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

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

Ganita Prakash Class 7 Chapter 7 Solutions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(ii) Here ∠B = 150° and ∠C = 15°.

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

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

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

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

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

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

InText Questions

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

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

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

A Tale of Three Intersecting Lines Class 7 Extra Questions

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

Question 1.
In the adjoining figure:

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

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

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

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

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

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

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

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

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

A Tale of Three Intersecting Lines Class 7 Short Question Answer

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

A Tale of Three Intersecting Lines Class 7 Long Question Answer

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Class 8 Maths Chapter 6 We Distribute Yet Things Multiply Solutions

Ganita Prakash Class 8 Chapter 6 Solutions

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Case-III

The pattern holds for fractions also.

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

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

Case-III

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer:
Case – I

6	7
13	14

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

Case – II

9	10
16	17

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

We Distribute Yet Things Multiply Class 8 Extra Questions

Multiple Choice Questions

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

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

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

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

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

Assertion and Reasoning

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

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

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

Case Based Questions

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

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

6	7
13	14

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

(iii) Let us consider

2	3
9	10

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

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

Class 7 Maths Chapter 8 Working with Fractions Solutions

Ganita Prakash Class 7 Chapter 8 Solutions

Class 7 Maths Ganita Prakash Chapter 8 Solutions Working with Fractions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

InText Questions

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

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

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

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

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

Working with Fractions Class 7 Extra Questions

Working with Fractions Class 7 Very Short Question Answer

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

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

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

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

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

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

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

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

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

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

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

Working with Fractions Class 7 Short Question Answer

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Working with Fractions Class 7 Long Question Answer

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

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

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

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

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

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

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

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

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

(vi) 10 × \(\frac{1}{5}\) = 2 × 1 = 2

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

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

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

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

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

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

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

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

Class 7 Maths Chapter 2 Arithmetic Expressions Solutions

Ganita Prakash Class 7 Chapter 2 Solutions

Class 7 Maths Ganita Prakash Chapter 2 Solutions Arithmetic Expressions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

InText Questions

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

Solution:
(i)
∴ 245 + 289 > 246 + 285

(ii)
∴ 273 – 145 = 272 – 144

(iii)
∴ 364 + 587 < 363 + 589

(iv)
∴ 142 + 245 < 129 + 245

(v)
∴ 213 – 77 < 214 – 76

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

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

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

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

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

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

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

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

Arithmetic Expressions Class 7 Extra Questions

Arithmetic Expressions Class 7 Very Short Question Answer

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

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

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

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

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

Arithmetic Expressions Class 7 Short Question Answer

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Arithmetic Expressions Class 7 Long Question Answer

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

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

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

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

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

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

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

Arithmetic Expressions Class 7 Case Based Questions

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

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

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

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

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

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

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

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

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

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

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

Class 8 Science Curiosity Chapter 7 Question Answer

Class 8 Science Ch 7 Particulate Nature of Matter Question Answer

Class 8 Science Chapter 7 Particulate Nature of Matter Question Answer

Probe and Ponder Questions

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

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

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

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

Question 5.
Share your questions?
Answer:

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

InText Questions

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

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

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

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

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

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

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

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

Particulate Nature of Matter Class 8 Questions and Answers

Keep the Curiosity Alive (Pages 113-114)

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

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

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

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

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

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

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

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

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

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

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

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

Ice (Solid)

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

Liquid Water

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

Water Vapor (Gas)

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

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

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

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

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

Class 8 Science Chapter 7 Question Answer

Activity 1

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

Procedure:

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

Observations:

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

Inferences:

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

Activity 2.

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

Procedure:

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

Observations:

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

Inference:

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

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

Activity 3.

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

Procedure:

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

Observations:

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

Inference:

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

Activity 4.

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

Procedure:

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

Observations:

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

Inference:

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

Activity 5.

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

Procedure:

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

Observations:

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

Inference:

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

Activity 6.

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

Procedure:

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

Observations :

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

Inference:

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

Activity 7.

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

Procedure:

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

Observations :

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

Inference:

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

Activity 8.

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

Procedure:

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

Observations:

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

Inference:

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

Activity 9.

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

Procedure:

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

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

Particulate Nature of Matter Class 8 Extra Questions and Answers

Short Answer Type Questions

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

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

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

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

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

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

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

Long Answer Type Questions

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

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

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

Case-Study Based Questions

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

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

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

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

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

Picture Based Questions

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

Answer:
A → Evaporation
B → Boiling

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

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

Particulate Nature of Matter Class 8 MCQ

Multiple Choice Questions (Mcqs)

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

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

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

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

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

Assertion and Reasoning

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

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

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

Fill in the blanks

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

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

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

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

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

True or False

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

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

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

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

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

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

Class 8 Maths Chapter 5 Number Play Solutions

Ganita Prakash Class 8 Chapter 5 Solutions

Class 8 Maths Ganita Prakash Chapter 5 Solutions Number Play

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Figure it Out (Page 131) :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Number Play Class 8 Extra Questions

Multiple Choice Questions

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

(d) 4 × 347 × 3

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

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

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

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

Assertion and Reasoning

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

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

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

Case Based Questions

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

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

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

For the cryptarithm,

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

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

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

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

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

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

Class 8 Science Curiosity Chapter 6 Question Answer

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

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

Probe and Ponder Questions

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

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

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

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

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

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

InText Questions

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

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

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

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

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

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

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

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

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

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

Keep the Curiosity Alive (Pages 94-96)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Procedure:

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

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

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

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

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

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

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

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

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

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

Class 8 Science Chapter 6 Question Answer

Activity 1.

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

Procedure:

Case I: When two pipes are of different diameters.

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

Case II: When two pipes are of same diameters.

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

Observations:

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

Inferences:

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

Activity 2.

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

Procedure:

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

Observations:

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

Inference:

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

Activity 3.

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

Procedure:

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

Observations:

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

Inference:

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

Activity 4.

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

Procedure:

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

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

Inference:

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

Activity 5.

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

Procedure:

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

Observations:

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

Inference:

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

Activity 6.

Let us observe

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

Procedure:

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

Observations:

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

Inference:

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

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

Short Answer Type Questions

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

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

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

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

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

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

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

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

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

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

Long Answer Type Questions

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

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

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

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

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

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

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

Case-Study Based Questions

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

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

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

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

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

Picture Based Questions

I. Observe the picture and answer the questions.

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

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

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

Pressure, Winds, Storms, and Cyclones Class 8 MCQ

Multiple Choice Questions (Mcqs)

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

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

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

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

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

Assertion and Reasoning

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

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

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

Fill in the blanks

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

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

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

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

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

True or False

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

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

3. Atmospheric pressure decreases with altitude.
Answer:
True

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

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

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

Class 8 Maths Chapter 4 Quadrilaterals Solutions

Ganita Prakash Class 8 Chapter 4 Solutions

Class 8 Maths Ganita Prakash Chapter 4 Solutions Quadrilaterals

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

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

Solution:

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

(ii) The given rectangle is PSRQ.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Solution:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Quadrilaterals Class 8 Extra Questions

Multiple Choice Questions

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

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

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

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

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

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

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

Assertion and Reasoning

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

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

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

Case Based Questions

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

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

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

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

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

Class 8 Science Curiosity Chapter 5 Question Answer

Class 8 Science Ch 5 Exploring Forces Question Answer

Class 8 Science Chapter 5 Exploring Forces Question Answer

Probe and Ponder Questions

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

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

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

Question 4.
Share your questions
Answer:

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

InText Questions

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

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

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

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

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

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

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

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

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

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

Exploring Forces Class 8 Questions and Answers

Keep the Curiosity Alive (Pages 25-26)

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

Answer:

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

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

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

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

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

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

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

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

Question 7.
Why do we sometimes slip on smooth surfaces like ice or polished floors? Explain.
Answer:
When we walk on surfaces like ice, we often slip, which means we lose our balance and fall. This happens because the force that helps us stay upright and move forward (friction) is not enough. These surfaces have fewer irregularities. Minimizing the contact area and the force of friction between the surface and our shoes makes it easier to slide instead of grip. A layer of water, even a thin one on ice, can further reduce friction by acting as a lubricant, making the surface even more slippery.

Question 8.
Is any force being applied to an object in a non-uniform motion?
Answer:
Yes, for an object to be in non-uniform motion, a force must be acting upon it. Nonuniform motion, also known as accelerated motion, means the object’s velocity is changing, either in speed or direction, or both. This change in velocity requires a force to be applied.

Question 9.
The weight of an object on the Moon becomes one-sixth of its weight on the Earth. What causes this change? Does the mass of the object also become one-sixth of its mass on the Earth?
Answer:
The change in an object’s weight on the moon compared to Earth is due to the difference in gravitational force. The moon’s gravity is significantly weaker than Earth’s, roughly one-sixth as strong. However, the mass of an object remains the same regardless of location; only weight changes with gravitational pull.

Question 10.
Three objects 1,2 and 3 of the same size and shape but made of different materials are placed in the water. They dip to different depths as shown in fig. If the weights of the three objects 1,2 and 3 are w₁, w₂, and w₃, respectively, then

(i) w₁=w₂=w₃
(ii) w₁>w₂>w₃
(iii) w₂>w₃>w₁
(iv) w₃>w₁>w₂
Answer:
(ii) The relationship between the weights of the objects is w₁>w₂>w₃.
Object 1 is the deepest meaning it displaces the most water.
Object 2 is less deep than object 1 but deeper than object 3.
Object 3 is the least deep, meaning it displaces the least amount of water.
Since the objects have the same size and shape, the greater the depth, the greater the weight of the object (assuming they are all made of the same material).
Hence, the object with the greatest weight will sink the deepest, and the object with the least weight will be the closest to the surface.

Class 8 Science Chapter 5 Question Answer

Activity 1.

Let us explore
Aim: To experience the push or pull.
Materials Required: A large cardboard box, a rope.

Procedure :

• Take a heavy and large cardboard box.
• Try to push or pull it by yourself.
• Try to move the box in as many different ways as you can do.
• Are you able to move the box in any other way than shown in fig?

Answer:
It is really very difficult to move.
Observations: In every method you are either pushing or pulling the box.

Inferences:

• No matter how you chose to move the box – by dragging, sliding, rolling, lifting etc. Some form of push or pull is always involved.
• Generally, the push or pull applied on an object is called force.

Activity 2.

Let us analyse
Aim: To analyse and study the effect of force.

Procedure:

• Think of different examples and situations where a force either push or pull is applied.
• List them in the table given below.
• Write the effect of the force in the table also.
• Some of them are already listed for you.

Observations :
Different actions and their effects

Answer:

Inferences:
We have noticed that the effect of the force are :

• A force can make an object move from rest.
• A force can change the speed of an object if it is moving. Either increase or decrease.
• A force can change the direction of motion of an object.
• A force can bring about a change in the shape of an object.
• A force can cause some or all of these effects.

Activity 3.

Let us investigate
Aim : To show that friction always opposes the relative motion between the two bodies, irrespective of the direction of motion.
Materials Required : A book.

Procedure :

• Gently push a book on a table.
• It stops after moving for some distance.
• Repeat this activity pushing the book from the opposite direction. [Fig. (b).
• Does the book stop this time, too? Yes, the book stop this time too.
• Can you think of an explanation ?
  The book slides for some time and then stops. The reason is that

Inference:
The force acting along the two surfaces in contact which opposes the motion of one body over the other, is called the force of friction.

Activity 4.

Let us explore
Aim: To study that force of friction depends upon the nature of the two surfaces in contact.
Materials Required: A wooden board, a pencil cell, bricks or books, a piece of cloth, sand.

Procedure:

• Make an inclined plane on a smooth floor, or on a table. (You may use a wooden board supported by bricks or books).
• Put a mark with a pen at any point ‘A’ on the inclined plane.
• Now let a pencil cell move down from this point.
• How far does it move on the plane before coming to rest ?
  It moves up to the end of the inclined plane and comes to rest on the table.
• Note down the distance.
• Now spread a piece of cloth over the table. Make sure that there are no wrinkles in the cloth.
  Try this activity again [Fig. (b).
• Repeat this activity by spreading a thin layer of sand over the table. Maintain the same slope throughout the activity.
  Inference: Friction depends upon the nature of surfaces in contact.

Activity 5.

Let us test
Aim : To show that a magnet can exert force on another magnet without being in contact with it.
Materials Required : Two ring magnets, a wooden stick.

Procedure :

• First of all we will take two ring magnets and a wooden stick.
• Now, hold the stick in a vertical position on a wooden table.
• Insert one ring magnet into the stick. (See fig.)
• Now, take another ring magnet and insert in such a way that like poles of two magnets face each other.

Observations:

• Second magnet stay floating above the first magnet. It means that they repel each other.
• The second magnet still remains floating after reversing the poles of both magnets.

Inference:

• A magnet can exert force on another magnet without being in contact with it.
• The force exerted by a magnet on another magnet or a magnetic material is called magnetic force. Since a magnet can exert a force from a distance without being in contact it is called a non-contact force.

Activity 6.

Let us experiment
Aim: To show electrostatic force.
Materials Required : A plastic straw, a plastic scale, a piece of polythene and small piece of paper.

Procedure :

• Take a plastic scale/straw and rub it vigorously with polythene.
• It is advised that never touch the rubbed part with your hand or any other metal object.
• Bring the rubbed straw or scale close to the small pieces of paper which is kept on the table.
• Do not touch the paper.
• Write down your observations.

Observations:
The paper pieces get pulled towards the plastic scale/straw and stick to it when it is brought close to paper pieces.

Inference:

• When two objects are rubbed, electrical charges are build up on their surfaces which is called static charges.
• A charged object attract an uncharged objects like small pieces of paper.
• This is called electrostatic force and comes into play even when the two bodies are not in contact.

Activity 7.

Let us experiment
Aim: To show that like charges repel and unlike charges attract each other.
Materials Required: Two balloons, a length of thread, and a woollen cloth.

Procedure:

• First of all we will take two balloons, a thread and a woollen cloth.
• Now, inflate the two balloons and hang them.
• We should be careful that the two balloons do not touch each other. (See fig. (a))
• Now, rub both balloons with the woollen cloth and release them. Remember that these two balloons should not be touched with your fingers. Note down your observation.
• Now, bring the woollen cloth used for rubbing the balloons close to one of the rubbed balloons. Note down your observation.

Observations :

• Case-I: The balloons move away from each other i.e., repelling each other.
• Case-II: They attract each other.

Inference:

• The force exerted by a charged body on another charged body or an uncharged body is called electrostatic force. It is a non-contact force.
• We can infer that like or similar charged body repel each other and unlike (opposite) charged bodies attract each other.
• The two kinds of static charges are said to be ‘positive’ and ‘negative’ charges.

Activity 8.

Let us observe
Aim: To demonstrate the gravitational force.
Materials Required: A ball, a stone or any object around you.

Procedure :

• Take a ball or a stone or your eraser and throw it vertically upwards.
• Now, throw it again, but this time harder than previous.
• Not down your observations in both the cases.
• You may think of different situations around you and can throw any object in any direction.

Observations :

• Case-I: The ball or stone come down.
• Case-II: Even in this case also the ball still fall back down to the ground but takes little more time.

Inference:

• Finally all objects falls or comes back to the ground or floor.
• This shows that some force is acting in downward direction. This is called the gravitational force or force of gravity or simply gravity.

Activity 9.

Let us explore
Aim : To show that the earth pulls different objects with different forces and it depends on their masses.
Materials Required : A spring, a few objects of different masses e.g., a pencil box, a tiffin box, a small stone.

Procedure :

• Hang one end of the spring from a nail and on the other end, hang an object and observe the spring.
• When you suspend an object from a spring, the spring stretches due to the force applied by the earth.
• Now hang the other objects, one by one and notice the stretch in the spring in each cases.

Observations: If you hang objects of different mass one after another, each time causes a different amount of stretching.
Inference: This proves that heavier objects are pulled with more force i.e., the weight of an object depends on its mass; heavier objects have greater weight.

Activity 10.

Let us observe
Aim : To measure maximum weight by a spring balance.
Materials Required : A spring balance.

Procedure :

• Take a spring balance and hang it as shown in fig.
• Now, look at the spring balance carefully.
• What is the maximum weight it can measure ?

Observations :

• The maximum weight a spring balance can measure is 10 N.
• Range of spring balance is the maximum weight it can measure.

Inference : If a spring balance shows values from 0 to 10 N , its range is 0.10 N . means that it can measure weight upto 10 N.

Activity 11.

Let us calculate
Aim: To determine the smallest readable value (Least count) of and close-up of its scale a spring balance.
Materials Required: A spring balance.

Procedure :

• Note down the weight difference indicated between the two bigger marks.
• The weight difference between 0 and 01 N or between 01 N and 02 N is 1 N.
• Count the number of divisions between these two bigger marks.

Demonstration :

1. Weight different between 0 and 1 N or between 01 N and 2 N =1 N Number of divisions between 0 and 1 or 1 and 2=5
∴ Least count = \(\frac{1}{5}\)=0.2 N

Inference:

• The least count of a spring balance is the smallest difference in weight it can measure.
• So, the least count of a spring balance is 0.2 N.

Activity 12.

Let us measure
Aim: To measure the weight and mass of an object using a spring balance.
Materials Required: Pencil box, water bottle filled with water partially.

Procedure :

• Hang a spring balance from the hook of a spring balance (without exceeding its maximum range).
• The pointer or reading on the scale shows the object’s weight in newtons (N).
• This method can be repeated for many objects and results should be recorded in a table.
• The mass scale (in g/kg) on a spring balance assuming earth’s gravity.
• The mass reading is only correct on earth where gravitational acceleration is standard.
• You can repeat the above Activities-10 to 12 for the mass scale which is shown on the left side on the spring balance (Fig. 13) to measure the mass of an object.

Observations :
Measuring weight using a spring balance

S.No.	Object	Weight (N)
1.	Pencil Box
2.	Partially filled water bottle

Inference: We can measure the weight and mass of an object using a spring balance.

Activity 13.

Let us investigate
Aim: To experience a buoyant force or upthrust by a liquid.
Materials Required: A bucket, an empty plastic bottle.

Procedure:

• Take an empty plastic bottle (with its lid closed tightly) and a bucket full of water.
• Now, push the bottle in the water. (See fig.)
• You will feel an upward force or push.
• Does the bottle bounce up and comes on top of water ?

Observation:

• Yes, we feel an upward push on our hand.
• The bottle bounces back up when released.

Inference:

• Gravity (gravitational force acts on downward direction i.e., pulls downward. The buoyant force (upthrust) pushes it upwards.
• If the gravitational force (weight of the object) is greater than buoyant force, the object sinks. i.e. W>U, sink
• If the two forces are equal i.e., W=U, the object floats.
• The density of the liquid affects the buoyant force.

Exploring Forces Class 8 Extra Questions and Answers

Short Answer Type Questions

Question 1.
What is electrostatic force ? Why is it called non-contact force?
Answer:
The force exerted by a charged body on another charged or uncharged body is called electrostatic force. This force comes into play even when the bodies are not in contact, so it is called non-contact force.

Question 2.
What is Muscular Force ? Why is the force called contact force?
Answer:
Muscular Force : The force resulting due to the action of muscles is called Muscular Force. Since muscular force can be applied on an object when muscle is in contact with the object, so it is called a contact force.

Question 3.
What causes friction ?
Answer:
Friction occurs when two bodies move on each other. Each surface has some irregularities on it. When two such objects move on each other their irregularities get interlocked and friction arises.

Question 4.
You might have noticed that when used for a long time, slippers with rubber soles become slippery. Explain the reason.
Answer:
After using a rubber soles for a long time, their surfaces become smooth. Hence, the friction between the sole and the floor decreases. So, the slippers become slippery.

Question 5.
A blacksmith hammers a hot piece of iron while making a tool. How does the force due to hammering affect the piece of iron?
Answer:
The force due to hammering causes the change in the shape of the iron and iron can be moulded in the shape of the required tool.

Question 6.
Iqbal has to push a lighter box and Seema has to push a similar heavier box on the same floor. Who will have to apply a larger force and why ?
Answer:
The heavy object will be pressed hard against the opposite surface and produces more friction. So Seema will have to apply a larger force due to excess friction.

Long Answer Type Questions

Question 1.
In the following situations identify the agent exerting a force and the object on which it acts. State the effect of the force in each case.
(a) Squeezing a piece of lemon between the fingers to extract its juice.
(b) Taking out paste from a toothpaste tube.
(c) A load suspended from a spring while its other end is on a hook fixed to a wall.
(d) An athlete making a high jump to clear the bar at a certain height.
Answer:
(a) Agent are fingers, object is lemon, effect of force can be observable in form of lemon juice being expelled by squeezing.
(b) Agent is hand of the person squeezing the tube, object is toothpaste tube and effect of the force can be observed as the paste coming out of the tube.
(c) Agent is the load suspended, object is the spring and effect can be seen in the form of elongation of spring on suspension of load.

Question 2.
What are the different types of forces ? Give an example of each.
Answer:
Types of forces : The following are the different types of forces :
(a) Muscular force
(b) Magnetic force
(c) Electrostatic force
(d) Gravitational force
(e) Frictional force
(a) Muscular Force : The force applied by a living being with its muscles is known as muscular force e.g., bullocks apply muscular force to draw a cart.
(b) Magnetic Force : The force exerted by a magnet is called magnetic force. Example: A magnet attracts nails and pins made fron iron even from some distance.
(c) Electrostatic Force : The force exerted by an electrified body is called electrostatic force. Example : Pieces of paper get attracted towards a charged comb.
(d) Gravitational Force : The force of attraction between any two objects possessing mass is called force gravitation. An object dropped from a certain height falls on the earth due to the gravitational force.
(e) Frictional Force : The force which always opposes the motion of one body over another body is called frictional force. Example, a marble rolled on the ground stops after some time due to the frictional force.

Question 3.
An archer shoots an arrow in the air horizontally. However, after moving some distance, the arrow falls to the ground. Name the initial force that sets the arrow in motion. Explain why the arrow ultimately falls down.
Answer:
When an archer shoots an arrow he stretches the string of the bow by applying muscular force. In this process the shape of the bow changes. As soon as the string is released, it regains its original position which provides the initial force to set the arrow in motion. The force of gravity that acts on the arrow in the downward direction which finally brings it to the ground.

Case-Study Based Questions

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

Generally, to apply a force on an object, your body has to be in contact with the object. The contact may also be with the help of a stick or a piece of rope. This force is caused by the action of muscles in our body. The force resulting due to the action of muscles is known as the muscular force. Animals also make use of muscular force to carry out their physical activities and other tasks. Animals like bullocks, horses, donkeys and camels are used to perform various tasks for us. In per-forming these tasks they use muscular force. Since muscular force can be applied only when it is in contact with an object, it is also called a contact force.

Like Poles of two magnets repel each other and unlike poles attract each other. A straw is said to have acquired electrostatic charge after it has been rubbed with a sheet of paper. Such a straw is an example of a charged body. The force exerted by a charged body on another charged or uncharged body is known as electrostatic force. This force comes into play even when the bodies are not in contact. The electrostatic force, therefore, is another example of a non-contact force.

Similarly, the force exerted by a magnet on a piece of iron is also a noncontact force. Objects or things fall towards the earth because it pulls them. This force is called the force of gravity, or just gravity. This is an attractive force. The force of gravity acts on all objects. The force of gravity acts on all of us all the time without our being aware of it. Water begins to flow towards the ground as soon as we open a tap. Water in rivers flows downward due to the force of gravity.

(i) Gravity is : …………..
(a) Repulsive
(b) Attraction + Repulsive force
(c) Attractive force
(d) Not a force
Answer:
(c) Attractive force

(ii) An example of a non-contact force is :
(a) force exerted by us to lift a bucket
(b) pushing a stationary car
(c) force exerted by magnet
(d) hitting a circket ball for a 6 run
Answer:
(c) force exerted by magnet

(iii) The force exerted by a charged body on another charged body can :
(a) attract each other
(b) both (a) and (c)
(c) repel each other
(d) none of these
Answer:
(b) both (a) and (c)

(iv) Two objects repel each other. This repulsion could be due to charged or uncharged body is :
(a) magnetic force
(b) frictional force
(c) musculer force
(d) electrostatic force
Answer:
(d) electrostatic force

Picture Based Questions

I. Observe the picture and answer the following questions :

(i) Why do soles of shoes wear out?
(a) Due to friction
(b) Due to
old shoes
(c) Due to bad quality
(d) None of
Answer:
(a) Due to friction

(ii) Can it is possible to reduce friction upto zero ?
(a) Yes
(b) No
(c) Some times
(d) 50-50
Answer:
(b) No

Exploring Forces Class 8 MCQ

Multiple Choice Questions (Mcqs)

Question 1.
What is the SI unit of force ?
(a) Newton
(b) Joule
(c) Watt
(d) Kilogram
Answer:
(a) Newton

Question 2.
What is force described as in science ?
(a) Pressure
(b) Push or pull
(c) Friction
(d) Energy
Answer:
(b) Push or pull

Question 3.
Which of these is a non-contact force ?
(a) Friction
(b) Electrostatic force
(c) Tension
(d) Muscular force
Answer:
(b) Electrostatic force

Question 4.
What causes an object thrown upwards to return to earth ?
(a) Air pressure
(b) Gravity
(c) Electrostatics
(d) Friction
Answer:
(b) Gravity

Question 5.
Which of the following changes can a force cause ?
(a) Director change in
(b) Shape change in
(c) Speed change in
(d) All of these
Answer:
(d) All of these

Assertion and Reasoning

These questions consist of two statements, each printed as Assertion (A) and Reason (R). While answering these questions, you are required to choose any one of the following four responses.

(a) Assertion and reason both are correct and reason is correct explanation for assertion.
(b) Assertion and reason both are correct and reason is not correct explanation for assertion.
(c) Assertion is correct but the reason is wrong.
(d) Assertion is wrong but the reason is correct.

1. Assertion (A): A push or a pull on an object is called a force.
Reason (R): Forces applied on an object in the same direction add to one another.
Answer:
(b) Assertion and reason both are correct and reason is not correct explanation for assertion.

2. Assertion (A): A change in either the speed of an object, or its direction of motion, or both, is described as a change in its state of motion.
Reason (R): A force may bring a change in the state of motion of an object.
Answer:
(a) Assertion and reason both are correct and reason is correct explanation for assertion.

Fill in the blanks

1. To strecth the bow, the archer applies a force that causes a change in its ………….
Answer:
shape

2. The force applied by the archer to stretch the bow is an example of ………… force.
Answer:
muscular

3. The type of force responsible for a change in the state of motion of the arrow is an example of a ………… force.
Answer:
contact

4. While the arrow moves towards its target, the forces acting on it are due to ………… and that due to ………… of air.
Answer:
gravity and friction

5. A force arises due to ………… between two objects.
Answer:
interaction

True or False

1. A cyclist exerts a force of pull on the paddles of bicycle.
Answer:
False

2. Magnetic force is a contact force.
Answer:
False

3. A force can change the direction of motion of a moving object.
Answer:
True

4. The S.I. unit of pressure is newton per square metre.
Answer:
True

5. Pressure due to a liquid contained in a vessel is equal to all the points on its surface.
Answer:
True

Students can refer to BSE Odisha Class 8 Math Solution and Ganita Prakash Chapter 3 A Story of Numbers Class 8 Question Answer to understand textbook questions step by step.

Class 8 Maths Chapter 3 A Story of Numbers Solutions

Ganita Prakash Class 8 Chapter 3 Solutions

Class 8 Maths Ganita Prakash Chapter 3 Solutions A Story of Numbers

1. REEMA’S CURIOSITY
Page : 51

Question 1.
How do we ensure that all cows have returned safely after grazing?
Answer:
To ensure all cows return safely after grazing, farmers should use a combination of proper fencing and herd management, including counting cows as they enter and leave paddocks, using incentives like feed to encourage return to the holding area, and having a routine that minimizes the risk of lost or separated animals. Regular observation, clear containment through effective fencing, and understanding the animals’ natural behaviors are crucial for a safe herd return.

Question 2.
Do we have fewer cows than our neighbour?
Answer:
How to solve the problem
1. Identify the variables – Look for two specific numbers:
(i) The number of cows you have.
(ii) The number of cows your neighbor has.

2. Compare the numbers – Using the greater than (> is greater than >) and less than (< is less than <) symbols and compare the two numbers.
(i) If your cows < neighbor’s cows, then you have fewer cows.
(ii) If your cows > neighbor’s cows, then you do not have fewer cows than your neighbours.

3. State your conclusion – Answer the question based on the comparison you made

Question 3.
If there are fewer, how many more cows would we need so that we have the same number of cows as our neighbour?
Answer:
To find out how many more cows you need, you simply subtract the number of cows you have from the number of cows your neighbor has.

Formula : Number of more cows needed = (Number of your neighbor’s cows) – (Number of your cows)

Figure it Out : Page : 54

Question 1.
Suppose you are using the number system that uses sticks to represent numbers, as in Method 1. Without using either the number names or the numerals of the Hindu number system, -give a method for adding, subtracting, multiplying and dividing two numbers or two collections of sticks.
Answer:
Method 1 : Addition (Putting Together)

• Collect sticks representing the first quantity.
• Collect another set of sticks representing the second quantity.
• Combine both sets into a single group.
• The total number of sticks in the combined group represents the sum.
  Example:
  Group A: | | | | (4 sticks)
  Group B: | | | (3 sticks)
  Total: | | | | | | | (7 sticks)

Method 2: Subtraction (Taking Away)

• Take the Part with the group of sticks representing the larger collection.
• Remove or take away sticks equal to the number of the smaller quantity.
• The remaining sticks show me result of subtraction.
  Example:
  Start with: | | | | | | | (7 sticks)
  Take away: | | | (3 sticks)
  Left: | | | | (4 sticks)

Method 3: Multiplication (Repeated Addition)

• Make several groups of sticks, each containing the same number.
• Count all the sticks across all groups together.
• The total number of sticks represents the product.
  Example:
  Multiply 3 groups of | | | (3 sticks each):
  Group 1: | | |
  Group 2: | | |
  Group 3: | | |
  Total: (| | |) (| | |) (| | |) (9 sticks)

Method 4: Division (Equal Sharing or Grouping)

• Take the total number of sticks.
• Split them into equal groups equal to the given number by which we need to divide
  Either:
  • Count how many sticks are in each group (equal sharing), or
  • Count how many such groups can be made (repeated subtraction).
    Example (Equal Sharing):
    Total: | | | | | | (6 sticks), divide into 2 groups → | | | and | | | (3 sticks each)
    Example (Grouping):
    How many groups of | | (2 sticks) can be made from | | | | | | (6 sticks)?
    3 groups.

Question 2.
One way of extending the number system in Method 2 is by using strings with more than one letter for example, we could use ‘aa’ for 27. How can you extend this system to represent all the numbers? There are many ways of doing it!
Answer:
Treat it like a base-26 system using letters.
Each letter acts like a number and we treat sequences like base-26 numbers, where:
‘a’ = 1, ‘b’ = 2, …, ‘z’ = 26

After ‘z’, we continue with:
‘aa’ = 27
‘ab’ = 28 …
‘az’ = 52
‘ba’ = 53 …,
bb = 54,…, bz = 78, ca = 79, cb = 80,…, cz = 104,…

Question 3.
Try making your own number system.
Answer:
My Own Number System: The “ABC Number System”

• In this number system, I use the letters A, B, C, D and E instead of normal digits.
• Each letter stands for a number: A = 0, B = 1, C = 2, D = 3 and E = 4.
  This means I can count using only these five letters, just like we normally count with digits 0 to 9 in the usual number system.
• I also follow place value – the rightmost letter is worth Is, then 5s, then 25s, and so on (because this is a base-5 system).
  For example, the code BD means B = 1 (in 5s place) and D = 3 (in Is place). So BD = (1 × 5) + 3 = 8.
  This system is fun and feels like a secret code!
• I can count and do Maths using only letters, which helps me understand how numbers can be written in many different ways.

2. SOME EARLY NUMBER SYSTEMS
Figure it Out : Page : 59

Question 1.
Represent the following numbers in the Roman system.
(i) 1222
(ii) 2999
(iii) 302
(iv) 715
Answer:
(i) 1222
Break into parts:
1000 + 200 + 20 + 2 = M + CC + XX + II = MCCXXII

(ii) 2999
Break into parts:
1000 + 1000 + 900 + 90 + 9
= M + M + CM + XC + IX
= MMCMXCIX

(iii) 302
Break into parts:
300 + 2 = CCC + II = CCCII

(iv) 715
Break into parts:
700 + 10 + 5 = DCC + X + V = DCCXV

Figure it Out : Page : 60 – 61

Question 1.
A group of indigenous people in a Pacific island use different sequences of number names to count different objects. Why do you think they do this?
Answer:
Indigenous people on a Pacific island might use different sequences of number names for different objects because their language and culture are closely connected to daily life and nature. They may count coconuts, fish, people or days differently because each object is important in a different way and may follow different traditions.

For example, they might use one type of number word for living things and another for non-living things or they may count pairs of items (like eyes or shoes) instead of single pieces. Using different number systems helps them understand, group and remember things more easily in their own way.

Question 2.
Consider the extension of the Gumulgal number system beyond 6 in the same way of counting by 2s. Come up with ways of performing the different arithmetic operations (+, ×, ÷) for numbers occurring in this system, without using Hindu numerals. Use this to evaluate the following:
(i) (ukasar-ukasar-ukasar-ukasar- urapon) + (ukasar-ukasarukasar- urapon)
(ii) (ukasar-ukasar-ukasar-ukasar- urapon) – (ukasar-ukasarukasar)
(iii) (ukasar-ukasar-ukasar-ukasar- urapon) × (ukasar-ukasar)
(iv) (ukasar-ukasar-ukasar-ukasar- ukasar-ukasar-ukasar-ukasar) ÷ (ukasar- ukasar)
Answer:
Understanding the Gumulgal number system, which counts in groups of 2 using the following number names:
urapon = 1
ukasar = 2
ukasar-urapon = 3 (2 + 1)
ukasar-ukasar = 4 (2 + 2)
ukasar-ukasar-urapon = 5 (2 + 2 + 1)
ukasar-ukasar-ukasar = 6 (2 + 2 + 2) and so on…
► Converting Gumulgal terms to Hindu numerals:
(ukasar-ukasar-ukasar-ukasar-urapon) → 2 + 2 + 2 + 2 + 1 = 9 (ukasar-ukasar-ukasar-urapon) → 2 + 2 + 2 + 1 = 7
(ukasar-ukasar) → 2 + 2 = 4
(ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) → 8 ukasar = 8 × 2 = 16
► Performing Arithmetic Operations:
(i) Addition : 9 + 7 = 16
→ Convert 16 back into Gumulgal style:
8 ukasar → ukasar-ukasar-ukasar-ukasar- ukasar-ukasar-ukasar-ukasar

(ii) Subtraction : 9 – 6 = 3 → 2 + 1 = ukasar-urapon

(iii) Multiplication : 9 × 4 = 36 → Break 36 as 2 + 2 + 2 + … 18 times =18 ukasar → ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukaar- ukasar-ukasar-ukasar-ukasar-ukasar-ukasar

(iv) Division : 16 ÷ 4 = 4 → 2 + 2 = ukasar-ukasar

Question 3.
Identify the features of the Hindu number system that make it efficient when compared to the Roman number system.
Answer:
Features that Make the Hindu Number System Efficient than Roman number system:

• Uses a place value system where the position of a digit determines its value.
• Includes zero (0) as both a digit and a placeholder.
• Needs only 10 symbols (0 – 9) to represent any number.
• Allows for easy and quick arithmetic operations (addition, subtraction; etc.).
• Unambiguous and compact representation of even very large numbers.
• Forms the base for modern mathematics and science.
• Globally accepted and used in all fields today.

Question 4.
Using the ideas discussed in this section, try refining the number system you might have made earlier.
Answer:
After learning from this chapter, I improved my number system as follows:

• I improved my number system by making it a base-5 system.
• It uses five symbols: A, B, C, D, E (where A = 0, B = 1, …, E = 4).
• Each position from right to left represents powers of 5 (1, 5, 25, 125…).
• I added place value, so the same symbol has different values based on its position.
• Including A as zero helps avoid confusion and allows writing large numbers easily.
• This system is now more compact, clear and good for calculations, just like the Hindu number system.

3. THE IDEA OF A BASE
Figure it Out : Page : 62

Question 1.
Represent the following numbers in the Egyptian system: 10458, 1023, 2660, 784, 1111, 70707.
Answer:

Question 2.
What numbers do these numerals stand for?

Answer:
(i) 100 + 100 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 1 + 1 + 1 + 1 + 1 + 1
= 200 + 70 + 6 = 276

(ii) 1000 + 1000 + 1000 + 1000 + 100 + 100 + 100 + 10 + 10 + 1 + 1 = 4000 + 300 + 20 + 2
= 4322

Figure it Out : Page: 63

Question 1.
Write the following numbers in the above base-5 system using the symbols in Table 2: 15, 50,137, 293, 651.
Answer:

Question 2.
Is there a number that cannot be represented in our base-5 system above? Why or why not?
Answer:
No, there is no number that cannot be represented in our base-5 system.

Because:

• A base-5 system uses digits A, B, C, D, E (which stand for 0 to 4).
• Any number, no matter how big, can be written using combinations of these symbols and place values based on powers of 5 (1,5,25,125…).
• Since there is no upper limit on how many places we can use, we can represent every whole number.

Question 3.
Compute the landmark numbers of a base-7 system. In general, what are the landmark numbers of a base-n system?
Answer:
The landmark numbers of a base-7 system:
In a base-7 system, the landmark numbers are powers of 7:
7⁰ = 1
7¹ = 7
7² = 49
7³ = 343
7⁴ = 2401
7⁵ = 16807 …and so on.
The landmark numbers of a base-n number system are the powers of n starting from n⁰ = 1, n, n², n³,…

Figure it Out : Page : 65

Question 1.
Add the following Egyptian-numerals:

Answer:

Question 2.
Add the following numerals that are in the base-5 system that we created:

Answer:

Figure it Out : Page : 69 – 70

Question 1.
Can there be a number whose representation in Egyptian numerals has one of the symbols occurring 10 or more times? Why not?
Answer:
No, there cannot be a number whose representation in Egyptian numerals has one of the symbols occurring 10 or more times.
Because :

• The Egyptian number system is an additive system, not a place value system.
• Symbols are repeated to add up values, but each symbol is used at most 9 times.
• The system had only a limited number of symbols, each of which was repeated no more than 9 times in writing any number.
  So, whenever a symbol would be needed 10 times or more, Egyptians would move to the next higher symbol instead of repeating it, making their writing more compact and systematic.

Question 2.
Create your own number system of base 4, and represent numbers from 1 to 16.
Answer:
I have created my own number system called the Quad-Code System, which is based on base-4. In this system, I use four special symbols instead of regular digits:
A = 0 B = 1 C = 2 D = 3
In base-4, the place values increase as powers of
4. So, the rightmost place is 4⁰ = 1 = 1, the next is 4¹ = 4 and then 4² = 16 and so on. Using this system, I can write any number using just these four symbols.

Here is how I write numbers from 1 to 16 :

• 1 is written as B
• 2 is written as C
• 3 is written as D
• 4 is written as BA
• 5 is written as BB
• 6 is written as BC
• 7 is written as BD
• 8 is written as CA
• 9 is written as CB
• 10 is written as CC
• 11 is written as CD
• 12 is written as DA
• 13 is written as DB
• 14 is written as DC
• 15 is written as DD
• 16 is written as ABA

Question 3.
Give a simple rule to multiply a given number by 5 in the base-5 system that we created.
Answer:
The simple rule to multiply a number by 5 in the base-5 system I created using symbols (A = 0, B = 1, C = 2, D = 3, E = 4):
Rule : Add a zero (A) at the end of the number In base-5, multiplying any number by 5 is the same as shifting its digits one place to the left and adding A (zero) at the right end – just like adding a zero in base-10 when multiplying by 10.
Example :
Let’s take the number BC (which is 1 × 5 + 2 = 7 in decimal)

Now multiply by 5, just add A at the end to get BCA. BCA in base-5 = 7 × 5 = 35 in decimal
Because in base-5, the digits shift just like in base-10. Adding a zero (A) multiplies the number by the base itself, i.e., 5.

4. PLACE VALUE REPRESENTATION
Figure it Out : Page : 73

Question 1.
Represent the following numbers in the Mesopotamian system –
(i) 63
(ii) 132
(iii) 200
(iv) 60
(v) 3605
Answer:

Page : 76

Question 1.
Represent the following numbers using the Mayan system:
(i) 77
(ii) 100
(iii) 361
(iv) 72.1
Answer:
Mayan System Representation of Numbers
1. Convert to base-20: Divide the number by 20.

2. Determine the remainder and quotient : The quotient becomes the next level, and the remainder the value for the current level.

3. Represent with symbols : Convert the quotient and remainder into Mayan symbols.
(i) For 77:
(i) 77 ÷ 20 = 3 remainder 17
(ii) The remainder 17 would be at the bottom (ones place).
(iii) The quotient 3 would be in the next level (twenties place).
(iv) Result : This would be written with a bar and two dots for 3 at the top, and three bars and two dots for 17 at the bottom.

(ii) For 100 :

• 100 ÷ 20 = 5 remainder 0.
• Result: A bar (5) at the top and a shell (0) at the bottom.

(iii) For 361 :

• 361 ÷ 20 = 18 remainder 1.
• 18 ÷ 20 = 0 remainder 18.
• Result: This would be written with two bars and three dots for 18 at the top, and two bars and three dots for 18 at the bottom.

(iv) For 721 :

• 721 ÷ 20 = 36 remainder 1.
• 36 ÷ 20 = 1 remainder 16.
• 1 ÷ 20 = 0 remainder 1.
• Result: This would be written with two bars and three dots for 18 at the bottom, a bar and one dot for 16 in the middle, and two dots for 1 at the top.

Figure it Out : Page : 80

Question 1.
Why do you think the Chinese alternated between the Zong and Heng symbols? If only the Zong symbols were to be used, how would 41 be represented? Could this numeral be interpreted in any other way if there is no significant space between two successive positions?
Answer:
Using Zong and Heng symbols:

• The Chinese number system used Zong (vertical) and Heng (horizontal) symbols to show place value clearly.
• They alternated the direction of the symbols at each place (units, tens, hundreds, etc.) to avoid confusion when reading the number.
• This made it easier to know which digit belonged to which place even when spaces were small or missing.

► If only Zong symbols were used, how would 41 be written?
In Chinese system : 41 = 4 tens and 1 unit. Using only Zong symbols, it would be written as: (Zong for 4) followed by (Zong for 1) → looks like: IIII IWithout alternating the symbol direction or keeping proper spacing: IIII I could be misread as 5 (i.e., 1 five) instead of 41.
The lack of direction or spacing removes the clue that one part is “tens” and the other is “units”.

► The Chinese alternated between Zong and Heng symbols to make place values visually clear and easy to read, especially in handwritten or closely packed texts. Without this, numbers like 41 could easily be misunderstood.

Question 2.
Form a base-2 place value system using ‘ukasar’ and ‘urapon’ as the digits. Compare this system with that of the Gumulgal’s.
Answer:
To form a base-2 place value system using ‘ukasar’ and ‘urapon’, we assign:

• ‘ukasar’ = 0
• ‘urapon’ = 1
  This system works just like the binary number system, where each position from right to left represents increasing powers of 2. For example:
• The first place is 2⁰ = 1
• The next is 2¹ = 2
• Then 2² = 4 and so on.

So, we can represent numbers like this :

• The number 1 is written as urapon
• The number 2 is written as urapon ukasar
• The number 3 is urapon urapon
• The number 4 becomes urapon ukasar ukasar Each position tells us how many of that power of 2 we have and we use ukasar for 0 and urapon for 1.

Comparison: If we compare this to the Gumulgal number system, there’s a big difference. The Gumulgal system doesn’t use place value. Instead, it adds groups of 2s (ukasar) and Is (urapon) to build numbers. For example, to make 7, they would say something like ukasar-ukasar- ukasar-urapon (2 + 2 + 2 + 1).

So the main difference is the base-2 system with ukasar and urapon is a place value system – more efficient and better for large numbers. The Gumulgal system is group-based and additive, which is fine for small numbers but becomes confusing as numbers get bigger.

Question 3.
Where in your daily lives, and in which professions, do the Hindu numerals, and 0, play an important role? How might our lives have been different if our number system and 0 hadn’t been invented or conceived of?
Answer:
Hindu numerals and the digit 0 are used every day in our lives-for telling time, counting money, reading prices, doing math in school and writing phone numbers. Many professions like banking, teaching, engineering and science rely heavily on this number system. Zero plays a key role in place value and calculations, making big numbers easy to write and understand.

If zero and the Hindu number system had not been invented, life would be very difficult. We would struggle to calculate, trade or even write dates properly. Modern technology like comput¬ers and calculators would not exist, slowing down progress in every field.

Question 4.
The ancient Indians likely used base 10 for the Hindu number system because humans have 10 fingers, and so we can use our fingers to count. But what if we had only 8 fingers? How would we be writing numbers then? What would the Hindu numerals look like if we were using base 8 instead? Base 5? Try writing the base- 10 Hindu numeral 25 as base-8 and base-5 Hindu numerals, respectively. Can you write it in base-2?

Answer:
If humans had only 8 fingers :
We would probably have developed a base-8 number system instead of base-10. Just like we now count from 0 to 9 in base-10, we would count from 0 to 7 in base-8. All our numerals, math and calculations would be based on powers of 8.

Conversion of the base-10 number 25 :
► In base-8:
25 ÷ 8 = 3 remainder 1
3 ÷ 8 = 0 remainder 3
So, 25 in base-8 = 31

► In base-5:
25 ÷ 5 = 5 remainder 0
5 ÷ 5 = 1 remainder 0
1 ÷ 5 = 0 remainder 1
So, 25 in base-5 = 100

► In base-2 (binary):
25 ÷ 2 = 12 remainder 1
12 ÷ 2 = 6 remainder 0
6 ÷ 2 = 3 remainder 0
3 ÷ 2 = 1 remainder 1
1 ÷ 2 = 0 remainder 1

A Story of Numbers Class 8 Extra Questions

Multiple Choice Questions

Question 1.
What numeral is represented by ‘v’ in Roman System?
(a) 1
(b) 5
(c) 10
(d) 100
Solution:
In Roman system ‘v’ represents 5.
(b) 5

Question 2.
Which of the following is incorrectly matched?

Answer:
(d) M 250, is incorrectly matched as M in Roman System represents 1000.

Question 3.
Which of the following Roman Numbers represents the number 1222?
(a) MIIXXII
(b) MCCXII
(c) MCCXXII
(d) MCCCXI
Solution:
Here, 1222 = 1000 + 100 + 100 + 20 + 2
= 1000 + 100 + 100 + 10 + 10 + 1 + 1
= M + C + C + X + X + I + I
= MCCXXII
(c) MCCXXII

Question 4.
Which of the following Roman Numbers represents the number 2999?
(a) MMCMXCIX
(b) IMMM
(c) CMMIC
(d) MMCMCIX
Solution:
Here, 2999 = 1000 + 1000 + 900 + 90 + 9
= MMCMXCIX.
(a) MMCMXCIX

Question 5.
Which of the following Roman Numerals is represented by 2362?
(a) MMCCCLXII
(b) MMMCCLXII
(c) MMCCCLLII
(d) MMCCLCXII
Solution:
Here,
2362 = 1000 + 1000 + 100 + 100 + 100 + 50 + 10 + 1 + 1 = MMCCCLXII
(a) MMCCCLXII

Assertion and Reasoning

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

Question 1.
Assertion (A) : In Roman numerals LXII represents 62 of Hindu numerals.
Reason (R) : In Roman Numerals L represents 50, X represents 10 and I represents 1 of Hindu System.
Answer:
(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).

Question 2.
Assertion (A) : Roman Numeral CDXIII = Hindu Numeral 413
Reason (R) : ∵, C = 100, D = 500, XIII = 13.
∵, So CD = 400 and CDXIII = 413.
Answer:
(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).

Case Based Questions

Question 1.
Aarya was writing some numbers on a white board. She had put some operation sign in between the numbers.

Based on the above answer the following:
(i) In which numeral these numbers are written?
(ii) What will come in place of blank space in (A)?
(iii) What will come in place of blank space in (B)?
(iv) What will come in place of blank space in (C)?
(v) What will come in place of blank space in (D)?
Answer:
(i) These numbers are of Egyptian number system.