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Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 2 Lines and Angles Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 2 Lines and Angles Solutions

Ganita Prakash Class 6 Chapter 2 Solutions

Class 6 Maths Ganita Prakash Chapter 2 Solutions Lines and Angles

Question 1.
Can you find the angles in the given pictures? Draw the rays forming any one of the angles and name the vertex of the angle.

Solution:
Yes,

Question 2.
Mark any four points on your paper so that no three of them are on one line. Label them A, B, C and D. Draw all possible lines going through pairs of these points. How many lines do you get? Name them. How many angles can you name using A, B, C and D? Write them all down and mark each of them with a curve.
Solution:
A, B, C and D are four points.

The six possible lines are \(\overleftrightarrow{A B}\), \(\overleftrightarrow{B C}\), \(\overleftrightarrow{C D}\), \(\overleftrightarrow{A D}\), \(\overleftrightarrow{A C}\) and \(\overleftrightarrow{B D}\).
And, 12 possible angles arc ∠BAC, ∠CAD, ∠BAD, ∠ADB, ∠BDC, ∠ADC, ∠DCA. ∠ACB, ∠DCB, ∠ABD, ∠DBC, and ∠ABC.

Question 3.
Find out the number of acute angles in each of the figures below.

What will be the next figure and how many acute angles will it have? Do you notice any pattern in the numbers?
Solution:
First figure: There are three acute angles in the first figure.
Second figure: There are 12 acute angles in second figure (each of the 4 smaller triangles has 3 acute angles).
Third figure: There are 21 acute angles in the third figure (each of the 7 smaller triangles has 3 acute angles).

The next figure is given alongside. And, it has 30 acute angles.
The number of acute angles: 3, 12, 21, 30, …
Pattern: number of acute angles increase by 9 in each step.

Question 4.
Angles in a clock
(a) The hands of a clock make different angles at different times. At 1 o’clock, the angle between the hands is 30°. Why?
(b) What will be the angle at 2 o’clock? And at 4 o’clock? 6 o’clock?
(c) Explore other angles made by the hands of a clock.

Solution:
(a) The clock is divided into 12 hours, so each hour mark is 30° apart, (\(\frac{360^{\circ}}{12}\) = 30°)
Therefore, at 1 o’clock the hour hand is at 1 and the minute hand is at 12, forming a 30° angle.

(b) At 2 o’clock, it is 60° (i.e. 30° × 2 = 60°),
At 4 o’clock, it is 120° (i.e. 30° × 4 = 120°)
At 6 o’clock, it is 180° (i.e. 30° × 6 = 180°)

(c) The angle increases by 30° for each hour. Other angle includes 90° at 3 o’clock, 150° at 5 o’clock and so on. ‘ Thus, on multiplying the hour by 30°, we can find the angle at any hour.

Question 5.
Vidya is enjoying her time on the swing. She notices that the greater the angle with which she starts the swinging, the greater is the speed she achieves on her swing. But where is the angle? Are you able to see any angle?

Solution:
Yes, the angle is visible, and it is formed between the rope and the tree branch.

Question 6.
Observe the images below where there is an insect and its rotated version. Can angles be used to describe the amount of rotation?

How? What will be the arms of the angle and the vertex?
Solution:
Yes, angles can he used to show the amount of rotation.

One arm of the angle is the horizontal line that the insects are on, and the other arm is the line that meets it at a corner point. The point where the two lines meet is the vertex of the angle.

Question 7.
In this figure, ∠TER = 80°. What is the measure of ∠BET? What is the measure of ∠SET?

Solution:
Given, ∠TER = 80° and ∠SER = 90°
Now, ∠SET = ∠SER – ∠TER = 90° – 80° = 10°; ∠BET = ∠BES + ∠SET = 90° + 10° = 100°

Question 8.
Draw the letter ‘M’ such that the angles on the sides are 40° each and the angle in the middle is 80°.
Solution:

Note: In NCERT, the middle angle is given as 60°, which is not possible with side angles as 40°.

Question 9.
Draw the letter Y such that the three angles formed are 150°, 60° and 150°.
Solution:

Question 10.
The Ashoka Chakra has 24 spokes. What is the degree measure of the angle between two spokes next to each other? What is the largest acute angle formed between two spokes?

Solution:
The angle between two spokes is 15° (i.e. \(\frac{360^{\circ}}{24}\)) because the Ashoka Chakra is a circle and dividing 360° by the number of spokes 24 gives the angle between two adjacent spokes.
Now, each pair of spokes can form different angles depending on how many spokes apart they are:
1 spoke apart → 15°;
2 spokes apart → 30°;
3 spokes apart → 45° ;
4 spokes apart → 60°;
5 spokes apart → 75°;
6 spokes apart → 90° → Not acute
7 spokes apart → 105° → Not acute (but obtuse)
So, the largest acute angle formed between two spokes = 75°.

InText Questions

Question 1.
In given figure, we have ∠AOB = ∠BOC = ∠COD = ∠DOE = ∠EOF = ∠FOG = ∠GOH = ∠HOI = ______. Why?

Solution:
In the figure, it is shown that a straight angle is divided into 8 equal angles:
∠AOB, ∠BOC, ∠COD, ∠DOE, ∠EOF, ∠FOG, ∠GOH, and ∠HOI.
A straight angle around a point measures 180°.
Since all 8 angles are equal, we divide 180° equally among them:
∠AOB = ∠BOC = … = ∠HOI = \(\frac{180^{\circ}}{8}\) = 22.5°

Lines and Angles Class 6 Extra Questions

Lines and Angles Class 6 Very Short Question Answer

Question 1.
In the figure, 8 points are given. Join the points A to E, E to F, F to B, B to G, G to C, C to H, H to D and D to C. How many line segments are formed?

Solution:
While joining the points A to E, E to F, F to B, B to G, G to C, C to H, H to D and D to C, we get the figure given alongside.

From the figure, the line segments are \(\overline{A E}, \overline{E F}, \overline{F B}, \overline{B G}, \overline{G C}, \overline{C H}, \overline{H D}\) and \(\overline{D C}\).
So, 8 line segments are formed.

Question 2.
In the given figure, write concurrent lines and their points of concurrence.

Solution:
We know that three or more lines that pass through the same point are called concurrent lines.

Concurrent Lines	Point of Concurrence
line n, line q, line r	A
line l, line p, line q	B
line l, line m, line r	D

Question 3.
Complete the statements given in Column I using the appropriate statements from Column II.

	Column I		Column II
(i)	Three or more points are collinear	(a)	part of line having two endpoints.
(ii)	Through a point	(b)	can be either parallel or intersecting.
(iii)	Line segment is a	(c)	more than one line can pass.
(iv)	Through two points	(d)	if they lie on the same line.
(v)	Two lines in a plane	(e)	only one line can pass.

Solution:
We know that,
Three or more points are collinear if they lie on the same line.
Through a point more than one line can pass.
Line segment is a part of line having two endpoints. Through two points only one line can pass.
Two lines in a plane can be either parallel or intersecting.
∴ (i) – (d), (ii) – (c), (iii) – (a), (iv) – (e), (v) – (b)

Question 4.
Match the expressions in Column I with their correct descriptions from Column II.

	Column I		Column II
(i)	\(\overrightarrow{P Q}\)	(a)	Line PQ
(ii)	\(\overrightarrow{Q P}\)	(b)	Line segment PQ
(iii)	\(\overleftrightarrow{P Q}\)	(c)	A ray with starting point P
(iv)	\(\overline{P Q}\)	(d)	A ray with starting point Q

Solution:
Here,
(i) \(\overrightarrow{P Q}\) represents a ray with starting point P.
(ii) \(\overrightarrow{Q P}\) represents a ray with starting point Q.
(iii) \(\overleftrightarrow{P Q}\) represents a line PQ.
(iv) \(\overline{P Q}\) represents a line segment PQ.
∴ (i) – (c), (ii) – (d), (iii) – (a), (iv) – (b)

Question 5.
Write the name of the angles in the given figure.

Solution:
In the given figure, the possible angles are as follows:

Vertex	Arms	Name of angle
O	\(\overrightarrow{O P}\) and \(\overrightarrow{O Q}\)	∠POQ or ∠QOP
O	\(\overrightarrow{O P}\) and \(\overrightarrow{O R}\)	∠POR or ∠ROP
O	\(\overrightarrow{O Q}\) and \(\overrightarrow{O R}\)	∠ROQ or ∠QOR

Question 6.
How many acute angles are present in the given figure? Also, name them.

Solution:
We know that acute angles measure less than 90°.
In the given figure,
Acute angles: ∠ABC, ∠BCD, ∠DEF, ∠EFG and ∠GHI
So, 5 acute angles are present in the given figure.

Question 7.
In the figure, find ∠AOC.

Solution:
Given, ∠AOB = 51° and ∠BOC = 46°
Now, ∠AOC = ∠AOB + ∠BOC = 51° + 46° = 97°

Question 8.
In the figure, find the missing angle.

Solution:
From figure, ∠ACB is a straight angle.
∴ ∠ACB = 180°
⇒ ∠ACD + ∠BCD = 180° ,
⇒ 123° + ∠BCD = 180°
⇒ ∠BCD = 180°- 123° = 57°

Question 9.
Find the measure of the angle POQ given below using a protractor.

Solution:
Using protractor, we get ∠POQ = 137°.

Question 10.
Find the degree measures of ∠POR, ∠QOR and ∠QOS in the below figure.

Solution:
Using protractor, we get
∠POR = 98°, ∠QOR = 50°, ∠QOS = 76°

Question 11.
In the figure, find the missing angle.

Solution:
From the figure, ∠POQ is a straight angle.
∴ ∠POQ = 180°
⇒ ∠QOR + ∠ROS + ∠SOP = 180°
⇒ 26° + ∠ROS + 32° = 180°
⇒ ∠ROS + 58° = 180°
⇒ ∠ROS = 180°-58° = 122°

Question 12.
In the figure given below, find the value of x°.

Solution:
We know that the complete angle at a point measures 360°
∴ ∠AOB + ∠BOC + ∠COD + ∠AOD = 360°
⇒ 98° + 23° + 76° + x° = 360°
⇒ 197° + x° = 360°
⇒ x° = 360°- 197° = 163°

Lines and Angles Class 6 Short Question Answer

Question 1.
What is the maximum and minimum number of points of intersection of four lines in a plane?

Solution:
When four lines intersect each other in such a way that each pair of lines meet at a different point, we get the maximum number of intersecting points.

Let the four lines be p, q, r and s. Then, the lines can intersect as shown in the figure to give the maximum number of points of intersection.

So, the maximum number of points of intersection of four lines in a plane is six.

Formula to confirm the answer:
Maximum number of points of intersection of n lines = \(\frac{n(n-1)}{2}\)
If all four lines are parallel, then they will not intersect at any point.
So, the minimum number of points of intersection of four lines in a plane is zero.

Question 2.
Lines l, m and n are concurrent. Also, lines p, m and n are concurrent. State whether lines p, l, m and n are concurrent or not.
Solution:
Three or more lines that pass through one common point is called concurrent lines. We are told that:

Lines l, m, and n are concurrent. So, they all meet at a point, let’s call it point A. Also, lines p, m, and n are concurrent. That means these lines also meet at point A. Now, let’s look at all the lines:
Line l goes through point A.
Line p goes through point A.
Line m goes through point A.
Line n goes through point A.
So, all four lines (l, m, n, and p) go through the same point A.
That means they are all concurrent.

Question 3.
Name all the line segments in each of the following figures.

Solution:
All the line segments in figure (i) are \(\overline{A B}\), \(\overline{A C}\) and \(\overline{A D}\).
All the line segments in figure (ii) are \(\overline{P Q}\), \(\overline{P T}\), \(\overline{P S}\), \(\overline{P R}\), \(\overline{Q T}\), \(\overline{S R}\) and \(\overline{S T}\).
All the line segments in figure (iii) are \(\overline{L M}\), \(\overline{M N}\), \(\overline{N O}\), \(\overline{O P}\), \(\overline{P Q}\) and \(\overline{Q L}\).

Question 4.
There are four non-collinear points as shown in the figure. Draw lines through these points taking two at a time. Name the lines. How many such different lines can be drawn?

Solution:
The lines through the given points taking two at a time are \(\overleftrightarrow{P Q}, \overleftrightarrow{P R}, \overleftrightarrow{P S}, \overleftrightarrow{Q R}, \overleftrightarrow{Q S}\) and \(\overleftrightarrow{R S}\)

So, 6 different lines can be drawn from the given four non-collinear points.

Question 5.
In the given figure, name

(i) lines containing the point P.
(ii) lines passing through the point Q.
(iii) line on which H lies.
(iv) three pairs of intersecting lines.
Solution:
(i) Lines containing the Point P are \(\overleftrightarrow{P Q}\), \(\overleftrightarrow{A X}\) and \(\overleftrightarrow{P G}\).
(ii) Lines through the point Q are \(\overleftrightarrow{A Y}\) and \(\overleftrightarrow{P Q}\).
(iii) Line on which lines is \(\overleftrightarrow{X Y}\).
(iv) Three pairs of intersecting hues are \(\overleftrightarrow{P Q}\) and \(\overleftrightarrow{P G}\); \(\overleftrightarrow{A X}\) and \(\overleftrightarrow{X Y}\); \(\overleftrightarrow{A Y}\) and \(\overleftrightarrow{X Y}\).

Question 6.
Identify and name the line segments and rays in each of the following figures.

Solution:
We know that the shortest path from point A to point B (including A and B) is called the line segment AB. It is denoted by either \(\overline{A B}\) or \(\overline{B A}\).
And a ray is a portion of a line that starts at one point (called the starting point or initial point of the ray) and goes on endlessly in a direction.

Figure Number	Line Segment	Ray
(i)	\(\overline{A C}\), \(\overline{D E}\), \(\overline{A B}\) and \(\overline{C D}\)	\(\overrightarrow{A B}\) and \(\overrightarrow{D E}\)
(ii)	\(\overline{R T}\), \(\overline{R P}\), \(\overline{T Q}\), \(\overline{R S}\) and \(\overline{T S}\)	\(\overrightarrow{R P}\), \(\overrightarrow{T Q}\), \(\overrightarrow{R S}\) and \(\overrightarrow{T S}\)
(iii)	\(\overline{O N}\), \(\overline{Q L}\), \(\overline{Q P}\), \(\overline{L P}\), \(\overline{M N}\), \(\overline{N O}\) and \(\overline{M O}\)	\(\overrightarrow{Q L}\), \(\overrightarrow{N M}\), \(\overrightarrow{N O}\) and \(\overrightarrow{Q P}\)

Question 7.
In the given figure, name the point(s):

(i) in the interior of ∠POQ.
(ii) in the interior of ∠POR.
(iii) in the exterior of ∠QOR.
(iv) in the exterior of ∠QOS. ..
(v) on ∠POS.
(vi) on ∠ROS.
Solution:
(i) Point A is in the interior of ∠POQ.
(ii) Points A, B, Q and C are in the interior of ∠POR.
(iii) Points A, P, D. E and S are in the exterior of ∠QOR.
(iv) Points A and P arc in the exterior of ∠QOS.
(v) Points P, O, E and S are on ∠POS.
(vi) Points R, O, E and S are on ∠ROS.

Question 8.
Compare two angles given below using superimposition and find which angle is smaller?

Solution:
Suppose we have a transparent circular paper which can be moved and placed from one figure to another figure.

Let us place the circular paper on the ∠PQR

The circular paper is placed in such a way that its centre is on the vertex of the angle i.e. Q. Mark the points A and B on the edge of circular paper at the points where the arms of the angle PQR pass through the circular paper.

After that put it on the other given angle ∠LMN, such that B lies on the arm MN and check which is smaller.

Hence, ∠LMN is smaller than ∠PQR.

Question 9.
In the figure, write another name for:

(i) ∠1
(ii) ∠2
(iii) ∠3
(iv) ∠4
(v) ∠5
Solution:
(i) Arms of’ the ∠1 are BA and BC and its vertex is B.
So. another name for ∠1 is ∠ABG or ∠GBA.

(ii) Aims of tIme ∠2 are GB amid GC and its vertex is G.
So, another name for ∠2 is ∠BGC or ∠CGB.

(iii) Arms of the ∠3 are CG and CE and its vertex is C.
So, another name for ∠3 is ∠GCE or ∠ECG.

(iv) Arms of the ∠4 arc ED and EC and its vertex is E.
So. another name for ∠4 is ∠CED or ∠DEC.

(v) Arms of the ∠5 are FE and FG and its vertex is F.
So, another name for ∠5 is ∠EFG or ∠GFE.

Question 10.
In each case, determine which angle is greater and why?
(i) ∠PQR or ∠XYZ
(ii) ∠XYZ or ∠LMN
(iii) ∠PQR or ∠LMN

Solution:
On comparing by superimposition, we get

(i) ∠PQR is greater than ∠XYZ because the amount of rotation of ∠PQR is greater than the amount of rotation of ∠XYZ.
(ii) ∠XYZ is greater than ∠LMN because the amount of rotation of ∠XYZ is greater than the amount of rotation of ∠LMN.
(iii) ∠PQR is greater than ∠LMN because the amount of rotation of ∠PQR is greater than the amount of rotation of ∠LMN.

Question 11.
How many acute and obtuse angles are present in the given figure? Also, name them.

Solution:
We know that acute angles measure less than 90° and obtuse angles measure more than 90° but less than 180°.
In the given figure,
Acute angles: ∠OLP, ∠MLP, ∠LMQ, ∠NMQ, ∠MNR, ∠ONR, ∠NOS and ∠LOS
Obtuse angles: ∠LPS, ∠LPQ, ∠MQP, ∠MQR, ∠NRQ, ∠NRS, ∠OSR and ∠OSP
So, 8 acute angles and 8 obtuse angles are present in the given figure.

Question 12.
Classify the following angles (acute, obtuse, right, straight, complete):
(i) 80°
(ii) 90°
(iii) 180°
(iv) 175°
(v) 360°
Solution:
The given angles can be classified as follows:
(i) 80°: Acute angle
(ii) 90°: Right angle
(iii) 180°: Straight angle
(iv) 175°: Obtuse angle
(v) 360°: Complete angle

Question 13.
In the figure, if ∠POQ = 23° and ∠POR = 62°, then find ∠QOR.

Solution:
Given,
∠POQ = 23° and ∠POR = 62°
From the figure, ∠POR = ∠POQ + ∠QOR
⇒ 62° = 23° + ∠QOR
Subtracting 23° from both sides, we get
62° – 23° = 23° + ∠QOR – 23°
⇒ ∠QOR = 39°

Question 14.
Construct an angle of 115° using a protractor.
Solution:

To draw an angle of 115° using a protractor, we follow the steps laid down below:
Step 1: Draw a ray \(\overrightarrow{O A}\).
Step 2: Place the protractor in such a way that its centre is exactly on the point 0 and the base line lies along \(\overrightarrow{O A}\).
Step 3: Starting from 0° on the right, move and look for the mark 115° mark on the protractor.
Step 4: Mark a point B against this 115° mark.
Step 5: Remove the protractor and draw the ray \(\overrightarrow{O B}\).
Thus, ∠AOB is the required angle whose measure is 115°.

Question 15.
How many degrees are there in:
(i) \(\frac{1}{2}\) of a straight angle?
(ii) \(\frac{3}{4}\) of a right angle?
Solution:
(i) \(\frac{1}{2}\) of a straight angle = \(\frac{1}{2}\) × 180° = 90°
(ii) \(\frac{3}{4}\) of right angle = \(\frac{3}{4}\) × 90°
= 3 × 22.5° = 67.5°

Question 16.
Measure the following angles with the help of a protractor and write the measure in degrees.

Solution:
(i) Using protractor, we get ∠ABC = 68°.
(ii) Using protractor, we get ∠PQR = 109°.
(iii) Using protractor, we get ∠XYZ = 125°.

Question 17.
Name the different angles and write their degree measures.

Solution:
All the possible angles are as follows: ;
∠AOB, ∠AOC, ∠AOD, ∠AOE, ∠BOC, ∠BOD, ∠BOE, ∠COD, ∠COE, ∠DOE
Now, ∠AOB = 40°; ∠AOC = 65°; ∠AOD = 125°; ∠AOE = 180°
∠BOC = ∠AOC – ∠AOB = 65° – 40° = 25°
∠BOD = ∠AOD – ∠AOB = 125°- 40° = 85°
∠BOE = ∠AOE – ∠AOB = 180° – 40° = 140°
∠COD = ∠AOD – ∠AOC = 125° – 65° = 60°
∠COE = ∠AOE – ∠AOC = 180° – 65°= 115°
∠DOE = ∠AOE – ∠AOD = 180°- 125°= 55°

Question 18.
Draw a line segment PQ of length 8 cm. Take a point R on PQ such that PR = 5 cm. At point R, draw SR ⊥ PQ.

Solution:
To draw a required perpendicular line using a protractor, we follow the steps laid down below:
Step 1: Draw a line segment PQ of length 8 cm.
Step 2: Mark a point R on the line segment PQ such that PR = 5 cm.
Step 3: Draw SR perpendicular to PQ at R using protractor.
Thus, SR ⊥ PQ at point R.

Question 19.
State the type (obtuse, reflex, complete) of each of the following angles with their measure.

Solution:
Using protractor, we find
∠HOG = 125° (Obtuse angle)
∠EOE = 133°
⇒ Reflex of ∠EOE = 360°- 133° = 227° (Reflex angle)
∠NON = 360° (Complete angle)

Question 20.
Find the degree measures of ∠XOW, ∠XOY, ∠XOZ and ∠YOZ in the below figure.

Solution:
We need to find the measures of ∠XOW, ∠XOY, ∠XOZ and ∠YOZ.
So, we place the protractor in such-a way that its centre coincide with the vertex 0 of the given angles and the base line lies along \(\overrightarrow{O W}\).
Starting from 0° on the right on inner scale, \(\overrightarrow{O X}\) passes through the 26° mark, \(\overrightarrow{O Y}\) passes through the 65° mark and \(\overrightarrow{O Z}\) passes through the 106° mark.
So, ∠XOW = 26°, ∠WOY = 65° and ∠WOZ = 106°
Now, ∠XOW = 26°
∴ ∠XOY = ∠WOY – ∠WOX = 65° – 26° = 39° [∵ ∠XOW = ∠WOX]
Also, ∠XOZ =∠WOZ – ∠WOX = 106° – 26° = 80°
And, ∠YOZ = ∠WOZ – ∠WOY = 106° – 65° = 41°

Lines and Angles Class 6 Long Question Answer

Question 1.
In the given figure, name

(i) Five pairs of intersecting lines.
(ii) Four collinear points.
(iii) Three collinear points.
(iv) Three concurrent lines.
Solution:
(i) Five pairs of intersecting lines are \(\overleftrightarrow{A B}\) and \(\overleftrightarrow{I J}\) ; \(\overleftrightarrow{P Q}\) and \(\overleftrightarrow{I J}\) ; \(\overleftrightarrow{X Y}\) and \(\overleftrightarrow{I J}\) ; \(\overleftrightarrow{X Y}\) and \(\overleftrightarrow{L M}\) ; \(\overleftrightarrow{X Y}\) and \(\overleftrightarrow{A B}\).
(ii) Four collinear points are X, G, H and Y.
(iii) Three collinear points are P, K and Q. As X, G, H and Y are collinear points, we can select any three points from X, G, H and Y as well for three collinear points.
(iv) Three concurrent lines are \(\overleftrightarrow{A B}\), \(\overleftrightarrow{L M}\) and \(\overleftrightarrow{P Q}\).

Question 2.
Consider the line XV in the adjoining figure. Find whether the given statements are true or false.

(i) B is a point on \(\overrightarrow{G Y}\).
(ii) A is a point on \(\overrightarrow{H Y}\).
(iii) G, B and H are points on the line segment GC.
(iv) \(\overrightarrow{H Y}\) is same as \(\overrightarrow{G Y}\).
(v) \(\overrightarrow{B X}\) is different from \(\overrightarrow{B Y}\).
(vi) A, B, C, G, H, X and Y are points on the line XY.
Solution:
(i) True
From the figure, the point B lies between the points G and Y.
Therefore, B is a point on \(\overrightarrow{G Y}\).

(ii) False
From the figure, the point A does not lie between the points H and Y.
Therefore, A is not a point on \(\overrightarrow{H Y}\).

(iii) True
From the figure, the point G is one endpoint of the line segment GC and the points B and H lie between points G and C.
Therefore, G, B and H are points on the line segment GC.

(iv) False
From the figure, the initial points of \(\overrightarrow{H Y}\) and \(\overrightarrow{G Y}\) are different.
Therefore, \(\overrightarrow{H Y}\) is not the same as \(\overrightarrow{G Y}\).

(v) True
From the figure, the initial points of the \(\overrightarrow{B X}\) and \(\overrightarrow{B Y}\) are same but they are moving in different directions.
Therefore, \(\overrightarrow{B X}\) is different from \(\overrightarrow{B Y}\).

(vi) True
From the figure, the points A, B, C, G, H, X and F lie on the line \(\overrightarrow{X Y}\).
Therefore, A, B, C, G, H, X and Y are points on the line \(\overrightarrow{X Y}\).

Question 3.
Draw rough diagrams of two angles for the condition which is possible.
(i) Vertex in common.
(ii) One arm in common.
(iii) Two arms in common.
(iv) Three points in common.
Solution:
Rough diagrams of two angles for the given condition are as follows:
(i) In the figure, ∠POQ and ∠XOY have vertex O in common.

(ii) In the figure, ∠POO and ∠QOR have arm \(\overrightarrow{O Q}\) in common.

(iii) In the figure, ∠POQ and ∠XOY have two arms in common.

(iv) In the figure, ∠POQ and ∠QOR have three points O, S and Q in common.

Question 4.
In each of the grids, join A to other grid points in the figure by a straight line to get:
(i) An acute angle

Solution:
(i) We know that an acute angle is greater than 0° and less than 90°.

Here, ∠BAX, ∠BAY, ∠DCX and ∠DCY are acute angles.

(ii) We know that an obtuse angle is greater than 90° and less than 180°.

Here, ∠BAX, ∠BAY, ∠DCX and ∠DCY are obtuse angles.

(iii) We know that an obtuse angle is greater than 180° and less than 360°.

Here, ∠BAX, ∠BAY, ∠DCX and ∠DCY are reflex angles.

(iv) We know that a right angle is equal to 90°

Here, ∠BAX, ∠BAY, ∠DCX and ∠DCY are right angles.

Question 5.
Find, the degree measures of ∠AOB, ∠BOE, ∠BOD, ∠COE and ∠COD.

Solution:
For ∠AOB, \(\overrightarrow{O A}\) is at 0° mark on the right and \(\overrightarrow{O B}\) passes through the 35° mark from right.
Therefore, ∠AOB = 35°
For ∠BOE, \(\overrightarrow{O E}\) is at 0° mark on the left and \(\overrightarrow{O B}\) passes through the 145° mark from left.
Therefore, ∠BOE = 145°
For ∠COE, \(\overrightarrow{O E}\) is at 0° mark on the left and \(\overrightarrow{O C}\) passes through the 105° mark from left.
Therefore, ∠COE = 105°
For ∠BOD, we have
∠BOD = ∠BOE – ∠DOE …….. (i)
For ∠DOE, \(\overrightarrow{O E}\) is at 0° mark on the left and \(\overrightarrow{O D}\) passes through the 65° mark from left.
Therefore, ∠DOE = 65°
From (i), ∠BOD = 145° – 65° = 80° [∵ ∠BOE = 145° and ∠DOE = 65°]
For ∠COD, we have
∠COD = ∠COE – ∠DOE = 105°- 65° = 40°

Question 6.
Identify the types of angles represented by shaded portion in each:

Solution:
(i) As the angle is less than quarter of a full turn, it is an acute angle.
(ii) As the angle is greater than quarter of a full turn but less than half of a full turn, it is an obtuse angle.
(iii) As the angle is greater than half of a full turn but less than a full turn, it is a reflex angle.
(iv) As the angle is equal to quarter of a full turn, it is a right angle.
(v) As the angle is equal to half of a full turn, it is a straight angle.
(vi) As the angle is equal to a full turn, it is a complete angle.

Lines and Angles Class 6 Case Based Questions

Question 1.
The Ashoka Chakra, found at the centre of the Indian national flag, is a symbol of righteousness and progress. It consists of 24 equally spaced spokes.
Based on the above information, answer the following questions:
(i) What is the angle between any two consecutive spokes of the Ashoka Chakra?
(ii) What is the largest acute angle that can be formed between any two spokes of the Ashoka Chakra?
(iii) What is the largest obtuse angle that can be formed between any two spokes of the Ashoka Chakra?
(iv) What is the smallest reflex angle that can be formed between any two spokes of the Ashoka Chakra?
Solution:
There are 24 spokes in the Ashoka Chakra.
(i) Since the spokes divide the circle into 24 equal parts, the angle between any two consecutive spokes is calculated by dividing the complete angle of a circle (360°) by 24. Thus, the angle between two consecutive spokes is \(\frac{360^{\circ}}{24}\) = 15°.

(ii) The largest acute angle that can be formed between any two spokes would be the greatest multiple of 15° that is still less than 90°.
The largest multiple of 15° and less than 90° is 75°.
Therefore, the largest acute angle between any two spokes is 75°.

(iii) The largest obtuse angle that can he formed between two spokes is the greatest multiple of 15°, which is greater than 90° but less than 1.80°.
The multiples of 15° (between 90° and 180°) are 105°, 120°, 135°, 150°, 165°.
Thus, the largest obtuse angle between any two spokes is 165°.

(iv) The smallest reflex angle that can be formed between two spokes is the smallest angle multiple of 15° greater than 180°, which would be 195° (since 180° + 15° = 195°).

Question 2.
Priya drawn a figure in her notebook as shown.

Based on the above information, answer the following questions:
(i) Write the collinear points.
(ii) Write the concurrent lines.
(iii) How many lines are concurrent at point N?
(iv) How many lines have point Q as the point of intersection?
Solution:
(i) Three or more points that lie on the same straight line are known as collinear points.
From the figure, N, P and O are collinear points.
And, M, S, R and Q are collinear points.

(ii) We know that three or more lines in a plane that pass through one point are known as concurrent lines.
From the figure, \(\overleftrightarrow{M N}\), \(\overleftrightarrow{S N}\) and \(\overleftrightarrow{Q N}\) are concurrent lines.

(iii) From the figure, we can see that three lines \(\overleftrightarrow{M N}\), \(\overleftrightarrow{S N}\) and \(\overleftrightarrow{Q N}\) are concurrent at point N.

(iv) From the figure, we can see that two lines. \(\overleftrightarrow{M Q}\) and \(\overleftrightarrow{N Q}\) intersect at point Q.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 4 Expressions using Letter Numbers MCQ improves accuracy in objective exams.

MCQ on Expressions using Letter Numbers Class 7

Expressions using Letter Numbers MCQ Class 7

Class 7 Maths Expressions using Letter Numbers MCQ

Question 1.
The value of the expression 2x – 7, when x = 5, is:
(a) 7
(b) 5
(c) 3
(d) -4
Solution:
(c) 3
Putting x = 5 in the expression 2x – 7, we get
2 × 5 – 7 = 10 – 7 = 3

Question 2.
The value of the expression 4p + 8q – 10, when p = 3 and q = -1, is:
(a) 8
(b) -24
(c) 14
(d) -6
Solution:
(d) -6
Putting p = 3 and q = – 1 in the expression
4p + 8q – 10,
we get 4 × 3 + 8 × (- 1) – 10 = 12 – 8 – 10
= 12 – 18 = – 6

Question 3.
The simplified form of 9x + 2y + 3x – 7y is:
(a) 12x – 5y
(b) 7xy
(c) 11x – 4y
(d) -3xy
Solution:
(a) 12x – 5y
Given, 9x + 2y + 3x – 7y
Here, 9x and 3x are like terms; 2y and -7y are like terms.
So, 9x + 2y + 3x – 7y = 9x + 3x + 2y – 7y
= 12x – 5y

Question 4.
Shown alongside is a BALANCED hanger using circular and triangular cut-outs. The weight of each circular cut-out is 1 unit and of each triangular cut-out is p units.
Which of the following will help in finding the weight of triangular cut-out?

(a) 6p = 7p
(b) 2p + 4 = 5p + 2
(c) 2(p + 4) = 5(p + 2)
(d) 4p + 2 = 2p + 5
Solution:
(b) 2p + 4 = 5p + 2
Left side: 4 circles and 2 triangles
Right side: 2 circles and 5 triangles
Thus, total weight on left side
= 4 × 1 + 2 × p = (4 + 2p) units
And total weight on right side
= 2 × 1 + 5 × p = (2 + 5p) units
Since the hanger is balanced, both sides must have equal weight.
∴ Weight on left side = Weight on right side
⇒ 4 + 2p = 2 + 5p
⇒ 2p + 4 = 5p + 2

Question 5.
Niharika is printing posters for an event using her home printer. It takes 15 seconds to set up the printer before it starts printing. Once it starts, each poster takes 6 seconds to print. Which of the following expressions shows the total time (in seconds) needed to print4p’ posters, assuming the printer is off initially?
(a) 15 + 6 + p
(b) (15 + 6)p
(c) 15 × 6p
(d) 15 + 6p
Solution:
(d) 15 + 6p
Time taken to set up the printer = 15 seconds (this happens only once).
Time taken to print 1 poster = 6 seconds
Therefore, time taken to print ‘p’ posters = 6p seconds
Thus, total time needed to print ‘p’ posters
= Time taken to set up the printer + Time taken to print ‘p’ posters = (15 + 6p) seconds

Question 6.
The simplified form of 5m – 2n – 7m + 10n + 3 is:
(a) 3
(b) 3m + 3n + 3
(c) 2m + 8n
(d) -2m + 8n + 3
Solution:
(d) -2m + 8n + 3
Given, 5m – 2n – 7m + 10n + 3
Here, 5m and -7m are like terms; -2n and 10n are like terms.
So, 5m – 2n – 7m + 10n + 3
= 5m – 7m – 2n + 10n + 3
= -2m + 8n + 3

Question 7.
Shown below is a representation of one of the properties of whole numbers.

Which of the following represents the property shown?
(a) a(b + c) = a × b + a × c
(b) a + b = b + a
(c) a × (b × c) = a × (b × c)
(d) a + (b + c) = (a + b) + c
Solution:
(a) a(b + c) = a × b + a × c
From figure, we have
Area on left side = Area on right side
⇒ 16 × 8 = 12 × 8 + 4 × 8
[∵ Area of rectangle = Length × Breadth]
⇒ 8 × 16 = 8 × 12 + 8 × 4
⇒ 8 × (12 + 4) = 8 × 12 + 8 × 4
[∵ 16 = 12 + 4]
Suppose a = 8, b = 12 and c = 4, we get
a(b + c) = a × b + a × c

Question 8.
Amit sells cold drinks and sandwiches at a school canteen. One cold drink costs ₹25 and one sandwich costs ₹40.
If m cold drinks and n sandwiches were sold in a day, which of the following expressions shows the total amount earned (in rupees) that day?
(a) 25m + 40n
(b) (25 + 40) × (m + n)
(c) 40m + 25n
(d) (25 + 40) × m + n
Solution:
(a) 25m + 40n
Cost of 1 cold drink = ₹25
Cost of 1 sandwich = ₹40
Money earned from m cold drinks
= 25 × m = 25m
Money earned from n sandwiches
= 40 × n = 40n
Total earnings = 25m + 40M

Question 9.
The formula of the given number machine is:

(a) 2a + b
(b) a × b + 5
(c) a + 5 × b
(d) None of these
Solution:
(b) a × b + 5
We can see,
4 × 1 + 5 = 445 = 9, 7 × 0 + 5 = 0 + 5 = 5,
3 × 2 + 5 = 6 + 5 = 11, 5 × 3 + 5 = 15 + 5 = 20
Here, the rule followed is “5 more than the product of two numbers”.
Thus, if two numbers are a and b, the formula for the given machine is a × b + 5.

Question 10.
Consider the following number machine:

What is the missing input?
(a) 8
(b) 10
(c) 12
(d) 15
Solution:
(a) 8
We can see,
4 × 3 + 2 × 2 = 12 + 4 = 16,
5 × 3 + 2 × 2=15 + 4=19,
7 × 3 + 3 × 2 = 21 + 6 = 27,
9 × 3 + 3 × 2 = 27 + 6 = 33
So, the rule for the given machine in algebraic expression is,
Output = a × 3 + b × 2, where a and b are inputs.
When output = 34 and b = 5,
34 = a × 3 + 5 × 2
⇒ 34 = 3a + 10 ⇒ 3a = 34 – 10 = 24
⇒ a = \(\frac{24}{3}\) = 8 3
Thus, the missing input is 8.

Question 11.
The value of expression 6m – 7n is:
(i) 6, when m = 4 and n = 2.
(ii) 9, when m = 5 and n = 3.
(iii) -9, when m = 2 and n = 3.
(iv) 10, when m = 3 and n = 1.
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii) only
(c) (ii), (iii) and (iv)
(d) (i) and (iv)
Solution:
(b) (ii) and (iii) only
When m = 4 and n = 2;
6m – 7n = 6 × 4 – 7 × 2 = 24 – 14 = 10
When m = 5 and n = 3;
6m – 7n = 6 × 5 – 7 × 3 = 30 – 21 = 9
When m = 2 and n = 3;
6m – 7n = 6 × 2 – 7 × 3 = 12 – 21 = – 9
When m = 3 and n = 1;
6m – 7n = 6 × 3 – 7 × 1 = 18 – 7 = 11
So, (ii) and (iii) are correct.

Question 12.
If p is a number, then which of the following statements are correct?
(i) 2 more than 3 times the number is 2p + 3.
(ii) 4 less than 7 times the number is 7p – 4.
(iii) 4 more than 5 times the number is 5p + 4.
(iv) 8 less than 4 times the number is 8p – 4.
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii) only
(c) (ii), (iii) and (iv)
(d) (i) and (iv)
Solution:
(b) (ii) and (iii) only
(i) 2 more than 3 times the number is 2 + 3 × p
= 3p + 2.
(ii) 4 less than 7 times the number is 7 × p – 4
= 7p – 4.
(iii) 4 more than 5 times the number is 4 + 5 × p
= 5p + 4.
(iv) 8 less than 4 times the number is 4 × p – 8
= 4p – 8.
So, (ii) and (iii) are correct.

Question 13.
The expression 5x – 3y + 4 is equivalent to
(i) 7x – 2y – (2x + y + 2) + 6
(ii) 5x – y + 3 – 2(y + 2x) + 1
(iii) x – y – (2y – 4x) + 4
(iv) 2(2y – x – 2) + 7(1 – y + x) + 1
Choose the correct option from the following:
(a) (i) only
(b) (ii) and (iii) only
(c) (i), (iii) and (iv) only
(d) (i), (ii), (iii) and (iv)
Solution:
(c) (i), (iii) and (iv) only
(i) 7x – 2y – (2x + y + 2) + 6
= 7x – 2y – 2x – y – 2 + 6
= (7x – 2x) + (-2y – y) + 4
= 5x + (-3y) + 4 = 5x – 3y + 4

(ii) 5x – y + 3 – 2(y + 2x) + 1
= 5x – y + 3 – 2y – 4x + 1
= (5x – 4x) + (-y – 2y) + 3 + 1
= x + (-3y) + 4
= x – 3y + 4 ≠ 5x – 3y + 4

(iii) x – y – (2y – 4x) + 4
= x – y – 2y + 4x + 4
= (x + 4x) + (-y – 2y) + 4
= 5x + (-3y) + 4 = 5x – 3y + 4

(iv) 2(2y – x – 2) + 7(1 – y + x) + 1
= 4y – 2x – 4 + 7 – 7y + 7x + 1
= (-2x + 7x) + (4y – 7y) – 4 + 7 + 1
= 5x + (-3y) + 4 = 5x – 3y + 4
So, (i), (iii) and (iv) are correct.

Expressions using Letter Numbers Class 7 Fill in the Blanks

Question 1.
If n is a number, then:
(i) 7 more than the number is _____.
(ii) 3 less than the number is __.
(iii) 5 less than 7 times the number is _________.
(iv) 12 more than 9 times the number is _____.
Solution: n + 7, n – 3, 7n – 5, 9n + 12.
(i) 7 more than the number is n + 7.
(ii) 3 less than the number is n – 3.
(iii) 5 less than 7 times the number is 7n – 5.
(iv) 12 more than 9 times the number is 9n + 12.

Question 2.
If x = – 4, then 5 – x = ____.
Solution: 9
Putting x = – 4 in the expression 5 – x, we get
5 – (-4) = 5 + 4 = 9
∴ If x = – 4, then 5 – x = 9.

Question 3.
If t = 21, then 3t = _______ .
Solution: 63
Putting t = 21 in the expression 31, we get
3 × 21 = 63
∴ If t = 21, then 3t = 63.

Question 4.
If p = 7 and q = – 3, then 3p + 2q = _______.
Solution: 15
Putting p = 7 and q = – 3 in the expression 3p + 2q, we get
3 × 7 + 2 × (- 3) = 21 – 6 = 15 .
∴ If p = 7 and q = – 3, then 3p + 2q = 15.

Question 5.
If x = 12, then 4x + 5 = ______.
Solution: 53
Putting x = 12 in the expression 4x + 5, we get
4 × 12 + 5 = 48 + 5 = 53
∴ If x = 12, then 4x + 5 = 53.

Question 6.
If m = – 4, then 5(m + 1) = _______.
Solution: -15
Putting m = -4 in the expression 5(m + 1), we get
5(- 4 + 1) = 5 × (- 3) = -15
∴ If = -4, then 5(m + 1) = -15.

Question 7.
If p = – 10 and q = 5, then 5p + 10q = ______.
Solution: 0
Putting p = -10 and q = 5 in the expression 5p + 10q, we get
5 × (- 10) + 10 × 5 = – 50 + 50 = 0
∴ If p = – 10 and q = 5 then 5p + 10q = 0.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 3 A Peek Beyond the Point MCQ improves accuracy in objective exams.

MCQ on A Peek Beyond the Point Class 7

A Peek Beyond the Point MCQ Class 7

Class 7 Maths A Peek Beyond the Point MCQ

Question 1.
How many tenths make a unit?
(a) \(\frac{1}{10}\)
(b) 10
(c) 100
(d) 1
Solution:
(b) 10
We know, 1 unit = 10 one-tenths.

Question 2.
The value of \(4 \frac{4}{10}+7 \frac{3}{10}\) is:
(a) \(7 \frac{7}{10}\)
(b) \(11 \frac{7}{10}\)
(c) \(10 \frac{1}{10}\)
(d) \(7 \frac{17}{10}\)
Solution:
(b) \(11 \frac{7}{10}\)
We can write, \(4 \frac{4}{10}=\frac{44}{10} \text { and } 7 \frac{3}{10}=\frac{73}{10}\)
Now, \(4 \frac{4}{10}+7 \frac{3}{10}=\frac{44}{10}+\frac{73}{10}\)
= \(\frac{117}{10}=\frac{110}{10}+\frac{7}{10}=11+\frac{7}{10}=11 \frac{7}{10}\)

Question 3.
How is one-hundredth represented in decimal form?
(a) 0.01
(b) 0.001
(c) 1.0
(d) 0.1
Solution:
(a) 0.01
One-hundredth represents 100 parts of a unit, i.e. \(\frac{1}{100}=\frac{01}{100}\)
Here, number of zeros in denominator is 2.
So, putting decimal point in the numerator 2 places to the left, we get 0.01.
Thus, the decimal representation of one- hundredth is 0.01.

Question 4.
The place value of 5 in the decimal 11.485 is:
(a) Five-tenths
(b) Five-hundredths
(c) Five-thousandths
(d) Five-ones
Solution:
(c) Five-thousandths
The expanded form of 11.485 is:
11.485 = 10 × 1 + 1 × 1 + \(\left(4 \times \frac{1}{10}\right)\) + \(\left(8 \times \frac{1}{100}\right)+\left(5 \times \frac{1}{1000}\right)\)
Thus, the place value of 5 in the decimal 11.485 is five-thousandths, i.e. 0.005.

Question 5.
The decimal form of 729 hundredths is:
(a) 729
(b) 72.9
(c) 7.29
(d) None of these
Solution:
(c) 7.29
The decimal form of 729 hundredths is \(\frac{729}{100}\) = 7.29.

Question 6.
The decimal representation of \(\frac{3}{4}\) is:
(a) 0.75
(b) 0.85
(c) 0.65
(d) 0.95
Solution:
(a) 0.75
We have, \(\frac{3}{4}=\frac{3 \times 25}{4 \times 25}=\frac{75}{100}\) = 0.75

Question 7.
The place value of the digit 7 in the number 3.476 is:
(a) 7 ones
(b) 7 tenths
(c) 7 hundredths
(d) 7 thousandths
Solution:
(c) 7 hundredths
3.476 can he represented in the decimal place value chart as:

Tens	Ones	.	Tenths	Hundredths	Thousandths
	3	.	4	7	6

Hence, the place value of the digit 7 in the number 3.476 is 7 hundredths, i.e. 0.07.

Question 8.
If a number line is divided into 10 equal parts between 3 and 4, then the decimal number at the 7^th division is:
(a) 3.07
(b) 3.7
(c) 3.17
(d) 37
Solution:
(b) 3.7
The number line, divided into 10 equal parts between 3 and 4, is given as:

Thus, the decimal number at the 7^th division is 3.7.

Question 9.
The greatest decimal number among the following is:
(a) 1.2
(b) 1.02
(c) 1.22
(d) 1.21
Solution:
(c) 1.22
The given decimal numbers are: 1.2, 1.02, 1.22, 1.21
Converting them to like decimals, we get 1.20, 1.02, 1.22, 1.21

Now, as the whole number part of all the decimal numbers is the same, i.e. 1, we compare the digits after the decimal point from left to right.

As 1.02 has 0 at the tenths place while other decimals have 2, 1.02 is the smallest.
Now, comparing the hundredths place of 1.20, 1.22, 1.21, we get
1.22 > 1.21 > 1.20 (As 2 >1 > 0)
Thus, 1.22 is the greatest among the given numbers.

Question 10.
Which of the following sequence is decreasing?
(a) 4.5, 4.55, 4.6
(b) 2.3, 2.2, 2.1
(c) 3.1, 3.0, 3.1
(d) 1.01,2.001,3.0001
Solution:
(b) 2.3, 2.2, 2.1
A sequence is decreasing if each term is smaller
than the one before it.
2.3, 2.2, 2.1 → each term decreases by 0.1 i.e. (2.3 > 2.2 > 2.1).
Thus, 2.3, 2.2, 2.1 is the decreasing sequence.

Question 11.
How can we express 0.5 hours in minutes?
(a) 50 minutes
(b) 45 minutes
(c) 25 minutes
(d) 30 minutes
Solution:
(d) 30 minutes
We know, 1 hour = 60 minutes
Thus, 0.5 hours = 0.5 × 60 = \(\frac{5}{10}\) × 60
= 30 minutes

Question 12.
Which of the following pairs of decimal numbers are equivalent?
(i) 0.5 and 0.500
(ii) 0.05 and 0.0500
(iii) 1.06 and 1.060
(iv) 0.06 and 0.600
Choose the correct option from the following:
(a) (i) and (iv)
(b) (ii) and (iii) only
(c) (i), (ii) and (iii)
(d) (i) and (iii) only
Solution:
(c) (i), (ii) and (iii)
We know that adding trailing zeros after the last non-zero digit of a decimal does not change its value. Thus,
(i) 0.5 and 0.500 are equivalent.
(ii) 0.05 and 0.0500 are equivalent.
(iii) 1.06 and 1.060 are equivalent.
(iv) 0.06 and 0.600 are not equivalent as 0.06 represents 6-hundredths, while 0.600 represents 6-tenths.

Question 13.
Which of these are correct comparisons of decimals?
(i) 0.70 = 0.7
(ii) 0.701 > 0.7
(iii) 0.65 < 0.605
(iv) 0.506 < 0.56
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (ii) and (iv) only
(c) (i), (ii) and (iv)
(d) (i), (iii) and (iv)
Solution:
(c) (i), (ii) and (iv)
(i) 0.70 and 0.7 are equal in value, i.e. 7 tenths. Thus, this comparison is correct.
(ii) 0.701 is greater than 0.7 because it has a 1 in the thousandths place, while 0.7 has 0. Thus, this comparison is correct.
(iii) 0.65 is greater than 0.605, since 0.65 = 0.650. So, this comparison is incorrect.
(iv) 0.506 < 0.56, since 0.506 = 506 thousandths, and 0.56 = 560 thousandths. Thus, this comparison is correct.

Question 14.
Which of the following statements are true?
(i) The decimal number with the greater whole number part is greater.
(ii) Milligrams are used for measuring heavier weights like bags of rice, etc. while kilograms are used for very light weights like medicine dosage, etc.
(iii) 3.5 feet means 3 feet and 5 inches.
(iv) 2.4 hours means 2 hours 24 minutes.
Choose the correct option from the following:
(a) (ii) and (iv)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (i) and (iii)
Solution:
(c) (i) and (iv)
The decimal number with greater whole number part is greater. For example, 7.1 > 6.99 because 7 > 6.
Kilograms are used for heavier things (like bags of rice) while milligrams are used for very tiny weights (like medicine).
Since 1 foot =12 inches, 0.5 feet = 6 inches. So, 3.5 feet = 3 feet 6 inches.
0.4 hours = 0.4 × 60 = 24 minutes. So, 2.4 hours = 2 hours 24 minutes.
Thus, statements (i) and (iv) are true.

A Peek Beyond the Point Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): Fractional part of a number is always less than 1.
(R): Decimal places represent values like
\(\frac{1}{10}, \frac{1}{100}, \frac{1}{1000},\) and so on. These are all parts of a whole.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Decimal places represent values like \(\frac{1}{10}, \frac{1}{100}, \frac{1}{1000},\)
and so on. These are a parts of a whole, they are less than 1.
That’s why fractional part of a number is always less than 1.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): \(3 \frac{6}{100}\) is greater than \(3 \frac{6}{10}\).
(R): \(\frac{6}{100}\) has a larger denominator than \(\frac{6}{10}\).
Solution:
(d) A is false but R is true.
We can write \(3 \frac{6}{100}=\frac{306}{100}\) and \(3 \frac{6}{10}=\frac{36}{10}=\frac{360}{100}\)
Clearly, 360 > 306 ⇒ \(\frac{360}{100}>\frac{306}{100}\)
⇒ \(3 \frac{6}{10}>3 \frac{6}{100}\)
Thus, Assertion (A) is false.
Now, as 100 > 10, \(\frac{6}{100}\) has larger denominator than \(\).
Thus, Reason (R) is true.

Question 3.
(A): If we subtract 47.38 from 89.62, the whole number parts differ by 42, and the actual result is more than 41 and less than 43.
(R): When subtracting two decimal numbers, the result lies between one less than and one more than the difference of their whole number parts.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Whole number part of 89.62 = 89
Whole number part of 47.38 = 47
Difference of whole number parts
= 89 – 47 = 42
Actual difference: 89.62 – 47.38 = 42.24
Now, (42 – 1) < 42.24 < (42 + 1) i.e.
41 < 42.24 < 43.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 4.
(A): 38 paise = ₹0.38
(R): One rupee =100 paise
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that 1 rupee = 100 paise.
So, 1 paisa = \(\frac{1}{100}\) rupee
Therefore, 38 paise = ₹ \(\frac{38}{100}\) = ₹0.38
Hence, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

A Peek Beyond the Point Class 7 Fill in the Blanks

Question 1.
The length \(2\frac{7}{10}\) cm is read as _____.
Solution: two and seven- tenths centimetres
The length \(2\frac{7}{10}\) cm is read as two and seven- tenths centimetres.

Question 2.
Kashvi cuts a 1-metre ribbon into 10 equal pieces. The length of one piece is ______ metre.
Solution: \(\frac{1}{10}\) or one-tenth
Given, length of ribbon is 1 metre and number of equal pieces is 10.
∴ Length of each piece = \(\frac{1}{10}\) metre
Thus, the length of one piece is \(\frac{1}{10}\) or one-tenth metre.

Question 3.
Fill in the blanks:
(i) __________ mm = 42.9 cm
(ii) 826 cm = ________ m
(iii) _____ g = 9.5 kg
(iv) 1834 mm = _____ cm
(v) 1749 g = ____ kg
(vi) ₹7 = _____ paise
(vii) ₹ 18.75 = ______ paise
(viii) ₹ _______ = 4250 paise
Solution: 429, 8.26, 9500, 183.4, 1.749, 700, 1875, 42.50
(i) We know that 1 cm = 10 mm.
Therefore, 42.9 cm = 42.9 × 10 mm
= \(\frac{429}{10}\) × 10 mm = 429 mm

(ii) We know that 1 cm = \(\frac{1}{100}\) m
Therefore, 826 cm = \(\frac{826}{100}\) m = 0.86 m

(iii) We know that 1 kg = 1000 g.
Therefore, 9.5 kg = 9.5 × 1000 g 95
= \(\frac{95}{10}\) × 1000 g = 9500 g

(iv) We know that 1 mm = \(\frac{1}{10}\) cm.
Therefore, 1834 mm = \(\frac{1834}{10}\) cm = 183.4 cm

(v) We know that 1 g = \(\frac{1}{1000}\) kg.
Therefore, 1749 g = \(\frac{1749}{100}\) kg = 1.749 kg

(vi) We know that 1 rupee = 100 paise.
Therefore, 7 rupees = 7 × 100 paise
= 700 paise

(vii) We know that 1 rupee =100 paise.
Therefore, ₹ 18.75 = 18.75 × 100 paise
= \(\frac{1875}{100}\) × 100 = 1875 paise

(viii) We know that 1 paise = \(\frac{1}{100}\) rupee
4250 paise = \(\frac{4250}{100}\) rupees = 42.50 rupees

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

Class 8 Science Curiosity Chapter 9 Question Answer

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

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

Probe and Ponder Questions

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

Answer:

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

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

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

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

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

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

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

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

InText Questions

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

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

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

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

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

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

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

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

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

Keep the Curiosity Alive (Pages 149-151)

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

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

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

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

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

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

Question 2.
Fill in the blanks:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Class 8 Science Chapter 9 Question Answer

Activity 1

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

Materials Required: A glass tumbler, salt.

Procedure:

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

Observations:
Table: Dissolution of salt in water

Answer:

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

Inferences:

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

Activity 2.

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

Procedure:

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

Observations:

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

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

Activity 3.

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

Procedure:

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

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

Activity 4.

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

Observations:

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

Inference:

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

2. What is the smallest volume it can measure?

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

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

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

Activity 5.

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

Procedure:

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

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

Determining Volume of Solid Objects with Regular Shapes

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

Activity 6.

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

Procedure:

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

Observations:

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

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

Inference:

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

Activity 7.

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

Procedure:

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

Observations:

Table: Volume of irregular solids

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

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

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

Short Answer Type Questions

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

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

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

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

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

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

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

Long Answer Type Questions

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

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

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

Case-Study Based Questions

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

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

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

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

Picture Based Questions

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

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

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

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

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

Multiple Choice Questions (Mcqs)

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

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

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

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

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

Assertion and Reasoning

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

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

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

Fill in the blanks

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

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

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

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

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

True or False

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

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

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

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

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

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 2 Arithmetic Expressions MCQ improves accuracy in objective exams.

MCQ on Arithmetic Expressions Class 7

Arithmetic Expressions MCQ Class 7

Class 7 Maths Arithmetic Expressions MCQ

Question 1.
The value of expression 8 + 7 + 6 is:
(a) 21
(b) 19
(c) 20
(d) 17
Solution:
(a) 21
We have, 8 + 7 + 6 = 15 + 6 = 21

Question 2.
Compare and choose the correct option to fill the box: 24 + 16 ☐ 56 – 16.
(a) >
(b) <
(c) =
(d) None of these
Solution:
(c) =
Given, 24 + 16 ☐ 56 – 16
Left hand side (LHS) = 24 + 16 = 40
Right hand side (RHS) = 56 – 16 = 40
Clearly, LHS = RHS So, 24 + 16 = 56 – 16

Question 3.
Compare and choose the correct option to fill the box: 4 × 3 ☐ 78 ÷ 6.
(a) >
(b) <
(c) =
(d) None of these
Solution:
(b) <
Left hand side (LHS): 4 × 3 = 12
Right hand side (RHS): 78 ÷ 6 = 13
Clearly, 12 < 13 ⇒ 4 × 3 < 78 ÷ 6

Question 4.
In a sweet shop, Meena packed 12 laddoos in one box and 4 in another. Kunal packed 10 laddoos in one big box and none in the second. Who packed more laddoos?
(a) Meena
(b) Kunal
(c) Both packed the same number
(d) Depends on the sweetness
Solution:
(a) Meena
Laddoos packed by Meena = 12 + 4= 16
Laddoos packed by Kunal = 10 + 0= 10
Thus, Meena packed more laddoos.

Question 5.
The number of terms in the expression 40 – 7 × 3 + 8 ÷ 2 – 4is:
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(c) 4
We have, 40 – 7 × 3 + 8 ÷ 2 – 4

So, there are 4 terms in the given expression.

Question 6.
The value of the expression 100 – {25 – (10 × 5)} is:
(a) 125
(b) 115
(c) 120
(d) 130
Solution:
(a) 125
We have, 100 – {25 – (10 × 5)}
We know, on removing the brackets preceded by a minus sign, the signs of all the terms inside the brackets change.
Thus, 100 – {25 – (10 × 5)}
= 100 – 25 + (10 × 5)
= 100 – 25 + 50 = 150 – 25 = 125

Question 7.
The value of the expression 240 – [60 – (30 – 8 ÷ 2) + 10] is:
(a) 192
(b) 194
(c) 196
(d) 198
Solution:
(c) 196
240 – [60 – (30 – 8 ÷ 2) + 10]
= 240 – [60 – (30 – 4) + 10]
= 240 – [60 – 26 + 10]
= 240 – 44 = 196

Question 8.
Which of the following is equal to the expression ‘53 – 17 + 4’?
(a) 53 – (17 + 4)
(b) 53 + (17 – 4)
(c) 53 – (17 – 4)
(d) 4 + (17 – 53)
Solution:
(c) 53 – (17 – 4)
We know, on removing the brackets preceded by a minus sign, the signs of all the terms inside the brackets change.
So, 53 – (17 – 4) = 53 – 17 + 4

Question 9.
Which of the following expressions are the arithmetic expressions?
(i) 12 + (2 × 4 – 6)
(ii) 7x + 2y – 3
(iii) 18 ÷ 3 + 7
(iv) 4xy – 11
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (ii) and (iv) only
(c) (i), (ii), (iii) and (iv)
(d) None of these
Solution:
(a) (i) and (iii) only
We know, an arithmetic expression is a
mathematical sentence that contains numbers and operations like addition (+), subtraction (-), multiplication (×) or division (÷).
So, 12 + (2 × 4 – 6) and 18 ÷ 3 + 7 are arithmetic expressions.
And, 7x + 2y – 3 and 4xy – 11 are not arithmetic expressions because x and y are not numbers.

Question 10.
Which of the following are correct?
(i) The signs ‘>’, ‘=’ and ‘<’ are used to compare the values of two expressions.
(ii) 276- 19 < 275- 18 (iii) 366 – 20 = 380 – 34 (iv) 490 – 20 > 396 – 30
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (ii) and (iv) only
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Solution:
(d) (i), (iii) and (iv)
(i) Yes, the signs ‘>’, ‘=’ and ‘<’ are used to compare the values of two expressions.
So, (i) is correct.

(ii) Left-hand side (LHS): 276 – 19 = 257
Right-hand side (RHS): 275 – 18 = 257
Clearly, 276- 19 = 275- 18.
So, (ii) is incorrect.

(iii) Left-hand side (LHS): 366 – 20 = 346
Right-hand side (RHS): 380 – 34 = 346
Clearly, 366 – 20 = 380 – 34.
So, (iii) is correct.

(iv) Left-hand side (LHS): 490 – 20 = 470
Right-hand side (RHS): 396 – 30 = 366
Clearly, 470 > 366
⇒ 490 – 20 > 396 – 30
So, (iv) is correct.

Arithmetic Expressions Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): 110 – 27 > 112 – 27
(R): When the same number is subtracted from two different numbers, the difference is greater for the larger number.
Solution:
(d) A is false but R is true.
We know, when the same number is subtracted
from two different numbers, the difference is greater for the larger number.
Since 112 is larger than 110,
110 – 27 < 112 – 27 Therefore, Assertion (A) is false, but Reason (R) is true.

Question 2.
(A): 568 + 247 > 572 + 248
(R): If a < b and c < d, then a + c < b + d, where a, b, c and d are numbers.
Solution:
(d) A is false but R is true.
If a < b and c < d, then a + c < b + d, where a,
b, c and d are numbers.
Clearly, 568 < 572 and 247 < 248
So, 568 + 247 < 572 + 248
Therefore, Assertion (A) is false, but Reason (R) is true.

Question 3.
(A): 5 – 2 ≠ 5 + (-2)
(R): Subtracting a number is the same as adding its additive inverse.
Solution:
(d) A is false but R is true.
We know that subtracting a number is the same
as adding its additive inverse.
∴ 5 – 2 = 5 + (- 2) as – 2 is additive inverse of 2.
Therefore, Assertion (A) is false and Reason (R) is true.

Question 4.
(A) : 16 – (- 7 + 19) = 16 – 7 + 19
(R) : On removing the brackets preceded by a plus “ + ’ sign, the signs of all the terms inside the brackets remain same.
Solution:
(d) A is false but R is true.
We know, on removing the brackets preceded by a ‘+’ sign, the signs of all the terms inside the brackets remains same.
But in expression ‘16 – (-7 + 19)’, brackets are preceded by minus sign.
Therefore, 16 – (- 7 + 19) = 16 + 7 – 19
Therefore, Assertion (A) is false, but Reason (R) is true.

Arithmetic Expressions Class 7 Fill in the Blanks

Question 1.
Fill in the blanks with the correct sign: ‘<’ , ‘>’ or ‘=’:
(i) 111 – 28 _____ 85
(ii) 55 ÷ 11 ____ 5
Solution: <, =
(i) LHS: 111 – 28 = 83
RHS: 85
Clearly, 83 < 85 ⇒ 111 – 28 < 85

(ii) LHS: 55 ÷ 11 = 5
RHS: 5
Clearly, LHS = RHS ⇒ 55 – 11 = 5

Question 2.
Fill in the blanks to make the expressions equal on both sides of the ‘ = ’ sign:
(i) 19 + 6 = ____ + 9
(ii) 9 × _____ = 72 ÷ 2
Solution: 16, 4
(i) 19 + 6 = 16 + 9
[∵ 19 + 6 = 25 and 16 + 9 = 25]
(ii) 9 × 4 = 72 ÷ 2
[∵ 72 ÷ 2 = 36 and 9 × 4 = 36]

Question 3.
In the blanks below, fill ‘<’, ‘>’ or ‘=’ after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out, and not by evaluating the expressions.
(i) (12 – 7) × 36 _____ (7 – 12) × 36
(ii) 20 + 6 × 15 _____ (20 + 6) × 15
Solution: <, <
(i) LHS : (12 – 7) × 36 = Positive number × 36 → Positive number
RHS : (7 – 12) × 36 = Negative number × 36 → Negative number
We know, a positive number is always greater than a negative number.
Thus, LHS > RHS
Hence, (12 – 7) × 36 > (7 – 12) × 36

(ii) RHS : (20 + 6) × 15

Thus, LHS < RHS
Hence, 20 + 6 × 15 < (20 + 6) × 15

Question 4.
Fill in the blanks with numbers and boxes with operation signs such that the expressions on both sides are equal.
(i) 21 + (___ ☐ ____) = 21 + 12 – 5
(ii) 32 – (3 + 8) = 32 ☐ 3 – _____
(iii) 25 – (11 ☐ 7) = 25 – 11 + 7
(iv) 19 – (17 – 12) = 19 ☐ 17 ☐ 12
Solution: 12 – 5, 8, -, -, +
(i) 21 + ( 12 – 5) = 21 + 12 – 5
(ii) 32 – (3 + 8) = 32 – 3 – 8
(iii) 25 – (11 – 7) = 25 – 11 + 7
(iv) 19 – (17 – 12) = 19 – 17 + 12

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 1 Patterns in Mathematics Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 1 Patterns in Mathematics Solutions

Ganita Prakash Class 6 Chapter 1 Solutions

Class 6 Maths Ganita Prakash Chapter 1 Solutions Patterns in Mathematics

Question 1.
Why are 1, 3, 6, 10, 15, … called triangular numbers? Why are 1, 4, 9, 16, 25, … called square numbers or squares? Why are 1, 8, 27, 64, 125, … called cubes?
Solution:
As the dot representation of sequence 1,3, 6, 10, 15, … forms triangles, it is called triangular numbers sequence. As the dot representation of sequence 1,4, 9, 16, 25, … forms squares, it is called square numbers sequence.

As the dot representation of sequence 1,8, 27, 64, 125, … forms cubes, it is called cube numbers sequence.

Question 2.
You will have noticed that 36 is both a triangular number and a square number! That is, 36 dots can be arranged perfectly both in a triangle and in a square. Make pictures in your notebook illustrating this!
This shows that the same number can be represented differently and play different roles, depending on the context. Try representing some other numbers pictorially in different ways!
Solution:
Representation of 36 as a triangular number:

Representation of 36 as a square number:

Representation of 10 as even number:

Representation of 10 as triangular number:

Question 3.
Can you think of pictorial way to visualise the sequence of Powers of 2? Powers of 3?
Solution:
Powers of 2 can be visualised as:

Powers of 3 can be visualised as:

Question 4.
Can you find a similar pictorial explanation for why adding counting numbers up and down, i.e., 1, 1+2 + 1,1+2 + 3 + 2 + 1, … , gives square numbers?
Solution:
Yes,
As we can see, the dot representation of the addition of counting numbers up and down forms the dot representation of square numbers.

Question 5.
Which sequence do you get when you start to add the All 1’s sequence up? What sequence do you get when you add the All l’s sequence up and down?
Solution:
Adding all 1 ’s sequence up
1 = 1
1 + 1 = 2
1 + 1 + 1 = 3
1 + 1 + 1 + 1 = 4 and so on.
Here, we get a sequence of counting numbers i.e. 1, 2, 3, 4, … .

Adding all l’s sequence up and down 1 = I
1 + 1 + 1 = 3
1 + 1 + 1 + 1 + 1 = 5 and so on.
Here, we get a sequence of odd numbers.

Question 6.
What happens when you add up pairs of consecutive triangular numbers? That is, take 1 + 3, 3 + 6, 6 + 10, 10 + 15,… ? Which sequence do you get?
Solution:
Adding up pairs of consecutive triangular numbers, we get
1 + 3 = 4;
3 + 6 = 9;
6 + 10 = 16;
10 + 15 = 25 and so on.
Here, we get a sequence of square numbers.

Question 7.
What happens when you multiply the triangular numbers by 6 and add 1 ? Which sequence do you get?
Solution:
Triangular numbers are 1, 3, 6, 10, 15, …
On multiplying triangular numbers by 6 and add 1 to it, we get
1 × 6 + 1 = 7;
3 × 6 + 1 = 19;
6 × 6 + 1 = 37;
10 × 6 + 1 = 61;
15 × 6 + 1 = 91 and so on.
Hence, we get a sequence of hexagonal numbers.

InText Questions

Question 1.
Observe the pattern given below:
1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
Why does this happen? Do you think it will happen forever?
Solution:
The sequence of odd numbers is: 1, 3, 5, 7, 9, 11,…
Sum of first 1 odd number = 1 = 1²
Sum of first 2 odd numbers = 1 + 3 = 4 = 2²
Sum of first 3 odd numbers = 1 + 3 + 5 = 9 = 3²
Sum of first 4 odd numbers =1 + 3 + 5 + 7 = 16 = 4²
Sum of first 5 odd numbers =1+3 + 5 + 7 + 9 = 25 = 5²
Sum of first 6 odd numbers =1 + 3 + S + 7 + 9 + 11 = 36 = 6²
Each time we add another odd number, the total becomes a perfect square.
Since the sequence of odd numbers keeps going forever, and the sum of the first n odd numbers is always n², this pattern will continue forever.

Question 2.
Pictorially represent and find the sum of the first 10 odd numbers.
Solution:
From figure, it is clear that
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100 = 10²

Patterns in Mathematics Class 6 Extra Questions

Patterns in Mathematics Class 6 Very Short Question Answer

Question 1.
Find the next term of the sequence 1, 4, 9, 16,… .
Solution:
Given sequence is 1, 4, 9, 16, …, i.e. 1², 2², 3², 4², …, which is a sequence of squares.
So, next term, i.e. fifth term = 5² = 25

Question 2.
What is the sum of first 10 terms of the sequence of counting numbers?
Solution:
We know, the sequence of counting numbers is 1, 2, 3, 4, ….
Now, sum of first 10 terms of the sequence of counting numbers
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55

Question 3.
Identify the rule in the following number pattern and write the missing entries:
111 × 11 = 1221
121 × 11 = 1331
131 × 11 = 1441
141 × _____ = ______
_____ × 11 = 1661
161 × 11 = ______
Solution:
We observe that the first number on the left side increases by 10 each time: 111, 121, 131, 141, 151, 161,…
The result also increases by 110 each time: 1221, 1331, 1441, 1551, 1661, 1771, …
The first number is multiplied by 11 each time.
Thus, the missing numbers are as follows:
141 × 11 = 1551;
151 × 11 = 1661;
161 × 11 = 1771

Question 4.
Find the next term of the sequence 2 + 1, 2 + 2, 2 + 3, 2 + 4,… .
Solution:
Given sequence is 2 + 1, 2 + 2, 2 + 3, 2 + 4, ….
First term = 2 + 1
Second term = 2 + 2
Third term = 2 + 3
Fourth term = 2 + 4
So, next term, i.e. fifth term = 2 + 5

Question 5.
Find the next term of the sequence 2, 6, 12, 20, …
Solution:
C liven sequence is 2, 6, 12, 20, ….
Here, 2 = 1 × 2
6 = 2 × 3
12 = 3 × 4
20 = 4 × 5
So, next term = 5 × 6 = 30

Question 6.
Find the next term of the sequence 1, 3, 6, 10, 15,
Solution:
Given sequence is 1, 3, 6, 10, 15
First term = 1
Second term = 3 = 1 + 2 (First term + 2)
Third term = 6 = 3 + 3 (Second term + 3)
Fourth term = 10 = 6 + 4 (Third term + 4)
Fifth term = 15 = 10 + 5 (Fourth term + 5)
So, next term = Fifth term + 6 = 15 + 6 = 21

Question 7.
What is the sum of first 12 terms of the sequence of all 1s?
Solution:
We know, the sequence of all 1 s is 1, 1, 1, 1, ….
Now, sum of first 12 terms of the sequence of all 1s
= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12

Question 8.
What is the sum of first 6 terms of the sequence of odd numbers?
Solution:
The sequence of odd numbers is 1, 3, 5, 7
Now, sum of first 6 terms of the sequence of odd numbers = 1 + 3 + 5 + 7 + 9 + 11 = 36 = 6²

Question 9.
Write the first 5 square numbers.
Solution:
We know, the sequence of square numbers is 1, 4, 9, 16, 25, 36, …….
So, the first 5 square numbers are 1, 4, 9, 16 and 25.

Question 10.
Write the first 4 triangular numbers.
Solution:
We know, the sequence of triangular numbers is 1, 3, 6, 10, 15, 21, 28, … .
So, the first 4 triangular numbers are 1, 3, 6 and 10.

Question 11.
Which sequence do you get on adding the counting numbers up?
Solution:
We know, the sequence of counting numbers is 1. 2, 3, 4, 5, … .
Now, on adding the counting numbers up, we get the following sequence:
1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 = 10
1 + 2 + 3 + 4 + 5 = 15
So, the sequence is 1, 3, 6, 10, 15, …, which is the sequence of triangular numbers.

Patterns in Mathematics Class 6 Short Question Answer

Question 1.
Find the next term of the sequence 2, 16, 54, 128,
Solution:
Given sequence is 2, 16, 54, 128, ….
First term = 2 = 2 × 1 = 2 × 1³
Second term = 16 = 2 × 8 = 2 × 2³
Third term = 54 = 2 × 27 = 2 × 3³
Fourth term = 128 = 2 × 64 = 2 × 4³
So, next term, i.e. fifth term = 2 × 53 = 2 × 125 = 250

Question 2.
Can you identify numbers which are both triangular as well as square numbers? Find 2 such numbers.
Solution:
We know, the sequence of triangular numbers is 1, 3, 6, 10, 15, 21, 28, 36, 45, ….
And the sequence of squares is 1,4, 9, 16, 25, 36, 49, 64, 81, ….
So, 1 and 36 are both triangular as well as square numbers.

Question 3.
Observe the pattern shown below and write the next three steps:
1 × 9 + 2 = 11
12 × 9 + 3 = 111
123 × 9 + 4 = 1111
Solution:
We observe that

• On the left side, the first number starts at 1, then becomes 12, then 123, each time we add the next digit in order.
• We multiply by 9 each time.
• Then, we add next number (2, then 3, then 4)
• On the right side the answer is made of all 1 s and the number of Is is one more than the number of digits in starting number.

So, next three steps will be:
1234 × 9 + 5 = 11111
12345 × 9 + 6 = 111111
123456 × 9 + 7 = 1111111

Question 4.
Identify the pattern in the following number pattern and write the missing terms:
2178 × 4 = 8712
21978 × 4 = 87912
219978 × 4 = _______
_____ × 4 = 8799912
21999978 × 4 = ________
219999978 × ______ = 879999912
Solution:
In the first number, we start with 2178 and keep adding one more 9 before 78 in each step.
Then, we multiply the number by 4. The result is a number starting with 87, followed by the same number of 9s, and ending with 12.
The missing numbers are as follows:
2178 × 4 = 8712
21978 × 4 = 87912
219978 × 4 = 879912
2199978 × 4 = 8799912
21999978 × 4 = 87999912
219999978 × 4 = 879999912

Question 5.
Pairs of consecutive triangular numbers are added (i.e. 1 + 3, 3 + 6, …). Which sequence will you get on such addition? Write the 6^th term of the new obtained sequence.
Solution:
We know, the sequence of triangular numbers is 1, 3, 6, 10, 15, 21, 28, … .
Now, on adding pairs of consecutive triangular numbers we get the following sequence:
1 + 3 = 4
3 + 6 = 9
6 + 10 = 16
10 + 15 = 25
15 + 21 = 36
21 + 28 = 49
So, the sequence is 4, 9, 16, 25, 36, 49, …, which represents the square numbers starting with 4.
Now, 6^th term of the new obtained sequence is 49.

Question 6.
Which sequence do you get on adding the odd numbers up?
Solution:
We know, the sequence of odd numbers is 1, 3, 5, 7, 9, … .
Now, on adding the odd numbers up, we get the following sequence:
1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
So, the sequence is 1,4, 9, 16, 25, …, which is the sequence of square numbers.

Patterns in Mathematics Class 6 Long Question Answer

Question 1.
What will happen if you multiply the triangular numbers by 6 and add 1? Which sequence do you get?
Solution:
We know, the sequence of triangular numbers is 1, 3, 6, 10, 15, 21, 28, 36, 45, … .
Now, if we multiply the triangular numbers by 6 and add 1, we get the following sequence:
1 × 6 + 1 = 6 + 1 = 7
3 × 6 + 1 = 18 + 1 = 19
6 × 6 + 1 = 36 + 1 = 37
10 × 6 + 1 = 60 + 1 = 61
15 × 6 + 1 = 90 + 1 = 91
So, the required sequence is 7, 19, 37, 61, 91, …, which represents hexagonal numbers starting with 7.

Question 2.
Find the number of line segments connecting any two distinct vertices of the polygon as shown in the figure.

Solution:
We know, in a complete graph, every pair of vertices is connected by a unique line segment. The number of line segments in a complete graph with n vertices is given by \(\frac{n(n-1)}{2}\).
For K₇ (Heptagon):
Number of vertices = 7
∴ Number of line segments
= \(\frac{n(n-1)}{2}\) = \(\frac{7(7-1)}{2}\) = 7 × 3 = 21

Question 3.
Find the number of sides of a Koch Snowflake obtained after 5 iterations.
Solution:
To get from one shape to the next shape in the Koch Snowflake sequence, each line segment ‘_______’ is replaced by a speed bump ‘‘

Number of iterations	No. of sides
0	3 = 3 × 40
1	12 = 3 × 41
2	48 = 3 × 42
3	192 = 3 × 43
4	768 = 3 × 44

Thus,
After 5 iterations, number of sides = 3 × 4⁵ = 3 × 1024 = 3072.

Patterns in Mathematicse Class 6 Case Based Questions

Question 1.
Players wear different jersey numbers to help fans and broadcasters identify them, especially in games like cricket and football where they look similar on the field.

One day, Ashish went to the stadium to watch a cricket match. There, he observed the jersey numbers of some cricketers and found them to follow a sequence.
Based on the above information, answer the following questions:
(i) Write the rule for the sequence followed by the given numbers in words.
(ii) What will be the number on Captain’s jersey?
(iii) What will be the number on Vice-captain’s jersey?
Solution:
(i) Numbers on the jersey of players are given as
2 = 1² + 1;
5 = 2² + 1;
10 = 3² + 1;
17 = 4² + 1;
26 = 5² + 1
Rule of the sequence: n² + 1; n = 1, 2, 3, …
(ii) Since captain is at 6^th position, the number on his jersey is 6² + 1 = 36 + 1 = 37.
(iii) Since vice-captain is at 7^th position, the number on his jersey is 7² + 1 = 49 + 1 = 50

Question 2.
In Ms. Rina’s class, each student is given a unique roll number. One day, while arranging the books in the library, she noticed that the roll numbers of the students returning books followed a specific number pattern. She found that the roll numbers were: 1, 2, 3, 5, 8, 13, …
Based on the above information, answer the following questions:
(i) Write the rule for the sequence followed by the given numbers.
(ii) What is the name of the number sequence followed by the given numbers?
(iii) What will be the roll number of the 10^th student?
Solution:
The given numbers are 1, 2, 3, 5, 8, 13, …
(i) First number = 1
Second number = 2
Third number = 3 = 1 + 2
(First number + Second Number)
Fourth number = 5 = 2 + 3
(Second number + Third Number)
Fifth number = 8 = 3 + 5
(Third number + Fourth Number)
Sixth number = 13 = 5 + 8
(Fourth number + Fifth Number)
Therefore, the rule for the sequence is,
First number = 1, Second number = 2, any other number = sum of previous two numbers

(ii) The given numbers are known as Virahanka numbers.

(iii) Observing the pattern (from above)
Seventh number = Fifth number + Sixth number
= 8 + 13 = 21
Eighth number = Sixth number + Seventh number
= 13 + 21 = 34
Ninth number = Seventh number + Eighth number
= 21 + 34 = 55
Tenth number = Eighth number + Ninth number
= 34 + 55 = 89
Therefore, the roll number of 10^th student is 89.

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

Class 8 Maths Chapter 7 Proportional Reasoning 1 Solutions

Ganita Prakash Class 8 Chapter 7 Solutions

Class 8 Maths Ganita Prakash Chapter 7 Solutions Proportional Reasoning 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Figure it Out (Page 170 – 171) :

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

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

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

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

Figure it Out (Page 175) :

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

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

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

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

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

Figure it Out (Page 176 – 177) :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Proportional Reasoning 1 Class 8 Extra Questions

Multiple Choice Questions

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

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

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

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

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

Assertion and Reasoning

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

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

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

Case Based Questions

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

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

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

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

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

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