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 *x*_{1} *y*_{1} = *x*_{2} *y*_{2}

7 / 84 = *x* / 180

*x *= [ 7 × 180 ] / 84 = 15

*x* = 15 kg

**15 kgof the onions bought for **₹

**(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 *x*_{1}/*y*_{1} = *x*_{2}/*y*_{2}

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 *x*_{1} *y*_{1} = *x*_{2} *y*_{2}

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 *x*_{1} / *y*_{1} = *x*_{2} / *y*_{2}

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 *x*_{1} / *y*_{1} = *x*_{2} / *y*_{2}

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 *x*_{1} *y*_{1} = *x*_{2} *y*_{2}

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 *x*_{1} / *y*_{1} = *x*_{2} / *y*_{2}

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 *x*_{1} /*y*_{1} = *x*_{2} / *y*_{2}

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 *x*_{1} *y*_{1} = *x*_{2} *y*_{2}

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 *x*_{1} / *y*_{1} = *x*_{2} / *y*_{2}

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 *x*_{1} *y*_{1} = *x*_{2} *y*_{2}

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 *x*_{1} *y*_{1} = *x*_{2} *y*_{2}

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 *x*_{1} *y*_{1} = *x*_{2} *y*_{2}

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**

**ANSWERS**

**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

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

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