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# Exercise 4.2 (Inverse Proportion)

Fill in the blanks, Find the value of the following, Objective type questions, Text Book Back Exercises Questions with Answers, Solution

Exercise 4.2

1. Fill in the blanks.

(i) 16 taps can fill a petrol tank in 18 minutes. The time taken for 9 taps to fill the same tank will be 32 minutes.

(ii) If 40 workers can do a project work in 8 days, then 80 workers can do it in 4 days.

2. 6 pumps are required to fill a water sump in 1 hr 30 minutes. What will be the time taken to fill the sump if one pump is switched off?

Let x be the required time taken to fill the sump.

Number of Pumps                6                                     5

Time Taken                          1 hr 30 mts (90 mts)        x

When the number of pumps decrease. Time taken will be increased. So it is in inverse proportion

Hence x1 y1 = x2 y2

6 × 90 = 5 × x

x = [ 6 × 90 ] / 5 = 108 minutes

x = 108 mts = 1 hr 48 mts

Time taken to fill the sump if one pump is switched off = 1 hr 48 mts.

3. A farmer has enough food for 144 ducks for 28 days. If he sells 32 ducks, how long will the food last?

Let x be the required number of days.

Number of ducks          144       112

Number of days             28              x

When number of ducks decrease food last for days will be increased So, it is in inverse proportion

Hence x1 y1 = x2 y2

144 × 28 = 112 × x

x = ( 144 × 28) / 112 = 36 days

The food will last for 36 days.

4. It takes 60 days for 10 machines to dig a hole. Assuming that all machines work at the same speed, how long will it take 30 machines to dig the same hole?

Let x be the required number of days.

Number of Machines    10                  30

Number of days             60                 x

When number of machines increase, number of days will be decreased.

So, it is in inverse proportion.

Hence x1 y1 = x2 y2

10 × 60 = 30 × x

x = [ 10 × 60 ] / 30 = 20 days

30 machines will take 20 days to dig the same hole.

5. Forty students stay in a hostel. They had food stock for 30 days. If the students are doubled then for how many days the stock will last?

Let x be the required number of days

Number of students            40            80

Number of days                  30              x

When number of students increase, the number of days the stock last will be decreased. So, it is in inverse proportion.

Hence x1 y1 = x2 y2

40 × 30 = 80 × x

x = [ 40 × 30 ] / 80 = 15 days

The stock will last for 15 days.

6. Meena had enough money to send 8 parcels each weighing 500 grams through a courier service. What would be the weight of each parcel, if she has to send 40 parcels for the same money?

Let x be the required weight of each parcel.

Number of parcels              8              40

Weight in grams                 500              x

When number of parcels increase the weight will be decreased. So, it is in inverse proportion.

Hence x1 y1 = x2 y2

8 × 500 = 40 × x

x = [ 8 × 500 ] / 40 = 100 grams

The weight of each parcel would be 100 grams.

7. It takes 120 minutes to weed a garden with 6 gardeners If the same work is to be done in 30 minutes, how many more gardeners are needed?

Let x be the required number of gardeners.

Time Taken                   120 mts              30 mts

Number of gardeners         6                        x

When time taken decreases the number of gardeners will be increased So, it is in inverse proportion.

Hence x1 y1 = x2 y2

120 × 6 = 30 × x

x = [120 × 6] / 30 = 24

x = 24

Number of more gardeners needed = 24 – 6

= 18

18 more gardeners are needed

8. Neelaveni goes by bi-cycle to her school every day. Her average speed is 12km/hr and she reaches school in 20 minutes. What is the increase in speed, if she reaches the school in 15 minutes?

Let x be the required speed

Speed km/hr                        12                  x

Time taken minutes             20                 15

When Time Taken is decreased the speed will be increased.

So, it is in inverse proportion

Hence x1 y1 = x2 y2

12 × 20 = x × 15

x = [12 × 20] / 15 = 16

x = 16 km / hr

The increase in speed = (16 – 12)km/hr.

= 4 km /hr.

9. A toy company requires 36 machines to produce car toys in 54 days. How many machines would be required to produce the same number of car toys in 81 days?

Let x be the required number of machines.

Number of machines         36              x

Number of days                 54             81

When number of days increase, number of machines will be decreased So, it is in inverse proportion.

Hence x1 y1 = x2 y2

36 × 54 = x × 81

x = [36 × 54]  / 81 = 24

x = 24

24 machines w ould be required to produce the same number of car toys in 81 days.

Objective type questions

10. 12 cows can graze a field for 10 days. 20 cows can graze the same field for_____ days.

(i) 15

(ii) 18

(iii) 6

(iv) 8

Answer : (iii) 6

11. 4 typists are employed to complete a work in 12 days. If two more typists are added, they will finish the same work in _______ days.

(i) 7

(ii) 8

(iii) 9

(iv) 10

Answer : (ii) 8

Exercise 4.2

1. (i) 32

(ii) 80

2. 1 hour 48 minutes

3. 36 days

4. 20 days

5. 15 days

6. 100 gm

7. 18

8. 4 km/hr

9. 24

Objective Type Questions

10. (iii) 6

11. (ii) 8

Tags : Questions with Answers, Solution | Term 1 Chapter 4 | 7th Maths , 7th Maths : Term 1 Unit 4 : Direct and Inverse Proportion
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7th Maths : Term 1 Unit 4 : Direct and Inverse Proportion : Exercise 4.2 (Inverse Proportion) | Questions with Answers, Solution | Term 1 Chapter 4 | 7th Maths