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# Exercise 4.3

7th Maths : Term 1 Unit 4 : Direct and Inverse Proportion : Exercise 4.3 : Miscellaneous Practice problems, Challenge Problems, Text Book Back Exercises Questions with Answers, Solution

Exercise 4.3

Miscellaneous Practice problems

1. If the cost of 7kg of onions is ₹ 84 find the following

(i) Weight of the onions bought for ₹ 180 (ii) The cost of 3 kg of onions

(i) Weight of the onions bought for ₹180

Let x be the required weight of the onions bought

Weight of onions (kg)              7            x

Cost (₹)                                   84          180

When cost increases weight also increases. So it is in direct proportion.

Hence x1 y1 = x2 y2

7 / 84 = x / 180

x = [ 7 × 180 ] / 84 = 15

x = 15 kg

15 kgof the onions bought for 180

(ii) The cost of 3 kg of onions

Let x be the required cost

Weight of onions (kg)           7            3

Cost (₹)                              84             x

When weight decreases cost also decrease. So, it is in direct proportion

Hence x1/y1 = x2/y2

7 / 84 = 3 / x

x = [ 3 × 84 ] / 7  = 36

x = ₹ 36

The cost of 3 kg of onions = ₹ 36

2. If C =kd, (i) what is the relation between C and d?

(ii) find k when C = 30 and d = 6 (iii) find C, when d = 10

(i) What is the relation between C and d?

C = kd

Direct proportion

(ii) Find k when C = 30 and d = 6

C = 30, d = 6

C = kd

30 = 6k

K = 30 / 6 = 5

K = 5

(iii) Find C, when d = 10

C = Kd

K = 5,

d= 10

C = 5 × 10 = 50

C = 50

3. Every 3 months Tamilselvan deposits ₹ 5000 as savings in his bank account. In how many years he can save ₹ 1,50,000.

Let x be the required number of months

Months            3                x

Savings ₹     5000               150000

When saving increases months also will be increased. So, it is direct proportion.

Hence x1 y1 = x2 y2

3 / 5000 = x / 150000

x = [ 3 × 150000 ] / 5000 = 90

x = 90 months

In 90 months he can save ₹ 1,50,000

4. A printer, prints a book of 300 pages at the rate of 30 pages per minute. Then, how long will it take to print the same book if the speed of the printer is 25 pages per minute?

Let x be the required time taken

Number of pages                 25         300

Time Taken (mts)                 1              x

When number of pages increase time taken also will be increased. So, it is in direct proportion.

Hence x1 / y1 = x2 / y2

25 / 1 = 300 / x

x = [ 300 × 1 ] / 25 = 12

x = 12 mts

It will take 12 mts to print the same book.

5. If the cost of 6 cans of juice is ₹ 210, then what will be the cost of 4 cans of juice?

Let the required cost is ₹ x

Number of cans                 6              4

Cost ₹                             210               x

When number of cans decrease the cost also will be decreased. So it is direct proportion.

Hence x1 / y1 = x2 / y2

6 / 210 = 4 / x

x = [4 × 210] / 6 = 140

x = ₹ 140

The cost of 4 cans of juice ₹ 140

6. x varies inversly as twice of y. Given that when y = 6, the value of x is 4. Find the value of x when y = 8.

x                        4                    x

y                        6                    8

Given x varies inversly

Hence x1 y1 = x2 y2

4 × 6 = x × 8

x = [4 × 6] / 8 = 3

x = 3

When y = 8, the value of x = 3

7. A truck requires 108 liters of diesel for covering a distance of 594km. How much diesel will be required to cover a distance of 1650km?

Let x be the required amount of diesel.

Amount of diesel                108 l                x

Distance (km)                       594 km         1650 km

When distance increase the amount of diesel required also will be increased. So, it is direct proportion.

Hence x1 / y1 = x2 / y2

108 / 594 = x / 1650

x = [ 108 × 1650 ] / 594 = 300 l

x = 300 liters

300 liters of diesel will be required to cover a distance of 1650 km.

Challenge Problems

8. If the cost of a dozen soaps is ₹ 396, what will be the cost of 35 such soaps?

Let x be the required cost

Number of soaps             12                    35

Cost (₹)                        396                        x

When number of soaps increase the cost also will be increased. So, it is direct proportion.

Hence x1 /y1 = x2 / y2

12 / 396 = 35 / x

12 × x = 35 × 396

x = [ 35 × 396 ] / 12 = 1155

x = ₹ 1155

The cost of 35 soaps = ₹ 1155

9. In a school, there is 7 period a day each of 45 minutes duration. How long each period is, if the school has 9 periods a day assuming the number of hours to be the same?

Let x be the required minutes

Number of periods          7              9

Time duration (mts)        45             x

When number of periods increase the time duration for each period will be decreased. So, it is in indirect proportion.

Hence x1 y1 = x2 y2

7 × 45 = 9 × x

x = [7 × 45] / 9 = 35 mts

Each period is 35 mts, if the school has 9 periods a day

10. Cost of 105 notebooks is ₹ 2415. How many notebooks can be bought for ₹ 1863?

Let x be the required number of notebooks

Number of note books          105                x

Cost(₹)                                 2415             1863

When cost decreases, number of note books also will be decreased. So, it is in direct proportion.

Hence x1 / y1 = x2 / y2

105 / 2415 = x / 1863

x = [ 105 × 1863 ] / 2415 = 81

x = 81

81 note books can be bought for = ₹ 1863

11. 10 farmers can plough a field in 21 days. Find the number of days reduced if 14 farmers ploughed the same field?

Let x be the required number of days.

Number of farmers          10             14

N umber of days              21                x

When number of farmers increase days will be decreased. So, it is in indirect proportion.

Hence x1 y1 = x2 y2

10 × 21 = 14 × x

x = [ 10 × 21 ] / 14= 15

x = 15

Number of days = 15 days

The number of days reduced = (21 – 15) days = 6 days

12. A flood relief camp has food stock by which 80 people can be benefited for 60 days.

After 10days 20 more people have joined the camp. Calculate the number of days of food shortage due to the addition of 20 more people?

Let x be the required number of days.

Number of people                         80                100

Number of days                            50                    x

When number of people increase days will be decreased. So. it is in indirect proportion.

Hence x1 y1 = x2 y2

80 × 50 = 100 × x

x = [80 × 50] / 100 = 40

x = 40 days

Number of days of food shortage = (50 – 40) days = 10 days.

13. Six men can complete a work in 12 days. Two days later, 6 more men joined them. How many days will they take to complete the remaining work?

Let x be the required number of days.

Number of men              6                 12

Number of days            10                  x

When men increases, days will be decreased. So, it is in indirect proportion.

Hence x1 y1 = x2 y2

6 × 10 = 12 × x

x = [ 6 × 10 ] / 12 = 5

x = 5 days

Number of days they will take to complete the remaining work = 5 days

Exercise 4.3

1. (i) 15 kg.

(ii) ₹ 36

2. (i) Direct proportion (ii) k = 5 (iii) C = 50

3. 90 months

4. 12 minutes

5. ₹ 140

6. 3

7. 300 litres

Challenge Problems

8. ₹ 1155

9. 35 minutes

10. 81

11. 6 days

12. 10 days

13. 5 days

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7th Maths : Term 1 Unit 4 : Direct and Inverse Proportion : Exercise 4.3 | Questions with Answers, Solution | Direct and Inverse Proportion | Term 1 Chapter 4 | 7th Maths