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Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 5 Parallel and Intersecting Lines Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 5 Parallel and Intersecting Lines Solutions

Ganita Prakash Class 7 Chapter 5 Solutions

Class 7 Maths Ganita Prakash Chapter 5 Solutions Parallel and Intersecting Lines

Question 1.
List all the linear pairs and vertically opposite angles you observe in the given figure:

Solution:
We know that adjacent angles formed by two intersecting lines, are called linear pairs. Linear pairs always add up to 180°.
And opposite angles formed by two intersecting lines, are called vertically opposite angles. Vertically opposite angles are always equal to each other.

Linear pairs	∠a and ∠b, ∠b and ∠c, ∠c and ∠d, ∠d and ∠a
Pairs of Vertically Opposite Angles	∠b and ∠d, ∠a and ∠c

 

Question 2.
Using your sense of how parallel lines look, try to draw lines parallel to the line segments on this dot paper.
(a) Did you find it challenging to draw sorne of them?
(b) Which ones?
(c) How did you do ii?

Solution:

(a) Yes, some line segments are a litle more difficult to draw than others.
(b) Line segments e,f. h and g.
(c) Lines parallel to a given line segment are drawn by keeping thern equidistant from the given line segment.

Question 3.
In the figure, which line is parallel to line a — line b or line c? How do you decide this?

Solution:
Line c is parallel to line a because the corresponding dots on these lines are equidistant from each other. So, they do not intersect, no matter how far they are extended.

Question 4.
Can you draw a line parallel to l, that goes through point A? How will you do it with the tools from your geometry box? Describe your method.

Solution:
Tools needed: Ruler, Set-squares (right-angled triangle), Pencil Steps of Construction:
Step 1: Place the set square so that one side is along the line l.
Step 2: Hold the ruler against the other side of the set square (the ruler won’t move).
Step 3: Slide the set square along the ruler until one side reaches point A.
Step 4: Draw a line along the edge of the set square through point A.
Step 5: This new line is parallel to line l and passes through point A.

Question 5.
Find the angles marked below.


Solution:
(i) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain a = 48°.

(ii) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain b = 52°.

(iii) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain c = 81°.

(iv) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain d = 99°.

(v) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain e = 69°.

(vi) Since the sum of interior angles on the same side of a transversal intersecting a pair of parallel lines is always equal to 180°,f + 132° = 180° ⇒ f = 180°- 132° ⇒ f = 48°

(vii) Since corresponding angles formed by a transversal intersecting a pair of parallel sides are equal, we obtain g = 122°.

(viii) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain h = 75°.

(ix) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain i = 54°.

(x) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain j = 97°.

Question 6.
In the figures below, what angles do x and y stand for?

Solution:
(i) The lines and angles are marked as shown in the figure.

Line m is parallel to line n and line a is a transversal.
∴ ∠2 = 65° + ∠1 [∵ Corresponding angles]
⇒ 90° = 65° + ∠1
⇒ ∠1 = 90°- 65°
⇒ ∠1 = 25°
And, ∠1 = x = 25° [∵ Vertically opposite angles]
Also, line m is parallel to line n and line b is a transversal.
∴ ∠1 +y = 180° [∵ Sum of co-interior angles = 180°]
⇒ 25° + y = 180° [∵ ∠1 = 25°]
⇒ y = 180° – 25° ⇒ y = 155°
Thus, the values of x and y are 25° and 155°, respectively.

(ii) The lines and angles are marked as shown in the figure.

Line a is parallel to line b and line d is a transversal.
∴ ∠2 = 53° [∵ Alternate interior angles]
Also, line a is parallel to line b and c is a transversal.
∴ ∠1 + ∠2 = 78° [∵ Alternate interior angles]
⇒ ∠1 + 53° = 78° [∵ ∠2 = 53°]
⇒ ∠1 = 78°- 53°
⇒ ∠1 = 25°
Therefore, ∠1 = x = 25° [∵ Vertically opposite angles]

Question 7.
What is the measure of ∠NOP in the figure?

[Hint: Draw lines parallel to LM and PQ through points N and O.]
Solution:
Lines parallel to LM and PQ through points N and O are drawn and angles are marked as shown in the figure.

Since LM is parallel to EF and MN is a transversal,
∠1 = 40° [∵ Alternate interior angles]
Now, ∠1 + ∠2 = 96°
⇒ 40° + ∠2 = 96° [∵ ∠1 = 40°]
⇒ ∠2 = 96° – 40°
⇒ ∠2 = 56°
Since EF is parallel to GH and AT is a transversal,
∠2 = ∠3 = 56° [∵ Alternate interior angles]
Also, GH is parallel to PQ and OP is a transversal.
∴ ∠4 = 52° [∵ Alternate interior angles]
So, a = ∠3 + ∠4
⇒ a = 56° + 52° ⇒ a = 108°
Thus, ∠NOP = 108°.

InText Questions

Question 1.
Can two straight lines intersect at more than one point?
Solution:
No, two straight lines cannot intersect at more than one point. If two lines intersect at more than one point, then they are coincident lines.

Question 2.
Take a plain square sheet of paper (use a newspaper for this).

(i) How would you describe the opposite edges of the sheet? They are ________ to each other.
(ii) How would you describe the adjacent edges of the sheet? The adjacent edges are ________ to each other. They meet at a point. They form right angles.
(iii) Fold the sheet horizontally in half. A new line is formed (see figure).
How many parallel lines do you see now?
(iv) Make one more horizontal fold in the folded sheet. How many parallel lines do you see now?
(v) What will happen if you do it once more? How many parallel lines will you get? Is there a pattern? Check if the pattern extends further, if you make another horizontal fold.
(vi) Make a vertical fold in the square sheet. This new vertical line is _________ to the previous horizontal lines.
(vii) Fold the sheet along a diagonal. Can you find a fold that creates a line parallel to the diagonal line?
Solution:
(i) They are parallel to each other.
(ii) The adjacent edges are perpendicular to each other.
(iii) We see three parallel horizontal lines — the top edge, the fold and the bottom edge. The new horizontal line is perpendicular to the vertical edges of paper.
(iv) On folding the paper horizontally one more time, we see five parallel horizontal lines.
(v) On folding the paper horizontally once more, we will get nine parallel lines. The number of horizontal parallel lines follow the sequence:
1^st fold → 3 lines
2^nd fold → 5 lines
3^rd fold → 9 lines and so on
So, after each fold, the number of horizontal lines increases as folding doubles the sections and add extra fold lines. The pattern continues as we fold more.

(vi) This new vertical line is perpendicular to the previous horizontal lines.

(vii) Yes, we can make a fold parallel to the diagonal by folding the sheet in the same slanting direction at equal angles or by folding a smaller triangle inside the square.

Parallel and Intersecting Lines Class 7 Extra Questions

Parallel and Intersecting Lines Class 7 Very Short Question Answer

Question 1.
In the given figure, write the vertically opposite angles of ∠AOD and ∠AOC.

Solution:
We know that when two lines intersect, the angles opposite to each other are called vertically opposite angles. They are formed without sharing a common arm and are always equal.
Thus, in the given figure, ∠BOC is vertically opposite angle of ∠AOD and ∠BOD is vertically opposite angle of ∠AOC.

Question 2.
Find the complement of each of the following angles:
(i) 28°
(ii) 66°
(iii) 75°
(iv) 80°
Solution:
We know that complement of angle x° is (90° – x°).
(i) Complement of 28° = 90° – 28° = 62°
(ii) Complement of 66° = 90° – 66° = 24°
(iii) Complement of 75° = 90° – 75° = 15°
(iv) Complement of 80° = 90° – 80° = 10°

Question 3.
Find the supplement of each of the following angles:
(i) 28°
(ii) 95°
(iii) 130°
(iv) 155°
Solution:
We know that supplement of angle x° is (180° – x°).
(i) Supplement of 28° = 180° – 28° = 152°
(ii) Supplement of 95° = 180° – 95° = 85°
(iii) Supplement of 130° = 180° – 130° = 50°
(iv) Supplement of 155° = 180° – 155° = 25°

Question 4.
Find the complement of each of the following angles:
(i) 32°
(ii) 72°
(iii) 68°
(iv) 20°
Solution:
We know that complement of angle x° is (90 – x)°.
(i) Complement of 32° = 90° – 32° = 58°
(ii) Complement of 72° = 90° – 72° = 18°
(iii) Complement of 68° = 90° – 68° = 22°
(iv) Complement of 20° = 90° – 20° = 70°

Question 5.
Find the supplement of each of the following angles:
(i) 25°
(ii) 92°
(iii) 142°
(iv) 165°
Solution:
We know that supplement of angle x° is (180 – x)°.
(i) Supplement of 25° = 180° – 25° = 155°
(ii) Supplement of 92° = 180° – 92° = 88°
(iii) Supplement of 142° = 180° – 142° = 38°
(iv) Supplement of 165° =180° – 165° = 15°

Question 6.
In the given figure, write the angles that form a linear pair with ∠AOD and with ∠BOC.

Solution:
We know that the adjacent angles formed by two lines intersecting each other, are called linear pair of angles. Linear pairs always add up to 180°.
In the given figure, ∠AOC and ∠BOD form a linear pair of angles with ∠AOD.
∠AOC and ∠BOD form a linear pair of angles with ∠BOC.

Question 7.
In the given figure, find the value of a.

Given,
∠SOR = 3a + 20° and ∠ROT = a
From the given figure, we can say that ∠SOR and ∠ROT form a linear pair.
∴ ∠SOR + ∠ROT = 180°
⇒ 3a + 20° + a = 180°
⇒ 4a = 160° ⇒ a = 40°

Question 8.
Find the value of x in the following figure, if AB is parallel to CD.

Solution:
Given, AB is parallel CD.
Since x and 3x are co-exterior angles on the same side of transversal, they add up to 180°.
∴ x + 3x = 180° ⇒ 4x = 180°
⇒ x = \(\frac{180^{\circ}}{4}\) ⇒ x = 45°

Question 9.
Find the value of x in the following figure, if AB is parallel to CD.

Solution:
Given, AB is parallel CD.
Since x and 2x are interior angles on the same side of transversal, they add up to 180°.
∴ x + 2x = 180°
⇒ 3x = 180° ⇒ x = 60°

Parallel and Intersecting Lines Class 7 Short Question Answer

Question 1.
Identify the complementary and supplementary pairs of angles from the following pairs:
(i) 42°, 48°
(ii) 85°, 95°
(iii) 30°, 60°
(iv) 135°, 45°
Solution:
We know that if the sum of the measures of two angles is 90°, then the angles are called complementary angles and if the sum of the measures of two angles is 180°, then the angles are called supplementary angles.
(i) The given angles are 42° and 48°.
Now, sum of given angles = 42° + 48° = 90°
Thus, the given angles are complementary angles.

(ii) The given angles are 85° and 95°.
Now, sum of given angles = 85° + 95° = 180°
Thus, the given angles are supplementary angles.

(iii) The given angles are 30° and 60°.
Now, sum of given angles = 30° + 60° = 90°
Thus, the given angles are complementary angles.

(iv) The given angles are 135° and 45°.
Now, sum of given angles = 135° + 45° = 180°
Thus, the given angles are supplementary angles.

Question 2.
In the given figure, find the value of x.

Solution:
Given, ∠AOB = 47°, ∠BOC = x, ∠COD = 83°, ∠DOE = 92° and ∠EOA = 75°
Now, as ∠AOB, ∠BOC, ∠COD, ∠DOE and ∠EOA are angles at a point, they add up to 360°.
∴ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°
⇒ 47° + x + 83° + 92° + 75°= 360°
⇒ 297° + x = 360°
⇒ x = 360° – 297° = 63°
Thus, the value of x is 63°.

Question 3.
In the given figure, line AB is parallel to line DG. Find the value of x + y.

Solution:
Given, line AB is parallel to line DG.
∴ ∠ACE = ∠CEG
[∵ Alternate interior angles are equal]
⇒ x = 80° [∵ ∠ACE = 80° and ∠CEF = x]
Since ∠GFH and ∠EFH form linear pair, they add up to 180°.
∴ ∠GFH + ∠EFH = 180°
⇒ 150° + y = 180° [∵ ∠GFH = 150°]
⇒ y = 180° – 150° = 30°
∴ x + y = 80° + 30° = 110°

Question 4.
In the given figure, l is parallel to m. Find the value of x and y.

Solution:
Given, l is parallel to m and n is transversal to l and m.
Since 2x and y are vertically opposite angles, they are equal.
∴ y = 2x …(i)
Since 4x and y are interior angles on the same side of the transversal, they add up to 180°.
∴ 4x + y = 180°
⇒ 4x + 2x = 180° [From (i)]
⇒ 6x = 180° ⇒ x = 30°
Substituting the value of x in (i), we get
y = 2 × 30° = 60°

Parallel and Intersecting Lines Class 7 Long Question Answer

Question 1.
Which lines appear to be perpendicular to each other in the given figure?

Solution:
When two lines intersect and the angles formed are 90° (i.e. all four angles are equal), the lines are said to be perpendicular to each other.

In the given figure, since line i is parallel to line d, perpendicular to any of these lines is also perpendicular to other. Therefore, lines b, c, h and /are perpendicular to lines i and d.

Similarly, lines d and i are perpendicular to lines b, c, h andf.

Question 2.
In the given figure, AB is parallel to CD and AD is parallel to BC. Find the values of x, y and z.

Solution:
Given, AB || CD, AD ||
BC and ∠BAD = 70°.
As AD || BC and AB is transversal, ∠ABC and ∠BAD are co-interior angles.
∴ ∠ABC + ∠BAD = 180°
⇒ x + 70° = 180° [∵ ∠BAD = 70°]
⇒ x = 180°- 70° = 110°
Now, as AB || DC and AD is transversal, ∠ADC and ∠BAD are co-interior angles.
∴ ∠ADC + ∠BAD = 180°
⇒ z + 70° = 180° [∵ ∠BAD = 70°]
⇒ z = 180°- 70° = 110°
Also, as AD || BC and DC is transversal, ∠ADC and ∠BCD are co-interior angles.
∴ ∠ADC + ∠BCD = 180°
⇒ 110°+ y = 180° [∵ ∠ADC =110°]
⇒ y = 180°- 110° = 70°

Question 3.
In the following figure, find the value of each marked angle.

Solution:
Given, line l || m || n.
Let point P lie on line l, point R lie on line p and the point of intersection of lines p and l be Q, as shown in the figure.

Since ∠PQR and 32° are vertically opposite angles, they are equal.
∴ ∠PQR = 32°
As l || m, ∠b and ∠PQR are interior angles on the same side of transversal p.
∴ ∠b + ∠PQR = 180°
⇒ ∠b + 32° = 180° [∵ ∠PQR = 32°]
⇒ ∠b = 180°- 32° = 148°
As l || n, ∠a and ∠PQR are corresponding angles.
∴ ∠a = ∠PQR
⇒ ∠a = 32° [∵ ∠PQR = 32°]

Question 4.
In the following figure, AB is parallel to CD and AD is parallel to BC. Find the values of x, y and z.

Solution:
Given, AB || CD, AD || BC, ∠DAC = 45° and ∠BAC = 30°.
As AB || CD anddC is transversal, ∠DAC and ∠ACB are alternate interior angles.
∴ ∠ACB = ∠DAC ⇒ x = 45°
[∵ ∠DAC = 45° and ∠ACB = x]
As AB || CD and AC is transversal, ∠BAC and ∠ACD are alternate interior angles.
∴ ∠ACD = ∠BAC ⇒ y = 30°
[∵ ∠BAC = 30° and ∠ACD = y]
As AD || BC and AB is transversal. ∠DAB and ∠ABC are co-interior angles.
∴ ∠DAB +∠ABC = 180°
⇒ ∠DAC + ∠CAB + ∠ABC = 180°
[∵ ∠DAB = ∠DAC + ∠CAB]
⇒ 45° + 30° + z = 180°
[∵ ∠DAC = 45°, ∠CAB = 30° and ∠ABC = z]
⇒ z = 180°- 75° = 105°

Parallel and Intersecting Lines Class 7 Case Based Questions

Question 1.
In the given figure, two straight lines PQ and RS intersect each other at O such that ∠POT = 75°.

Based on the above information, answer the following questions:
(i) Find the value of b.
(ii) Find the value of a.
(iii) Find the value of c.
Solution:
(i) Given, ∠ROP = 4b, ∠POT = 75°, ∠TOS = b
Here, ∠ROS = 180°
[Since ∠ROS is a straight angle]
⇒ ∠ROP + ∠POT + ∠TOS = 180°
⇒ 4b + 75° + b = 180°
⇒ 5b + 75° = 180°
⇒ 5b = 180° – 75° = 105°
b = \(\frac{105^{\circ}}{5}\) = 21°
Thus, the value of b is 21°.

(ii) Since ∠ROP and ∠QOS are vertically opposite angles, they are equal.
∴ ∠QOS = ∠ROP
⇒ a = 4b
⇒ a = 4 × 21° [∵ b = 21°]
⇒ a = 84°
Thus, the value of a is 84°.

(iii) Since ∠QOS and ∠QOR form a linear pair, they add up to 180°.
∴ ∠QOS + ∠QOR = 180°
⇒ a + 2c = 180°
⇒ 84° + 2c =180° [∵ a = 84°]
⇒ 2c = 180° – 84° = 96°
c = \(\frac{96^{\circ}}{2}\) = 48°
Thus, the value of c is 48°.

Question 2.
In a class, a teacher asked a student to draw three lines on the board. The student draws the lines on the board as shown.

The line l is perpendicular to line n and ∠1 = 75°.
Based on the above information answer the following questions:
(i) What is the measure of ∠6?
(ii) What is the measure of ∠5?
(iii) What is the measure of ∠3?
(iv) What is the sum of the measure of ∠2 and ∠1?
Solution:
(i) Given, line l is perpendicular to line n.
∴ ∠1 + ∠6 = 90°
⇒ 75° + ∠6 = 90° [∵ ∠1 = 75°]
⇒ ∠6 = 90° – 75° = 15°

(ii) Given, line l is perpendicular to line n.
∴ ∠5 = 90°

(iii) We have, ∠6 = 15°
Since ∠3 and ∠6 are vertically opposite angles, they are equal.
∴ ∠3 = ∠6 = 15°

(iv) Given, line l is perpendicular to line n. So, ∠2 is equal to 90°.
∴ ∠1 + ∠2 = 75° + 90° = 165°

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 7 Math Solution and Ganita Prakash Chapter 4 Expressions using Letter Numbers Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 4 Expressions using Letter Numbers Solutions

Ganita Prakash Class 7 Chapter 4 Solutions

Class 7 Maths Ganita Prakash Chapter 4 Solutions Expressions using Letter Numbers

Question 1.
One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expressions(s) describe the total amount in rupees earned that day?
(a) 30x + 20y
(b) (30 + 20) × (x + y)
(c) 20x + 30)
(d) (20 + 30) × (x + y)
(e) 30x – 20y
Solution:
Cost of one plate of Jowar roti = ₹30
Cost of x plates of Jowar roti = ₹30x
Cost of one plate of Pulao = ₹20
Cost of y plate of Pulao = y
So, the expression for the total amount earned that day = 30x + 20y
Hence, the correct answer is option (a).

Question 2.
Write formulas for the perimeter of:
(i) triangle with all sides equal.
(ii) a regular pentagon.
(iii) a regular hexagon.
Solution:
(i) Let side length of triangle be a. Then,
Perimeter of triangle with all sides equal = a + a + a = 3a = 3 × side

(ii) Let side length of a regular pentagon be a. Then,
Perimeter of the regular pentagon = a + a + a + a + a = 5a = 5 × side

(iii) Let side length of a regular hexagon be a. Then,
Perimeter of the regular hexagon = a + a + a + a + a + a = 6a = 6 × side

Question 3.
Munirathna has a 20 m long pipe. However, he wants a longer watering pipe for his garden. He joins another pipe of some length to this one. Give the expression for the combined length of the pipe. Use the letter- number to denote the length in metres of the other pipe.
Solution:
Length of pipe Munirathna has = 20 m
Length of another pipe Munirathna wants to join = k m
∴ Combined length of the pipe = (20 + k) m

Question 4.
What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5?
Complete the following table:

Solution:

Number of ₹100 notes	Number of ₹20 notes	Number of ₹5 notes	Expression and Total amount (in ₹)
3	5	6	3 × 100 + 5 × 20 + 6 × 5 = 430
6	4	3	6 × 100 + 4 × 20 + 3 × 5 = 695
8	4	z	8 × 100 + 4 × 20 + z × 5 = 880 + 5z
x	y	z	x × 100 + y × 20 + z × 5 = 100x + 20y + 5z

Question 5.
In a calendar month, if any 2 × 3 grid full of dates is chosen as shown in the picture, write expressions for the dates in the blank cells if the bottom middle cell has date ‘w’.

Solution:
Date to the left to w will be 1 less than w.
Date to the right to w will be 1 more than w.
Since there are 7 clays in a week,
Date above w will be 7 less than w.
Date in the diagonally left cell to w will be 8 less than w.
Date in the diagonally right will to w will be 6 less than w.
Thus, the expressions in other five blank cells of the grid are as shown:

w – 1	w – 7	w – 6
w – 1	w	w + 1

Question 6.
A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights.
(1) Write an expression describing how far away the snail is from its starting position.
(ii) What can we say about the snail’s movement if d > u?
Solution:
(i) During the day the snail climbs up ‘u’ cm.
During the night the snail slips down ‘d’ cm.
So, the net distance covered in one day is (u – d) cm.
So, in 10 days and 10 nights the net distance covered by the snail = 10(u – d) cm.
Hence, the expression describing how far away the snail is from it starting position is 10(u – d).

(ii) If d > u, snail slips down more than it climbs.
It means the snail will never reach the top.

Question 7.
Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometres would Radha have cycled after 3 weeks?
Solution:
In first week, Radha cycles 5 km every day.
So, she cycled 5 × 7 = 35 km in first week.
In second week, Radha cycles (5 + z) km every day.
In third week, she cycles 5 + z + z = (5 + 2z) km every day.
So, she cycled (5 + 2z) × 7 = (35 + 14z) km in third week.
Thus, number of kilometres Radha cycled in 3 weeks
= 35 + (35 + 7z) + (35 + 14z)
= (35 + 35 + 35) + (7z + 14z) = (105 + 21z)km

Question 8.
In the following figure, observe how the expression it w + 2 becomes 4w + 20 along one path. Fill in the missing blanks on the remaining paths. The ovals contain expressions and the boxes contain operations.

Solution:
[(w + 2) – 5] × 3 = [w – 3] × 3 = 3w – 9
[(w + 2) – 8] – 4 = [w – 6] – 4 = w – 10
[(w + 2) – 4] × 3 = [w – 2] × 3 = 3w – 6

Question 9.
A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations.
(i) What is the algebraic expression for the time taken to travel from Yahapur to Yahapur?
(ii) If t = 4, what is the time taken to travel from Yahapur to Vahapur?
Solution:
(i)
Let the time taken to travel from one station to another station = t
So, time taken to travel from Yahanpur to Vahapur = 4t
As there are three stoppages between these two stations and the train stops for 2 minutes at each stoppage, total time taken during stoppages = 2 × 3 = 6 minutes.
So, the algebraic expression for total time taken (in minutes) is (4t + 6).

(ii) From (i), the algebraic expression for total time (in minutes) taken from Yahanpur to Vahapur is (4t + 6). So, the time taken to travel from Yahapur to Vahapur = 4 × 4 + 6 = 16 + 6 = 22 minutes.

Question 10.
Simplify the following expressions:
(i) 3a + 9b – 6 + 8a – 4b – 7a + 16
(ii) 3(3a – 3b) – 8a – 4b – 16
(iii) 2(2x – 3) + 8x + 12
(iv) 8x – (2x – 3) + 12
(v) 8h – (5 + 7h) + 9
(vi) 23 + 4(6m – 3n) – 8n – 3m – 18
Solution:
(i) 3a + 9b – 6 + 8a – 4b – 7a + 16 = (3a + 8a – 7a) + (9b – 4b) + (- 6 + 16) = 4a + 5b + 10
(ii) 3(3a – 3b) – 8a – 4b – 16 = 9a – 9b – 8a – 4b – 16 = (9a – 8a) + (-9b – 4b) – 16 = a – 13b – 16
(iii) 2(2x – 3) + 8x + 12 = 4x – 6 + 8x + 12 = (4x + 8x) + (-6 + 12) = 12x + 6
(iv) 8x – (2x – 3) + 12 = 8x – 2x + 3 + 12 = 6x + 15
(v) 8h – (5 + 7h) + 9 = 8h – 5 – 7h + 9 = 8h – 7h – 5 + 9 = h + 4
(vi) 23 + 4(6m – 3n) – 8n – 3m – 18 = 23 + 24m – 12n – 8n – 3m – 18
= 24m – 3m – 12w – 8n + 23 – 18 = 21m – 20n + 5

Question 11.
Add the expressions given below:
(i) 4d – 7c + 9 and 8c – 11 +9d
(ii) – 6f + 19 – 8s and – 23 + 13f + 12s
(iii) 8d – 14c + 9 and 16c – (11 + 9d)
(iv) 6f – 20 + 8s and 23 – 13f – 12s
(v) 13m – 12n and 12n – 13m
(vi) – 26m + 24n and 26m – 24n
Solution:
(i) (4d – 7c + 9) + (8c – 11 + 9d) = (4d + 9d) + (- 7c + 8c) + (9 – 11) = 13d + c – 2
(ii) (-6f + 19 – 8s) + (-23 + 13f + 12s) = (-6f + 13f) + (-8s + 12s) + (19 – 23) = 7f + 4s – 4
(iii) (8d – 14c + 9) + [16c – (11 + 9d)] = (8d – 14c + 9) + (16c – 11 – 9d)
= (8d – 9d) + (-14c + 16c) + (9 – 11) = -d + 2c – 2
(iv) (6f – 20 + 8s) + (23 – 13f – 12s) = (6f – 13f) + (8s – 12s) + (-20 + 23) = -7f – 4s + 3
(v) (13m – 12n) + (12n – 13m) = (13m – 13m) + (- 12n + 12n) = 0
(vi) (-26m + 24n) + (26m – 24n) = (-26m + 26m) + (24n – 24n) = 0

Question 12.
Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut?

Solution:
Step 1 (0 fold): We get 0 + 2 = 2 pieces
Step 2 (1 fold): We get 1 + 2 = 3 pieces
Step 3 (2 folds): We get 2 + 2 = 4 pieces
In the same way, if rope is folded 10 times and cut, we get 10 + 2 = 12 pieces.
In the same way, when the rope is folded r times and cut, we get r + 2 pieces.

Question 13.
Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares?

Solution:
Step 1: To make 1 square, we need 4 matchsticks.
Step 2: To make 2 squares, we need 4 + 3 = 7 matchsticks.
Step 3: To make 3 squares, we need 4 + 3 + 3 = 10 matchsticks.
So, to make w squares, we need —
4 + (w – 1) × 3 = 4 + 3 (w – 1) = 4 + 3w – 3 = (3w + 1) matchsticks.
To make 10 squares, substituting w = 10, we get
Number of required matchsticks = 3(10) + 1 = 30 + 1 = 31

Question 14.
Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares?

Solution:
Number of squares in step 1 = 5
Number of squares in step 2 = 5 + 4 = 9
Number of squares in step 3 = 5 + 4 + 4 = 5 + 2 × 4 = 5 + 8 = 13
Thus, number of squares in step 4 = 5 + 3 × 4 = 5 + 12 = 17
Number of squares in step 10 = 5 + 9 × 4 = 5 + 36 = 41
And, number of squares in step 50 = 5 + 49 × 4 = 5 + 196 = 201
So, the general formula for number of squares in step n = 5 + (n – 1) × 4 = 5 + 4 (n – 1) = 5 + 4n – 4 = 4n + 1
Number of vertices in step 1 = 16
Number of vertices in step 2 = 16 + 12 = 28
Number of vertices in step 3 = 16 + 12 + 12 = 16 + 2 × 12 = 16 + 24 = 40
Number of vertices in step n = 16 + (n – 1) × 12 = 16 + 12n – 12 = 12n + 4

InText Questions

Question 1.
Shabnam is 3 years older than Aftab. When Aftab’s age is 10 years, Shabnam’s age will be 13 years. Now, Aftab’s age is 18 years, what will Shabnam’s age be?
Solution:
Shabnam’s age = Aftab’s age + 3
When .Aftab’s age = 18 years,
Shabnam’s age = 18 + 3 = 21 years

Question 2.
Find the values of the following arithmetic expressions:
(i) 23 – 10 × 2
(ii) 83 + 28 – 13 + 32
(iii) 34 – 14 + 20
(iv) 42 + 15 – (8 – 7)
(v) 68 – (18 + 13)
(vi) 7 × 4 + 9 × 6
(vii) 20 + 8 × (16 – 6)
Solution:
(i) 23 – 10 × 2 = 23 – 20 = 3
(ii) 83 + 28 – 13 + 32
= (83 – 13) + (28 + 32)
= 70 + 60 = 130
(iii) 34 – 14 + 20 = (34 – 14) + 20 = 20 + 20 = 40
(vi) 42 + 15 – (8 – 7) = 42 + 15 – 1
= 42 + 14 = 56
(v) 68 – (18 + 13) = 68 – 31 = 37
(vi) 7 × 4 + 9 × 6 = 28 + 54 = 82
(vii) 20 + 8 × (16 – 6) = 20 + 8 × 10
= 20 + 80 = 100

Question 3.
Some simplifications are shown below where the letter-numbers are replaced by numbers and the value of the expression is obtained.
1. Observe each of them and identify if there is a mistake.
2. If you think there is a mistake, try to explain what might have gone wrong.
3. Then, correct it and give the value of the expression.

Solution:
(1) If a = – 4, then 10 – a = 6 is wrong.

(2) If d = 6, then 3d = 36 is wrong.
As 10 – a = 10 – (-4) = 10 + 4 = 14
As 3d = 3 × d = 3 × 6 = 18
So, if a = -4, then 10 – a = 14 is correct.
So, if d = 6, then 3d = 18 is correct.

(3) Ifs = 7, then 3s – 2 = 15 is wrong.

(4) If r = 8, then 2r + 1 = 29 is wrong.
As 3s – 2 = 3 × 7 – 2 = 21 – 2 = 19
As 2r + 1 = 2 × 8 + 1 = 16 + 1 = 17
So, if s = 7, then 3s – 2 = 19 is correct.
So, if r = 8, then 2r + 1 = 17 is correct.

(5) If j = 5, then 2j = 10 is correct.

(6) If m = – 6, then 3(m + 1) = 19 is wrong.
As 2j = 2 × 5 = 10
As 3(m + 1) = 3 × (- 6 + 1) = 3 × (- 5) = – 15
So, if m = – 6, then 3(m + 1) = – 15 is correct.

(7) If f = 3, g = 1, then 2f – 2g = 2 is wrong.

(8) If t = 4, b = 3, then 2t + b = 24 is wrong.
As 2f – 2g = 2 × 3 – 2 × 1 = 6 – 2 = 4
As 2t + b = 2 × 4 + 3 = 8 + 3 = 11
So, if f = 3, g = 1, then 2f – 2g = 4 is correct.
So, if t = 4, b = 3, then 2t + b = 11 is correct,

(9) If h = 5, n = 6, then h – (3 – n) = 4 is wrong.
As h – (3 – n) = 5 – (3 – 6) = 5 – (- 3) = 5 + 3 = 8
So, if h = 5, n = 6, then h – (3 – n) = 8 is correct.

Question 4.
Here is a table showing the number of pencils and erasers sold in a shop. The price per pencil is c, and the price per eraser is d. Find the total money earned by the shopkeeper during these three days.

(i) If c = ₹50, find the total amount earned by the sale of pencils.
(ii) Write the expression for the total money earned by selling erasers. Then, simplify the expression.
Solution:
Total money earned by the shopkeeper
= Money earned on day 1 + Money earned on day 2 + Money earned on day 3
= 5c + 4 d + 3c + 6d + 10c + d = 18c + 11d

(i) Total amount earned by the sale of pencils
= 5c + 3c + 10c = 18c
= 18 × ₹50 = ₹900

(ii) Given, the price per eraser is d.
Total money earned by selling erasers
= 4d + 6d + d = 11 d

Question 5.
Fill in the blanks below by replacing the letter- numbers by numbers; an example is shown. Then compare the values that 5u and 5 + u take.

Solution:

u	5u	5 + u
11	5 × 11 = 55	5 + 11 = 16
8	5 × 8 = 40	5 + 8 = 13
5	5 × 5 = 25	5 + 5 = 10

We see that the values of 5u and 5 + u are not equal for different values of u. So, the expressions 5w and 5 + u are not equal.

Question 6.
Are the expressions 10y – 3 and 10(y – 3) equal?
After filling in the two diagrams (given below), do you think the two expressions are equal?

Solution:

u	10y – 3	10(y – 3)
0	10 × 0 – 3 = -3	10(0 – 3) = -30
7	10 × 7 – 3 = 67	10(7 – 3) = 40
10	10 × 10 – 3 = 97	10(10 – 3) = 70

Since the values of 10v – 3 and 10(y – 3) are not equal for different values of y, the expressions 10y – 3 and 10(y – 3) are not equal.

Question 7.
Find the formulas of the number machines below and write the expression for each set of inputs.

Solution:
(i) The formula for the number machine is “sum of first number and second number minus two” and the expression is a + b – 2.
The expression for each set of inputs are:
5 + 2 – 2 = 5, 8 + 1 – 2 = 7,
9 + 11 – 2 = 18, 10 + 10 – 2 = 18
and a + b – 2

(ii) The formula for the number machine is “product of first number and second number plus one” and the expression is a × b + 1.
The expression for each set of inputs are:
4 × 1 + 1= 5, 6 × 0 + 1 = 1,
3 × 2 + 1 = 7, 10 × 3 + 1 = 31
and a × b + 1 = ab + 1

Question 8.
Somjit noticed a repeating pattern along the border of a saree.

Use this to find what design appears at positions 99, 122 and 148.
Solution:
We can see the design A, B and C appear at the positions 3n – 2, 3n – 1 and 3n, respectively.
For 99, the remainder on division by 3 is 0,’i.e. it is a multiple of 3.
So, at position 99, design C will appear.
For 122, the remainder on division by 3 is 2, i.e. it is 1 less than a multiple of 3, i.e. 3n – 1.
So, at position 122, design B will appear.
For 148 , the remainder on division by 3 is 1, i.e. it is 2 less than a multiple of 3, i.e. 3n – 2.
So, at position 148, design A will appear.

Expressions using Letter Numbers Class 7 Extra Questions

Expressions using Letter Numbers Class 7 Very Short Question Answer

Question 1.
Simplify the expression 7(u – 2v) + 2v.
Solution:
Using the distributive property, this expression can be simplified as
7(u – 2v) + 2v = 7u – 7 × 2v + 2v
= 7u – 14v + 2v = 7u – 12v
[Adding like terms together]

Question 2.
What is the sum of the numbers in the given picture (unknown values are denoted by letter- numbers)?

Solution:
Adding row wise, we get Sum of numbers
= (5 × 4) + (2g + 2h) + (2g + 2h) + (5 × 4)
= 20 + 2g + 2 h + 2g + 2h + 20
= (20 + 20) + (2g + 2g) + (2h + 2 h)
= 40 + 4g + 4 h

Expressions using Letter Numbers Class 7 Short Question Answer

Question 1.
Add the numbers in each picture below. Write their corresponding expressions and simplify them.

Solution:
(i) (x + y + y + x + y + y + x + y + y + x + y + y) + (4 × 6)
= 4x + 8y + 24

(ii) 2 × 3y + 4 × 2 + 2 × (-4x) + 4 × (-8)
= 6y + 8 – 8x – 32
= 6y – 8x – 24

(iii) 4 × (- 3n) + 6 × 7m
= -12m + 42m
= 42m – 12n

Question 2.
Simplify each of the following expressions:
(i) e + e + e + f + f
(ii) m + m – (m – n) – n – n
(iii) l – 7m + l – (m – 1)
(iv) 2x – x – (x + x)
(v) (3u – 2v) – (2u – 3v)
Solution:
(i) e + e + e + f + f = 3e + 2f
(ii) m + m – (m – n) – n – n = 2m – (m – n) – 2n
= 2m – m + n – 2n
= m – n

(iii) l – m + l – (m – l) = l + l – m – (m – l)
= 2l – m – m + l
= 3l – 2m

(iv) 2x – x – (x + x) = 2x – x – 2x = -x

(v) (3u – 2v) – (2u – 3v) = 3u – 2v – 2u + 3v
= 3u – 2v – 2v + 3u
= u + v

Expressions using Letter Numbers Class 7 Long Question Answer

Question 1.
Simplify the following expressions:
(i) 2x – 3y + 7x + y – 12
(ii) 4(x – 2y) + 3x – 5y + 1
(iii) 7 + 3h – 2g – (5h – 4g)
(iv) 3g + 9h – (5 + 6h – 2g)
(v) 5(2u + 7v – 2) – 3(8u – 10v + 4)
(vi) 17 – 13a + 19b – 2(9 + 7a – 3b)
Solution:
(i) 2x – 3y + 7x + y – 12
= 9x – 2y – 12

(ii) 4(x – 2y) + 3x – 5y + 1
= 4x – 8y + 3x – 5y + 1
= 7x – 13y + 1

(iii) 7 + 3h – 2g – (5h – 4g)
= 7 + 3h – 2g – 5h + 4g
= 7 – 2h + 2g

(iv) 3g + 9h – (5 + 6h – 2g)
= 3g + 9h – 5 – 6h + 2g
= 5g + 3h – 5

(v) 5(2u + 7v – 2) – 3(8u – 10v + 4)
= 10u + 35v – 10 – 24u + 30v – 12
= – 14u + 65v – 22

(vi) 17 – 13a + 19b – 2(9 + 7a – 3b)
= 17 – 13a + 19b – 18 – 14a + 6b
= – 27a + 25b – 1

Question 2.
Add the expressions given below:
(i) 3d+ 6c – 7 and 3c – 4d + 2
(ii) -8g + 3h + 2 and 6g – 4h – 12
(iii) 11k + l + 10 and k – (11l – 10)
(iv) 21m – 7n and -17m + 10n + 13
(v) 5(x + 7y) – 3 and 2x + 3y – 2
(vi) 7a + 2b + 9 and 3a + 8b + 1
Solution:
(i) (3d + 6c – 7) + (3c – 4d + 2)
= 6c + 3c + 3d – 4d – 7 + 2 = 9c – d – 5

(ii) (-8g + 3h + 2) + (6g – 4h – 12)
= -8g + 6g + 3h – 4h + 2 – 12
= – 2g – h – 10

(iii) (11k + l + 10) + {k – (11l – 10)}
= 11k + l + 10 + k = 11l + 10
= 11k + k + l – 11l + 10 + 10
= 12k – 10l + 20

(iv) (21m – 7n) + (-17m + 10n + 13)
= 21m – 17m – 7n + 10n + 13
= 4m + 3n + 13

(v) 5(x + 7y) – 3 + (2x + 3y – 2)
= 5x + 35y – 3 + 2x + 3y – 2
= 5x + 2x + 35y + 3y – 3 – 2
= 7x + 38y – 5

(vi) (7a + 2b + 9) + (3a + 8b + 1)
= 7a + 3a + 2b + 8b + 9 + 1
= 10a + 10b + 10

Expressions using Letter Numbers Class 7 Case Based Questions

Question 1.
A movie theatre has 6 columns of seats (labelled A to F) and the seats are arranged in endless rows, starting from the front. Each seat is numbered sequentially row wise, beginning with Seat 1 in Column A of Row 1, moving left to right.

Based on the above information, answer the following questions:
(i) In the seating layout, imagine selecting any 2 × 3 block of seats (2 rows and 3 columns), like the one shown in the picture.

	x

If the seat in the top middle of selected block is numbered ‘x’ write expressions to represent the seat numbers in the other five blank positions of the block.

(ii) Look at the group of seats arranged in the shape.

4		6
	11
16		18

Find the sum of all the seat numbers in this shape. Then compare this total with the number
at the centre i.e. 11.
Try this again with a different set of numbers arranged in the same shape and write your observation.
Solution:
(i) Number to the left of Y will be 1 less than x.
Number to the right of V will be 1 more than x.
Since there are 6 seats in one row,
Number below x will be 6 more than x.
Number in the diagonally left cell to ‘x’ will be 5 more than x for one less than x + 6).
Number in the diagonally right cell to ‘x’ will be 7 more thain x (or one more than x + 6).
Thus, the expressions in other five blank s positions of the block are as shown:

x – 1	x	x + 1
x + 5	x + 6	x + 7

(ii) Sum of all numbers
= 4 + 6 + 11 + 16 + 18 = 55 = 5 × 11
The sum is 5 times the number in the centre.
Now, let the number at the centre be 14, then the shape is given below:

7		9
	17
19		21

Sum of all the numbers
= 7 + 9 + 14 + 19 + 21
= 70 = 5 × 14
Again, the sum is 5 times the number in the centre.
Now, let the number at the centre be 9, then the shape is given below:

2		4
	9
14		16

Sum of all the numbers
= 2 + 4 + 9 + 14 + 16 = 45 = 5 × 9
Again, the sum is 5 times the number in the centre.

Question 2.
Observe the picture below. It shows a growing pattern of huts made using toothbrushes.
In Step 1, there is 1 hut.
In Step 2, there are 2 huts.
In Step 3, there are 3 huts and this pattern continues.

Based on the above information, answer the following questions:
(i) What is the general rule to find the number of toothbrushes in step n?
(ii) How many toothbrushes will be there in step 9, step 37 and step 54?
Solution:
(i) We can see that the number of toothbrushes increases by 3 at each step.

Step No.	Number of Toothbrushes
1	4 = 4 + 0 × 3 = 4 + (4 – 1) × 3
2	4 + 3 = 4 + 1 × 3 = 4 + (2 – 1) × 3
3	4 + 3 + 3 = 4 + 2 × 3 = 4 + (3- 1) × 3
4	4 + 3 + 3 + 3 = 4 + 3 × 3 = 4 + (4 – 1) × 3

We can write general rule to find the number of toothbrushes in step n as:
4 + (n – 1) × 3 = 4 + 3n – 3 = 3n + 1,
where n = 1, 2, 3, …

(ii) Number of toothbrushes in step 9
= 3 × 9 + 1 = 27 + 1 = 28
Number of toothbrushes in step 37
= 3 × 37 + 1 = 111 + 1 = 112
Number of toothbrushes in step 54
= 3 × 54 + 1 = 162 + 1 = 163

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 6 Number Play Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 6 Number Play Solutions

Ganita Prakash Class 7 Chapter 6 Solutions

Class 7 Maths Ganita Prakash Chapter 6 Solutions Number Play

Question 1.
Arrange the stick figure cutouts in order of their heights, ensuring that each child (from left to right) states the number of children standing ahead who are taller than them.


(i) 0, 1 , 1 , 2, 4, 1 , 5
(ii) 0, 0, 0, 0, 0, 0, 0
(iii) 0, 1, 2, 3, 4, 5, 6
(iv) 0, 1,0, 1, 0, 1, 0
(v) 0, 1, 1, 1, 1, 1, 1
(vi) 0,0,0,3,3,3,3
Solution:
(i) The required arrangement is FCBGADF:

(ii) The required arrangement is AECGBDF.

(iii) The required arrangement is FDBGCEA

(iv) The required arrangement is EAGCDBF

(v) The required arrangement is FAECGBD

(vi) The required arrangement is BDFAECG

Question 2.
We know that:
(i) even + even = even
(ii) 0dd + 0dd = 0dd
(iii) even + odd = odd
Similarly, find out the parity for the below scenarios:
(iv) even – even = ___________
(v) 0dd – 0dd = _______
(vi) even – odd = ________
(vii) odd – even = ________
Solution:
(iv) even – even
Example: 6 – 2 = 4 → even
8 – 4 = 4 → even
Parity of result = even
∴ even – even = even

(v) odd – odd
Example: 7 – 3 = 4 → even
9 – 5 = 4 = 4 → even
Parity of result = even
∴ old – old = even

Question 3.
How many different magic squares can be made using numbers 1-9?
Solution:
Using the numbers 1-9, there is exactly one unique magic square (excluding rotations and reflections).

8	1	6
3	5	7
4	9	2

If transformations like rotations are allowed, then there are 8 variations of this magic square.

Question 4.
Create a magic square using numbers 2-10. What strategy would you use for this? Compare it with the magic squares made using 1-9.
Solution:
The numbers 2-10 are 9 consecutive numbers, just like 1-9, but increased up by 1.
Strategy: Start with the classic 1-9 magic square, and add 1 to each number.

8	1	6
3	5	7
4	9	2

Original: After adding 1 to each.
The magic square for numbers 2-10 would have a different magic sum (18) compared to numbers 1-9 (15). The structure remains similar, but the values are shifted up to 1.

9	2	7
4	6	8
5	10	3

Question 5.
Take a magic square, and
(i) increase each number by 1
(ii) double each number
In each case, is the resulting grid also a magic square?
How do the magic sums change in each case?
Solution:
Original:

8	1	6
3	5	7
4	9	2

(i) After increasing each number by 1:

9	2	7
4	6	8
5	10	3

This is still a magic square.
New magic sum = 15 + 3 × 1 = 18

(ii) After doubling each number

16	2	12
6	10	14
8	18	4

Still a magic square
New magic sum = 15 × 2 = 30

In case (i), adding a constant to every number → magic sum (for 3×3 grid) is increased by three times of that constant.
In case (ii), multiplying all by a constant → magic sum multiplied by that constant.

Question 6.
Using the generalised form, find a magic square if the centre number is 25.
Solution:
The generalised form of a magic square is given below.

m + 3	m – 4	m + 1
m – 2	m	m + 2
m – 1	m + 4	m – 3

If the centre number is 25, then m = 25.
Substituting m = 25 in generalised form of a magic square, we get

Question 7.
Write the result obtained by—
(i) adding 1 to every term in the generalised form.
(ii) doubling every term in the generalised form.
Solution:
The generalised form of a magic square is given below.

m + 3	m – 4	m + 1
m – 2	m	m + 2
m – 1	m + 4	m – 3

(i) Adding one to every term:

m + 4	m – 3	m + 2
m – 1	m + 1	m + 3
m	m + 5	m – 2

(ii) Doubling every term:

2m + 6	2m – 8	2m + 2
2m – 4	2m	2m + 4
2m – 2	2m + 8	2m – 6

Question 8.
Create a magic square whose magic sum is 60.
Solution:
A 3 × 3 magic square’s sum is 3 × middle element.
So, for a sum is 60, the middle element should be \(\frac{60}{3}\) = 20 .
To get a magic sum of 60, we will multiply the original magic square by 4 i.e.

Question 9.
Is it possible to get a magic square by filling nine non-consecutive numbers?
Solution:
Yes, it is possible.
Justification: Let us consider the two magic squares with magic sum 45.

18	11	16
13	15	17
14	19	12

9 consecutive numbers
and

24	3	18
9	15	21
12	27	6

9 non-consecutive numbers

Question 10.
A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why?
Solution:
Dorjee toggles the switch 77 times.

Each toggle changes the state of the bulb (ON to OFF or OFF to ON). Starting from ON.

An odd number of toggles will leaves the bulb OFF and an even number of toggles will leave the bulb ON. Since 77 is odd, after 77 toggles, the bulb will be OFF.

Question 11.
Here is a 2 × 3 grid. For each row and column, the parity of the sum is written in the circle; ‘e’ for even and ‘o’ for odd. Fill the 6 boxes with 3 odd numbers (‘o’) and 3 even numbers (‘e’) to satisfy the parity of the row and column sums.

Solution:
Let’s label the cells as:

A	B	C
D	E	F

So the constraints are:

• Row 1 (A, B, C): sum is odd
• Row 2 (D, E, F): sum is even
• Column 1 (A, D): sum is even
• Column 2 (B, E): sum is even
• Column 3 (C, F): sum is odd

We’ll track parities only (o or e), not actual numbers.
Row 1: A = o, B = e, C = e, then o + e + e = odd
Column 1 (A, D) – e means A must be paired with D as odd to get the sum as even.
So, if A = o, D = o, then o + o = even
Similarly, if B = e, E = e, then e + e = even
Again, if C = e, F = o, then e + o = odd
So, the 6 boxes with 3 odd numbers and 3 even numbers can be filled as follows:

Question 12.
Make a 3 × 3 magic square with 0 as the magic sum. All numbers can not be zero. Use negative numbers, as needed.
Solution:
It is given that

• The magic square is 3 × 3.
• The magic sum is 0.
• All numbers in the square cannot be zero, we can use negative numbers as needed.

So, we will use the numbers (- 4) to 4 to create a magic square whose magic sum is 0.
The required magic square is given below.

-3	2	1
4	0	-4
-1	-2	3

Question 13.
Two consecutive numbers in the Virahanka sequence are 987 and 1597. What are the next 2 numbers in the sequence? What are the previous 2 numbers in the sequence?
Solution:
Given numbers are 987 and 1597.
In the Virahanka sequence, each number is the sum of the two preceding numbers.
The next two numbers are:
987 + 1597 = 2584 and 1597 + 2584 = 4181
The previous two numbers are:
1597 – 987 = 610 and 987 – 610 = 377
The sequence is …, 377, 610, 987, 1597, 2584, 4181,…

Question 14.
What is the parity of the 20th term of the Virahanka sequence.
Solution:
Consider the Virahanka sequence given below:
1,2, 3, 5, 8, 13, 21,34, 55, 89, 144, 233, 377, 610, 987, …
Let us observe the pattern of odd/even in Virahanka sequence.
Here 1 → odd;
2 → even;
3 → odd
5 → odd;
8 → even;
13 → odd
21 → odd;
34 → even;
55 → odd
So parity cycle: odd, even, odd, (repeats every 3 terms)
So the parity of 20th term in Virahanka sequence is even.

Question 15.
Solve the following cryptarithm:

Solution:
Here, in TAT, letter T is a hundreds place.
So, T = 1.
⇒ A = 0 and U = 9.
So, we have U = 9, T = 1 and A = 0.

InText Questions

Question 1.
Kishor has a set of number cards and 5 boxes. If each box must contain exactly one number card, suggest an arrangement to help him distribute the cards, so that 5 cards add to 30? Is it possible?

Solution:
No, it is not possible, as the sum of 5 odd numbers is always odd and 30 is an even number.

Question 2.
In a 3 × 3 grid, there are 9 small squares, which is an odd number. Meanwhile, in a 3 × 4 grid, there are 12 small squares, which is an even number.
Given the dimensions of a grid, can you tell the parity of the number of small squares without calculating the product?

Solution:
Yes, we can determine the parity of the number of small squares in a grid without directly calculating the full product, simply by observing the parity of the dimensions.
Rule: The product of two numbers is:

• Even if at least one of the numbers is even.
• Odd if both numbers are odd.

Question 3.
We can describe how the numbers within the magic square are related to each other, i.e. the structure of the magic square.
Choose any magic square that you have made so far using consecutive numbers. If m is the letter- number of the number in the centre, express how other numbers are related to m, how much more or less than m.
Solution:
Consider the magic square

8	1	6
3	5	7
4	9	2

We can express it using the letter- number m for the number in the centre as:

m + 3	m – 4	m + 1
m – 2	m	m + 2
m – 1	m + 4	m – 3

Question 4.
Write the next 3 numbers in the sequence:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, , , , …
If you have to write one more number in the Sequence above, can you tell whether it will be an odd number or an even number (without adding the two previous numbers)?
Solution:
The next 3 terms in the sequence are:
55 + 89 = 144; 89 + 144 = 233; 144 + 233 = 377
Yes, we can determine the parity without adding the two previous number.
Since odd + odd = even
Hence, parity of next number in sequence is even.

Question 5.
Find out what each letter stands for.

Solution:
(i) YY is a two-digit number where both digits are the same. So it can be 99, 88, …
But Z is a 1-digit number and ZOO is a 3-digit number.
So, Y = 9, Z = 1 and O = 0.

\(\begin{array}{r}
99 \\
+\quad 1 \\
\hline 1 \quad 00 \\
\hline
\end{array}\)

(ii)

(iii) Here, KP is a 2-digit number and PRR is a 3-digit number. Basically, 2 × (KP) = PRR.
If P = 1, then R = 2.
Hence, K = 6, P = 1 and R = 2.

Here, C + 1 is a two-digit number i.e. 10.

Number Play Class 7 Extra Questions

Number Play Class 7 Very Short Question Answer

Question 1.
A bag holds 100 balls, each marked with an odd number. If two balls are picked at random and their numbers are added, what will be the parity of the result?
Solution:
We know that odd number + odd number = even number.
Since each ball is marked with an odd number. If two balls are picked at random and their numbers are added, the parity of the sum will be even.

Question 2.
A number has odd parity. What will be the parity of:
(i) Number + 7
(ii) Number – 2
(iii) Number + Number
Solution:
Given, a number has odd parity.
(i) Since the parity of the sum of two odd numbers is even, the parity of (Number + 7) will be even.
(ii) Since the parity of the difference of an even and an odd number is odd, parity of (N umber – 2) will be odd.
(iii) Since the parity of the sum of two odd numbers is even, the parity of (Number + Number) will be even.

Question 3.
Neha wants to climb a staircase with 7 steps. She has a simple rule: She can take either 1 step or 2 steps at a time. For example, one possible way to climb is: 2, 1, 2, 2.
In how many different ways can Neha climb to the top of the 7-step staircase?
Solution:
The number of different ways in which Neha can climb to the top of the 7-step staircase taking either 1 step or 2 steps at a time, is the 7^th element of Virahanka sequence.
Virahanka sequence: 1, 2, 3, 5, 8, 13, 21, …
7^th element of Virahanka sequence = 21
∴ Neha can climb to the top of the 7-step staircase in 21 different ways.

Number Play Class 7 Short Question Answer

Question 1.
During a maths quiz, a student says, “Two consecutive numbers add up to 98.” Is this claim correct? Justify your answer with reasoning.
Solution:
(i) The claim is not correct.
(ii) The counting numbers 1, 2, 3, 4, 5, … alternate between even and odd numbers.
In any two consecutive numbers, one will always be even and the other will always be odd.
We know that the parity of the sum of an even number and an odd number is odd.
Since 98 is an even number, his claim is incorrect.

Question 2.
Form two expressions, one that always gives odd parity and another that always gives even parity.
Solution:
We know that the parity of the product is even if at least one of them is even.
Also, in any two consecutive numbers, one is alwavs even and the other is always odd.
∴ For any number n, the expression n(n + 1) = n² + n always has even parity.
Since even number + odd number = odd number.
∴ The expression n² + n + 1 always has odd parity.

Question 3.
Solve the following cryptarithms:
(i)
\(\begin{array}{r}
\mathrm{B} 2 \\
+6 \mathrm{C} \\
\hline \mathrm{EC} 6
\end{array}\)
(ii)
\(\begin{array}{r}
\mathrm{PQ} \\
+\mathrm{QR} \\
\hline \mathrm{QRQ}
\end{array}\)
Solution:
We know that in cryptarithms:

• Each letter stands for unique digit and different letters represent different digits.
• The same letter always means the same digit.

(i) Here, 2 + C = 6 ⇒ C = 4
Now, B + 6 = EC = E4 [Since C = 4]
⇒ B = 8 and E = 1

(ii) Here, Q + R = Q ⇒ R = 0
Now, P + Q = QR
⇒ P + Q = Q0
⇒ P = 9 and Q = 1

Number Play Class 7 Long Question Answer

Question 1.
Aarav stores his stamp collection in small boxes. He has:

• an odd number of boxes with 5 stamps each,
• an odd number of boxes with 7 stamps each, and
• an even number of boxes with 10 stamps each. He counts all his stamps and says the total is 175.

Did Aarav make a mistake? Explain your reasoning.
Solution:
We know that the parity of the product of two odd numbers is odd and the parity of the product of two even numbers is even. Thus,

• An odd number of boxes with 5(odd) stamps each gives an odd number of stamps.
• An odd number of boxes with 7(odd) stamps each also gives an odd number of stamps.
• An even number of boxes with 10(even) stamps each gives an even number of stamps.

Since, odd number + odd number = even number And even number + even number = even number, the final total must be even number. But Aarav claims the total is 175, which is odd. Therefore, he must have made a mistake in his counting.

Question 2.
Two consecutive numbers in the Virahanka sequence are 377 and 610. What are the previous 2 numbers in the sequence?
Solution:
We know that in Virahanka sequence, a number is the sum of previous two numbers.
Let the previous two numbers be x and y respectively.
The sequence will be as: …, x,y, 377, 610, …
Then, y + 377 = 610
⇒ v = 610-377 = 233
Now, x + v = 377 ,
⇒ x + 233 = 377
⇒ x = 377 – 233 = 144
Thus, the previous two numbers in the Virahanka sequence are 144 and 233.

Question 3.
Solve the following cryptarithms:
(i)
\(\begin{array}{r}
\mathrm{Tl} \\
+1 \mathrm{~T} \\
\hline 66
\end{array}\)
(ii)
\(\begin{array}{r}
\mathrm{D} 3 \\
+3 \mathrm{D} \\
\hline \mathrm{PPQ}
\end{array}\)
Solution:
(i) T + 1 = 6 ⇒ T = 5
Thus,
\(\begin{array}{r}
51 \\
+\quad 15 \\
\hline 66
\end{array}\)

(ii) A three-digit number abc can be written as 100a + 10b + c.
D3 = 10D + 3, 3D = 3 × 10 + D = 30 + D and PPQ = 100P + 10P + Q = 110P + Q
Now, D3 + 3D = (10D + 3) + (30 + D)
= 11 D + 33
When D ranges from 1 to 6, the expression 11 D + 33, gives a two-digit number.
At D = 7, 11 × 7 + 33 = 77 + 33 = 110
Putting P = 1 and Q = 0, we get
110P + Q = 110 × 1 + 0 = 110
∴ P = 1, Q = 0 and D = 7

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 7 A Tale of Three Intersecting Lines Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines Solutions

Ganita Prakash Class 7 Chapter 7 Solutions

Class 7 Maths Ganita Prakash Chapter 7 Solutions A Tale of Three Intersecting Lines

Question 1.
Use the points on the circle and/or the centre to form isosceles triangles.

Solution:
Select any two points on the circle and connect them with the centre of the circle.
Also, join these points to each other. This will form an isosceles triangle as the two radii are equal in length.

Question 2.
Can we say anything about the existence of a triangle for each of the following sets of lengths?
(i) 10 km, 10 km and 25 km
(ii) 5 mm. 10 mm and 20 mm
(iii) 12 cm. 20 cm and 40 cm
Solution:
(i) When we take direct path = 25 km.
Then roundabout path =10 km + 10 km = 20 km.
Since direct path is longer than the roundabout path.
So, existence of a triangle is not possible.

(ii) When we take direct path = 20 mm.
Then roundabout path = 10 mm + 5 mm = 15 mm.
Since direct path is longer than the roundabout path.
So, existence of a triangle is not possible.

(iii) When we take direct path = 40 cm.
Then roundabout path =12 cm + 20 cm = 32 cm.
Since direct path is longer than the roundabout path.
So, existence of a triangle is not possible.

Question 3.
Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? Justify your answer.
Solution:
Yes, an equilateral triangle with sides 50, 50, 50 exists because the sum of two sides is greater than the third side. For any positive number say x > 0, x + x > x. So, an equilateral triangle with all side lengths ‘x’ exists.

Question 4.
For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as sidelengths (decimal values can also be chosen): ;
(i) 1, 100
(ii) 5, 5
(iii) 3, 7
Solution:
(i) 5 possible values for the third length would be 99.5. 99.8, 100, 100.5, 100.9
Since, 100 < 1 + 99.5, 100 < 1 + 99.8, 100 < 1 + 100, 100.5 < 1 + 100, and 100.9 < 1 + 100

(ii) 5 possible values for the third length would be 1, 3.5, 5, 7.5, 8.9
Since, 5 < 1 + 5, 5 < 5 + 3.5, 5 < 5 + 5, 7.5 < 5 + 5, and 8.9 < 5 + 5

(iii) 5 possible values for the third length would be 4.5, 5, 6.9, 8, 9.8
Since, 7 < 3 + 4.5, 7 < 5 + 3, 7 < 3 + 6.9, 8 < 3 + 7, 9.8 < 3 + 7

Question 5.
Construct triangles for the measurement, 3 cm, 120°, 8 cm, where the angle is included between the sides.
Solution:

Steps of construction are given below:
Step 1: Construct a side AB of length 8 cm.
Step 2: Construct ∠d = 120° by drawing the other arm of-the angle.
Step 3: Mark the point C on the other arm such that AC, = 3 cm.
Step 4: Join BC to get the required triangle.

Question 6.
Construct triangles for the measurements, 25°, 3 cm, 60° where the side is included between the angles.
Solution:

Steps of construction are given below:
Step 1: Draw the base AB of length 3 cm.
Step 2: Draw ∠d and ∠B of measure 25° and 60° respectively.
Step 3: Make the point of intersection of the two new arms of ∠d and ∠B as point C to get the required triangle.

Question 7.
Determine which of the following pairs can be the angles of a triangle and which cannot:
(i) 35°, 150°
(ii) 70°, 30°
Solution:
(i) The sum of the given angles 35° + 150° = 185°.
This is not possible because sum of the angles of the triangle exceeds 180°.

(ii) The sum of the given angles 70° + 30° = 100°.
Possible third angle 180°- 100° = 80°.
Since the possible third angle comes out positive (80°), the given angles can be angles of a triangle.

Question 8.
Find the third angle of a triangle (using a parallel line) when two of the angles are:
(i) 36°, 72°
(ii) 150°, 15°
Solution:
(i) Here ∠B = 36° and ∠C = 72°.

Since the line BC is parallel to AT.
So, ∠XAB = ∠B = 36° [Alternate angles] ………. (i)
and ∠YAC = ∠C = 72 [Alternate angles] ………. (i)
Also, ∠XAB + ∠B AC + ∠YAC = 180° [∠XAY is a straight angle]
⇒ 36° + ∠BAC + 72° = 180° [Using (i) and (ii)]
⇒ ∠BAC = 180°- 108° = 72°.

(ii) Here ∠B = 150° and ∠C = 15°.

Since the line BC is parallel to AY.
So, ∠XAB = ∠B = 150° [Alternate angles] ………. (i)
and ∠YAC = ∠C = 15° [Alternate angles] ………. (ii)
Also, ∠XAB + ∠BAC + ∠YAC = 180° [∠XAY is a straight angle]
⇒ 150° + ∠BAC + 15° = 180° [Using (i) and (ii)]
⇒ ∠BAC = 180° – 165° = 15°.

Question 9.
Here is a triangle in which we know ∠B = ∠C and ∠A = 50°, Can you find ∠B and ∠C?

Solution:
Given, ∠A = 50° and ∠B – ∠C.

Draw a line Xy which is parallel to BC.
Now, ∠XA B = ∠B and ∠YAC = ∠C [Alternate angles] ………… (i)
Also, ∠XAB + ∠BAC + ∠YAC = 180°
⇒ ∠B + 50° + ∠C = 180° [Using (i)]
⇒ ∠B + ∠C = 180° – 50° = 130°
⇒ 2 ∠B = 130° [∵ ∠B = ∠C]
⇒ ∠B = 65° = ∠C.

Question 10.
Construct a triangle TRY with RY = 4 cm, TR = 7 cm, ∠R = 140°. Construct an altitude from T to RY.
Solution:
Steps of construction are given below:

Step 1: Construct a side TR of length 7 cm.
Step 2: Construct ∠R = 140° by drawing the other arm of the angle.
Step 3: Mark the point 1 on the other arm such that RY = 4 cm.
Step 4: Join TY to get the required triangle.
Step 5: Keep the ruler aligned to RY. Place the set square along the ruler such that one of the edges of the right angle touches the ruler.
Step 6: Slide the set square along the ruler till the perpendicular edge of the set square touches the vertex T.
Step 7: Extend the line YR and then draw the altitude through T on extended YR using the perpendicular edge of the set square.

InText Questions

Question 1.
What happens when the three vertices lie on a straight line? Will these points form a triangle?
Solution:
When the three vertices lie on a straight line, they become collinear. This means they no longer form a triangle because the three points do not enclose any area.

Question 2.
Construct a triangle with sidelengths 3 cm, 4 cm, and 8 cm. What is happening? Are you able to construct the triangle?
Solution:
Let the triangle be ABC, where AB = 8 cm.

Since, the arcs from the points A and B do not meet. So, we are not able to construct the triangle with side lengths 3 cm, 4 cm, and 8 cm.

A Tale of Three Intersecting Lines Class 7 Extra Questions

A Tale of Three Intersecting Lines Class 7 Very Short Question Answer

Question 1.
In the adjoining figure:

(i) Name the vertex opposite to side PQ.
(ii) Name the side opposite to vertex Q.
(iii) Name the angle opposite to side QR.
(iv) Name the side opposite to ∠R.
Solution:
In the given figure,
(i) The vertex opposite to side PQ is R.
(ii) The side opposite to vertex Q is PR.
(iii) The angle opposite to side QR is ∠P.
(iv) The side opposite to ∠R is PQ.

Question 2.
In the given triangle ABC, find the value of x.

Solution:
Given, ∠T = 7x – 3°, ∠B = 130°, ∠C = 6x + 1°
We know that sum of all the angles in a triangle is 180°.
∴ ∠A + ∠B + ∠C = 180°
⇒ 7x – 3° + 130° + 6x + 1° = 180°
⇒ 13x + 128° = 180° ⇒ 13x = 180°- 128° = 52°
⇒ x = \(\frac{52^{\circ}}{13},\) = 4°
Thus, the value of x is 4°.

Question 3.
Can a triangle be formed for the following set of angles?
(i) 80°, 70° and 50°
(ii) 56°, 64° and 60°
Solution:
We know that the sum of all angles in a triangle is 180°.
(i) Given set of angles are 80°, 70° and 50°.
Now, sum of the angles = 80° + 70° + 50° = 200° ≠ 180°
Thus, the given set of angles cannot form a triangle.

(ii) Given set of angles are 56°, 64° and 60°.
Now, sum of the angles = 56° + 64° + 60° = 180°
Thus, the given set of angles can form a triangle.

Question 4.
In the following figure, find the value of p.

Solution:
In the given figure, ∠SPR is an exterior angle, and ∠PRQ and ∠PQR are two interior opposite angles to ∠SPR.
∴ ∠SPR = ∠PRQ + ∠PQR [Exterior angle property]
⇒ p = 105° + 45° = 150°
Thus, the value of p is 150°.

Question 5.
In ∆XYZ, YX is extended to O. If ∠Y = 57° and XY = XZ, then find ∠ZXO.

Solution:
Given, ∠Y = 57° and XY = XZ
We know that in an isosceles triangle, angles opposite to equal sides are equal.
∴ ∠Z = ∠Y = 57° ……… (i)
In the given figure, ∠ZXO is an exterior angle.
And, ∠Y and ∠Z are two interior opposite angle to ∠ZXO.
We know, Exterior angle = Sum of two interior opposite angles
∴ ∠ZXO = ∠Y + ∠Z
⇒ x – 57° + 57° = 114° [Using (i)]
Thus, the value of ∠ZXO is 114°.

A Tale of Three Intersecting Lines Class 7 Short Question Answer

Question 1.
Construct a triangle PQR such that PQ =3.5 cm, QR = 6.5 cm and PR = 4 cm.
Solution:
Given, PQ = 3.5 cm, QR = 6.5 cm and PR = 4 cm

To construct the triangle PQR of given sides, we follow the steps laid down below:
Step 1: Using ruler, draw a line segment PQ, 3.5 cm long.
Step 2: With P as centre, draw an arc with radius equal to 4 cm.
Step 3: With Q as centre, draw an arc with radius equal to 6.5 cm to cut the previous arc at R.
Step 4: Join R to P and Q. Thus, ∆PQR is required triangle.

Question 2.
Construct a triangle PQR such that PQ = 7 cm, QR = 8.5 cm and PR = 9 cm.
Solution:
Given, PQ = 7 cm, QR = 8.5 cm and PR = 9 cm.

To construct the triangle PQR of given side lengths, follow the steps laid down below:
Step 1: Using ruler, draw a line segment PQ, 7 cm long.
Step 2: With P as centre, draw an arc with radius equal to 9 cm.
Step 3: With Q as centre, draw an arc with radius equal to 8.5 cm to cut the previous arc at R.
Step 4: Join R to P and Q. Thus, ∆PQR is required triangle.

Question 3.
In the given figure, PQ is parallel to RS. Find the value of y.

Solution:
In ∆PQR, we have
∠QPR = 82°, ∠PRQ = 42° and ∠PQR = x
We know that the sum of all angles in a triangle is 180°. ‘
∴ ∠QPR + ∠PRQ + ∠PQR = 180°
⇒ 82° + 42° + x = 180°
⇒ 124° + x = 180°
⇒ x = 180°-124° = 56°
Since PQ is parallel to RS and QR is a transversal, ∠PQR and ∠QRS are alternate interior angles, they are equal.
⇒ ∠QRS = ∠PQR
⇒ Y = ,x = 56°
Thus, the value of y is 56°.

Question 4.
In the given figure, find the value of x.

Solution:
In ∆DBE, ∠DEF = 150° is an exterior angle.
We know that the exterior angle of a triangle is equal to the sum of opposite interior angles.
∴ ∠DEF = ∠ERD + ∠EDB
⇒ 150° = 50° + ∠EDB
⇒ ∠EDB = 150° – 50° = 100°
Again, in ADAG, ∠EDB is an exterior angle.
∴ ∠EDB = ∠DAG + ∠AGD
⇒ 100° = ∠DAG + 70°
[∵ ∠EDB = 100°, ∠AGD = 70°]
⇒ ∠DAG = 100° – 70° = 30°
Also, ∠DAG and ∠GAH form a linear pair of angles.
∴ ∠DAG + ∠GAH = 180°
[∵ ∠DAG = 100°, ∠GAH = x]
⇒ 30° + x = 180°
⇒ x = 180°-30° = 150°

Question 5.
In ∆IJK, if IJ = IK, ∠IKJ = 50°, ∠JIK = m, find the value of m.

Solution:
Given, IJ = IK, ∠IKJ = 50°, ∠JIK = m
We know that in an isosceles triangle, angles opposite to equal sides are equal.
∴ ∠IKJ = ∠IJK = 50°
Now, sum of all the angles = 180°
⇒ ∠JIK + ∠IKJ + ∠IJK = 180°
⇒ m + 50° + 50° = 180°
⇒ m + 100° = 180°
⇒ m = 180°- 100° = 80°
Thus, the value of m is 80°.

Question 6.
In the given figure, if ∠NMO = 84° and LM = MN, then find the values of ∠N and ∠L.

Solution:
Given, ∠NMO = 84° and LM = MN
We know that in an isosceles triangle, angles opposite to equal sides are equal.
∴ ∠N = ∠L ………. (i)
In the given figure, ∠NMO = 84° is an exterior angle.
And, ∠N and ∠L are two interior opposite angles to ∠NMO.
We know,
Exterior angle = Sum of two interior opposite angles
∴ ∠NMO = ∠N + ∠L
⇒ 84° = ∠N + ∠N [Using (i)]
⇒ 2∠N = 84° ⇒ ∠N = \(\frac{84^{\circ}}{2}\) = 42°
Thus, ∠N = ∠L = 42°

Question 7.
In the given figure, find the value of y.

Solution:
In the given figure, ∠BCD is an exterior angle.
And, ∠BAC and ∠ABC are two interior opposite angles to ∠BCD.
We know, Exterior angle = Sum of two interior opposite angles
∴ ∠BCD = ∠BAC + ∠ABC
⇒ 7y + 6° = 50° + 96°
⇒ 7v + 6° = 146°
⇒ 7y = 146° – 6° = 140°
⇒ y = \(\frac{140^{\circ}}{7}\) = 20°
Thus, the value of y is 20°.

Question 8.
In ∆PQR, ZQ is thrice of ∠P and ∠R is twice of sum of ∠P and ∠Q. Find the angles.
Solution:
Let ∠P be x. Then,
∠Q = 3 x, ∠P = 3x
and ∠R = 2(∠P + ∠Q) = 2(x + 3x) = 2 × 4x = 8x
We know that sum of all the angles in a triangle is 180°.
∴ ∠P + ∠Q + ∠R = 180°
⇒ x + 3x + 8x = 180°
⇒ 12x = 180°
⇒ x = \(\frac{180^{\circ}}{12}\) = 15°
Thus, ∠P = x = 15°, ∠Q = 3x = 3 × 15° = 45° and ∠R = 8x = 8 × 15° = 120°

A Tale of Three Intersecting Lines Class 7 Long Question Answer

Question 1.
Mrs. Roy gave straws of length 21 cm to her students and asked them to cut the straw to get three pieces, which can be used to form each of the three types of triangles. The lengths of the three pieces were supposed to be whole numbers (1, 2, 3, …) only. Based on these criteria, write any one possible combination of straw lengths in Column 3 of the below table.

Column 1	Column 2	Column 3
	Type of triangle	Side lengths (cm)
(a)	Scalene triangle
(b)	Isosceles triangle
(c)	Equilateral triangle

Solution:
We know that if a given set of three lengths satisfies the triangle inequality (each length < sum of the other two lengths), then a triangle exists having those as side lengths.
(a) If all three sides of a triangle are different in length, then it is called a scalene triangle.
Thus, the possible sets of side lengths are:
(i) 2 cm, 9 cm and 10 cm
(ii) 3 cm, 8 cm and 10 cm
(iii) 4 cm, 7 cm and 10 cm
(iv) 4 cm, 8 cm and 9 cm
(v) 5 cm, 6 cm and 10 cm
(vi) 5 cm, 7 cm and 9 cm
(vii) 6 cm, 7 cm and 8 cm

(b) If any two sides of a triangle are equal in length, then it is called an isosceles triangle.
Thus, the possible sets of side lengths are:
(i) 6 cm, 6 cm and 9 cm
(ii) 8 cm, 8 cm and 5 cm
(iii) 9 cm, 9 cm and 3 cm
(iv) 10 cm, 10 cm and 1 cm

(c) If all three sides of a triangle are equal in length, then it is called an equilateral triangle. Thus, the possible set of side lengths is 7 cm, 7 cm, 7 cm.

Question 2.
In the given figure, find ∠EAB + ∠ABC + ∠BCD + ∠CDE + ∠DEA.

Solution:
We know that the sum of angles of a triangle is 180°.
∴ In ∆ABC, we have
∠CAB + ∠ABC + ∠BCA = 180° ………… (i)
In ∆ACD, we have
∠ACD + ∠CDA + ∠DAC = 180° ………. (ii)
And in AADE, we have
∠ADE + ∠DEA + ∠EAD = 180° ……….. (iii)
Adding (i), (ii) and (iii), we get
∠CAB + ∠ABC + ∠BCA + ∠ACD + ∠CDA + ∠DAC + ∠ADE + ∠DEA + ∠EAD = 180° + 180° + 180°
⇒ (∠EAD + ∠DAC + ∠CAB) + ∠ABC + (∠BCA + ∠ACD) + (∠CDA + ∠ADE) + ∠DEA = 540°
⇒ ∠EAB + ∠ABC + ∠BCD + ∠CDE + ∠DEA = 540°
[∵ ∠EAD + ∠DAC + ∠CAB = ∠EAB, ∠BCA + ∠ACD = ∠BCD and ∠CDA + ∠ADE = ∠CDE]

Question 3.
Two line segments PS and QR intersect at O. Joining PQ and SR, we get two triangles, ∆POQ and ∆ROS as shown in the figure. Find the values of ∠P and ∠Q.

Solution:
In ∆ROS, ∠R = 35° and ∠S = 70°
We know that sum of all the angles in a triangle is 180°.
∴ ∠R + ∠ROS + ∠S = 180°
⇒ 35° + ∠ROS + 70° = 180°
⇒ ∠ROS + 105° = 180°
⇒ ∠ROS = 180° – 105° = 75°
Since ∠ROS and ∠POQ are vertically opposite angles, ∠ROS = ∠POQ = 75°.
Now, in ∆POQ, ∠P = 5x, ∠POQ =75° and ∠Q = 4x + 6°
We know that sum of all the angles in a triangle is 180°.
∴ ∠P + ∠POQ + ∠Q = 180°
⇒ 5i + 75° + 4x + 6° = 180°
⇒ 9x + 81° = 180°
⇒ 9x = 180° – 81° = 99°
⇒ x = \(\frac{99^{\circ}}{9}\) = 11°
Thus, ∠P = 5x = 5 × 11° = 55° and ∠Q = 4x + 6° = 4 × 11° + 6° = 44° + 6° = 50°

Question 4.
Construct a triangle ∆XYZ with XY = 4 cm, YZ = 7 cm and ∠Y = 108°.
Solution:
Given, in ∆XYZ, XY = 4 cm, YZ = 7 cm and ∠Y = 108°
Steps of construction of ∆XYZ are as follows:
Step 1: Using a ruler, draw a line segment XY, 4 cm long.
Step 2: Using protractor, draw the given vertex angle, ∠Y = 108°.
Step 3: On the new arm created from point F, mark a point Z such that YZ = 7 cm using a ruler and a compass.
Step 4: Using a ruler, connect points X and Z to complete the required triangle XYZ.

Question 5.
Construct a triangle IJK if IJ = 5 cm, JK = 6 cm and included angle between sides IJ and JK is 37°.
Solution:
Given, in ∆IJK, IJ = 5 cm, JK = 6 cm and included angle between sides IJ and JK is 37°.
Steps of construction of ∆IJK are as follows:
Step 1: Using a ruler, draw a line segment IJ of 5 cm length.
Step 2: Using protractor, draw the given vertex angle, ∠J = 37°.

Step 3: On the new arm created from point K mark a point K such that JK = 6 cm using a ruler anti a compass.
Step 4: Using a ruler, connect points I and K to complete the triangle IJK.

Question 6.
Construct a triangle XYZ with ∠Z = 30°, ∠Y = 115° and YZ = 5 cm.
Solution:
Given, in ∆XYZ, ∠Z = 30°, ∠Y = 115° and YZ = 5 cm
Steps of construction of ∆XYZ are as follows:
Step 1: Using a ruler, draw a line segment YZ of 5 cm length.
Step 2: Using protractor, draw the given vertex angle, ∠Y = 115°.

Step 3: Using protractor, draw the given vertex angle, ∠Z = 30°.
Step 4: Mark the intersecting point of two new lines (arms of angles) as X. Thus, the required triangle, ∆XFZ is constructed.

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

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 8 Working with Fractions Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 8 Working with Fractions Solutions

Ganita Prakash Class 7 Chapter 8 Solutions

Class 7 Maths Ganita Prakash Chapter 8 Solutions Working with Fractions

Question 1.
A team of workers can make 1 km of a water canal in 8 days. So, in one day, the team can make _______ km of the water canal. If they work 5 days a week, they can make _______ km of the water canal in a week.
Solution:
Water canal made by team of workers in 8 days = 1 km
So, water canal made by team of workers in 1 dav = \(\frac{1}{8}\) km
The length of the water canal made by the team of workers in 5 days = 5 × \(\frac{1}{8}\) km = \(\frac{5}{8}\) km
Hence, the team can make \(\frac{5}{8}\) km of the water canal in one week.

Question 2.
Safia saw the Moon setting on Monday at 10 pm. Her mother, who is a scientist, told her that every day the Moon sets \(\frac{5}{6}\) hour later than the previous day. How many hours after 10 pm will the moon set on Thursday?
Solution:
There are 3 days from Monday to Thursday. Since the Moon sets \(\frac{5}{6}\) hours later than previous day, the number of hours the Moon will set later on Thursday than Monday = 3 × \(\frac{5}{6}\) hours = \(\frac{15}{6}\) hours = \(\frac{5}{2}\) hours.
We know that 1 hour = 60 minutes
⇒ \(\frac{5}{2}\) hours = \(\frac{5}{2}\) × 60 minutes = \(\frac{300}{2}\) minutes = 150 minutes
Now, 150 minutes = 120 minutes + 30 minutes = 2 hours 30 minutes
Hence, on Thursday the Moon will set 2 hours 30 minutes after 10 pm.

Question 3.
Find the following products. Use a unit square as a whole for representing the fractions and carrying out the operations.
(i) \(\frac{2}{3}\) × \(\frac{4}{5}\)
(ii) \(\frac{1}{4}\) × \(\frac{2}{3}\)
Solution:
(i) Each row represents \(\frac{1}{5}\) and each column represents \(\frac{1}{3}\). The whole is divided into 5 rows and 3 columns creating 5 x 3 = 15 equal parts and 2 × 4 = 8 of the parts is double-shaded, that is \(\frac{8}{15}\) of the whole is double-shaded.
Therefore, \(\frac{2}{3}\) × \(\frac{4}{5}\) = \(\frac{8}{15}\)

(ii) Each row represents \(\frac{1}{3}\) and each column represents \(\frac{1}{4}\). The whole is divided into 3 rows and 4 columns, creating 3×4=12 equal parts and 2 of the parts are double-shaded, that is \(\frac{2}{12}\) of the whole is double-shaded.
Therefore, \(\frac{1}{4}\) × \(\frac{2}{3}\) = \(\frac{2}{12}\) = \(\frac{1}{6}\)

Question 4.
A water tank is filled from a tap. If the tap is open for 1 hour, \(\frac{7}{10}\) of the tank gets filled. How much of the tank is filled if the tap is open for
(i) \(\frac{1}{3}\) hours
(ii) \(\frac{7}{10}\) hours
Solution:
In 1 hour, part of the tank gets filled = \(\frac{7}{10}\)
(i) In \(\frac{1}{3}\) hours, part oLthe tank gels filled = \(\frac{1}{3}\) × \(\frac{7}{10}\) = \(\frac{1 \times 7}{3 \times 10}\) = \(\frac{7}{30}\)
Therefore, in \(\frac{1}{3}\) hours, \(\frac{7}{30}\) of the tank gets filled.

(ii) In \(\frac{7}{10}\) hours, part of the tank gets filled = \(\frac{7}{10}\) × \(\frac{7}{10}\) = \(\frac{7 \times 7}{10 \times 10}\) = \(\frac{49}{100}\)
Therefore, in \(\frac{7}{10}\) hours, \(\frac{49}{100}\) of the tank gets filled.

Question 5.
Tsewang plants four saplings in a row in his garden. The distance between two saplings is \(\frac{3}{4}\) m. Find the distance between the first and last sapling.
Solution:

The distance between the first and the last sapling = \(\frac{3}{4}\) + \(\frac{3}{4}\) + \(\frac{3}{4}\) = 3 × \(\frac{3}{4}\) = \(\frac{9}{4}\) m

Question 6.
Which is heavier: \(\frac{12}{15}\) of 500 grams or \(\frac{3}{20}\) of 4 kg?
Solution:
We have \(\frac{12}{15}\) of 500 g = \(\frac{12}{15}\) × 500 g = \(\frac{12 \times 500}{15}\) = 400 g
And \(\frac{3}{20}\) × 4000 g = \(\frac{3 \times 4000}{20}\) g = 600 g [∵ 1 kg = 1000 g]
∵ 600 g is heavier than 400 g.
∵ \(\frac{3}{20}\) of 4 kg is heavier than \(\frac{12}{15}\) of 500 grams.

Question 7.
(i) 3 ÷ \(\frac{7}{9}\)
(ii) \(\frac{14}{6}\) \(\frac{7}{3}\)
(iii) \(\frac{1}{6}\) ÷ \(\frac{11}{12}\)
(iv) 3\(\frac{2}{3}\) ÷ 1\(\frac{3}{8}\)
Solution:
(i) 3 ÷ \(\frac{7}{9}\) = 3 × \(\frac{9}{7}\) = \(\frac{9}{7}\) = \(\frac{3 \times 9}{7}\)
= 3\(\frac{6}{7}\)

(ii) \(\frac{14}{6}\) \(\frac{7}{3}\) = \(\frac{14}{6}\) × \(\frac{3}{7}\) = \(\frac{14 \times 3}{6 \times 7}\)
= \(\frac{42}{42}\) = 1

(iii) \(\frac{1}{6}\) ÷ \(\frac{11}{12}\) = \(\frac{1}{6}\) × \(\frac{12}{11}\) = \(\frac{1 \times 12}{6 \times 11}\)
= \(\frac{2}{11}\)

(iv) 3\(\frac{2}{3}\) ÷ 1\(\frac{3}{8}\) = \(\frac{11}{3}\) ÷ \(\frac{11}{8}\) = \(\frac{11}{3}\) × \(\frac{11}{8}\)
= \(\frac{11 \times 8}{3 \times 11}\)
= \(\frac{8}{3}\) = 2\(\frac{2}{3}\)

Question 8.
Patiganita a book written by Sridharacharya in the 9^th century CE, mentions this problem: “Friend, after thinking, what sum will be obtained by adding together 1÷ \(\frac{1}{6}\), 1 ÷ \(\frac{1}{10}\), 1 ÷ \(\frac{1}{13}\), 1 ÷ \(\frac{1}{9}\) and 1 ÷ \(\frac{1}{2}\). What should the friend say?
Solution:
1÷ \(\frac{1}{6}\) = 1 × 6 = 6, 1 ÷ \(\frac{1}{10}\) = 1 × 10 = 10, 1 ÷ \(\frac{1}{13}\) = 1 × 13 = 13, 1 ÷ \(\frac{1}{9}\) = 1 × 9 = 9 and 1 ÷ \(\frac{1}{2}\) = 1 × 2 = 2
Therefore, the sum obtained by adding together
1÷ \(\frac{1}{6}\), 1 ÷ \(\frac{1}{10}\), 1 ÷ \(\frac{1}{13}\), 1 ÷ \(\frac{1}{9}\) and 1 ÷ \(\frac{1}{2}\)
= 6 + 10 + 13 + 9 + 2 = 40
Thus, the friend should say the ‘sum’ is 40.

Question 9.
Amritpal decides on a destination for his vacation. If he takes a train, it will take him 5\(\frac{1}{6}\) hours to get there. If he takes a plane, it will take him \(\frac{1}{2}\) hour. How many hours does the plane save?
Solution:
The difference between the two durations = 5\(\frac{1}{6}\) – \(\frac{1}{2}\) = \(\frac{31}{6}\) – \(\frac{1}{2}\) [∵ 5\(\frac{1}{6}\) = \(\frac{31}{6}\)]
= \(\frac{31}{6}\) – \(\frac{3}{6}\) = \(\frac{31-3}{6}\) = \(\frac{28}{6}\)
= \(\frac{14}{3}\) = 4\(\frac{2}{3}\) hours
Hence, the plane saves 4\(\frac{2}{3}\) hours.

Question 10.
What fraction of the whole square is shaded?

Solution:
In the given figure, the big square is divided into 4 identical squares. So, one small square occupies \(\frac{1}{4}\) of the area of the big square. Now, consider the smaller square.
The small square in the figure is divided into 8 identical triangles in which 3 are shaded.

So, the shaded part is \(\frac{3}{8}\) of the small square.
But the small square is \(\frac{3}{8}\) of the big square.
∴ The shaded part is \(\frac{1}{4}\) × \(\frac{3}{8}\) = \(\frac{3}{32}\) of the big square.
Hence, \(\frac{3}{32}\) of the whole square is shaded.

Question 11.
A colony of ants set out in search of food. As they search, they keep splitting equally at each point (as shown in the Fig. 8.7) and reach two food sources, one near a mango tree and another near a sugarcane field. What fraction of the original group reached each food source?

Solution:
At first point ants split in two ways. So, fraction of ants at each way is 1 ÷ \(\frac{1}{2}\) = \(\frac{1}{2}\).
At the second point. ants split in two ways.
So the fraction of ants at each war \(\frac{1}{2}\) ÷ 2 = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

At the third point, ants split in four ways.
So, fraction of ants at each way
= \(\frac{1}{4}\) ÷ 4 = \(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\)
At the fourth point, ants split in 2 wars.
So fraction of ants at each way
= \(\frac{1}{16}\) ÷ 2 = \(\frac{1}{16}\) × \(\frac{1}{2}\) = \(\frac{1}{32}\)
Hence, fraction of ants at mango tree = \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{16}\) + \(\frac{1}{16}\) + \(\frac{1}{32}\)
= \(\frac{16}{32}\) + \(\frac{8}{32}\) + \(\frac{2}{32}\) + \(\frac{2}{32}\) + \(\frac{1}{32}\) = \(\frac{16+8+2+2+1}{32}\)
= \(\frac{29}{32}\)
Fraction of ants near sugarcane field = \(\frac{1}{32}\) + \(\frac{1}{16}\) = \(\frac{1}{32}\) + \(\frac{2}{32}\)
= \(\frac{1+2}{32}\) = \(\frac{3}{32}\)

Question 12.
What is (1 – \(\frac{1}{2}\))?
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\) ?
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) ?
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) × (1 – \(\frac{1}{6}\)) × (1 – \(\frac{1}{7}\)) × (1 – \(\frac{1}{8}\)) × (1 – \(\frac{1}{9}\)) × (1 – \(\frac{1}{10}\)) ?
Make a general statement and explain.
Solution:
1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\) = \(\frac{1}{2}\) × \(\frac{2}{3}\) = \(\frac{1}{3}\);
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) = \(\frac{1}{2}\) × \(\frac{2}{3}\) × \(\frac{3}{4}\) × \(\frac{4}{5}\) = \(\frac{1}{5}\);
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) × (1 – \(\frac{1}{6}\)) × (1 – \(\frac{1}{7}\)) × (1 – \(\frac{1}{8}\)) × (1 – \(\frac{1}{9}\)) × (1 – \(\frac{1}{10}\))
= \(\frac{1}{2}\) × \(\frac{2}{3}\) × \(\frac{3}{4}\) × \(\frac{4}{5}\) × \(\frac{5}{6}\) × \(\frac{6}{7}\) × \(\frac{7}{8}\) × \(\frac{8}{9}\) × \(\frac{9}{10}\) = \(\frac{1}{10}\)
Here, we observe that in this pattern, denominator of each term cancels the numerator of the next term, and found that the final product has the numerator of the first term and the denominator of the last term.
In general, (1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × ………… × (1 – \(\frac{1}{n}\)) = \(\frac{1}{n}\)

InText Questions

Question 1.
In each of the figures given below, find the fraction of the big square that the shaded region occupies.

Solution:
(i) In the figure. the smaller square in the up- right corner is divided into 4 smaller squares and each square within it is further divided into 2 triangles.

∴ Total number of triangles in whole square
= (4 × 2) × 4 = 32
Out of these 32 triangles. 12 triangular regions are shaded.
∴ Fraction of shaded region = \(\frac{12}{32}\) = \(\frac{3}{8}\)
Thus, the shaded region occupies \(\frac{3}{8}\) of the area of the whole square.

(ii) In the figure, the smaller square is divided into 4 triangles and a square, which can be divided into 4 triangles having the same area. Therefore, total triangle in 1 smaller square is 8.

∴ Total number of triangles in whole square = 8 × 4 = 32
Out of these 32 triangles, 2 triangular regions are shaded.
∴ Fraction of shaded region = \(\frac{2}{32}\) = \(\frac{1}{16}\)
Thus, the shaded region occupies \(\frac{1}{16}\) of the area of the whole square.

Working with Fractions Class 7 Extra Questions

Working with Fractions Class 7 Very Short Question Answer

Question 1.
A library has 2400 books, and \(\frac{5}{8}\) of them are placed on the ground floor. The rest are kept on the first floor. How many books are on the first floor?
Solution:
Given, total number of books in library = 2400
Number of books on ground floor
= \(\frac{5}{8}\) of total books = \(\frac{5}{8}\) × 2400 = 5 × 300 = 1500
Hence, number of books on the first floor
= 2400 – 1500 = 900.

Question 2.
A water tank can hold 1250 litres of water. If 2\(\frac{1}{5}\) of such tanks are filled, how much water is used in total?
Solution:
Given, capacity of one tank = 1250 litres
And, 2\(\frac{1}{5}\) of such tanks are filled.
∴ Total water used = 2\(\frac{1}{5}\) × capacitv of one tank
= \(\frac{11}{5}\) × capacity of one tank
= \(\frac{11}{5}\) × 125o = 11 × 250
= 2750 litres

Question 3.
Divide:
(i) 25 by \(\frac{1}{3}\)
(ii) 48 by 3\(\frac{3}{4}\)
Solution:
(i) 25 ÷ \(\frac{1}{3}\) = 25 × \(\frac{3}{1}\)
= \(\frac{25 \times 3}{1}\) = 75

(ii) 48 ÷ 3\(\frac{3}{4}\) = 48 ÷ \(\frac{15}{4}\)
= 48 × \(\frac{4}{15}\) [∵ 3\(\frac{3}{4}\) = \(\frac{15}{4}\)]
= \(\frac{16 \times 4}{5}\) = \(\frac{64}{5}\) = 12\(\frac{4}{5}\)

Question 4.
Solve the following:
(i) \(\frac{3}{4}\) ÷ \(\frac{2}{5}\)
(ii) 2\(\frac{1}{2}\) ÷ \(\frac{5}{8}\)
Solution:
(i) \(\frac{3}{4}\) ÷ \(\frac{2}{5}\) = \(\frac{3}{4}\) ÷ \(\frac{5}{2}\)
= \(\frac{3 \times 5}{4 \times 2}\) = \(\frac{15}{8}\)

(ii) 2\(\frac{1}{2}\) ÷ \(\frac{5}{8}\) = \(\frac{5}{2}\) ÷ \(\frac{5}{8}\)
= \(\frac{5}{2} \times \frac{8}{5}\) = \(\frac{5 \times 8}{2 \times 5}\) = 4

Question 5.
Solve:
(i) \(\frac{5}{6}\) ÷ 3
(ii) 1\(\frac{2}{3}\) ÷ 4
Solution:
(i) \(\frac{5}{6}\) ÷ 3 = \(\frac{5}{6}\) × \(\frac{1}{3}\) = \(\frac{5 \times 1}{6 \times 3}\) = \(\frac{5}{18}\)

(ii) 1\(\frac{2}{3}\) ÷ 4 = \(\frac{5}{3}\) ÷ 4 = \(\frac{5}{3}\) × \(\frac{1}{4}\) = \(\frac{5 \times 1}{3 \times 4}\) = \(\frac{5}{12}\) [∵ 1\(\frac{2}{3}\) = \(\frac{5}{3}\)]

Question 6.
During a community health program, each participant was provided \(\frac{2}{5}\) litres of clean water per day. If 40 litres were distributed on a particular day, calculate the number of participants who received the water.
Solution:
Quantity of water for I participant = \(\frac{2}{5}\) litres
Total quantity oÍ waler = 40 litres
Now, number of participants
= \(\frac{\text { Total quantity of water }}{\text { Quantity of water for } 1 \text { participant }}\)
= 40 ÷ \(\frac{2}{5}\) = 40 × \(\frac{5}{2}\) = 100

Question 7.
What number should be multiplied by 5\(\frac{3}{4}\) to get 3\(\frac{3}{5}\)?
Solution:
Let the required number be x. Then,
x × 5\(\frac{3}{4}\) = 3\(\frac{3}{5}\)
⇒ x = \(\frac{3 \frac{3}{5}}{5 \frac{3}{4}}\)
⇒ x = \(\frac{\frac{18}{5}}{\frac{23}{4}}\) = \(\frac{15}{5}\) × \(\frac{4}{23}\) = \(\frac{18 \times 4}{5 \times 23}\)
= \(\frac{72}{115}\)

Question 8.
Each guest is served 1\(\frac{3}{5}\) litres of juice. If 40 litres of juice is available, how many guests can be served?
Solution:
Given, juice served to each guest.
= 1\(\frac{3}{5}\) litres = \(\frac{8}{5}\)litres
Total quantity of juice available = 40 litres
Now, number of guests that can be served
= \(\frac{\text { Total quantity of juice available }}{\text { Juice served to each guest }}\)
= \(\frac{40}{\frac{8}{5}}\) = 40 × \(\frac{5}{8}\) = 5 × 5 = 25

Working with Fractions Class 7 Short Question Answer

Question 1.
Multiply the following fractions and express as mixed fraction:
(i) 1\(\frac{3}{4}\) × 2\(\frac{2}{5}\)
(ii) 4\(\frac{1}{3}\) × \(\frac{3}{7}\)
(iii) 5\(\frac{2}{9}\) × 1\(\frac{1}{6}\)
(iv) 3\(\frac{4}{5}\) × 2\(\frac{3}{8}\)
Solution:
(i) 1\(\frac{3}{4}\) × 2\(\frac{2}{5}\) = \(\frac{7}{4}\) × \(\frac{12}{5}\)
= \(\frac{7 \times 3}{1 \times 5}\) = \(\frac{21}{5}\)
= 4\(\frac{1}{5}\)

(ii) 4\(\frac{1}{3}\) × \(\frac{3}{7}\) = \(\frac{13}{3}\) × \(\frac{3}{7}\)
= \(\frac{13 \times 1}{1 \times 7}\) = \(\frac{13}{7}\)
= 1\(\frac{6}{7}\)

(iii) 5\(\frac{2}{9}\) × 1\(\frac{1}{6}\) = \(\frac{47}{9}\) × \(\frac{7}{6}\)
= \(\frac{47 \times 7}{9 \times 6}\) = \(\frac{329}{54}\)
= 6\(\frac{5}{54}\)

(iv) 3\(\frac{4}{5}\) × 2\(\frac{3}{8}\) = \(\frac{19}{5}\) × \(\frac{19}{8}\)
= \(\frac{19 \times 19}{5 \times 8}\) = \(\frac{361}{40}\)
= 9\(\frac{1}{4}\)

Question 2.
Meena invited \(\frac{2}{7}\) of her students to a weekend workshop. If \(\frac{3}{5}\) of the invited students were girls, how many boys were present at the workshop if her class has 105 students?
Solution:
Given, total students in the class = 105
∴ Nunber of students invited
= \(\frac{2}{7}\) × Total students =
= \(\frac{2}{7}\) × 105 = 2 × 15 = 30
Now, girls among invited students
= \(\frac{3}{5}\) of the invited students
= \(\frac{3}{5}\) × 30 = 3 × 6 = 18
Hence, number of boys present at the workshop
= 30 – 18 = 12.

Question 3.
A library donated \(\frac{3}{8}\) of its books to a village school. If \(\frac{2}{3}\) of the donated books were story books, how many non-story books were donated to the school if the library originally had 1280 books?
Solution:
Given, total number of books in library = 1280
Number of books donated = \(\frac{3}{8}\) of total books
= \(\frac{3}{8}\) × 1280 = 3 × 160 = 480
Story books among donated books = \(\frac{2}{3}\) × of donated books = \(\frac{2}{3}\) × 480 = 2 × 160 = 320
Hence, number of non-story books among donated books = 480 – 320 = 160

Question 4.
The area of a rectangular field is 63\(\frac{3}{5}\) m². If its length is 7\(\frac{1}{2}\)m, find its breadth?
Solution:
Given area of rectangular field = 63\(\frac{3}{5}\) = \(\frac{318}{5}\) m²
length of rectangular field = 7\(\frac{1}{2}\) = \(\frac{15}{2}\) m
We know.
Area of rectangle length × Breadth
⇒ Breadth = \(\frac{\text { Area of rectangle }}{\text { Length }}\) = \(\frac{318}{5}\) ÷ \(\frac{15}{2}\)
= \(\frac{318}{5}\) × \(\frac{2}{15}\) = \(\frac{106 \times 2}{5 \times 5}\)
= \(\frac{212}{25}\) = 8\(\frac{12}{25}\) m

Question 5.
Determine the number by which 5\(\frac{4}{7}\) must be multiplied to obtain 4\(\frac{3}{5}\).
Solution:
We have. 5\(\frac{4}{7}\) = \(\frac{39}{7}\) and 4\(\frac{3}{5}\) = \(\frac{23}{5}\)
Let the number to be found b x. Then,
\(\frac{39}{7}\) × x = \(\frac{23}{5}\)
Multiplying by the reciprocal of \(\frac{39}{7}\) on both sides, we get
\(\frac{7}{39}\) × \(\frac{39}{7}\) × x = \(\frac{7}{39}\) × \(\frac{23}{5}\) [∵ Reciprocal of \(\frac{39}{7}\) is \(\frac{7}{39}\)]
⇒ x = \(\frac{7 \times 23}{39 \times 5}\) = \(\frac{161}{195}\)

Question 6.
A bookstore sells journals at ₹ 6\(\frac{1}{2}\) per copy. If the shop’s revenue from journal sales totalled ₹ 975, how many dozens of journals were sold?
Solution:
Given, total revenue from journal sales = ₹ 975
Price of 1 journal = ₹ 6\(\frac{1}{2}\) = ₹ \(\frac{13}{2}\)
Now, number of journals sold
= \(\frac{\text { Total revenue from journal sales }}{\text { Price of } 1 \text { journal }}\)
⇒ Number of journals sold = \(\frac{975}{\frac{13}{2}}\) = 975 × \(\frac{2}{13}\)
= \(\frac{75 \times 2}{1}\) = 150
Now, the number of journals sold (in dozens)
= \(\frac{150}{12}\) = \(\frac{25}{2}\) = 12\(\frac{1}{2}\) [∵ 1 dozen = 12]
Thus, 12\(\frac{1}{2}\) dozens of journals were sold.

Question 7.
Divide:
(i) 36 by \(\frac{3}{8}\)
(ii) 54 by 2\(\frac{2}{5}\)
(iii) 42 by \(\frac{7}{6}\)
Solution:
(i) 36 ÷ \(\frac{3}{8}\) = 36 × \(\frac{8}{3}\) = 12 × 8 = 96

(ii) 54 ÷ 2\(\frac{2}{5}\)
= 54 ÷ \(\frac{12}{5}\) = 54 × \(\frac{5}{12}\)
= \(\frac{9 \times 5}{2}\) = \(\frac{45}{2}\) = 22\(\frac{1}{2}\) [∵ 2\(\frac{2}{5}\) = \(\frac{12}{5}\)]

(iii) 42 ÷ \(\frac{2}{5}\) = 42 × \(\frac{6}{7}\)
= 6 × 6 = 36

Question 8.
How many square tiles with a side length of \(\frac{1}{2}\) m are required to cover an area of 12\(\frac{1}{4}\) m²?
Solution:
Given, the side length of a square tile is \(\frac{1}{2}\) m.
Therefore, area of one tile = Side × Side
= \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\) m²
Area to be covered by the tiles = 12\(\frac{1}{4}\) m² = \(\frac{49}{4}\) m²
Now, number of required tiles = Total area that need to be covered ÷ Area of one tile
= \(\frac{49}{4}\) ÷ \(\frac{1}{4}\) = \(\frac{49}{4}\) × \(\frac{4}{1}\) = 49
Thus, the number of required tiles is 49.

Question 9.
A fruit seller earns ₹ 4\(\frac{1}{2}\) per apple. He earns ₹864. How many dozens of apples did he sell?
Solution:
Given, total earnings = ₹864
Amount earned per apple = ₹4\(\frac{1}{2}\) = ₹\(\frac{9}{2}\)
Now, number of apples sold = \(=\frac{\text { Total earnings }}{\text { Amount earned per apple}}\)
= \(\frac{864}{\frac{9}{2}}\) = 864 × \(\frac{2}{9}\) = 96 × 2 = 192
So, number of apples sold (in dozens)
= \(\frac{192}{12}\) = 16 [∵ 1 dozen = 12]

Working with Fractions Class 7 Long Question Answer

Question 1.
Fill in the boxes using <, > or = without actually finding the product:
(i) \(\frac{3}{8}\) × \(\frac{2}{7}\) ☐ \(\frac{2}{7}\)
(ii) \(\frac{5}{9}\) × \(\frac{4}{11}\) ☐ \(\frac{5}{9}\)
(iii) 4\(\frac{1}{2}\) × 3\(\frac{3}{4}\) ☐ 4\(\frac{1}{2}\)
(iv) \(\frac{6}{13}\) × \(\frac{15}{4}\) ☐ \(\frac{15}{4}\)
Solution:
(i) We know, the product of two proper fractions is less than each of them.
Here, \(\frac{3}{8}\) and \(\frac{2}{7}\) are proper tractions.
∴ \(\frac{3}{8}\) × \(\frac{2}{7}\) < \(\frac{2}{7}\)

(ii) We know, the product of two proper fractions is less than each of them.
Here. \(\frac{5}{9}\) and \(\frac{4}{11}\) are proper fractions.
∴ \(\frac{5}{9}\) × \(\frac{4}{11}\) < \(\frac{5}{9}\)

(iii) We know, the product of two improper fractions is greater than both the fractions or equal to either of them.
Here, 4\(\frac{1}{2}\) and 3\(\frac{3}{4}\) are mixed tractions and mixed tractions can he converted into improper fractions.
∴ 4\(\frac{1}{2}\) × 3\(\frac{3}{4}\) > 4\(\frac{1}{2}\)

(iv) We know, the product of a proper fraction and an improper fraction (greater than 1) lies between both the fractions.
Here. \(\frac{6}{13}\) is a proper traction and \(\frac{15}{4}\) is an improper traction (greater than 1).
∴ \(\frac{6}{13}\) × \(\frac{15}{4}\) < \(\frac{15}{4}\)

Question 2.
Solve the following and write the result in simplest form:
(i) 8 × \(\frac{5}{6}\)
(ii) 4 × \(\frac{2}{5}\)
(iii) 7 × \(\frac{3}{8}\)
(iv) 5 × \(\frac{4}{3}\)
(v) \(\frac{11}{4}\) × 3
(vi) 10 × \(\frac{1}{5}\)
Solution:
(i) 8 × \(\frac{5}{6}\) = \(\frac{48 \times 5}{6}\) = \(\frac{4 \times 5}{3}\)
= \(\frac{20}{3}\)

(ii) 4 × \(\frac{2}{5}\) = \(\frac{4 \times 2}{5}\)
= \(\frac{8}{5}\)

(iii) 7 × \(\frac{3}{8}\) = \(\frac{7 \times 3}{8}\)
= \(\frac{21}{8}\)

(iv) 5 × \(\frac{4}{3}\) = \(\frac{5 \times 4}{3}\)
= \(\frac{20}{3}\)

(v) \(\frac{11}{4}\) × 3 = \(\frac{11 \times 3}{4}\)
= \(\frac{33}{4}\)

(vi) 10 × \(\frac{1}{5}\) = 2 × 1 = 2

Question 3.
Solve the following and express as a mixed fraction:
(i) 3\(\frac{5}{8}\) × 2\(\frac{1}{4}\)
(ii) 7\(\frac{2}{3}\) × 4\(\frac{3}{5}\)
(iii) 5\(\frac{3}{10}\) × 3\(\frac{4}{7}\)
(iv) \(\frac{4}{9}\) × 6\(\frac{2}{5}\)
(v) \(\frac{7}{10}\) × 9\(\frac{1}{6}\)
(vi) \(\frac{3}{8}\) × 8\(\frac{5}{9}\)
Solution:
(i) 3\(\frac{5}{8}\) × 2\(\frac{1}{4}\) = \(\frac{29}{8}\) × \(\frac{9}{4}\) = \(\frac{29 \times 9}{8 \times 4}\)
= \(\frac{261}{32}\) = 8\(\frac{5}{32}\)

(ii) 7\(\frac{2}{3}\) × 4\(\frac{3}{5}\) = \(\frac{23}{3}\) × \(\frac{23}{5}\) = \(\frac{23 \times 23}{3 \times 5}\)
= \(\frac{529}{15}\) = 35\(\frac{4}{15}\)

(iii) 5\(\frac{3}{10}\) × 3\(\frac{4}{7}\) = \(\frac{53}{10}\) × \(\frac{25}{7}\)
= \(\frac{265}{14}\) = 18\(\frac{13}{14}\)

(iv) \(\frac{4}{9}\) × 6\(\frac{2}{5}\) = \(\frac{4}{9}\) × \(\frac{32}{5}\) = \(\frac{4 \times 32}{9 \times 5}\)
= \(\frac{128}{45}\) = 2\(\frac{38}{45}\)

(v) \(\frac{7}{10}\) × 9\(\frac{1}{6}\) = \(\frac{7}{10}\) × \(\frac{55}{6}\)
= \(\frac{77}{12}\) = 6\(\frac{5}{12}\)

(vi) \(\frac{3}{8}\) × 8\(\frac{5}{9}\) = \(\frac{3}{8}\) × \(\frac{77}{9}\) = \(\frac{1 \times 77}{8 \times 3}\)
= \(\frac{77}{24}\) = 3\(\frac{5}{24}\)

Question 4.
A car travels 5\(\frac{1}{4}\) km north, then \(\frac{1}{2}\) km west and 4\(\frac{3}{8}\) km north. Find total distance travelled (in km)?
What fraction of the journey was travelled in the north direction?
Solution:
Given: Distance travelled in north direction
d₁ = 5\(\frac{1}{4}\) km = \(\frac{21}{4}\) km
Distance travelled in west direction, d₂ = \(\frac{1}{2}\) km
Distance travelled in north direction again.
d₃ = 4\(\frac{3}{8}\)km = \(\frac{35}{8}\) km
∴ Total distance travelled = d₁ + d₂ + d₃
= \(\frac{21}{4}\) + \(\frac{1}{2}\) + \(\frac{35}{8}\) = \(\frac{42}{8}\) + \(\frac{4}{8}\) + \(\frac{35}{8}\)
= \(\frac{42+4+35}{8}\) = \(\frac{81}{8}\) = 10\(\frac{1}{8}\) km
Total distance travelled in north direction
= d₁ + d₃ = \(\frac{21}{4}\) + \(\frac{35}{8}\) = \(\frac{42+35}{8}\) = \(\frac{77}{8}\) km
∴ Fraction of the journey travelled in the north direction
= \(\frac{d_1+d_3}{d_1+d_2+d_3}\) = \(\frac{\frac{77}{8}}{\frac{81}{8}}\)
= \(\frac{77}{8}\) × \(\frac{8}{81}\) = \(\frac{77}{81}\)

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 2 Arithmetic Expressions Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 2 Arithmetic Expressions Solutions

Ganita Prakash Class 7 Chapter 2 Solutions

Class 7 Maths Ganita Prakash Chapter 2 Solutions Arithmetic Expressions

Question 1.
Fill in the blanks to make the expressions equal on both sides of the ‘=’ sign:
(i) 13 + 4 = _____ + 6
(ii) 22 + = 6 × 5
(iii) 8 × = 64 4 2
(iv) 34 – _____ = 25
Solution:
(i) 13 + 4 = 11 + 6 [∵ 13 + 4 = 17 and 11 + 6 = 17]
(ii) 22 + 8 = 6 × 5 [∵ 6 × 5 = 30 and 22 + 8 = 30]
(iii) 8 × 4 = 64 ÷ 2 [∵ 64 ÷ 2 = 32 and 8 × 4 = 32]
(iv) 34 – 9 = 25

Question 2.
Arrange the following expressions in ascending (increasing) order of their values.
(i) 67 – 19
(ii) 67 – 20
(iii) 35 × 25
(iv) 5 × 11
(v) 120 ÷ 3
Solution:
(i) 67 – 19 = 48
(ii) 67 – 20 = 47
(iii) 35 + 25 = 60
(iv) 5 × 11 = 55
(v) 120 ÷ 3 = 40
Clearly, 40 < 47 < 48 < 55 < 60
∴ 120 ÷ 3 < 67 – 20 < 67 – 19 < 5 × 11 < 35 4 25
Hence, (v) < (ii) < (i) < (iv) < (iii).

Question 3.
For each of the following situations, write the expression describing the situation, identify its terms and find the value of the expression.
(i) Queen Alia gave 100 gold coins to Princess Elsa and 100 gold coins to Princess Anna last year. Princess Elsa used the coins to start a business and doubled her coins. Princess Anna bought jewellery and has only half of the coins left. Write an expression describing how many gold coins Princess Elsa and Princess Anna together have.
(ii) A metro train ticket between two stations is ₹40 for an adult and ₹20 for a child. What is the total cost of tickets:
(a) for four adults and three children?
(b) for two groups having three adults each?
(iii) Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture.

Solution:
(i) Number of gold coins Princess Elsa got = 100
Number of gold coins Princess Anna got = 100
Princess Elsa used the coins to start the business and double her coins.
So, the number of coins Princess Elsa has = 2 × 100
Princess Anna bought jewellery and has only half of the coins left.
So, the total number of coins Princess Anna has = \(\frac{100}{2}\)
Therefore, the total number of gold coins Princess Elsa and Princess Anna have together
= 2 × 100 + \(\frac{100}{2}\) = 200 + 50 = 250
Thus, the expression describing the above situation is, 2 × 100 + \(\frac{100}{2}\).
Terms: 2 × 100, \(\frac{100}{2}\)

(ii) (a) Fare of metro train ticket for an adult = ₹40
So, the fare of metro train ticket for four adults (in ₹) = 4 × 40
Fare of metro train ticket for a child = ₹20
So, the fare of metro train ticket for three children (in ₹) = 3 × 20
Therefore, the expression describing the total cost of tickets (in ₹) for four adults and three children is 4 × 40 + 3 × 20.
Total fare = 4 × ₹40 + 3 × ₹20 = ₹ 160 + ₹60 = ₹220
Terms: 4 × 40, 3 × 20

(b) Fare of metro train ticket for an adult = ₹40
So, the fare of metro train ticket for a group of three adults (in ₹) = 3 × 40
Therefore, the expression describing the total cost of tickets (in ₹) for the two groups having three adults each is 2 × (3 × 40).
Total fare = 2 × (3 × ₹ 40) = 2 × ₹120 = ₹240
Terms: 2 × (3 × 40)

(iii) By observing the given picture, the total height of the window
= Number of gaps × 5 cm + Number of grills × 2 cm + Number of borders × 3 cm
Here, total number of gaps = 7; Total number of grills = 6; Total number of borders = 2
∴ Total height of window (in cm) = 7 × 5 + 6 × 2 + 2 × 3 = 35 + 12 + 6 = 47 + 6 = 53
Terms: 7 × 5, 6 × 2, 2 ×3

Question 4.
Remove the brackets and write the expression having the same value.
(i) 14 + (12 + 10)
(ii) 14 – (12 + 10)
(iii) 14 + (12 – 10)
(iv) 14 – (12 – 10)
(v) -14 + 12 – 10
(vi) 14 – (-12 – 10)
Solution:
(i) 14 + (12 + 10) = 14 + 12 + 10 = 14 + 22 = 36
(ii) 14 – (12 + 10) = 14 – 12 – 10 = 14 – 22 = -8
(iii) 14 + (12 – 10) = 14 + 12 – 10 = 14 + 2 = 16
(iv) 14 – (12 – 10) = 14 – 12 + 10 = 14 – 2 = 12
(v) – 14 + 12 – 10 = – 14 + 2 = – 12
(vi) 14 – (-12 – 10) = 14 + 12 + 10 = 14 + 22 = 36
Here, expressions given in (i) and (vi) have same value.

Question 5.
Find the values of the following expressions. For each pair, first try to guess whether they have the same value. When are the two expressions equal?
(i) (6 + 10) – 2 and 6 + (10 – 2)
(ii) 16 – (8 – 3) and (16 – 8) – 3
(iii) 27 – (18 + 4) and 27 + (-18 – 4)
Solution:
(i) (6 + 10) – 2 = 16 – 2 = 14 and 6 + (10 – 2) = 6 + 8 = 14
Clearly, (6 + 10) – 2 = 6 + (10 – 2)
Hence, both the expressions have the same value.

(ii) 16 – (8 – 3) = 16 – 5 = 11 and (16 – 8) – 3 = 8 – 3 = 5
Clearly, 16 – (8 – 3) ≠ (16 – 8) – 3
Hence, both the expressions do not have the same value.

(iii) 27 – (18 + 4) = 27 – 22 = 5 and 27 + (-18 – 4) = 27 + (- 22) = 27 – 22 = 5
Clearly, 27 – (18 + 4) = 27 + (-18 – 4)
Hence, both the expressions have the same value.

Question 6.
Add brackets at appropriate places in the expressions such that they lead to the values indicated.
(i) 34 – 9 + 12 = 13
(ii) 56 – 14 – 8 = 34
(iii) – 22 – 12 + 10 + 22 = – 22
Solution:
(i) 34 – (9 + 12) = 34 – 21 = 13
(ii) (56 – 14) – 8 = 42 – 8 = 34
(iii) – 22 – (12 + 10) + 22 = – 22 – 22 + 22 = – 22

Question 7.
Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions.

Solution:
For I : 5 × 4 + 4 × 8 = 20 + 32 = 52 or 4 × (4 + 8) + 4 = 4 × 12 + 4 = 52
For II : 8 × (5 + 6) = 8 × 11 = 88 or 8 × 5 + 8 × 6 = 40 + 48 = 88

Question 8.
Read the situations given below. Write appropriate expressions for each of them and find their values.
(i) The district market in Begur operates on all seven days of a week. Rahim supplies 9 kg of mangoes each day from his orchard and Shyam supplies 11 kg of mangoes each day from his orchard to this market. Find the amount of mangoes supplied by them in a week to the local district market.
(ii) Binu earns ₹20,000 per month. She spends ₹5,000 on rent, ₹5,000 on food and ₹2,000 on other expenses every month. What is the amount Binu will save by the end of a year?
(iii) During the daytime a snail climbs 3 cm up a post, and during the night while asleep, accidentally slips down by 2 cm. The post is 10 cm high, and a delicious treat is on its top. In how many days will the snail get the treat?
Solution:
(i) Amount of mangoes supplied by Rahim each day = 9 kg
Amount of mangoes supplied by Shyam each day = 11 kg
Total supplies of mangoes in the market on each day = (9 + 11) kg
∴ Total supplies of mangoes in the market in a week (7 days) = 7 × (9 + 11) = 7 × 20 = 140 kg

(ii) Binu’s per month earning = ₹20,000
Binu’s total monthly expenditures
= ₹5,000 on rent + ₹5,000 on food + ₹ 2,000 on other expenses
= ₹(5,000 + 5,000 + 2,000)
Therefore, Binu’s monthly savings = ₹20,000 – ₹(5,000 + 5,000 + 2,000) = ₹20,000 – ₹12,000 = ₹8,000
Thus, Binu’s total yearly savings = 12 × 8000 = ₹96000
Hence, Binu will save ₹96000 by the end of the year.

(iii) Since the snail climbs 3 cm up the post in daytime and slips down by 2 cm at night.
The distance climbed by the snail in a day = 3 – 2 = 1 cm
∴ The distance climbed in 7 days = 7 cm
The height of the post is 10 cm.
The distance climbed on the 8th day before slipping = 7 + 3=10 cm
So, the snail will take 8 days to reach the top of the post and get the delicious treat.

Question 9.
Find different ways of evaluating the following expressions:
(i) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
(ii) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
Solution:
(i) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 = (1 + 3 + 5 + 7 + 9) + (-2 – 4 – 6 – 8 – 10) = 25 + (- 30) = – 5
OR
1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 = (1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) + (9 – 10)
= (-1) + (-1) + (-1) + (-1) + (-1) = -5

(ii) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 = (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1)
= 0 + 0 + 0 + 0 + 0 = 0
OR
1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1= (1 + 1 + 1 + 1 + 1) + (- 1 – 1 – 1 – 1 – 1)
= 5 + (- 5) = 0

Question 10.
Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets. There can be more than one expression which is equal to the given expression.
(i) 83 – 37 – 12
(a) 84 – 38 – 12
(b) 84 – (37 + 12)
(c) 83 – 38 – 13
(d) -37 + 83 – 12
(ii) 93 + 37 × 44 + 76
(a) 37 + 93 × 44 + 76
(b) 93 + 37 x× 76 + 44
(c) (93 + 37) × (44 + 76)
(d) 37 × 44 + 93 + 76
Solution:
(i) 83 – 37 – 12 = 83 – 37 – 12 + (1 – 1) = (83 + 1) – 37 – 1 – 12 = 84 – 38 – 12
Also, 83 – 37 – 12 = – 37 + 83 – 12
Hence, (a) and (d) are equal to the given expression 83 – 37 – 12.

(ii) 93 + 37 × 44 + 76
Rearranging the terms, we get 37 × 44 + 93 + 76, which is equal to the given expression in option (d). Hence, (d) is equal to the given expression 93 + 37 × 44 + 76.

InText Questions

Question 1.
Use ‘>’ or ‘<’ or ‘=’ in each of the following expressions to compare them. Can you do it without complicated calculations? Explain your thinking in each case.

Solution:
(i)
∴ 245 + 289 > 246 + 285

(ii)
∴ 273 – 145 = 272 – 144

(iii)
∴ 364 + 587 < 363 + 589

(iv)
∴ 142 + 245 < 129 + 245

(v)
∴ 213 – 77 < 214 – 76

Question 2.
Can you explain why subtracting a number is the same as adding its inverse, using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
Solution:
In the token model:
Subtracting a positive number (e.g. subtracting 3) means removing 3 positive tokens.
Adding a negative number (e.g. adding- 3) meaning adding 3 negative tokens. These 3 negative tokens cancel out 3 existing positive tokens (by forming zero pairs), which is equivalent to removing 3 positive tokens.
Since both actions result in removing the same number of positive tokens, the final value is the same.

Question 3.
Does adding the terms of an expression in any order give the same value? Take some expressions and check. Consider expressions with more than 3 terms also.
Solution:
Yes
(-2) + (-3) + (-4) + (-5)
= (-2) + (-3) + (-4) + (-5)
= (- 5) + (-4) + (-5)
= (-9) + (-5) = -14

(-2) + (-3) + (-4) + (-5)
= (-2) + (-3) + (-4) + (-5)
= (-2) + (-7) + (-5)
= (-2) + (-12) = -14

(-2) + (-3) + (-4) + (-5)
= (-2) + (-3) + (-4) + (-5)
= (-2) + (-3) + (-9)
= (-5) + (- 9) = -14
Note: Students can do on their own by taking different numbers.

Question 4.
5 × 4 + 3 ≠ 5 × (4 + 3). Can you explain why?
Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5?
Solution:
Expression 5 × 4 + 3 means 3 more than 5 × 4, which is equal to 23, but 5 × (4 + 3) means 5 times the sum of 3 and 4 which is equal to 35.
Hence, 5 × 4 + 3 + 5 ×(4 + 3)
Now, 5 × (4 + 3), 5 × (3 + 4), and (3 + 4) × 5 have the same meaning, which is 5 times the sum of 3 and 4 and give the same value.
Hence, 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5

Question 5.
Use distributive property to find the following products:
(i) 95 × 8
(ii) 104 × 15
(iii) 49 × 50
Is this quicker than the multiplication procedure you use generally?
Solution:
(i) 95 × 8
= (100 – 5) × 8
= (100 × 8) – (5 × 8)
= 800 – 40 = 760

(ii) 104 × 15
= (100 + 4) × 15
= (100 × 15) + (4 × 15)
= 1500 + 60 = 1560

(iii) 49 × 50
= (50 – 1) × 50
= (50 × 50) – (50 × 1)
= 2500 – 50 = 2450
Yes, this procedure is quicker than the general multiplication procedure.

Arithmetic Expressions Class 7 Extra Questions

Arithmetic Expressions Class 7 Very Short Question Answer

Question 1.
Riya buys 5 notebooks per day for 4 days and 7 notebooks per day for the remaining 3 days of a week. Form an expression to represent the total number of notebooks she buys in that week.
Solution:
For 4 days: Riya buys 5 notebooks per day
For 3 days: Riya buys 7 notebooks per day
Thus, expression for total number of notebooks
= 4 × 5 + 3 × 7

Question 2.
Compare which is greater: 79 – 96 or 117 – 130.
Solution:
The value of 79 – 96 is – 17.
The value of 117 – 130 is – 13.
Clearly, – 17 < -13
Hence, 117 – 130 is greater than 79 – 96.

Question 3.
Anaya is preparing for a temple festival. She decorates 5 pillars, placing 7 marigold garlands on each. However, 2 garlands fall and cannot be used. Find the total number of garlands finally used.
Solution:
Total garlands before any fall = 5 × 7 = 35
Garlands that cannot be used = 2
Now, total garlands finally used = 35 – 2 = 33

Question 4.
Priya buys 2 packs of biscuits for ₹15 each and a juice for ₹20. What is the total cost that Priya needs to pay?
Solution:
Given, Priya buys 2 packs of biscuits for ₹15 each and a juice for ₹20.
Cost of biscuits = 2 × ₹15
Cost of juice = ₹ 20
Now, total cost = 2 × ₹ 15 + ₹20 = ₹30 + ₹20
= ₹50

Question 5.
Add brackets at appropriate places in the expressions such that they lead to the values indicated.
(i) 25 – 7 + 12 = 6
(ii) 42 – 15 – 7-9 = 29
(iii) – 32 – 18 + 14 + 32 = – 32
(iv) 35 – 18 – 5 + 10 = 32
Solution:
(i) 25 – (7 + 12) = 6
(ii) 42 – 15 – (7 – 9) = 29
(iii) – 32 – (18 + 14) + 32 = – 32
(iv) 35 – (18 – 5) + 10 = 32

Arithmetic Expressions Class 7 Short Question Answer

Question 1.
Ankit is planning a birthday party. He buys:

• 5 party hats, each costing ₹ 60
• 2 big balloons, each costing ₹90
• If the total cost exceeds ₹400, he receives a discount of ₹50.

Write the expression that shows the amount (in ₹) Ankit has to pay.
Solution:
Given, cost of five party hats = 5 × ₹ 60
Cost of two big balloons = 2 × ₹90
Discount = ₹50 [If total cost > ₹400]
Now, total cost (in ₹) = 5 × 60 + 2 × 90 = 300 + 180 = 480
As 480 > 400, Ankit gets a discount of ₹50.
So, the expression that shows the amount (in ₹) Ankit has to pay (after applying the discount) is
5 × 60 + 2 × 90 – 50.

Question 2.
Simplify:
(i) (-40) × (-1) + 28 ÷ 7
(ii) 7 – [13 – 2{4 × (- 4)}]
(iii) 81 × [59 – {7 × 8 + (13 – 2 × 5)}]
Solution:
(i) (-40) × (- 1) + 28 ÷ 7
= (40 × 1) + 28 ÷ 7
= 40 + 4 = 44 [∵ (-) × (-) = ( + )]

(ii) 7 – [13 – 2{4 × (- 4)}]
= 7 – [13 – 2 × {-16}] [∵ (+) × (-) = (-)]
= 7 – [13 + 32] = 7 – 45 = -38

(iii) 81 × [59 – {7 × 8 + (13 – 2 × 5)}]
= 81 × [59 – {7 × 8 + (13 – 10)}]
= 81 × [59 – {56 + 3}]
= 81 × [59 – 59] = 81 × 0 = 0

Question 3.
Find the following products using the distributive property.
(i) 107 × 12
(ii) 98 × 14
Solution:
(i) Let’s break 107 as 100 + 7.
Now, 107 × 12 = (100 + 7) × 12
= 100 × 12 + 7 × 12 [Using distributive property]
= 1200 + 84 = 1284
Hence, 107 × 12 = 1284

(ii) Let’s break 98 as 100 – 2.
Now, 98 × 14 = (100 – 2) × 14
= 100 × 14 – 2 × 14
= 1400 – 28 = 1372
Hence, 98 × 14 = 1372

Question 4.
Simplify: 659 – [219 – (750 — 255 ÷ 5 × 9)]
Solution:
We know, on removing the brackets preceded by a minus sign, the signs of all the terms inside the brackets change.
Thus, 659 – [219 – (750 – 255 ÷ 5 × 9)]
= 659 – 219 + (750 – 255 ÷ 5 × 9)
= 659 – 219 + (750 – 51 × 9) [As 255 ÷ 5 = 51]
= 659 – 219 + (750 – 459) [As 51 × 9 = 459]
= 1409-678
[As 659 + 750 = 1409 and – 219 – 459 = – 678]
= 731
Hence, 659 – [219 – (750 – 255 ÷ 5 × 9)] = 731

Question 5.
Simplify:
63 – (- 3){- 2 – 8 – 3} ÷ {5 + (- 2)(- 1)}
Solution:
63 – (-3){-2 – 8 – 3} ÷ {5 + (-2)(-1)}
= 63 – (- 3){-2 – 5} ÷ {5 + (- 2)(- 1)}
[Removal of bar]
= 63 + 3{-2 – 5} ÷ {5 + 2}
[As – (-3) = 3 and (-2) (-1) = 2]
= 63 + 3{-7} ÷ 7
= 63 – 21 ÷ 7[As 3(-7) = -21]
= 63 – 3 = 60[As 21 ÷ 7 = 3]

Question 6.
Ravi took part in a painting competition and got scores 34, 37 and 31 from three judges. Later, it was found that second judge had mistakenly given 37 instead of the correct score 35. What are Ravi’s initial and updated total scores?
Solution:
Given, Ravi got scores 34, 37 and 31 from three judges.
So, the initial total score = 34 + 37 + 31 = 102
Later, the second judge’s score was corrected from 37 to 35, which is 2 less than the initial score.
Since only the second score changed, the updated total score = 102 – 2 = 100
Hence, Ravi’s updated total score is 100.

Question 7.
Remove the brackets and write the expression having the same value.
(i) 18 + (12 + 6)
(ii) 18 + (12 – 6)
(iii) 18 – (12 + 6)
(iv) 18 – (12 – 6)
(v) 18 – (-12 – 6)
(vi) 18 – (-12 + 6)
Solution:
On removing the brackets preceded by a plus sign, the signs of all the terms inside the brackets remain same.
(i) 18 + (12 + 6) = 18 + 12 + 6
(ii) 18 + (12 – 6) = 18 + 12 – 6
On removing the brackets preceded by a minus sign, the signs of all the terms inside the brackets change.
(iii) 18 – (12 + 6) = 18 – 12 – 6
(iv) 18 – (12 – 6) = 18 – 12 + 6
(v) 18 – (-12 – 6) = 18 + 12 + 6
(vi) 18 – (-12 + 6) = 18 + 12 – 6

Question 8.
Leela is organizing candles for a festival. She places 36 candles in one box and 27 in another. She gives away 8 candles from the second box to her neighbour.
Write an expression for the number of candles Leela is left with.
Solution:
Candles in the first box = 36
Candles in the second box = 27
Candles given to neighbour = 8
Now, the number of candles left with Leela
= Candles in the first box + Candles in the second box – Candles given to neighbour
= 36 + 27 – 8, which is the required expression.

Question 9.
In three sections of a school library, 64, 93 and 81 books were recorded respectively.
Later, the librarian found that there were mistakes in two sections:

• The first section actually had 84 books, not 64.
• The second section actually had 73 books, not 93.

What is the difference between total number of books initially and after correcting the record?
Solution:
The error in the number of books in the first section was + 20 (i.e., 84 – 64 = +20).
The error in the second section was -20
(i.e., 73 – 93 = – 20).
Since these two errors cancel each other out, the overall total remains unchanged.
Therefore, there is no difference in the total number of books.

Arithmetic Expressions Class 7 Long Question Answer

Question 1.
Find the following products using the distributive property.
(i) 125 × 16
(ii) 87 × 13
(iii) 106 × 104
(iv) 107 × 91
Solution:
(i) Let’s break 125 as 100 + 25.
Now, 125 × 16 = (100 + 25) × 16
= 100 × 16 + 25 × 16 = 1600 + 400 = 2000
Hence, 125 × 16 = 2000

(ii) Let’s break 87 as 100 -13.
Now, 87 × 13 = (100 – 13) × 13
= 100 × 13 – 13 × 13 = 1300 – 169 = 1131
Hence, 87 × 13 = 1131

(iii) Let’s break 106 as 100 + 6.
Now, 106 × 104 = (100 + 6) × 104
= 100 × 104 + 6 × 104 = 10400 + 6 × (100 + 4)
[As 104 = 100 + 4]
= 10400 + 6 × 100 + 6 × 4
= 10400 + 600 + 24 = 11024
Hence, 106 × 104 = 11024

(iv) Let’s break 107 as 100 + 7.
Now, 107 × 91 = (100 + 7) × 91
= 100 × 91 + 7 × 91 = 9100 + 7 × (100 – 9)
[As 91 = 100 – 9]
= 9100 + 7 × 100 – 7 × 9
= 9100 + 700 – 63
= 9800 – 63 = 9737
Hence, 107 × 91 = 9737.

Question 2.
Simplify:
(i) 121 ÷ [17 – {15 – 3(7 – 4)}]
(ii) 32 ÷ [32 + {32 – (32 + 32 – 32 )}]
(iii) 15 – (-3) × [{4- 7 – 3} + 3 × {5 + (-3) × (-6)}]
Solution:
(i) 121 ÷ [17 – {15 – 3(7 – 4)}]
= 121 ÷ [17 – {15 – 3 × 3}]
= 121 ÷ [17 – {15 – 9}]
= 121 ÷ [17 – 6]
= 121 ÷ 11 = 11

(ii) 32 ÷ [32 + {32 – (32 + 32 – 32 )}]
= 32 ÷ [32 + {32 – (32 + 0)}]
= 32 ÷ [32 + {32 – 32}]
= 32 ÷ [32 + 0] = 32 + 32 = 1

(iii) 15 – (-3) × [{4 – 7 – 3 } ÷ 3 × {5 + (-3) × (-6)}]
= 15 + 3 × [{4 – 4} ÷ 3 × {5 + 18}]
= 15 + 3 × [0 ÷ 3 × 23]
= 15 + 3 × [0 × 23] [As 0 ÷ 3 = 0]
= 15 + 3 × 0 = 15

Arithmetic Expressions Class 7 Case Based Questions

Question 1.
At Surya Vidya Mandir, the annual Art Festival is being celebrated. Ishan is participating in the event and earning performance points based on his art scenes.
Here’s how the points are awarded:

• In the 1^st scene, he puts in a solid effort and earns 22 points.
• In the 2^nd scene, he gets more confident and scores 8 more points than in the 1^st scene.
• In the 3^rd scene, he is slightly exhausted and earns half the points he got in the 2^nd scene.

Based on the above information, answer the following questions:
(i) Write arithmetic expressions to represent the number of points earned in the 2nd and 3rd scenes respectively.
(ii) What is the total number of points earned in all three scenes?
(iii) If each point is worth ₹50, how much money can Ishan claim?
Solution:
Points earned in 1^st scene = 22
Points earned in 2^nd scene
= 8 more points than in the 1^st scene
= 22 + 8 = 30
Points earned in 3^rd scene
= \(\frac{1}{2}\) × (Points earned in 2^nd scene)
= \(\frac{1}{2}\) × 30 = 15

(i) Arithmetic expression to represent the number of points earned in 2^nd scene = 22 + 8
Arithmetic expression to represent the number of points earned in 3^rd scene = \(\frac{1}{2}\) × 30

(ii) Total number of points earned in all three scenes = 22 + 30 + 15 = 67

(iii) Given, 1 point = ₹50
Total money that Ishan can claim
= ₹50 × 67 = ₹3,350

Question 2.
Ria buys 5 packs of pencils for her art class.
Each pack has 6 coloured pencils and 4 graphite pencils. But she loses 3 coloured pencils on the way.
Based on the above information, answer the following questions

(i) How many coloured pencils does Ria have?
(ii) How many graphite pencils does she have?
(iii) What is the total number of pencils Ria finally has?
Solution:
(i) In 1 pack, there are 6 coloured pencils.
So, number of coloured pencils in 5 packs
= 5 × 6 = 30
Given, Ria lost 3 coloured pencils on the way.
Now, number of remaining coloured pencils
= 30 – 3 = 27

(ii) In 1 pack, there are 4 graphite pencils.
So, number of graphite pencils in 5 packs
= 4 × 5 = 20

(iii) Total number of pencils with Ria
= Number of coloured pencils + Number of graphite pencils
= 27 + 20 = 47

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