body is called electrostatic force. It is a non-contact force.
• We can infer that like or similar charged body repel each other and unlike (opposite) charged bodies attract each other.
• The two kinds of static charges are said to be ‘positive’ and ‘negative’ charges.

Activity 8.

Let us observe
Aim: To demonstrate the gravitational force.
Materials Required: A ball, a stone or any object around you.

Procedure :

• Take a ball or a stone or your eraser and throw it vertically upwards.
• Now, throw it again, but this time harder than previous.
• Not down your observations in both the cases.
• You may think of different situations around you and can throw any object in any direction.

Observations :

• Case-I: The ball or stone come down.
• Case-II: Even in this case also the ball still fall back down to the ground but takes little more time.

Inference:

• Finally all objects falls or comes back to the ground or floor.
• This shows that some force is acting in downward direction. This is called the gravitational force or force of gravity or simply gravity.

Activity 9.

Let us explore
Aim : To show that the earth pulls different objects with different forces and it depends on their masses.
Materials Required : A spring, a few objects of different masses e.g., a pencil box, a tiffin box, a small stone.

Procedure :

• Hang one end of the spring from a nail and on the other end, hang an object and observe the spring.
• When you suspend an object from a spring, the spring stretches due to the force applied by the earth.
• Now hang the other objects, one by one and notice the stretch in the spring in each cases.

Observations: If you hang objects of different mass one after another, each time causes a different amount of stretching.
Inference: This proves that heavier objects are pulled with more force i.e., the weight of an object depends on its mass; heavier objects have greater weight.

Activity 10.

Let us observe
Aim : To measure maximum weight by a spring balance.
Materials Required : A spring balance.

Procedure :

• Take a spring balance and hang it as shown in fig.
• Now, look at the spring balance carefully.
• What is the maximum weight it can measure ?

Observations :

• The maximum weight a spring balance can measure is 10 N.
• Range of spring balance is the maximum weight it can measure.

Inference : If a spring balance shows values from 0 to 10 N , its range is 0.10 N . means that it can measure weight upto 10 N.

Activity 11.

Let us calculate
Aim: To determine the smallest readable value (Least count) of and close-up of its scale a spring balance.
Materials Required: A spring balance.

Procedure :

• Note down the weight difference indicated between the two bigger marks.
• The weight difference between 0 and 01 N or between 01 N and 02 N is 1 N.
• Count the number of divisions between these two bigger marks.

Demonstration :

1. Weight different between 0 and 1 N or between 01 N and 2 N =1 N Number of divisions between 0 and 1 or 1 and 2=5
∴ Least count = \(\frac{1}{5}\)=0.2 N

Inference:

• The least count of a spring balance is the smallest difference in weight it can measure.
• So, the least count of a spring balance is 0.2 N.

Activity 12.

Let us measure
Aim: To measure the weight and mass of an object using a spring balance.
Materials Required: Pencil box, water bottle filled with water partially.

Procedure :

• Hang a spring balance from the hook of a spring balance (without exceeding its maximum range).
• The pointer or reading on the scale shows the object’s weight in newtons (N).
• This method can be repeated for many objects and results should be recorded in a table.
• The mass scale (in g/kg) on a spring balance assuming earth’s gravity.
• The mass reading is only correct on earth where gravitational acceleration is standard.
• You can repeat the above Activities-10 to 12 for the mass scale which is shown on the left side on the spring balance (Fig. 13) to measure the mass of an object.

Observations :
Measuring weight using a spring balance

S.No.	Object	Weight (N)
1.	Pencil Box
2.	Partially filled water bottle

Inference: We can measure the weight and mass of an object using a spring balance.

Activity 13.

Let us investigate
Aim: To experience a buoyant force or upthrust by a liquid.
Materials Required: A bucket, an empty plastic bottle.

Procedure:

• Take an empty plastic bottle (with its lid closed tightly) and a bucket full of water.
• Now, push the bottle in the water. (See fig.)
• You will feel an upward force or push.
• Does the bottle bounce up and comes on top of water ?

Observation:

• Yes, we feel an upward push on our hand.
• The bottle bounces back up when released.

Inference:

• Gravity (gravitational force acts on downward direction i.e., pulls downward. The buoyant force (upthrust) pushes it upwards.
• If the gravitational force (weight of the object) is greater than buoyant force, the object sinks. i.e. W>U, sink
• If the two forces are equal i.e., W=U, the object floats.
• The density of the liquid affects the buoyant force.

Exploring Forces Class 8 Extra Questions and Answers

Short Answer Type Questions

Question 1.
What is electrostatic force ? Why is it called non-contact force?
Answer:
The force exerted by a charged body on another charged or uncharged body is called electrostatic force. This force comes into play even when the bodies are not in contact, so it is called non-contact force.

Question 2.
What is Muscular Force ? Why is the force called contact force?
Answer:
Muscular Force : The force resulting due to the action of muscles is called Muscular Force. Since muscular force can be applied on an object when muscle is in contact with the object, so it is called a contact force.

Question 3.
What causes friction ?
Answer:
Friction occurs when two bodies move on each other. Each surface has some irregularities on it. When two such objects move on each other their irregularities get interlocked and friction arises.

Question 4.
You might have noticed that when used for a long time, slippers with rubber soles become slippery. Explain the reason.
Answer:
After using a rubber soles for a long time, their surfaces become smooth. Hence, the friction between the sole and the floor decreases. So, the slippers become slippery.

Question 5.
A blacksmith hammers a hot piece of iron while making a tool. How does the force due to hammering affect the piece of iron?
Answer:
The force due to hammering causes the change in the shape of the iron and iron can be moulded in the shape of the required tool.

Question 6.
Iqbal has to push a lighter box and Seema has to push a similar heavier box on the same floor. Who will have to apply a larger force and why ?
Answer:
The heavy object will be pressed hard against the opposite surface and produces more friction. So Seema will have to apply a larger force due to excess friction.

Long Answer Type Questions

Question 1.
In the following situations identify the agent exerting a force and the object on which it acts. State the effect of the force in each case.
(a) Squeezing a piece of lemon between the fingers to extract its juice.
(b) Taking out paste from a toothpaste tube.
(c) A load suspended from a spring while its other end is on a hook fixed to a wall.
(d) An athlete making a high jump to clear the bar at a certain height.
Answer:
(a) Agent are fingers, object is lemon, effect of force can be observable in form of lemon juice being expelled by squeezing.
(b) Agent is hand of the person squeezing the tube, object is toothpaste tube and effect of the force can be observed as the paste coming out of the tube.
(c) Agent is the load suspended, object is the spring and effect can be seen in the form of elongation of spring on suspension of load.

Question 2.
What are the different types of forces ? Give an example of each.
Answer:
Types of forces : The following are the different types of forces :
(a) Muscular force
(b) Magnetic force
(c) Electrostatic force
(d) Gravitational force
(e) Frictional force
(a) Muscular Force : The force applied by a living being with its muscles is known as muscular force e.g., bullocks apply muscular force to draw a cart.
(b) Magnetic Force : The force exerted by a magnet is called magnetic force. Example: A magnet attracts nails and pins made fron iron even from some distance.
(c) Electrostatic Force : The force exerted by an electrified body is called electrostatic force. Example : Pieces of paper get attracted towards a charged comb.
(d) Gravitational Force : The force of attraction between any two objects possessing mass is called force gravitation. An object dropped from a certain height falls on the earth due to the gravitational force.
(e) Frictional Force : The force which always opposes the motion of one body over another body is called frictional force. Example, a marble rolled on the ground stops after some time due to the frictional force.

Question 3.
An archer shoots an arrow in the air horizontally. However, after moving some distance, the arrow falls to the ground. Name the initial force that sets the arrow in motion. Explain why the arrow ultimately falls down.
Answer:
When an archer shoots an arrow he stretches the string of the bow by applying muscular force. In this process the shape of the bow changes. As soon as the string is released, it regains its original position which provides the initial force to set the arrow in motion. The force of gravity that acts on the arrow in the downward direction which finally brings it to the ground.

Case-Study Based Questions

1. Read the following passage carefully and answer the questions that follow :

Generally, to apply a force on an object, your body has to be in contact with the object. The contact may also be with the help of a stick or a piece of rope. This force is caused by the action of muscles in our body. The force resulting due to the action of muscles is known as the muscular force. Animals also make use of muscular force to carry out their physical activities and other tasks. Animals like bullocks, horses, donkeys and camels are used to perform various tasks for us. In per-forming these tasks they use muscular force. Since muscular force can be applied only when it is in contact with an object, it is also called a contact force.

Like Poles of two magnets repel each other and unlike poles attract each other. A straw is said to have acquired electrostatic charge after it has been rubbed with a sheet of paper. Such a straw is an example of a charged body. The force exerted by a charged body on another charged or uncharged body is known as electrostatic force. This force comes into play even when the bodies are not in contact. The electrostatic force, therefore, is another example of a non-contact force.

Similarly, the force exerted by a magnet on a piece of iron is also a noncontact force. Objects or things fall towards the earth because it pulls them. This force is called the force of gravity, or just gravity. This is an attractive force. The force of gravity acts on all objects. The force of gravity acts on all of us all the time without our being aware of it. Water begins to flow towards the ground as soon as we open a tap. Water in rivers flows downward due to the force of gravity.

(i) Gravity is : …………..
(a) Repulsive
(b) Attraction + Repulsive force
(c) Attractive force
(d) Not a force
Answer:
(c) Attractive force

(ii) An example of a non-contact force is :
(a) force exerted by us to lift a bucket
(b) pushing a stationary car
(c) force exerted by magnet
(d) hitting a circket ball for a 6 run
Answer:
(c) force exerted by magnet

(iii) The force exerted by a charged body on another charged body can :
(a) attract each other
(b) both (a) and (c)
(c) repel each other
(d) none of these
Answer:
(b) both (a) and (c)

(iv) Two objects repel each other. This repulsion could be due to charged or uncharged body is :
(a) magnetic force
(b) frictional force
(c) musculer force
(d) electrostatic force
Answer:
(d) electrostatic force

Picture Based Questions

I. Observe the picture and answer the following questions :

(i) Why do soles of shoes wear out?
(a) Due to friction
(b) Due to
old shoes
(c) Due to bad quality
(d) None of
Answer:
(a) Due to friction

(ii) Can it is possible to reduce friction upto zero ?
(a) Yes
(b) No
(c) Some times
(d) 50-50
Answer:
(b) No

Exploring Forces Class 8 MCQ

Multiple Choice Questions (Mcqs)

Question 1.
What is the SI unit of force ?
(a) Newton
(b) Joule
(c) Watt
(d) Kilogram
Answer:
(a) Newton

Question 2.
What is force described as in science ?
(a) Pressure
(b) Push or pull
(c) Friction
(d) Energy
Answer:
(b) Push or pull

Question 3.
Which of these is a non-contact force ?
(a) Friction
(b) Electrostatic force
(c) Tension
(d) Muscular force
Answer:
(b) Electrostatic force

Question 4.
What causes an object thrown upwards to return to earth ?
(a) Air pressure
(b) Gravity
(c) Electrostatics
(d) Friction
Answer:
(b) Gravity

Question 5.
Which of the following changes can a force cause ?
(a) Director change in
(b) Shape change in
(c) Speed change in
(d) All of these
Answer:
(d) All of these

Assertion and Reasoning

These questions consist of two statements, each printed as Assertion (A) and Reason (R). While answering these questions, you are required to choose any one of the following four responses.

(a) Assertion and reason both are correct and reason is correct explanation for assertion.
(b) Assertion and reason both are correct and reason is not correct explanation for assertion.
(c) Assertion is correct but the reason is wrong.
(d) Assertion is wrong but the reason is correct.

1. Assertion (A): A push or a pull on an object is called a force.
Reason (R): Forces applied on an object in the same direction add to one another.
Answer:
(b) Assertion and reason both are correct and reason is not correct explanation for assertion.

2. Assertion (A): A change in either the speed of an object, or its direction of motion, or both, is described as a change in its state of motion.
Reason (R): A force may bring a change in the state of motion of an object.
Answer:
(a) Assertion and reason both are correct and reason is correct explanation for assertion.

Fill in the blanks

1. To strecth the bow, the archer applies a force that causes a change in its ………….
Answer:
shape

2. The force applied by the archer to stretch the bow is an example of ………… force.
Answer:
muscular

3. The type of force responsible for a change in the state of motion of the arrow is an example of a ………… force.
Answer:
contact

4. While the arrow moves towards its target, the forces acting on it are due to ………… and that due to ………… of air.
Answer:
gravity and friction

5. A force arises due to ………… between two objects.
Answer:
interaction

True or False

1. A cyclist exerts a force of pull on the paddles of bicycle.
Answer:
False

2. Magnetic force is a contact force.
Answer:
False

3. A force can change the direction of motion of a moving object.
Answer:
True

4. The S.I. unit of pressure is newton per square metre.
Answer:
True

5. Pressure due to a liquid contained in a vessel is equal to all the points on its surface.
Answer:
True

Students can refer to BSE Odisha Class 8 Math Solution and Ganita Prakash Chapter 3 A Story of Numbers Class 8 Question Answer to understand textbook questions step by step.

Class 8 Maths Chapter 3 A Story of Numbers Solutions

Ganita Prakash Class 8 Chapter 3 Solutions

Class 8 Maths Ganita Prakash Chapter 3 Solutions A Story of Numbers

1. REEMA’S CURIOSITY
Page : 51

Question 1.
How do we ensure that all cows have returned safely after grazing?
Answer:
To ensure all cows return safely after grazing, farmers should use a combination of proper fencing and herd management, including counting cows as they enter and leave paddocks, using incentives like feed to encourage return to the holding area, and having a routine that minimizes the risk of lost or separated animals. Regular observation, clear containment through effective fencing, and understanding the animals’ natural behaviors are crucial for a safe herd return.

Question 2.
Do we have fewer cows than our neighbour?
Answer:
How to solve the problem
1. Identify the variables – Look for two specific numbers:
(i) The number of cows you have.
(ii) The number of cows your neighbor has.

2. Compare the numbers – Using the greater than (> is greater than >) and less than (< is less than <) symbols and compare the two numbers.
(i) If your cows < neighbor’s cows, then you have fewer cows.
(ii) If your cows > neighbor’s cows, then you do not have fewer cows than your neighbours.

3. State your conclusion – Answer the question based on the comparison you made

Question 3.
If there are fewer, how many more cows would we need so that we have the same number of cows as our neighbour?
Answer:
To find out how many more cows you need, you simply subtract the number of cows you have from the number of cows your neighbor has.

Formula : Number of more cows needed = (Number of your neighbor’s cows) – (Number of your cows)

Figure it Out : Page : 54

Question 1.
Suppose you are using the number system that uses sticks to represent numbers, as in Method 1. Without using either the number names or the numerals of the Hindu number system, -give a method for adding, subtracting, multiplying and dividing two numbers or two collections of sticks.
Answer:
Method 1 : Addition (Putting Together)

• Collect sticks representing the first quantity.
• Collect another set of sticks representing the second quantity.
• Combine both sets into a single group.
• The total number of sticks in the combined group represents the sum.
  Example:
  Group A: | | | | (4 sticks)
  Group B: | | | (3 sticks)
  Total: | | | | | | | (7 sticks)

Method 2: Subtraction (Taking Away)

• Take the Part with the group of sticks representing the larger collection.
• Remove or take away sticks equal to the number of the smaller quantity.
• The remaining sticks show me result of subtraction.
  Example:
  Start with: | | | | | | | (7 sticks)
  Take away: | | | (3 sticks)
  Left: | | | | (4 sticks)

Method 3: Multiplication (Repeated Addition)

• Make several groups of sticks, each containing the same number.
• Count all the sticks across all groups together.
• The total number of sticks represents the product.
  Example:
  Multiply 3 groups of | | | (3 sticks each):
  Group 1: | | |
  Group 2: | | |
  Group 3: | | |
  Total: (| | |) (| | |) (| | |) (9 sticks)

Method 4: Division (Equal Sharing or Grouping)

• Take the total number of sticks.
• Split them into equal groups equal to the given number by which we need to divide
  Either:
  • Count how many sticks are in each group (equal sharing), or
  • Count how many such groups can be made (repeated subtraction).
    Example (Equal Sharing):
    Total: | | | | | | (6 sticks), divide into 2 groups → | | | and | | | (3 sticks each)
    Example (Grouping):
    How many groups of | | (2 sticks) can be made from | | | | | | (6 sticks)?
    3 groups.

Question 2.
One way of extending the number system in Method 2 is by using strings with more than one letter for example, we could use ‘aa’ for 27. How can you extend this system to represent all the numbers? There are many ways of doing it!
Answer:
Treat it like a base-26 system using letters.
Each letter acts like a number and we treat sequences like base-26 numbers, where:
‘a’ = 1, ‘b’ = 2, …, ‘z’ = 26

After ‘z’, we continue with:
‘aa’ = 27
‘ab’ = 28 …
‘az’ = 52
‘ba’ = 53 …,
bb = 54,…, bz = 78, ca = 79, cb = 80,…, cz = 104,…

Question 3.
Try making your own number system.
Answer:
My Own Number System: The “ABC Number System”

• In this number system, I use the letters A, B, C, D and E instead of normal digits.
• Each letter stands for a number: A = 0, B = 1, C = 2, D = 3 and E = 4.
  This means I can count using only these five letters, just like we normally count with digits 0 to 9 in the usual number system.
• I also follow place value – the rightmost letter is worth Is, then 5s, then 25s, and so on (because this is a base-5 system).
  For example, the code BD means B = 1 (in 5s place) and D = 3 (in Is place). So BD = (1 × 5) + 3 = 8.
  This system is fun and feels like a secret code!
• I can count and do Maths using only letters, which helps me understand how numbers can be written in many different ways.

2. SOME EARLY NUMBER SYSTEMS
Figure it Out : Page : 59

Question 1.
Represent the following numbers in the Roman system.
(i) 1222
(ii) 2999
(iii) 302
(iv) 715
Answer:
(i) 1222
Break into parts:
1000 + 200 + 20 + 2 = M + CC + XX + II = MCCXXII

(ii) 2999
Break into parts:
1000 + 1000 + 900 + 90 + 9
= M + M + CM + XC + IX
= MMCMXCIX

(iii) 302
Break into parts:
300 + 2 = CCC + II = CCCII

(iv) 715
Break into parts:
700 + 10 + 5 = DCC + X + V = DCCXV

Figure it Out : Page : 60 – 61

Question 1.
A group of indigenous people in a Pacific island use different sequences of number names to count different objects. Why do you think they do this?
Answer:
Indigenous people on a Pacific island might use different sequences of number names for different objects because their language and culture are closely connected to daily life and nature. They may count coconuts, fish, people or days differently because each object is important in a different way and may follow different traditions.

For example, they might use one type of number word for living things and another for non-living things or they may count pairs of items (like eyes or shoes) instead of single pieces. Using different number systems helps them understand, group and remember things more easily in their own way.

Question 2.
Consider the extension of the Gumulgal number system beyond 6 in the same way of counting by 2s. Come up with ways of performing the different arithmetic operations (+, ×, ÷) for numbers occurring in this system, without using Hindu numerals. Use this to evaluate the following:
(i) (ukasar-ukasar-ukasar-ukasar- urapon) + (ukasar-ukasarukasar- urapon)
(ii) (ukasar-ukasar-ukasar-ukasar- urapon) – (ukasar-ukasarukasar)
(iii) (ukasar-ukasar-ukasar-ukasar- urapon) × (ukasar-ukasar)
(iv) (ukasar-ukasar-ukasar-ukasar- ukasar-ukasar-ukasar-ukasar) ÷ (ukasar- ukasar)
Answer:
Understanding the Gumulgal number system, which counts in groups of 2 using the following number names:
urapon = 1
ukasar = 2
ukasar-urapon = 3 (2 + 1)
ukasar-ukasar = 4 (2 + 2)
ukasar-ukasar-urapon = 5 (2 + 2 + 1)
ukasar-ukasar-ukasar = 6 (2 + 2 + 2) and so on…
► Converting Gumulgal terms to Hindu numerals:
(ukasar-ukasar-ukasar-ukasar-urapon) → 2 + 2 + 2 + 2 + 1 = 9 (ukasar-ukasar-ukasar-urapon) → 2 + 2 + 2 + 1 = 7
(ukasar-ukasar) → 2 + 2 = 4
(ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) → 8 ukasar = 8 × 2 = 16
► Performing Arithmetic Operations:
(i) Addition : 9 + 7 = 16
→ Convert 16 back into Gumulgal style:
8 ukasar → ukasar-ukasar-ukasar-ukasar- ukasar-ukasar-ukasar-ukasar

(ii) Subtraction : 9 – 6 = 3 → 2 + 1 = ukasar-urapon

(iii) Multiplication : 9 × 4 = 36 → Break 36 as 2 + 2 + 2 + … 18 times =18 ukasar → ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukaar- ukasar-ukasar-ukasar-ukasar-ukasar-ukasar

(iv) Division : 16 ÷ 4 = 4 → 2 + 2 = ukasar-ukasar

Question 3.
Identify the features of the Hindu number system that make it efficient when compared to the Roman number system.
Answer:
Features that Make the Hindu Number System Efficient than Roman number system:

• Uses a place value system where the position of a digit determines its value.
• Includes zero (0) as both a digit and a placeholder.
• Needs only 10 symbols (0 – 9) to represent any number.
• Allows for easy and quick arithmetic operations (addition, subtraction; etc.).
• Unambiguous and compact representation of even very large numbers.
• Forms the base for modern mathematics and science.
• Globally accepted and used in all fields today.

Question 4.
Using the ideas discussed in this section, try refining the number system you might have made earlier.
Answer:
After learning from this chapter, I improved my number system as follows:

• I improved my number system by making it a base-5 system.
• It uses five symbols: A, B, C, D, E (where A = 0, B = 1, …, E = 4).
• Each position from right to left represents powers of 5 (1, 5, 25, 125…).
• I added place value, so the same symbol has different values based on its position.
• Including A as zero helps avoid confusion and allows writing large numbers easily.
• This system is now more compact, clear and good for calculations, just like the Hindu number system.

3. THE IDEA OF A BASE
Figure it Out : Page : 62

Question 1.
Represent the following numbers in the Egyptian system: 10458, 1023, 2660, 784, 1111, 70707.
Answer:

Question 2.
What numbers do these numerals stand for?

Answer:
(i) 100 + 100 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 1 + 1 + 1 + 1 + 1 + 1
= 200 + 70 + 6 = 276

(ii) 1000 + 1000 + 1000 + 1000 + 100 + 100 + 100 + 10 + 10 + 1 + 1 = 4000 + 300 + 20 + 2
= 4322

Figure it Out : Page: 63

Question 1.
Write the following numbers in the above base-5 system using the symbols in Table 2: 15, 50,137, 293, 651.
Answer:

Question 2.
Is there a number that cannot be represented in our base-5 system above? Why or why not?
Answer:
No, there is no number that cannot be represented in our base-5 system.

Because:

• A base-5 system uses digits A, B, C, D, E (which stand for 0 to 4).
• Any number, no matter how big, can be written using combinations of these symbols and place values based on powers of 5 (1,5,25,125…).
• Since there is no upper limit on how many places we can use, we can represent every whole number.

Question 3.
Compute the landmark numbers of a base-7 system. In general, what are the landmark numbers of a base-n system?
Answer:
The landmark numbers of a base-7 system:
In a base-7 system, the landmark numbers are powers of 7:
7⁰ = 1
7¹ = 7
7² = 49
7³ = 343
7⁴ = 2401
7⁵ = 16807 …and so on.
The landmark numbers of a base-n number system are the powers of n starting from n⁰ = 1, n, n², n³,…

Figure it Out : Page : 65

Question 1.
Add the following Egyptian-numerals:

Answer:

Question 2.
Add the following numerals that are in the base-5 system that we created:

Answer:

Figure it Out : Page : 69 – 70

Question 1.
Can there be a number whose representation in Egyptian numerals has one of the symbols occurring 10 or more times? Why not?
Answer:
No, there cannot be a number whose representation in Egyptian numerals has one of the symbols occurring 10 or more times.
Because :

• The Egyptian number system is an additive system, not a place value system.
• Symbols are repeated to add up values, but each symbol is used at most 9 times.
• The system had only a limited number of symbols, each of which was repeated no more than 9 times in writing any number.
  So, whenever a symbol would be needed 10 times or more, Egyptians would move to the next higher symbol instead of repeating it, making their writing more compact and systematic.

Question 2.
Create your own number system of base 4, and represent numbers from 1 to 16.
Answer:
I have created my own number system called the Quad-Code System, which is based on base-4. In this system, I use four special symbols instead of regular digits:
A = 0 B = 1 C = 2 D = 3
In base-4, the place values increase as powers of
4. So, the rightmost place is 4⁰ = 1 = 1, the next is 4¹ = 4 and then 4² = 16 and so on. Using this system, I can write any number using just these four symbols.

Here is how I write numbers from 1 to 16 :

• 1 is written as B
• 2 is written as C
• 3 is written as D
• 4 is written as BA
• 5 is written as BB
• 6 is written as BC
• 7 is written as BD
• 8 is written as CA
• 9 is written as CB
• 10 is written as CC
• 11 is written as CD
• 12 is written as DA
• 13 is written as DB
• 14 is written as DC
• 15 is written as DD
• 16 is written as ABA

Question 3.
Give a simple rule to multiply a given number by 5 in the base-5 system that we created.
Answer:
The simple rule to multiply a number by 5 in the base-5 system I created using symbols (A = 0, B = 1, C = 2, D = 3, E = 4):
Rule : Add a zero (A) at the end of the number In base-5, multiplying any number by 5 is the same as shifting its digits one place to the left and adding A (zero) at the right end – just like adding a zero in base-10 when multiplying by 10.
Example :
Let’s take the number BC (which is 1 × 5 + 2 = 7 in decimal)

Now multiply by 5, just add A at the end to get BCA. BCA in base-5 = 7 × 5 = 35 in decimal
Because in base-5, the digits shift just like in base-10. Adding a zero (A) multiplies the number by the base itself, i.e., 5.

4. PLACE VALUE REPRESENTATION
Figure it Out : Page : 73

Question 1.
Represent the following numbers in the Mesopotamian system –
(i) 63
(ii) 132
(iii) 200
(iv) 60
(v) 3605
Answer:

Page : 76

Question 1.
Represent the following numbers using the Mayan system:
(i) 77
(ii) 100
(iii) 361
(iv) 72.1
Answer:
Mayan System Representation of Numbers
1. Convert to base-20: Divide the number by 20.

2. Determine the remainder and quotient : The quotient becomes the next level, and the remainder the value for the current level.

3. Represent with symbols : Convert the quotient and remainder into Mayan symbols.
(i) For 77:
(i) 77 ÷ 20 = 3 remainder 17
(ii) The remainder 17 would be at the bottom (ones place).
(iii) The quotient 3 would be in the next level (twenties place).
(iv) Result : This would be written with a bar and two dots for 3 at the top, and three bars and two dots for 17 at the bottom.

(ii) For 100 :

• 100 ÷ 20 = 5 remainder 0.
• Result: A bar (5) at the top and a shell (0) at the bottom.

(iii) For 361 :

• 361 ÷ 20 = 18 remainder 1.
• 18 ÷ 20 = 0 remainder 18.
• Result: This would be written with two bars and three dots for 18 at the top, and two bars and three dots for 18 at the bottom.

(iv) For 721 :

• 721 ÷ 20 = 36 remainder 1.
• 36 ÷ 20 = 1 remainder 16.
• 1 ÷ 20 = 0 remainder 1.
• Result: This would be written with two bars and three dots for 18 at the bottom, a bar and one dot for 16 in the middle, and two dots for 1 at the top.

Figure it Out : Page : 80

Question 1.
Why do you think the Chinese alternated between the Zong and Heng symbols? If only the Zong symbols were to be used, how would 41 be represented? Could this numeral be interpreted in any other way if there is no significant space between two successive positions?
Answer:
Using Zong and Heng symbols:

• The Chinese number system used Zong (vertical) and Heng (horizontal) symbols to show place value clearly.
• They alternated the direction of the symbols at each place (units, tens, hundreds, etc.) to avoid confusion when reading the number.
• This made it easier to know which digit belonged to which place even when spaces were small or missing.

► If only Zong symbols were used, how would 41 be written?
In Chinese system : 41 = 4 tens and 1 unit. Using only Zong symbols, it would be written as: (Zong for 4) followed by (Zong for 1) → looks like: IIII IWithout alternating the symbol direction or keeping proper spacing: IIII I could be misread as 5 (i.e., 1 five) instead of 41.
The lack of direction or spacing removes the clue that one part is “tens” and the other is “units”.

► The Chinese alternated between Zong and Heng symbols to make place values visually clear and easy to read, especially in handwritten or closely packed texts. Without this, numbers like 41 could easily be misunderstood.

Question 2.
Form a base-2 place value system using ‘ukasar’ and ‘urapon’ as the digits. Compare this system with that of the Gumulgal’s.
Answer:
To form a base-2 place value system using ‘ukasar’ and ‘urapon’, we assign:

• ‘ukasar’ = 0
• ‘urapon’ = 1
  This system works just like the binary number system, where each position from right to left represents increasing powers of 2. For example:
• The first place is 2⁰ = 1
• The next is 2¹ = 2
• Then 2² = 4 and so on.

So, we can represent numbers like this :

• The number 1 is written as urapon
• The number 2 is written as urapon ukasar
• The number 3 is urapon urapon
• The number 4 becomes urapon ukasar ukasar Each position tells us how many of that power of 2 we have and we use ukasar for 0 and urapon for 1.

Comparison: If we compare this to the Gumulgal number system, there’s a big difference. The Gumulgal system doesn’t use place value. Instead, it adds groups of 2s (ukasar) and Is (urapon) to build numbers. For example, to make 7, they would say something like ukasar-ukasar- ukasar-urapon (2 + 2 + 2 + 1).

So the main difference is the base-2 system with ukasar and urapon is a place value system – more efficient and better for large numbers. The Gumulgal system is group-based and additive, which is fine for small numbers but becomes confusing as numbers get bigger.

Question 3.
Where in your daily lives, and in which professions, do the Hindu numerals, and 0, play an important role? How might our lives have been different if our number system and 0 hadn’t been invented or conceived of?
Answer:
Hindu numerals and the digit 0 are used every day in our lives-for telling time, counting money, reading prices, doing math in school and writing phone numbers. Many professions like banking, teaching, engineering and science rely heavily on this number system. Zero plays a key role in place value and calculations, making big numbers easy to write and understand.

If zero and the Hindu number system had not been invented, life would be very difficult. We would struggle to calculate, trade or even write dates properly. Modern technology like comput¬ers and calculators would not exist, slowing down progress in every field.

Question 4.
The ancient Indians likely used base 10 for the Hindu number system because humans have 10 fingers, and so we can use our fingers to count. But what if we had only 8 fingers? How would we be writing numbers then? What would the Hindu numerals look like if we were using base 8 instead? Base 5? Try writing the base- 10 Hindu numeral 25 as base-8 and base-5 Hindu numerals, respectively. Can you write it in base-2?

Answer:
If humans had only 8 fingers :
We would probably have developed a base-8 number system instead of base-10. Just like we now count from 0 to 9 in base-10, we would count from 0 to 7 in base-8. All our numerals, math and calculations would be based on powers of 8.

Conversion of the base-10 number 25 :
► In base-8:
25 ÷ 8 = 3 remainder 1
3 ÷ 8 = 0 remainder 3
So, 25 in base-8 = 31

► In base-5:
25 ÷ 5 = 5 remainder 0
5 ÷ 5 = 1 remainder 0
1 ÷ 5 = 0 remainder 1
So, 25 in base-5 = 100

► In base-2 (binary):
25 ÷ 2 = 12 remainder 1
12 ÷ 2 = 6 remainder 0
6 ÷ 2 = 3 remainder 0
3 ÷ 2 = 1 remainder 1
1 ÷ 2 = 0 remainder 1

A Story of Numbers Class 8 Extra Questions

Multiple Choice Questions

Question 1.
What numeral is represented by ‘v’ in Roman System?
(a) 1
(b) 5
(c) 10
(d) 100
Solution:
In Roman system ‘v’ represents 5.
(b) 5

Question 2.
Which of the following is incorrectly matched?

Answer:
(d) M 250, is incorrectly matched as M in Roman System represents 1000.

Question 3.
Which of the following Roman Numbers represents the number 1222?
(a) MIIXXII
(b) MCCXII
(c) MCCXXII
(d) MCCCXI
Solution:
Here, 1222 = 1000 + 100 + 100 + 20 + 2
= 1000 + 100 + 100 + 10 + 10 + 1 + 1
= M + C + C + X + X + I + I
= MCCXXII
(c) MCCXXII

Question 4.
Which of the following Roman Numbers represents the number 2999?
(a) MMCMXCIX
(b) IMMM
(c) CMMIC
(d) MMCMCIX
Solution:
Here, 2999 = 1000 + 1000 + 900 + 90 + 9
= MMCMXCIX.
(a) MMCMXCIX

Question 5.
Which of the following Roman Numerals is represented by 2362?
(a) MMCCCLXII
(b) MMMCCLXII
(c) MMCCCLLII
(d) MMCCLCXII
Solution:
Here,
2362 = 1000 + 1000 + 100 + 100 + 100 + 50 + 10 + 1 + 1 = MMCCCLXII
(a) MMCCCLXII

Assertion and Reasoning

(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).
(b) Assertion (A) and Reason (R) both are true but Reason (R) is not the correct explanation for the Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.

Question 1.
Assertion (A) : In Roman numerals LXII represents 62 of Hindu numerals.
Reason (R) : In Roman Numerals L represents 50, X represents 10 and I represents 1 of Hindu System.
Answer:
(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).

Question 2.
Assertion (A) : Roman Numeral CDXIII = Hindu Numeral 413
Reason (R) : ∵, C = 100, D = 500, XIII = 13.
∵, So CD = 400 and CDXIII = 413.
Answer:
(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).

Case Based Questions

Question 1.
Aarya was writing some numbers on a white board. She had put some operation sign in between the numbers.

Based on the above answer the following:
(i) In which numeral these numbers are written?
(ii) What will come in place of blank space in (A)?
(iii) What will come in place of blank space in (B)?
(iv) What will come in place of blank space in (C)?
(v) What will come in place of blank space in (D)?
Answer:
(i) These numbers are of Egyptian number system.

Students can refer to BSE Odisha Class 8 Math Solution and Ganita Prakash Chapter 2 Power Play Class 8 Question Answer to understand textbook questions step by step.

Class 8 Maths Chapter 2 Power Play Solutions

Ganita Prakash Class 8 Chapter 2 Solutions

Class 8 Maths Ganita Prakash Chapter 2 Solutions Power Play

Page : 19

Question 1.
How many times can you fold it over and over?
Estu says “I heard that a sheet of paper can’t be folded more than 7 times”.
Roxie replies “What if we use a thinner paper, like a newspaper or a tissue paper?”
Try it with different types of paper and see what happens.

Answer:
Any paper, how big or small, thick or thin, cannot be folded for more than 7 times.

Question 2.
Say you can fold a sheet of paper as many times as you wish. What would its thickness be after 30 folds? Make a guess.
Let us find out how thick a sheet of paper will be after 46 folds. Assume that the thickness of the sheet is 0.001 cm.
Answer:
We know that,
Thickness after 1^st fold = 2t, if t is the initial thickness.
∴, Thickness after 2nd fold = 2²t.
In this way, we can find that after 30 folds, the thickness = 2³⁰t.

Page : 20

Question 1.
The following table lists the thickness after each fold. Observe that the thickness doubles after each fold.

(We use the sign ‘≈’ to indicate ‘approximately equal to’.)
After 10 folds, the thickness is just above 1 cm (1.024 cm).
After 17 folds, the thickness is about 131 cm (a little more than 4 feet).
Answer:

Fold	Thickness
1	0.002 cm
2	0.004 cm
3	0.008 cm
4	0.016 cm
5	0.032 cm
6	0.064 cm
7	0.128 cm
8	0.256 cm
9	0.512 cm
10	1.024 cm
11	2.048 cm
12	4.096 cm
13	8.192 cm
14	16.384 cm
15	32.768 cm
16	65.536 cm
17	≈ 131 cm

Question 2.
Now, what do you think the thickness would be after 30 folds? 45 folds? Make a guess.
Answer:
After 30 folds :
Thickness = 0.002 × 2³⁰
= 0.002 × 536,870,912
= 1073741824

After 45 folds :
Thickness = 0.002 × 2⁴⁵
= 0.002 × 35184372088832
= 3.518 × 10¹³ m

Question 3.
Fill the table below.

After 26 folds, the thickness is approximately 670 m. Burj Khalifa in Dubai, the tallest building in the world, is 830 m tall.

After 30 folds, the thickness of the paper is about 10.7 km, the typical height at which planes fly. The deepest point discovered in the oceans is the Mariana Trench, with a depth of 11 km.

It might be hard to digest the fact that after just 46 folds, the thickness is more than 7,00,000 km. This is the power of multiplicative growth, also called exponential growth. Let us analyse the growth. We have seen that the thickness doubles after every fold.

Notice the change in thickness after two folds. By how much does it increase?
After any 3 folds, the thickness increases 8 times (= 2 × 2 × 2). Check if that is true. Similarly, from any point, the thickness after 10 folds increases by 1024 times (= 2 multiplied by itself 10 times), as shown in the table below.

Answer:

Fold	Thickness
21	20.8 m
22	41.6 m
23	83.2 m
24	166.4 m
25	332.8 m
26	665.6 m
27	≈ 1.3 km
28	2.6 km
29	5.2 km
30	10.4 km
31	20.8 km
32	41.6 km
33	83.2 km
34	166.4 km
35	332.8 km
36	665.6 km
37	1331.2 km
38	2662.4 km
39	5324.8 km
40	10,649.6 km
41	21,299.2 km
42	42,598.4 km
43	85,196.8 km
44	170,393.6 km
45	340,786.6 km
46	681,572.4 km
47	1,363,144.8 km

We can notice here that with just 30 folds of a paper, which is just 0.002 cm thick, the thickness of the folded paper will be approximately 10.4 km, which is the typical height at which planes fly This thickness is nearly another important milestone of the civilization which is the deepest point discovered in the ocean, i.e., the Mariana Trench, with a depth of about 11 km.

Page : 22

Question 1.
Which expression describes the thickness of a sheet of paper after it is folded 10 times? The initial thickness is represented by the letter-number υ.
(i) 10υ
(ii) 10 + υ
(iii) 2 × 10 × υ
(iv) 2¹⁰
(v) 2¹⁰υ
(vi) 10²υ
Some more examples of exponential notation:
4 × 4 × 4 = 4³ = 64.
(-4) × (-4) × (-4) = (-4)³ = -64.
Similarly,
a × a × a × b × b can be expressed as a³b² (read as a cubed b squared).
a × a × b × b × b × b can be expressed as a²b⁴ (read as a squared b raised to the power 4).
Remember that 4 + 4 + 4 = 3 × 4 = 12, whereas 4 × 4 × 4 = 4³ = 64.
Solution:
Here the initial thickness is υ.
Thickness after 1st fold = 2υ
Thickness after 2nd fold = 2²υ
Thickness after 3rd fold = 2³υ …….
Thickness after 10th fold = 2¹⁰υ
So, (υ) 2¹⁰υ describes the thickness of a sheet of paper after it is folded 10 times.

Question 2.
Express the number 32400 as a product of its prime factors and represent the prime factors in their exponential form.
Solution:
Prime factors of 32400 :

So, 32400 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5
= 2⁴ × 3⁴ × 5², which is the required exponential form.

Question 3.
What is (-)⁵? Is it positive or negative? What about (-1)⁵⁶?
Solution:
(-1)⁵ = (-1) × (-1) × (-1) × (-1) × (-1)
= (-1)
It is negative.
(-1)⁵⁶ = 1, it is positive.

Question 4.
Is (-2)⁴ = 16? Verify.
Solution:
(-2)⁴ = (-2) × (-2) × (-2) × (-2)
= (+4) × (+4) = (+16) = 16

Figure it Out : Page : 22 – 23

Question 1.
E×press the following in exponential form:
(i) 6 × 6 × 6 × 6
(ii) y × y
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) a × a × a × c × c × c × c × d
Solution:
(i) 6 × 6 × 6 × 6 = (6)⁴
(ii) y × y = (y)²
(iii) b × b × b × b = (b)⁴
(iv) 5 × 5 × 7 × 7 × 7 = (5)² × (7)³
(v) 2 × 2 × a × a = (2)² × (a)²
(vi) a × a × a × c × c × c × c × d = (a)³ × (c)⁴ × (d)¹

Question 2.
Express each of the following as a product of powers of their prime factors in exponential form.
(i) 648
(ii) 405
(iii) 540
(iv) 3600
Solution:
(i) 648 Prime factors of 648 :

So, 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3
= 2³ × 3⁴

(ii) 405
Prime factors of 405 :

∴, 405 = 3 × 3 × 3 × 3 × 5
= (3)⁴ × (5) or simply 3⁴ × 5

(iii) 540
Prime factors of 540 :

∴, 540 = 2 × 2 × 3 × 3 × 3 × 5
= 2² × 3³ × 5¹

(iv) 3600
Prime factors of 3600 :

∴, 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 2⁴ × 3² × 5²

Question 3.
Write the numerical value of each of the following:
(i) 2 × 10³
(ii) 7² × 2³
(iii) 3 × 4⁴
(iv) (-3)² × (-5)²
(v) 3² × 10⁴
(vi) (-2)⁵ × (-10)⁶
Solution:
(i) 2 × 10³ = 2 × 10 × 10 × 10 = 2000

(ii) 7² × 2³ = 7 × 7 × 2 × 2 × 2 = 49 × 8 = 392

(iii) 3 × 4⁴ = 3 × 4 × 4 × 4 × 4 = 3 × 256 = 768

(iv) (-3)² × (-5)² = (-3) × (-3) × (-5) × (-5) = 9 × 25 = 225

(v) 3² × 10⁴ = 3 × 3 × 10 × 10 × 10 × 10
= 9 × 10,000 = 90,000

(vi) (-2)⁵ × (-10)⁶ = (-2) × (-2) × (-2) × (-2) × (-2) × (-10) × (-10) × (-10) × (-10) × (-10) × (-10)
= (-32) × 10,00,000 = -3,20,00,000

Page : 23

Question 1.
Three daughters with curious eyes,
Each got three baskets – a kingly prize.
Each basket had three silver keys,
Each opens three big rooms with ease.
Each room had tables – one, two, three,
With three bright necklaces on each, you see.
Each necklace had three diamonds so fine…
Can you count these stones that shine?
Hint : Find out the number of baskets and rooms.

Solution:
Number of daughters = 3
Since each daughter has got 3 baskets,
So number of baskets in total = 3 × 3 = 9
Each basket has three silver keys,
So, number of keys in total = 9 × 3 = 27
Since each silver key opens three big rooms,
Total number of rooms opened by these keys = 27 × 3 = 81
Each room has three tables,
Total number of tables = 81 × 3 = 243
Each table has three necklaces on it,
So, total number of necklaces = 243 × 3 = 729
Now, each necklace has three diamonds,
Hence, total number of diamonds = 729 × 3 = 2187

Question 2.
How many rooms were there altogether?
The information given can be visualised as shown below.

Solution:
From the diagram, the number of rooms is 3⁴. This can be computed by repeatedly multiplying 3 by itself,
3 × 3 = 9.
9 × 3 = 27.
27 × 3 = 81.
81 × 3 = 243.
There were 81 rooms altogether as depicted by the above solution.

Question 3.
How many diamonds were there in total? Can we find out by just one multiplication using the products above?
The number of diamonds is 3 × 3 × 3 × 3 × 3 × 3 × 3 = 3⁷.
We can write
3⁷ = (3 × 3 × 3 × 3) × (3 × 3 × 3)
We had computed till 3⁴. To find 3⁷, we can just multiply 3⁴ (= 81) with 3³(= 27).
= 3⁴ × 3³

= 81 × 27 = 2187
Solution:
Total number of diamonds = 2187
We can find this number by multiplying the number of necklaces by 3, i.e., 729 × 3 = 2187
Or by multiplying total number of tables by 9, i.e., 243 × 9 = 2187
Or by multiplying total number of rooms by 27, i.e., 243 × 27 = 2187
Note : There can be more ways.

Page : 24

Question 1.
3⁷ can also be written as 3² × 3⁵. Can you reason out why?
This can be easily extended to products where exponents are the same letter-numbers.
Solution:
3⁷ can also be written as 3² × 3⁵ as :
3² × 3⁵ = 3 × 3 × 3 × 3 × 3 × 3 × 3 (in expanded form)
= 3⁷ (in exponential form)

Question 2.
Write the product p⁴ × p⁶ in exponential form.
p⁴ × p⁶ = (p × p × p × p) × (p × p × p × p × p × p) = p¹⁰
We can generalise this to –
n^a n^b = n^{a + b} where a and b are counting numbers
Solution:
p⁴ × p⁶ = p × p × p × p × p × p × p × p × p × p_(expanded form)
= p¹⁰ (exponential form)

Question 3.
Use this observation to compute the following.
(i) 2⁹
(ii) 5⁷
(iii) 4⁶
Solution:
(i) 2⁹ = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)
= (8) × (8) × (8) = (64) × (8) = 512

(ii) 5⁷ = (5 × 5 × 5 × 5) × (5 × 5 × 5)
= 5⁴ × 5³ = 625 × 125 = 78125

(iii) 4⁶ = (4 × 4) × (4 × 4) × (4 × 4)
= 16 × 16 × 16 = 256 × 16 = 4096

Question 4.
Is 2¹⁰ also equal to (2⁵)²? Write it as a product.
Solution:
2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= (2 × 2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2) [Regrouping in group of 5]
= (2⁵) × (2⁵) = (2⁵)²
Note : (a^m)ⁿ = (aⁿ)^m = a^{m × n} = a^mn), where ‘n’ and ‘n’ are counting numbers.

Question 5.
Write the following expressions as a power of a power in at least two different ways:
(i) 8⁶
(ii) 7¹⁵
(iii) 9¹⁴
(iv) 5⁸
Solution:
(i) 8⁶ = (8²)³ Also, 8⁶ = (8²)³
(ii) 7¹⁵ = (7³)⁵ Also, 7¹⁵ = (7⁵)³
(iii) 9¹⁴ = (9²)⁷ Also, 9¹⁴ = (9⁷)²
(iv) 5⁸ = (5²)⁴ Also, 5⁸ = (5⁴)²

Page : 25

Question 1.
In the middle of a beautiful, magical pond lies a bright pink lotus. The number of lotuses doubles every day in this pond. After 30 days, the pond is completely covered with lotuses. On which day was the pond half full?
If the pond is completely covered by lotuses on the 30th day, how much of it is covered by lotuses on the 29th day?
Since the number of lotuses doubles every day, the pond should be half covered on the 29th day.

Solution:
Since the pond is fully filled with lotuses in 30 days.
Hence, the pond should be half covered on the 29th day as number of flowers doubles every day.

Question 2.
Write the number of lotuses (in exponential form) when the pond was –
(i) fully covered
(ii) half covered
Solution:
(i) The number of lotuses when the pond was fully covered = 2³⁰
(ii) The number of lotuses when the pond was half covered = 2²⁹

Question 3.
There is another pond in which the number of lotuses triples every day. When both the ponds had no flowers, Damayanti placed a lotus in the doubling pond. After 4 days, she took all the lotuses from there and put them in the tripling pond. How many lotuses will be in the tripling pond after 4 more days?
After the first 4 days, the number of lotuses is 1 × 2 × 2 × 2 × 2 = 2⁴.
After the next 4 days, the number of lotuses is 2 × 3 × 3 × 3 × 3 = 2⁴ × 3⁴.
Solution:
Number of lotuses in the doubling pond after 4 days = 2⁴ = 16
Number of lotuses in the tripling pond after 4 more days = 2⁴ × 3⁴ = 16 × 81 = 1296

Question 4.
What if Damayanti had changed the order in which she placed the flowers in the lakes? How many lotuses would be there?
1 × 3⁴ × 2⁴ = (3 × 3 × 3 × 3) ×(2 × 2 × 2 × 2).
Solution:
If Damayanti had changed the order then she will place the lotus in the tripling pond first and then after 4 days, she will place all lotuses in the doubling pond.
So, the number of lotuses after first 4 days = (3)⁴ = 81
and the number of lotuses after next 4 days = (3)⁴ × (2)⁴ = 81 × 16 = 1296

Question 5.
Can this product be expressed as an exponent mⁿ, where m and n are some counting numbers?
Solution:
Here the product is 3⁴ × 2⁴ which can also be written as
3 × 3 × 3 × 3 × 2 × 2 × 2 × 2
By regrouping we can write them as:
(3 × 2) × (3 × 2) × (3 × 2) × (3 × 2) = 6 × 6 × 6 × 6 = 6⁴, which is the required “mⁿ” form.
Note : a^m × b^m = (a × b)^m = (ab)^m, where ‘m is- a counting number.

Question 6.
Simplify \(\frac{10^4}{5^4}\) and write it in exponential form.
In general, we can show that \(\frac{m^a}{n^a}\) = \(\left(\frac{m}{n}\right)^a .\).
Solution:

= 2 × 2 × 2 × 2 = 2⁴
Note : \(\frac{a^m}{b^m}\) = (\(\frac{a}{b}\))^m
, where ‘m’ is a counting number.

Page : 26

Question 1.
Estu has 4 dresses and 3 caps. How many different ways can Estu combine the dresses and caps?
For each cap, he can choose any of the 4 dresses, so for 3 caps, 4 + 4 + 4 = 4 × 3 = 12 combinations are possthle. We can also look at it as – for each dress, Estu can choose any of the 3 caps, so for 4 outfits, 3 + 3 + 3 + 3 = 3 × 4 = 12 combinations are possible.

Solution:
Estu can combine 4 dresses and 3 caps in 4 × 3 ways = 12 ways.

Question 2.
Roxie has 7 dresses, 2 hats, and 3 pairs of shoes. How many different ways can Roxie dress up?

Solution:
Roxie has 7 dresses, 2 hats, and 3 pairs of shoes.
Number of different ways Roxie can dress up = 7 × 2 × 3 = 42.

Question 3.
Estu and Roxie came’across a safe containing old stamps and coins that their great-grandfather had collected. It was secured with a 5-digit password. Since nobody knew the password, they had no option except to try every password until it opened. They were unlucky and the lock only opened with the last password, after they had tried all possible combinations. How many passwords did they end up checking?

Solution:
Getting the 5-digit password is the same as filling 5 empty boxes by 10 different objects if it comes after all the possible combinations.
So,
We can fill the first box in 10 ways, 2nd box in 10 ways, 3rd box in 10 ways, 4th box in 10 ways, and 5th box in 10 ways.
So, total number of possible combinations:
= 10 × 10 × 10 × 10 × 10 = 10⁵

Page : 27

Question 1.
How many passwords are possible with such a lock?
Solution:
There will be 10⁵ passwords.

Question 2.
Think about how many combinations are possible in different contexts. Some examples are-
(i) Pincodes of places in India – The Pincode of Vidisha in Madhya Pradesh is 464001. The Pincode of Zemabawk in Mizoram is 796017.
(ii) Mobile numbers.
(iii) Vehicle registration numbers.
Try to find out how these numbers or codes are allotted/generated.
Solution:
(i) Pincodes of places in India contain 6 digits, but the first digit cannot be a zero, though there is no restriction on the rest of the digits. Total number of combinations :

Also, there are no PINs like 100000, 200000, 300000, 400000, 500000, 600000, 700000, 800000, 900000.
So, total number of combinations for the pincode = 9 × 10⁵ – 9.

(ii) Ignoring the actual system followed while framing a mobile number, we can get the number of combinations as:
= 9 × 10⁹, as there are 10 digits in a mobile number.

(iii) Vehicle registration numbers in Delhi can be like DL4SAG7336.
So, the required number of possible combinations for vehicle registration number can be:
26 × 26 × 9 × 26 × 26 × 10 × 10 × 10 × 10 – 1
= 26⁵ × 9 × 10⁴ – 1
1 is subtracted as there will not be a registration number in India which will have all four zeroes at the end.

Question 3.
What is 2¹⁰⁰ ÷ 2²⁵ in powers of 2?
In a generalised form,
n^a ÷ n^b = n^{a – b},
where n ≠ 0 and a and b are counting numbers and a > b.
Solution:
2¹⁰⁰ ÷ 2²⁵ = (\(\frac{2^{100}}{2^{25}}\)) = 2^{100 – 25} = 22⁷⁵
Note : a^m ÷ aⁿ = a^{m – n}, where a ≠ 0 and m and n are counting numbers and m > n.

Page : 28

Question 1.
Why can’t n be 0?
Solution:
n cannot be zero as division by zero is not defined.

Question 2.
We have not covered the case when the exponent is 0; for example, what is 2⁰?
Let us define 2⁰ in a way that the generalised form above holds true.
2⁰ = 2^{4 – 4} = 2⁴ ÷ 2⁴ = \(\frac{2 \times 2 \times 2 \times 2}{2 \times 2 \times 2 \times 2}\) = 1.
In fact for any letter number a
2⁰ = 2^{a – a} = 2^a ÷ 2^a 1.
In general,
x^a ÷ x^a = x^{a – a}, and so
1 = x⁰,
where x ≠ 0 and a is a counting number.
Solution:
Let us write 2⁰ = 2 ^{5 – 5}

Therefore, 2⁰ = 1

Page : 29

Question 1.
Can we write 10³ = \(\frac{1}{10^{-3}}\)?
We can write,
\(\frac{1}{10^{-3}}\) = \(\frac{1}{1 / 10^3}\) = 1 ÷ \(\frac{1}{10^{-3}}\) – 1 × 10³ = 10³.
Similarly, 7² \(\frac{1}{7^{-2}}
\) = and 4^a = \(\).
In a generalised form,
n^-a = \(\) and n^a = \(\), where n ≠ 0
Consider the following general forms we have identified.

na × nb = na + b

(na)b = (nb)a = na × b

na ÷ nb = na – b

Solution:
Yes, we can write 10³ = \(\frac{1}{10^{-3}}\)
Note:
(i) n^a × n^b = n^{a + b}
(ii) (n^a)^b = (n^b)^a = n^{a × b}
(iii) n^a ÷ n^b = n^{a – b}

Question 2.
We had required a and b to be counting numbers. Can a and b be any integers? Will the generalised forms still hold true?
Solution:
Yes. But for the case of division, it must be non-zero integer.

Question 3.
Write equivalent forms of the following.
(i) 2^-4
(ii) 10^-5
(iii) (-7)^-2
(iv) (-5)^-3
(v) 10^-100
Solution:
(i) 2^-4 = \(\frac{1}{2^4}\)
(ii) 10^-5 = \(\frac{1}{10^5}\)
(iii) (-7)^-2 = \(\frac{1}{(-7)^2}\)
(iv) (-5)^-3 = \(\frac{1}{(-5)^3}\)
(v) 10^-100 = \(\frac{1}{10^100}\)

Question 4.
Simplify and write the answers exponential form.
(i) 2^-4 × 2⁷
(ii) 3² × 3^-5 × 3⁶
(iii) p³ × p^-10
(iv) 2⁴ × (-4)^-2
(v) 8^p × 8^q
Solution:
(i) 2^-4 × 2⁷ = 2^{– 4 + 7} = 2³

(ii) 3² × 3^-5 × 3⁶ = 3^{2 + (-5) + 6} = 3³

(iii) p³ × p^-10 = p^{3 – 10} = p^-7

(iv) 2⁴ × (-4)^-2 = \(\frac{2^4}{(-4)^2}\) = \(\frac{2^4}{(-4) \times(-4)}\) = \(\frac{2^4}{16}\) = \(\frac{2^4}{2^4}\) = 1

(v) 8^p × 8^q = 8^{p + q}

Page : 30

Question 1.
Can we say that 16384 (4⁷) is 16 (4²) times larger than 1,024 (4⁵)?
Yes, since 4⁷ ÷ 4⁵ = 4².
Solution:
16384 = 4⁷ = 4² × 4⁵ = 16 × 1024
So, we can definitely say that 16384 is 16 times larger than 1024.

Question 2.
How many times larger than 4^-2 is 4²?
Solution:
4² = 4^{– 2 + 4} = 4^-2 × 4⁴
∴ 4² is 4⁴ times larger than 4^-2.

Question 3.
Use the power line for 7 to answer the following questions.

Solution:

Question 4.
Write these numbers in the same way:
(i) 172,
(ii) 5642,
(iii) 6374.
Solution:
(i) 172 = 1 × 10² + 7 × 10¹ + 2 × 10⁰
(ii) 5642 = 5 × 10³ + 6 × 10² + 4 × 10¹ + 2 × 10⁰
(iii) 6374 = 6 × 10³ + 3 × 10² + 7 × 10¹ + 4 × 10⁰.

Question 5.
How can we write 561.903?
561.903 = (5 × 100) + (6 × 10) + 1 + (9 × \(\frac{1}{10}\)) + (0 × \(\frac{1}{100}\)) + (3 × \(\frac{1}{1000}\)).
Writing it using powers of 10, we have
561.903 = (5 × 10²) + (6 × 10¹) + (1 × 10⁰) + (9 × 10^-1) + (0 × 10^-2) + (3 × 10^-3).
Solution:
561.903 = 5 × 10² + 6 × 10¹ + 1 × 10⁰ + 9 × 10^-1 + 0 × 10^-2 + 3 × 10^-3

Page : 31

Question 1.
Write the large-number facts we read just before in this form.
Solution:
In scientific notation or scientific form (also called standard form), we write numbers as x × 10^y where x ≥ 1 and x < 10 is the coefficient and ‘y’, the exponent, is any integer.
For example :
(i) The sun is located
30,00,00,00,00,00,00,00,00,000 m from the centre of our Milky Way galaxy, i.e., 3.0 × 10²⁰ m.

(ii) The number of stars in our galaxy is 1,00,00,00,00,000, i.e., 1.0 × 10¹¹.

(iii) The mass of the Earth is 59,76,00,00,00,00,00,00,00,00,00,000 kg,
i.e., 5.976 × 10²⁴ kg.

Page : 32

Question 1.
Can you say which of the three distances is the smallest?

Solution:
The distance between the Sun and Saturn is 14,33,50,00,00,000 m = 1.4335 × 10¹² m.
The distance between Saturn and Uranus is
14,39,00,00,00,000 m = 1.439 × 10¹² m.
The distance between the Sun and Earth is
1,49,60,00,00,000 m = 1.496 × 10¹¹ m.
Amongst the given distances, the distance between the Sun and the Earth is the smallest.

Question 2.
The number line below shows the distance between the Sun and Saturn (1.4335 × 10¹² m). On the number line below, mark the relative position of the Earth. The distance between the Sun and the Earth is 1.496 × 10¹¹ m.

Solution:

Question 3.
Express the following numbers in standard form.
(i) 59,853
(ii) 65,950
(iii) 34,30,000
(iv) 70,04,00,00,000
Solution:
(i) 59,853 = 5.9853 × 10⁴
(ii) 65,950 = 6.595 × 10⁴
(iii) 34,30,000 = 3.43 × 10⁶
(iv) 70,04,00,00,000 = 7.004 × 10¹⁰

Page : 33

Question 1.
What would be the worth (in rupees) of the donated jaggery? What would be the worth (in rupees) of the donated wheat?

Solution:
Worth of jaggery (₹) = Roxie’s weight in kg × cost of 1 kg jaggery.
Worth of wheat (₹) = Estu’s weight in kg × cost of 1 kg wheat.

Question 2.
Make necessary and reasonable assumptions for the unknowns and find the answers. Remember, Roxie is 13 years old and Estu is 11 years old.
Solution:
Assuming Roxie’s weight to be 45 kg and the cost of 1 kg of jaggery to be ₹70, the worth of the donated jaggery is 45 × 70 = ₹3,150.

Assuming Estu’s weight to be 50 kg and the cost of 1 kg of wheat to be ₹50, the worth of donated wheat is 50 × 50 = ₹2,500.

Question 3.
Roxie wonders, “Instead of jaggery if we use 1-rupee coins, how many coins are needed to equal my weight?”. How can we find out?
Solution:
Weight of a 1-rupee coin = 3.76 grams (approx.)
Let us assume Roxie’s weight be 45 kg.
So, the number of 1-rupee coins = \(\frac{45 \mathrm{~kg}}{3.76 \mathrm{gram}}\) = \(\frac{45000 \text { gram }}{3.76 \text { gram }}\) = 11,968.08511 ≈ 11,968, ₹1 coins

Page : 34

Question 1.
Would the number of coins be in hundreds, thousands, lakhs, crores, or even more? Make an instinctive guess.
Solution:
The number of coins will be in thousands.

Question 2.
Find the answer by making necessary and reasonable assumptions and approximations for the unknowns. Remember, we are not looking for an exact answer but a reasonably close estimate.

Estu asks, “What if we use 5-rupee coins or 10-rupee notes instead?
How much money could it be?”
Solution:
The number of coins = 11,968 (approx.)
Estu says, “When I become an adult, I would like to donate notebooks worth my weight every year”. Roxie says, “When I grow up, I would like to do annadana (offering grains or meals) worth my weight every year”.

Question 3.
How many people might benefit from each of these offerings in a year? Again, guess first before finding out.
Roxie and Estu overheard someone saying- “We did pādayātra for about 400 km to reach this place! We arrived early this morning.”
Solution:
Assuming Roxie’s weight to be 45 kg, she will donate 45 kg of grains. Further assuming that one person requires 15 kg of grains for a month, it will help 1 person for 3 months.

Assuming Estu’s weight to be 50 kg, he will donate 50 notebooks. Further assuming that one person requires 10 notebooks in a year for academic works, it will help 5 children.

Roxie and Estu overheard someone saying – “We did padayatra for about 400 km to reach this place! We arrived early this morning.”

Question 4.
How long ago would they have started their journey?
Solution:
Assuming that a person can walk at the speed of about 4 km/hour, we can find that they have walked for 100 hours to cover 400 km distance:

Assuming that they must have rested during their journey (Padayatra) for about 8 hours each day. So, they have travelled: (24 – 8) = 16 hours each day.
So, number of days they have travelled
= \(\frac{100}{16}\) = 6.25 days
Thus, we can conclude that they would have started their journey 6 days ago.

Page : 35

Question 1.
How many times can a person circumnavigate (go around the world) the Earth in their lifetime if they walk nonstop? Consider the distance around the Earth as 40,000 km.
Solution:
Assuming that a person can walk at the speed of 4 km/h.
If the person walks non-stop, then the person requires:
\(\frac{40,000}{4}\) hours = 10,000 hours to circumnavi-gate the Earth.

Again, let us assume that a person lives for 100 years. Then, in his lifetime he has 100 × 365 × 24 hours.
So, hypothetically the person will circumnavigate for:
= \(\frac{8760}{100}\) = 87.6 times.
Hence, the person will circumnavigate the Earth for 87 times approximately (hypothetically).

Roxie tells Estu about a science- fiction novel she is reading where they build a ladder to reach the moon,
“… I wonder if we actually had a ladder like that, how many steps would it have?”.

Question 2.
What do you think? Make an instinctive guess first.
Solution:
The average distance between the Earth and the Moon is approximately 384,400 km. If
we place steps at \(\frac{1}{2}\) m each, then there will be:
384,400 × 1000 × 2 steps
= 768,800,000 steps = 7.688 × 10⁸ steps.

Page : 36

Question 1.
We have to find out how many 20 cm make 3,84,400 km.
If we calculate the value, we get the result as 1,92,20,00,000 steps, which is 192 crore and 20 lakh steps or 1 billion 922 million steps. The fixed increase in the distance from the earth with each step (a 20 cm gain after each step) is called linear growth.

To cover the distance between the Earth and the Moon, it takes
1,92,20,00,000 steps with linear growth whereas it takes just 46 folds of
a piece of paper with exponential growth! Linear growth is additive,
whereas exponential growth is multiplicative.

Some examples of exponential growth we have seen earlier in this chapter are ‘The Stones that Shine’, ‘Magical Pond’, ‘How Many Combinations’. We shall explore more such interesting examples in a later chapter and also in the next grade.
Solution:
To find out how many 20 cm make 3,84,400 km

= 1.922 × 10⁹
So, 1.922 × 10⁹, 20 cm make 3,84,400 km.

Question 2.
Can you come up with some examples of linear growth and of exponential growth?
Solution:
Some examples of Linear Growth :
(i) Aarya deposits ₹10 everyday in a piggy bank. The money accumulated will be a linear growth as money in the piggy bank will have a sequence as ₹10, ₹20, ₹30, ₹40, ₹50, ₹60, ₹70, ….

(ii) Distance covered by a car which gives a mileage of 12 km by using every litre of fuel. The sequence will be 12 km, 24 km, 36 km, 48 km, 60 km, 72 km, etc.

Some examples of Exponential Growth:
(i) The spread of a virus generally follows exponential growth.
(ii) Savings with a bank increases exponentially when the interest is compounded using compound interest.

Page : 38 – 39

Question 1.
With a global human population of about 8 × 10⁹ and about 4 × 10⁵ African elephants, can we say that there are nearly 20,000 people for every African elephant?
Solution:
To get the number of people for each of the African elephant we divide global human population by the population of African elephants.
So, we obtain \(\frac{8 \times 10^9}{4 \times 10^5}\) = 2 × 10⁴ = 20,000
So, there are nearly 20,000 people for every African elephant.

Question 2.
Calculate and write the answer using scientific notation:
(i) How many ants are there for every human in the world?
(ii) If a flock of starlings contains 10,000 birds, how many flocks could there be in the world?
(iii) If each tree had about 10⁴ leaves, find the total number of leaves on all the trees in the world.
(iv) If you stacked sheets of paper on top of each other, how many would you need to reach the Moon?
Solution:
(i) Population of ants globally = 20 padma
= 20 quadrillion = 2 × 10¹⁶
Global population of humans = 8 × 10⁹
∴, Number of ants for each human being:

= 25 × 10^{14 – 9} = 25 × 10⁵ = 25,00,000
Therefore, there are 25 lakh ants for each human being.

(ii) The estimated global population of starlings is around 1.3 arab = 1.3 billion
= 1,300,000,000 = 1.3 × 10⁹
A flock of starlings contains 10,000 birds.
∴, Number of flocks globally = \(\frac{1.3 \times 10^9}{10^4}\) = 1.3 × 10⁵

(iii) Total estimated number of trees in the world = 3 × 10¹²
Number of leaves on each tree = 10⁴
So, the total number of leaves on all the trees in the world = 3 × 10¹² × 10⁴ = 3 × 10¹⁶

(iv) The distance between the Earth and the Moon = 3,84,400 km.
Assume that the thickness of the sheet of paper = 0.001 cm
So, number of sheets needed to be stacked to reach the moon: = \(\frac{384400 \times 1000 \times 100}{0.001}\)
= 38,440,000,000 × 1000 (∴, \(\frac{1}{0.001}\) = 1000)
= 3.844 × 10¹³

Question 3.
If you have lived for a million seconds, how old would you be?
Solution:
1 million seconds = 1,000,000 seconds
= \(\frac{1000000}{60}\) minutes = 16,666.67 minutes
= 277.78 hours = 11.574 days

Page : 40

Question 1.
10⁵ seconds ≈ 1.16 days and 10⁶ seconds ≈ 11.57 days. Think of some events or phenomena whose time is of the order of
(i) 10⁵ seconds and
(ii) 10⁶ seconds. Write them in scientific notation.
Solution:
(i) Lifespan of an adult mayfly is of the order of 10⁵ seconds.
(ii) The Commonwealth Games typically span 10⁶ seconds.

Page : 42

Question 1.
Calculate and write the answer using scientific notation:
(i) If one star is counted every second, how long would it take to count all the stars in the Universe? Answer in terms of the number of seconds using scientific notation.
(ii) If one could drink a glass of water (200 ml) every 10 seconds, how long would it take to finish the entire volume of water on Earth?
Solution:
(i) Number of stars in the Universe = 2 × 10²³
If one star is counted every second, then to count these stars:
= 2 × 10²³ seconds = 2.0 × 10²³ seconds
Note: for days = \(\frac{2 \times 10^{23}}{60 \times 60 \times 24}\) days = 2.3 × 10¹⁸ days

(ii) Volume of water a person can drink in 10 sec = 200 ml.
So, the volume of water a person can drink in 1 second = 20 ml
The entire volume of water on the Earth = 1.25 × 10²⁴ ml
∴, Time needed to finish the entire volume of
water on Earth = \(\frac{1.25 \times 10^{24}}{20}\) seconds
= 6.25 × 10²² seconds

Page : 43

Question 1.
What does the first part of each name denote?
Continuing this, a thousand trillion is a quadrillion (10¹⁵).
This pattern continues. Observe the names million (10⁶), billion (10⁹), trillion (10¹²), quadrillion (10¹⁵), quintillion (10¹⁸), sextillion (10²¹), septillion (10²⁴), octillion (10²⁷), nonillion (10³⁰), decillion (10³³).
Solution:
The first part of each name denotes: mi → one, bi → two, tri → three, quad —» four, quint → five, sext → six, sept → seven, oct → eight, noni → nine, deci → ten.

Figure it Out: Page : 44 – 45

Question 1.
Find out the units digit in the value of 2²²⁴ ÷ 4³²? [Hint: 4 = 2²]
Solution:
Here, given expression = 2²²⁴ ÷ 4³²
= 2²²⁴ ÷ (2²)³² = 2²²⁴ ÷ 2⁶⁴
= 2^{224 – 64} = 2¹⁶⁰
To find out the unit’s digit, we consider the following:
(2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, 2⁶ = 64, 2⁷ = 128, 2⁸ = 256, 2⁹ = 512, …
∴, We get 2 if the power is of the form ‘4n + 1’
We get 4 if the power is of the form ‘4n + 2’
We get 8 if the power is of the form ‘4n + 3’
We get 6 if the power is of the form ‘4n’, as its unit place digit.
Now, 160 = 4n for n = 40
So, the unit place digit of 2²²⁴ ÷ 4³² is 6.

Question 2.
There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days?
Solution:
Number of bottles in a container = 5
Number of bottles brought in each day = 1
In 40 days there will be 40 containers inside.
So, after 40 days there will be 40 × 5 bottles = 200 bottles.

Question 3.
Write the given number as the product of two or more powers in three different ways. The powers can be any integers.
(i) 64³
(ii) 192⁸
(iii) 32^-5
Solution:
(i) 64³ = (2⁶)³ = 2¹⁸
Now, 2¹⁸ can be written as 2^{4 + 14}, 2^{6 + 12}, 2^{10 + 8} etc.
So, 2⁴ × 2¹⁴, 2⁶ × 2¹², 2¹⁰ × 2⁸, etc.

(ii) 192⁸ = 192^{1 + 7} = 192¹ × 192⁷
192⁸ = 192^{4 + 4} = 192⁴ × 192⁴
192⁸ = 192^{6 + 2} = 192⁶ × 192²

(iii) 32^-5 = 32^{– 1 – 4} = 32^-1 × 32^-4
= 32^-5 = 32^{– 2 – 3} = 32^-2 × 32^-3
= 32^-5 = 32^{– 6 + 1} = 32^-6 × 32¹

Question 4.
Examine each statement below and find out if it is ‘Always True’, ‘Only Sometimes True’, or ‘Never True’. Explain your reasoning.
(i) Cube numbers are also square numbers.
(ii) Fourth powers are also square numbers.
(iii) The fifth power of a number is divisible by the cube of that number.
(iv) The product of two cube numbers is a cube number.
(v) q⁴⁶ is both a 4th power and a 6th power (q is a prime number).
Solution:
(i) Sometimes True : Cube numbers like 64, 512, 729, etc., are also square numbers, while numbers like 8, 27, 125, etc., are cube numbers but they are not squares.

(ii) Always True: Any fourth power is also a square number.
It happens because if we consider a⁴ which is the fourth power of a, then a⁴ = a × a × a × a
= (a × a) × (a × a) = (a² × a² = (a²)²
For example: Consider 5⁴ = 5 × 5 × 5 × 5
= (5 × 5) × (5 × 5) = (25) × (25) = (25)²
So, 5⁴ = (25)²

(iii) Always True:
Consider b⁵ ÷ b³ = = b × b
So, the fifth power of a number is divisible by the cube of that number.

(iv) Always True: Consider a³ and b³
∴, product of a³ × b³ = (a × b)³
So, the product of two cube numbers is a cube number.

(v) Never True: q⁴⁶ is neither a 4^th power nor a 6^th power, where q is a prime number, as 46 is neither divisible by 4 nor divisible by 6.

Question 5.
Simplify and write these in the exponential form.
(i) 10^-2 × 10^-5
(ii) 5⁷ ÷ 5⁴
(iii) 9^-7 ÷ 9⁴
(iv) (13^-2)^-3
(v) m⁵n¹²(mn)⁹
Solution:
(i) 10^-2 × 10^-5 =
10^{(-2) + (-5)} = 10^-7

(ii) 5^-7 ÷ 5⁴ = 5^{7 – 4} = 5³

(iii) 9^-7 ÷ 9⁴ = 9^-7-4 = 9^-11

(iv) (13^-2)^-3 ÷ 13^{(-2) × (-3)} = 13^{[(-2) × (-3)]} = 13⁶

(v) m⁵n¹²(mn)⁹ = m⁵n¹²m⁹n⁹ = m^{5 + 9} + n^{12 + 9} + 9 = m¹⁴n²¹

Question 6.
If 12² = 144 what is
(i) (1.2)²
(ii) (0.12)²
(iii) (0.012)²
(iv) 120²
Solution:
(i) (1.2)² = 1.44

(ii) (0.12)² = 0.0144

(iii) (0.012)² = 0.000144

(iv) (120)² = 14400

Question 7.
Circle the numbers that are the same-
2⁴ × 3⁶ 6⁴ × 3² 6¹⁰ 18² × 6² 6²⁴
Solution:

Question 8.
Identify the greater number in each of the following –
(i) 4³ or 3⁴
(ii) 2⁸ or 8²
(iii) 100² or 2¹⁰⁰
Solution:
(i) 4³ > 3⁴
(ii) 2⁸ > 8²
(iii) 2¹⁰⁰ > 100²

Question 9.
A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0 – 9, how many digits should the code consist of?
Solution:
8.5 billion = 8.5 × 10⁹
So, the code should consist of 10 digits.

Question 10.
64 is a square number (8²) and a cube number (4³). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general?
Solution:
There are many other numbers which are both squares and cubes, for example- a³ = 729 and 27² = 729, so 729 is both square and cube. 16³ = 4096 and 64² = 4096, so 4096 is both square and cube. We can describe such numbers as (a²)³ and (a³)²

Question 11.
A digital locker has an alphanumeric (it can have both digits and letters) passcode of length 5. Some example codes are G89P0, 38098, BRJKW, and 003AZ. How many such codes are possible?
Solution:
Length of passcode = 5
Since it can have both digits and letter so, there can be (36)⁵ as it will be equal to the ways we can fill 5 boxes by 26 alphabets and 10 digits (0 – 9).

Question 12.
The worldwide population of sheep in 2024 is about 10⁹, and that of goats is also about the same. What is the total population of sheep and goats?
(i) 20⁹
(ii) 10¹¹
(iii) 10¹⁰
(iv) 10¹⁸
(v) 2 × 10¹¹
(vi) 10⁹ ÷ 10⁹.
Solution:
Worldwide population of sheep (2024) = 10⁹
Worldwide population of goats (2024) = 10⁹
The total population of sheep and goats = 10⁹ + 10⁹ = 2 × 10⁹
So, (v) 2 × (10)⁹ and (vi) 10⁹ + 10⁹ are true.

Question 13.
Calculate and write the answer in scientific notation:
(i) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing.
(ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees.
(iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world.
(iv) Total time spent eating in a lifetime in seconds.
Solution:
(i) Each person in the world had 30 pieces of clothing.
Number of persons in the world = 8 × 10⁹
The total number of pieces of clothing
= 8 × 10⁹ × 30 = 24 × 10¹⁰ = 2.4 × 10¹¹

(ii) Number of bee colonies in the world
= 100 million = 100 × 10⁶
= Number of honeybees in each colony
= 50000
So, total number of honeybees
= 100 × 10⁶ × 50000 = 5 × 10¹²

(iii) Number of bacterial cells in the human body
= 38 trillion = 38 × 10¹² = 3.8 × 10¹³
Number of the humans in the world = 8 × 10⁹
The total bacterial population residing in all humans in the world = 3.8 × 10¹³ × 8 × 10⁹
= 30.4 × 10²² = 3.04 × 10²³

(iv) Assuming time spent eating in a day = 40 min = 40 × 60 seconds = 2400 seconds Appoximate number of days in human’s life of 100 years = 100 × 365
so, number a person spent eating in a lifetime:
= 100 × 365 × 2400 seconds
= 36.5 × 10,000 × 24
= 876,000,000 = 8.76 × 10⁷ seconds

Question 14.
What was the date 1 arab/1 billion seconds ago?
Solution:
1 arab second = 1,000,000,000 seconds
= 16,666,666.67 minutes
= 277,777.7778 hours
= 11,574.07407 days = 31.710 years
Assuming today’s date as 12 August 2024, then 1 arab seconds ago it was 12 August 1993.

Power Play Class 8 Extra Questions

Multiple Choice Questions

Question 1.
Which of the following is same as 2⁴ × 3⁶?
(a) 6⁴
(b) 6⁶
(c) 6⁴ × 3²
(d) 6² × 3⁴
Solution:
Here, 2⁴ × 3⁶ = 2⁴ × 3⁴ × 3²
= (2 × 3)⁴ × 3² = 6⁴ × 3²
(c) 6⁴ × 3²

Question 2.
If thickness of a paper is 0.001 cm, then its thickness after 8th fold is:
(a) 0.128 cm
(b) 0.256 cm
(c) 0.064 cm
(d) 0.032 cm
Solution:
After 1st fold, the thickness = 0.002 cm,
After 2nd fold, the thickness = 0.004 cm,
After 3rd fold, the thickness = 0.008 cm,
After 4th fold, the thickness = 0.016 cm,
After 5th fold, the thickness = 0.032 cm,
After 6th fold, the thickness = 0.064 cm,
After 7th fold, the thickness = 0.128 cm,
After 8th fold, the thickness = 0.256 cm.
(b) 0.256 cm

Question 3.
Which expression describes the thickness of a sheet of paper after it is folded 4 times? The initial thickness is represented by the letter ‘t’
(a) 4t
(b) t⁴
(c) t + 4
(d) 2⁴ t
Solution:
Initial thickness = t
After 1st fold, thickness = 2t
After 2nd fold, thickness = 2²t
After 3rd fold, thickness = 2³t
After 4th fold, thickness = 2⁴t
(d) 2⁴t

Question 4.
The sum of exponents of the prime factors of 32400 is:
(a) 10
(b) 9
(c) 8
(d) 11
Solution:
As 32400 = 2⁴ × 5² × 3⁴
Here, sum of exponents = 4 + 2 + 4 = 10.
(a) 10

Question 5.
2^-4 × 2¹⁰ =
(a) 2¹⁰
(b) 2^-4
(c) 2⁶
(d) 2^-6
Solution:
2^-4 × 2¹⁰ = 2^{– 4 + 10} = 2⁶ (∵, x^a × x^b = x^{a + b})
(c) 2⁶

Assertion and Reasoning

(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).
(b) Assertion (A) and Reason (R) both are true but Reason (R) is not the correct explanation for the Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.

Question 1.
Assertion (A) : 2³ × 2² = 2⁵
Reason (R) : a^m × aⁿ = a^{m + n}
Answer:
(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).

Question 2.
Assertion (A) : a^m × b^m = (ab)^m
Reason (R) : a^m ÷ aⁿ = a^{m – n}
Answer:
(b) Assertion (A) and Reason (R) both are true but Reason (R) is not the correct explanation for the Assertion (A).

Case Based Questions.

Question 1.
Aarya has some dresses, some caps and some pairs of shoes.
Based on the above, answer the following:

(i) If there are 6 dresses, 4 caps and 4 pairs of shoes, then how many combinations of wearing these items are there?
(ii) If there are in total 48 combinations and there are 6 dresses and 4 caps, then how many pairs of shoes are there?
(iii) If there are 10 dresses, 5 caps and some number of pairs of shoes and total number of combinations of dresses and others is 200, then how many pairs of shoes are there?
(iv) If there are 20 dresses, 10 caps and 15 pairs of shoes, then the number of combinations of these items in scientific notation?
Answer:
(i) Number of dresses = 6
Number of caps = 4
Number of shoes = 4
∴, possible combination = 6 × 4 × 4 = 96

(ii) Number of dresses = 6
Number of caps = 4
Total number of possible combination = 48
∴, the number of pairs of shoes = \(\frac{48}{6 \times 4}\) = \(\frac{48}{24}\) = 2

(iii) Number of dresses = 10
Number of caps = 5
Number of possible combinations = 200
∴, pairs of shoes = \(\frac{200}{10 \times 5}\) = \(\frac{200}{50}\) = 4

(iv) Number of dresses = 20
Number of caps = 10
Number of shoes = 15 pairs
∴, Total number of possible combinations
= 20 × 10 × 15 = 3000 = 3.0 × 10³