ML Aggarwal Direct and Inverse Variation Exe-9.1 Class 8 ICSE Ch-9 Maths Solutions. We Provide Step by Step Answer of  Exe-9.1 Questions for Direct and Inverse Variation as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

ML Aggarwal Direct and Inverse Variation Exe-9.1 Class 8 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 8th Chapter-9 Direct and Inverse Variation Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-9.1 Questions Edition 2023-2024

Direct and Inverse Variation Exe-9.1

ML Aggarwal Class 8 ICSE Maths Solutions

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Question 1. Observe the following tables and find if x and y are directly proportional:

(i)

 x 5 8 12 15 18 20 y 15 24 36 60 72 100

(ii)

 x 3 5 7 9 10 y 9 15 21 27 30

(i) x/y = 5/15 = 1/3

x/y = 8/24 = 1/3

x/y = 12/36 = 1/3

But, x/y = 15/60 = 1/4

x/y = 18/72 = 1/4

x/y = 20/100 = 1/5 which is not equal to 1/3

So, x/y is not constant.

Therefore, x and y are not directly proportional.

(ii) x/y = 3/9 = 1/3

x/y = 5/15 = 1/3

x/y = 7/21 = 1/3

x/y = 9/27 = 1/3

x/y = 10/30 = 1/3

so, x/y is constant.

Therefore, x and y are directly proportional.

Direct and Inverse Variation Exe-9.1

ML Aggarwal Class 8 ICSE Maths Solutions

Page-153

Question 2. If x and y are in direct variation, complete the following tables:

(i)

 x 3 5 … … 10 y 45 … 90 120 …..

(ii)

 x 4 8 … 20 28 y 7 … 21 …… …..

(i) x/y = 3/45 = 1/15

x/y is constant with 1/15.

Hence, x and y are directly proportional.

Now,

x1/y1 = 1/15 => 5/y1 = 1/15

y1 = 5 × 15 = 75

x2/y2 = x2/90 = 1/15

x2 = 90/15 = 6

x3/y3 = x3/120 = 1/15

x3 = 120/15 = 8

x4/y4 = 10/y4 = 1/15

y4 = 10 × 15 = 150

 x 3 5 6 8 10 y 45 75 90 120 150

(ii) x/y = 4/7

x/y is constant with 4/7.

Hence, x and y are directly proportional.

x1/y1 = 8/y1 = 4/7

y1 = (8 × 7)/4 = 14

x2/y2 = x2/21 = 4/7

x2 = (21 × 4)/7 = 12

x3/y3 = 21/y3 = 4/7

y3 = (21 × 7)/4 = 35

x4/y4 = 28/y4 = 4/7

y4 = (28 × 7)/4 = 49

 x 4 8 12 20 28 y 7 14 21 35 49

Question 3. If 8 metres cloth costs ₹250, find the cost of 5·8 metres of the same cloth.

Let the cost of 5.8 m cloth = Rs. x

length in m=8, 5.8

cost of cloth (Rs.) =250 , x

hence it is the case of direct variation

Therefore 8:250 = 5.8:x

8/250 = 5.8/x

x = (5.8 × 250)/8

= ₹181.25

Question 4. If a labourer earns ₹672 per week, how much will he earn in 18 days?

Let the labourer earns Rs. x in 18 days

days= 7, 18

Money earned (in Rs.)=672,x

Hence in the case of direct variation

So, 7: 672 = 18: x

7/672 = 18/x

x = (18 × 672)/7

= ₹1728

Question 5. If 175 dollars cost ₹7350, how many dollars can be purchased in ₹24024?

Let x dollars be purchased in Rs.24024

Cost (in Rs.)=7350,24024

Dollars=175,x

Hence the question is of direct variation.

So, 7350: 175 = 24024: x

7350/175 = 24024/x

x = (24024 × 175)/7350

= 572 Dollars

Question 6. If a car travels 67·5 km in 4·5 litres of petrol, how many kilometres will it travel in 26-4 litres of petrol?

Let car travels x km in 26.4 lts. of petrol.

Distance (in km) =x , 67.5

Petrol (in 1)=26.4,4.5

Hence the question is of direct variation .

So, x: 26.4 = 67.5: 4.5

x/26.4 = 67.5/4.5

x = (67.5 × 26.4)/4.5

= 396 km

Question 7. If the thickness of a pile of 12 cardboard sheets is 45 mm, then how many sheets of the same cardboard would be 90 cm thick?

Thickness of 12 cardboard = 45 mm

Let ‘x’ be the number of cardboard whose thickness is 90 cm = 900 m

So, 12: x = 45: 900

12/x = 45/900

x = (12 × 900)/45

= 240 mm

Thickness of 90 cardboard is 240 mm.

Question 8. In a model of a ship, the mast (flagstaff) is 6 cm high, while the mast of the actual ship is 9 m high. If the length of the ship is 33 m, how long is the model of the ship?

Height of a model of ship=6 cm.

But height of actual ship=9 m.

If length of ship=33 m

Let length of model =x

Therefore 6:x=9m :33

6/x = 9/33

x = (6 × 33)/9

∴ Length of model is 22 cm.

Question 9. The mass of an aluminum rod varies directly with its length. If a 16 cm long rod has a mass of 192 g, find the length of the rod whose mass is 105 g.

length of rod=16 cm and mass=192 kg.

if mass is 105 g,then

then let length of rod=x cm.

Therefore 16:x=192:105

16/x = 192/105

x = (16 × 105)/192

= 35/4

= 8.75 cm

∴ Length of rod = 8.75cm.

Question 10. Anita has to drive from village A to village B. She measures a distance of 3.5 cm between these villages on the map. What is the actual distance between the villages if the map scale is 1 cm = 20 km?

distance from village A to b on the map=3.5 cm.

scale of map 1 cm.=20 km.

therefore 1:3.5=20:x

1/3.5 = 20/x

x = (20 × 3.5)/1

x = 70 km

Distance between village A and village B is 70 km.

Question 11. A 23 m 75 cm high water tank casts a shadow 20 m long. Find at the same time:

(i) the length of the shadow cast by a tree 9 m 50 cm high.
(ii) the height of the tree if the length of the shadow is 12 m.

Height of a water tank = 23 m 75 cm

= 23 ¾ m = 95/4 m.

(i) If height of a tree = 9 m 50 cm

= 9 ½ m

= 19/2 m

Now, let its shadow be ‘x’ m

So, 95/4 : 19/2 :: 20: x

(95/4) × (2/19) = 20/x

5/2 = 20/x

x = (2 × 20)/5

= 8 m

∴ Its shadow = 8 m

(ii) Let the height of a tree be ‘x’ m

and its shadow = 12 m

95/4 : x :: 20: 12

95/4x = 20/12

4x = (95 × 12)/20

x = 57/4 m

x =14 m 25 cm

Height of the tree is14 m 25 cm.

Question 12. If 5 men or 7 women can earn ₹525 per day, how much would 10 men and 13 women will earn per day

Since in a day 5 men can earn Rs 525

Therefore in a day, 1 man can earn =Rs 525/5 =Rs. 105

Therefore in a day , 10 men will earn Rs 105  x 10 = Rs 1050

as in a day 7 women can earn Rs 525

Therefore in a day 1 woman can earn Rs 525

Therefore in a day , 1 women can earn  Rs 525/7=Rs.75

In a day 13 women will earn Rs 75 x 13 = rs 975

Hence, Total earning of 7 men and 13 women  in a day=Rs(1050+975)

=Rs 2025

— End of Direct and Inverse Variation Exe-9.1 Class 8 ICSE Maths Solutions :–