Class 8 RD Sharma Solutions – Chapter 10 Direct And Inverse Variations – Exercise 10.1 | Set 2

Chapter 10 Direct And Inverse Variations – Exercise 10.1 | Set 1

Question 11. A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 12 minutes.

Solution:

Distance covered by the car in 1 hour = 50 km 

Therefore, the distance covered by car in 60 minutes = 50 km (1 h = 60 min)

Now, 

Distance covered by the car in 1 minute = 50/60  (Taking 60 to RHS)

Distance covered by the car in 12 minutes = Distance covered by the car in 1 minutes x Required minutes

=>Distance covered by the car in 12 minutes = (50/60) x 12 

Solving the above equation, we get 

=> Distance covered by the car in 12 minutes = 10 km

Question 12. 68 boxes of a certain commodity require a shelf length of 13.6 m. How many boxes of the same commodity would occupy a shelf of 20.4 m?

Solution:

Given, 
Let the number of boxes required for length 20.4m be x. 

Boxes	68	x
Shelf length (in m)	13.6	20.4

We have the following equation, 

=> 68/13.6 = x/20.4

Solving for x, 

x = (68 / 13.6) x 20.4

x = 102 boxes

Therefore, 102 boxes are required to cover a shelf of 20.4m

Question 13. In a library, 136 copies of a certain book require a shelf length of 3.4 meter. How many copies of the same book would occupy a shelf-length of 5.1 meters?

Solution:

Given, 

Let the number of required shelves be x.

Copies	136	x
Shelf length(in m)	3.4	5.1

We have the following equation,

=> 136/3.4 = x/5.1

Solving for x,

x = (136 / 3.4) x 5.1

x = 204 books

Therefore, 204 books are required to cover a shelf of 5.1 m

Question 14. The second class railway fare for 240 km of journey is Rs. 15.00. What would be the fare for a journey of 139.2 km?

Solution:

Railway fare for 240 km = Rs 15.00

Let the railway fare for 139.2 km be x.

Distance (in km)	240	139.2
Price (in Rs)	Rs 15	x

We have the following equation,

=> 240 / 139.2 = 15 /x 

Solving for x, we get , 

=> x = 5 x 139.2 / 240 

=> x = Rs 87

Therefore, railway fare for a journey of 139.2 km = Rs 87.

Question 15. If the thickness of a pile of 12 cardboards is 35 mm, find the thickness of a pile of 294 cardboards.

Solution:

Thickness of 12 cardboards = 35 mm 

Thickness of 1 cardboard = 35/12 mm

Cardboards	12	294
Thickness (in mm)	35	x

Thickness of 294 cardboards = Thickness of 1 cardboard x 294

=> Thickness of 294 cardboards = (35/12) x 294

=> Thickness of 294 cardboards = 85.75 mm

Question 16. The cost of 97 meter of cloth is Rs. 242.50. What length of this can be purchased for Rs. 302.50?

Solution:

Length of cloth in Rs 242.50 = 97 m

Length of cloth in Rs 302.50 = x 

Cloth length(in m)	97	x
Price	242.50	302.50

Solving for x, we get ,

=> 97/x = 242.50/302.50

=> 97/x = 24250/30250

=> x = (97 × 30250)/24250

=> x = 121 m 

Question 17. 11 men can dig 6_^3/4 meter long trench in one day. How many men should be employed for digging 27-meter long trench of the same type in one day?

Solution:

Men	11	x
Length of trench (in m)	6+ 3/4 = 27/4	27

11 / (27/4) = x/27

Solving for x, we get, 

(11 × 27)/(27/4) = x

Cancelling the numerator and denominator, we get, 

x = (11 × 4) men

x = 44 men

Therefore, 

44men can dig 27m long trench in one day

Question 18. A worker is paid Rs. 210 for 6 days work. If his total income of the month is Rs. 875, for how many days did he work?

Solution:

Days for which Rs 210 paid to worker = 6

Days for which Rs 875 paid to worker = x

Days	6	x
Salary	210	875

Equating , we have, 

210/6 = 875/x 

Solving for x, we have, 

x = (875 × 6)/210

x= 25 days

Question 19. A worker is paid Rs. 200 for 8 days work. If he works for 20 days, how much will he get?

Solution:

Salary for 8 days of work = Rs 200

Salary for 20 days of work = x

Salary	200	x
Days	8	20

Solving for x, we have,

8 /200 = 20/x 

=> x =( 20 × 200) /8

=> x = (20 × 25)

=> x= Rs 500

Therefore, Salary for 20 days of work = Rs 500

Question 20. The amount of extension in an elastic string varies directly as the weight hung on it. If a weight of 150 gm produces an extension of 2.9 cm, then what weight would produce an extension of 17.4 cm?

Solution:

Weight required to produce extension of 2.9 cm = 150 gm

Weight required to produce extension of 17.4 cm = x

Extension (in cm)	2.9	17.4
weight (in gm)	150	x

Solving for x, we have, 

150 /2.9 = x/17.4

(150 /2.9) × 17.4 = x

=> x = 900gm

Therefore, Weight required to produce extension of 17.4 cm = 900 gm

Question 21. The amount of extension in an elastic spring varies directly with the weight hung on it. If a weight of 250 gm produces an extension of 3.5 cm, find the extension produced by the weight of 700 gm.

Solution:

Extension produced by 250gm of weight = 3.5 cm

Extension produced by 700gm of weight = x

Extension (in cm)	3.5	x
weight (in gm)	250	700

By the following data, we have, 

250 /3.5 = 700/x 

Solving for x, 

(700 × 3.5) /250 = x 

=> x = (7 × 35) /25

=> x = 9.8 cm

Extension produced by 700gm of weight = 9.8 cm

Question 22. In 10 days, the earth picks up 2.6 × 10⁸ pounds of dust from the atmosphere. How much dust will it pick up in 45 days.

Solution:

We have, 

Dust(in pounds)	2.6 × 108	x
Days	10	45

Pounds of dust picked up in 10 days = 2.6 × 10⁸

Pounds of dust picked up in 45 days = x

Solving for x, we have, 

10 / (2.6 × 10⁸) = 45 /x

x = (2.6 × 10⁸ × 45) /10

=> x = 11.7 × 10⁸ pounds

Therefore, earth will pick up 11.7 × 10⁸ pounds of dust in45 days.

Question 23. In 15 days, the earth picks up 1.2 x 108 kg of dust from the atmosphere. In how many days it will pick up 4.8 x 10s kg of dust?

Solution:

We have, 

Days	15	x
Dust	1.2 × 108	4.8 × 108

Solving for x, we get, 

1.2 × 10⁸ /15 = 4.8 × 10⁸ /x

x = (4.8 × 10⁸ × 15) / (1.2 × 10⁸)

=> x = (4.8 × 15) / 1.2 

=> x = 60 days

Therefore, earth will pick up 4.8 × 10⁸ kg of dust in 60 days.