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# Class 8 RD Sharma – Chapter 10 Direct And Inverse Variations – Exercise 10.2 | Set 2

• Last Updated : 07 Apr, 2021

### Question 10. A car can finish a certain journey in 10 hours at a speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance?

Solution:

As we know if the speed of the car increases, then it will take less time to reach the destination. So, the speed of the car and time vary inversely. Let us take a km/hr be the speed of the car

So, 48 × 10 = a × 8

=> a = 60 km/hr

So, the speed needed = 60 km/hr

And, the speed of car = 48 km/hr

So, increased speed of car = 60 – 48 = 12 km/hr

Hence, the speed of the car must be increased by 12 km/hr

### Question 11. 1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were transferred to another fort and thus the food lasted now for 32 more days. How many soldiers left the fort?

Solution:

As we know that the number of soldiers and the number of days vary inversely. Let a be the number of soldiers present in the fort

So, 1200 × 28 = a × 32

=> a = 900

Total soldiers that were present in the fort = 1200

So, the number of soldiers left the fort = 1200 – 900 = 300

Hence, 900 soldiers are present in the fort and 300 soldiers left the fort

### Question 12.Three spraying machines working together can finish painting a house in 60 minutes. How long will it take for 5 machines of the same capacity to do the same job?

Solution:

As we know if the number of spraying machines is increased, then they will take less time to paint the house. So, the number of spraying machines and time vary inversely. Let the machines take a minute to paint the house

So, 3 × 60 = 5 × a

=> a = 36

Hence, 5 spraying machines will take 36 minutes to paint the house.

### Question 13.A group of 3 friends staying together, consume 54 kg of wheat every month. Some more friends join this group, and they find that the same amount of wheat lasts for 18 days. How many new members are there in this group now?

Solution:

As we know that the number of friends and their food consumption varies inversely. Let currently there be a number of friends in the group

So, 3 × 30 = a × 18

=> a = 5

Number of friends initially = 3

Number of friends now = 5

So, the number of friends that joined the group = 5 – 3 = 2

Hence, 2 new members joined the group

### Question 14. 55 cows can graze a field in 16 days. How many cows will graze the same field in 10 days?

Solution:

As we know that the number of cows and time taken to graze the field var inversely. Let a cow will graze the field in 10 days

So, 55 × 16 = a × 10

=> a = 88

Hence, 88 cows will graze the field in 10 days

### Question 15. 18 men can reap a field in 35 days. For reaping the same field in 15 days, how many men are required?

Solution:

As we know that the number of men and time taken to reap the field var inversely. Let a men will reap the field in 15 days

So, 18 × 35 = a × 15

=> a = 42

Hence, 42 men will reap the field in 15 days

### Question 16.A person has money to buy 25 cycles worth Rs. 500 each. How many cycles he will be able to buy if each cycle is costing Rs. 125 more?

Solution:

As we know that if the prices of cycles are less, the person can buy more cycles and vice-versa. So, the number of cycles and their cost varies inversely. Let a be the number of cycles, the person can buy

So, 25 × 500 = a × 625

=> a = 20

Hence, the person can buy 20 cycles if each cycle is costing Rs. 125 more

### Question 17. Raghu has enough money to buy 75 machines worth Rs. 200 each. How many machines can he buy if he gets a discount of Rs. 50 on each machine?

Solution:

As we know that if the prices of machines are less, Raghu can buy more machines and vice-versa. So, the number of machines and their cost vary inversely. Let a be the number of machines, Raghu can buy

So, 75 × 200 = a × 150

=> a = 100

Hence, Raghu can buy 100 machines if he gets a discount of Rs. 50 on each machine

### Question 18. If x and y vary inversely as each other and

i) x = 3 when y = 8, find y when x = 4

Solution:

As we know x and y vary inversely

So, 3 × 8 = 4 × y1

=> y1 = 6

ii) x = 5, when y = 15, find x when y = 12

Solution:

As we know x and y vary inversely

So, 5 × 15 = x1 × 12

=> x1 = 25/4

iii) x = 30, find y when constant of variation = 900

Solution:

Constant of variation i.e k = 900

As we know x and y vary inversely, which means

x × y = k

So, 30 × y = 900

Hence, y = 30

iv) y = 35, find x when constant of variation = 7

Solution:

Constant of variation i.e k = 7

As we know x and y vary inversely, which means

x × y = k

So, x × 35 = 7

Hence, x = 1/5

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