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
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
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