Direct and Inverse Proportions

# Direct and Inverse Proportions Notes | Study Mathematics (Maths) Class 8 - Class 8

## Document Description: Direct and Inverse Proportions for Class 8 2022 is part of Direct and Inverse Proportions for Mathematics (Maths) Class 8 preparation. The notes and questions for Direct and Inverse Proportions have been prepared according to the Class 8 exam syllabus. Information about Direct and Inverse Proportions covers topics like and Direct and Inverse Proportions Example, for Class 8 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Direct and Inverse Proportions.

Introduction of Direct and Inverse Proportions in English is available as part of our Mathematics (Maths) Class 8 for Class 8 & Direct and Inverse Proportions in Hindi for Mathematics (Maths) Class 8 course. Download more important topics related with Direct and Inverse Proportions, notes, lectures and mock test series for Class 8 Exam by signing up for free. Class 8: Direct and Inverse Proportions Notes | Study Mathematics (Maths) Class 8 - Class 8
``` Page 1

Q u e s t i o n : 1
Observe the tables given below and in each one find whether x and y are proportional:
i
x 3 5 8 11 26
y 9 15 24 33 78
ii
x 2.5 4 7.5 10 14
y 10 16 30 40 42
iii
x 5 7 9 15 18 25
y 15 21 27 60 72 75
S o l u t i o n :
i
Clearly,
x
y
=
3
9
=
5
15
=
8
24
=
11
33
=
26
78
=
1
3
constant Therefore, x and y are proportional.
ii
Clearly,
x
y
=
2.5
10
=
4
16
=
7.5
30
=
10
40
=
1
4
, while
14
42
=
1
3
i. e. ,
2.5
10
=
4
16
=
7.5
30
=
10
40
is not equal to
14
42
. Therefore, x and y are not proportional.
iii
Clearly,
x
y
=
5
15
=
7
21
=
9
27
=
25
75
=
1
3
, while
15
60
=
18
72
=
1
4
i. e. ,
5
15
=
7
21
=
9
27
=
25
75
is not equal to
15
60
and
18
72
. Therefore, x and y are not proportional.
Q u e s t i o n : 2
If x and y are directly proportional, find the values of x
1
, x
2
and y
1
in the table given below:
x 3 x
1
x
2 10
y 72 120 192 y
1
S o l u t i o n :
Since x and y are directly propotional, we have:
3
72
=
x 1
120
=
x 2
192
=
10
y 1
Now,
3
72
=
x 1
120
? x
1
=
120×3
72
= 5
And,
3
72
=
x 2
192
? x
2
=
3 × 192
72
= 8
And,
3
72
=
10
y 1
? y
1
=
72×10
3
= 240
Therefore, x
1
= 5, x
2
= 8 and y
1
= 240
Q u e s t i o n : 3
A truck covers a distance of 510 km in 34 litres of diesel. How much distance would it cover in 20 litres of diesel?
S o l u t i o n :
Let the required distance be x km. Then, we have:
Quantity of diesel inlitres

34 20
Distance inkm 510 x
Clearly, the less the quantity of diesel consumed, the less is the distance covered.
So, this is a case of direct proportion.
Now,
34
510
=
20
x
?
1
15
=
20
x
? x ×1 = 20 ×15 = 300
Therefore, the required distance is 300 km.
Q u e s t i o n : 4
A taxi charges a fare of Rs 2550 for a journey of 150 km. How much would it charge for a journey of 124 km?
S o l u t i o n :
Let the charge for a journey of 124 km be x.
Pricein 2550 x
Distanceinkm 150 124
More is the distance travelled, more will be the price.
So, it is a case of direct proportion.
?
2550
150
=
x
124
? x =
2550×124
150
= 2108
Thus, the taxi charges 2,108 for the distance of 124 km.
Q u e s t i o n : 5
A loaded truck covers 16 km in 25 minutes. At the same speed, how far can it travel in 5 hours?
S o l u t i o n :
Let the required distance be x km. Then, we have:
1 h = 60 mini. e. , 5 h = 5 ×60 = 300 min
( )
Page 2

Q u e s t i o n : 1
Observe the tables given below and in each one find whether x and y are proportional:
i
x 3 5 8 11 26
y 9 15 24 33 78
ii
x 2.5 4 7.5 10 14
y 10 16 30 40 42
iii
x 5 7 9 15 18 25
y 15 21 27 60 72 75
S o l u t i o n :
i
Clearly,
x
y
=
3
9
=
5
15
=
8
24
=
11
33
=
26
78
=
1
3
constant Therefore, x and y are proportional.
ii
Clearly,
x
y
=
2.5
10
=
4
16
=
7.5
30
=
10
40
=
1
4
, while
14
42
=
1
3
i. e. ,
2.5
10
=
4
16
=
7.5
30
=
10
40
is not equal to
14
42
. Therefore, x and y are not proportional.
iii
Clearly,
x
y
=
5
15
=
7
21
=
9
27
=
25
75
=
1
3
, while
15
60
=
18
72
=
1
4
i. e. ,
5
15
=
7
21
=
9
27
=
25
75
is not equal to
15
60
and
18
72
. Therefore, x and y are not proportional.
Q u e s t i o n : 2
If x and y are directly proportional, find the values of x
1
, x
2
and y
1
in the table given below:
x 3 x
1
x
2 10
y 72 120 192 y
1
S o l u t i o n :
Since x and y are directly propotional, we have:
3
72
=
x 1
120
=
x 2
192
=
10
y 1
Now,
3
72
=
x 1
120
? x
1
=
120×3
72
= 5
And,
3
72
=
x 2
192
? x
2
=
3 × 192
72
= 8
And,
3
72
=
10
y 1
? y
1
=
72×10
3
= 240
Therefore, x
1
= 5, x
2
= 8 and y
1
= 240
Q u e s t i o n : 3
A truck covers a distance of 510 km in 34 litres of diesel. How much distance would it cover in 20 litres of diesel?
S o l u t i o n :
Let the required distance be x km. Then, we have:
Quantity of diesel inlitres

34 20
Distance inkm 510 x
Clearly, the less the quantity of diesel consumed, the less is the distance covered.
So, this is a case of direct proportion.
Now,
34
510
=
20
x
?
1
15
=
20
x
? x ×1 = 20 ×15 = 300
Therefore, the required distance is 300 km.
Q u e s t i o n : 4
A taxi charges a fare of Rs 2550 for a journey of 150 km. How much would it charge for a journey of 124 km?
S o l u t i o n :
Let the charge for a journey of 124 km be x.
Pricein 2550 x
Distanceinkm 150 124
More is the distance travelled, more will be the price.
So, it is a case of direct proportion.
?
2550
150
=
x
124
? x =
2550×124
150
= 2108
Thus, the taxi charges 2,108 for the distance of 124 km.
Q u e s t i o n : 5
A loaded truck covers 16 km in 25 minutes. At the same speed, how far can it travel in 5 hours?
S o l u t i o n :
Let the required distance be x km. Then, we have:
1 h = 60 mini. e. , 5 h = 5 ×60 = 300 min
( )
.
Distance inkm

16 x
Time inmin 25 300
Clearly, the more the time taken, the more will be the distance covered.
So, this is a case of direct proportion.
Now,
16
25
=
x
300
? x =
16×300
25
? x = 192
Therefore, the required distance is 192 km.
Q u e s t i o n : 6
If 18 dolls cost Rs 630, how many dolls can be bought for Rs 455?
S o l u t i o n :
Let the required number of dolls be x. Then, we have:

No of dolls 18 x
Cost of dolls
inrupees
630 455
Clearly, the less the amount of money, the less will be the number of dolls bought.
So, this is a case of direct proportion.
Now,
18
630
=
x
455
?
1
35
=
x
455
? x =
455
35
? x = 13
Therefore, 13 dolls can be bought for Rs 455.
Q u e s t i o n : 7
If 9 kg of sugar costs  238.50, how much sugar can be bought for  371?
S o l u t i o n :
Let the quantity of sugar bought for 371 be x kg.
Quantity inkg 9 x
Pricein 238.50 371
The price increases as the quantity increases. Thus, this is a case of direct proportion.
?
9
238.50
=
x
371
? x =
9×371
238.50
= 14
Thus, the quantity of sugar bought for 371 is 14 kg.
Q u e s t i o n : 8
The cost of 15 metres of a cloth is Rs 981. What length of this cloth can be purchased for Rs 1308?
S o l u t i o n :
Let the length of cloth be x m. Then, we have:
Length of cloth inmetres 15 x
Cost of cloth inrupees 981 1308
Clearly, more length of cloth can be bought by more amount of money.
So, this is a case of direct proportion.
Now,
15
981
=
x
1308
? x =
15×1308
981
? x = 20
Therefore, 20 m of cloth can be bought for Rs 1,308.
Q u e s t i o n : 9
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 15m high. If the length of the ship is 35 metres, how long is the model ship?
S o l u t i o n :
Let x m be the length of the model of the ship. Then, we have:
1 m = 100 cmTherefore, 15 m = 1500 cm35 m = 3500 cm

Length of the mast incm Length of the  ship incm
Actual ship 1500 3500
Model of the
ship
9 x
Clearly, if the length of the actual ship is more, then the length of the model ship will also be more.
So, this is a case of direct proportion.
Now,
1500
9
=
3500
x
? x =
3500×9
1500
? x = 21 cm
Therefore, the length of the model of the ship is 21 cm.
Q u e s t i o n : 1 0
In 8 days, the earth picks up (6.4 × 10
7
) kg of dust from the atmosphere. How much dust will it pick up in 15 days?
S o l u t i o n :
Let x kg be the required amount of dust. Then, we have:

No. of days 8 15
Dust inkg 6. 4 ×10
7
x
( )
Page 3

Q u e s t i o n : 1
Observe the tables given below and in each one find whether x and y are proportional:
i
x 3 5 8 11 26
y 9 15 24 33 78
ii
x 2.5 4 7.5 10 14
y 10 16 30 40 42
iii
x 5 7 9 15 18 25
y 15 21 27 60 72 75
S o l u t i o n :
i
Clearly,
x
y
=
3
9
=
5
15
=
8
24
=
11
33
=
26
78
=
1
3
constant Therefore, x and y are proportional.
ii
Clearly,
x
y
=
2.5
10
=
4
16
=
7.5
30
=
10
40
=
1
4
, while
14
42
=
1
3
i. e. ,
2.5
10
=
4
16
=
7.5
30
=
10
40
is not equal to
14
42
. Therefore, x and y are not proportional.
iii
Clearly,
x
y
=
5
15
=
7
21
=
9
27
=
25
75
=
1
3
, while
15
60
=
18
72
=
1
4
i. e. ,
5
15
=
7
21
=
9
27
=
25
75
is not equal to
15
60
and
18
72
. Therefore, x and y are not proportional.
Q u e s t i o n : 2
If x and y are directly proportional, find the values of x
1
, x
2
and y
1
in the table given below:
x 3 x
1
x
2 10
y 72 120 192 y
1
S o l u t i o n :
Since x and y are directly propotional, we have:
3
72
=
x 1
120
=
x 2
192
=
10
y 1
Now,
3
72
=
x 1
120
? x
1
=
120×3
72
= 5
And,
3
72
=
x 2
192
? x
2
=
3 × 192
72
= 8
And,
3
72
=
10
y 1
? y
1
=
72×10
3
= 240
Therefore, x
1
= 5, x
2
= 8 and y
1
= 240
Q u e s t i o n : 3
A truck covers a distance of 510 km in 34 litres of diesel. How much distance would it cover in 20 litres of diesel?
S o l u t i o n :
Let the required distance be x km. Then, we have:
Quantity of diesel inlitres

34 20
Distance inkm 510 x
Clearly, the less the quantity of diesel consumed, the less is the distance covered.
So, this is a case of direct proportion.
Now,
34
510
=
20
x
?
1
15
=
20
x
? x ×1 = 20 ×15 = 300
Therefore, the required distance is 300 km.
Q u e s t i o n : 4
A taxi charges a fare of Rs 2550 for a journey of 150 km. How much would it charge for a journey of 124 km?
S o l u t i o n :
Let the charge for a journey of 124 km be x.
Pricein 2550 x
Distanceinkm 150 124
More is the distance travelled, more will be the price.
So, it is a case of direct proportion.
?
2550
150
=
x
124
? x =
2550×124
150
= 2108
Thus, the taxi charges 2,108 for the distance of 124 km.
Q u e s t i o n : 5
A loaded truck covers 16 km in 25 minutes. At the same speed, how far can it travel in 5 hours?
S o l u t i o n :
Let the required distance be x km. Then, we have:
1 h = 60 mini. e. , 5 h = 5 ×60 = 300 min
( )
.
Distance inkm

16 x
Time inmin 25 300
Clearly, the more the time taken, the more will be the distance covered.
So, this is a case of direct proportion.
Now,
16
25
=
x
300
? x =
16×300
25
? x = 192
Therefore, the required distance is 192 km.
Q u e s t i o n : 6
If 18 dolls cost Rs 630, how many dolls can be bought for Rs 455?
S o l u t i o n :
Let the required number of dolls be x. Then, we have:

No of dolls 18 x
Cost of dolls
inrupees
630 455
Clearly, the less the amount of money, the less will be the number of dolls bought.
So, this is a case of direct proportion.
Now,
18
630
=
x
455
?
1
35
=
x
455
? x =
455
35
? x = 13
Therefore, 13 dolls can be bought for Rs 455.
Q u e s t i o n : 7
If 9 kg of sugar costs  238.50, how much sugar can be bought for  371?
S o l u t i o n :
Let the quantity of sugar bought for 371 be x kg.
Quantity inkg 9 x
Pricein 238.50 371
The price increases as the quantity increases. Thus, this is a case of direct proportion.
?
9
238.50
=
x
371
? x =
9×371
238.50
= 14
Thus, the quantity of sugar bought for 371 is 14 kg.
Q u e s t i o n : 8
The cost of 15 metres of a cloth is Rs 981. What length of this cloth can be purchased for Rs 1308?
S o l u t i o n :
Let the length of cloth be x m. Then, we have:
Length of cloth inmetres 15 x
Cost of cloth inrupees 981 1308
Clearly, more length of cloth can be bought by more amount of money.
So, this is a case of direct proportion.
Now,
15
981
=
x
1308
? x =
15×1308
981
? x = 20
Therefore, 20 m of cloth can be bought for Rs 1,308.
Q u e s t i o n : 9
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 15m high. If the length of the ship is 35 metres, how long is the model ship?
S o l u t i o n :
Let x m be the length of the model of the ship. Then, we have:
1 m = 100 cmTherefore, 15 m = 1500 cm35 m = 3500 cm

Length of the mast incm Length of the  ship incm
Actual ship 1500 3500
Model of the
ship
9 x
Clearly, if the length of the actual ship is more, then the length of the model ship will also be more.
So, this is a case of direct proportion.
Now,
1500
9
=
3500
x
? x =
3500×9
1500
? x = 21 cm
Therefore, the length of the model of the ship is 21 cm.
Q u e s t i o n : 1 0
In 8 days, the earth picks up (6.4 × 10
7
) kg of dust from the atmosphere. How much dust will it pick up in 15 days?
S o l u t i o n :
Let x kg be the required amount of dust. Then, we have:

No. of days 8 15
Dust inkg 6. 4 ×10
7
x
( )
Clearly, more amount of dust will be collected in more number of days.
So, this is a case of direct proportion.
Now,
8
6.4×10
7
=
15
x
? x =
15×6.4×10
7
8
? x = 12 ×10
7
Therefore, 12,00,00,000 kg of dust will be picked up in 15 days.
Q u e s t i o n : 1 1
A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 1 hour 12 minutes?
S o l u t i o n :
Let x km be the required distance. Then, we have:
1 h = 60 mini. e. , 1h 12 min = (60 +12) min = 72 min

Distance covered inkm 50 x
Time inmin 60 72
Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.
Now,
50
60
=
x
72
? x =
50×72
60
? x = 60
Therefore, the distance travelled by the car in 1 h 12 min is 60 km.
Q u e s t i o n : 1 2
Ravi walks at the uniform rate of 5 km/hr. What distance would he cover in 2 hours 24 minutes?
S o l u t i o n :
Let x km be the required distance covered by Ravi in 2 h 24 min.
Then, we have:
1 h = 60mini. e. , 2 h 24 min = (120 +24) min = 144 min

Distance covered
inkm
5 x
Time inmin 60 144
Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.
Now,
5
60
=
x
144
? x =
5×144
60
? x = 12
Therefore, the distance covered by Ravi in 2 h 24 min is 12 km.
Q u e s t i o n : 1 3
If the thickness of a pile of 12 cardboards is 65 mm, find the thickness of a pile of 312 such cardboards.
S o l u t i o n :
Let x mm be the required thickness. Then, we have:

Thickness of cardboard inmm 65 x
No. of cardboards 12 312
Clearly, when the number of cardboard is more, the thickness will also be more.
So, it is a case of direct proportion.
Now,
65
12
=
x
312
? x =
65×312
12
? x = 1690
Therefore, the thickness of the pile of 312 cardboards is 1690 mm.
Q u e s t i o n : 1 4
11 men can dig 6
3
4
-metre-long trench in one day. How many men should be employed for digging 27-metre-long trench of the same type in one day?
S o l u t i o n :
Let x be the required number of men.
Now, 6
3
4
m =
27
4
m
Then, we have:
Number of men 11 x
Length of trench inmetres
27
4
27
Clearly, the longer the trench, the greater will be the number of men required.
So, it is a case of direct proportion.
Now,
11
27
4
=
x
27
?
11×4
27
=
x
27
? x = 44
Therefore, 44 men should be employed to dig a trench of length 27 m.
Q u e s t i o n : 1 5
Page 4

Q u e s t i o n : 1
Observe the tables given below and in each one find whether x and y are proportional:
i
x 3 5 8 11 26
y 9 15 24 33 78
ii
x 2.5 4 7.5 10 14
y 10 16 30 40 42
iii
x 5 7 9 15 18 25
y 15 21 27 60 72 75
S o l u t i o n :
i
Clearly,
x
y
=
3
9
=
5
15
=
8
24
=
11
33
=
26
78
=
1
3
constant Therefore, x and y are proportional.
ii
Clearly,
x
y
=
2.5
10
=
4
16
=
7.5
30
=
10
40
=
1
4
, while
14
42
=
1
3
i. e. ,
2.5
10
=
4
16
=
7.5
30
=
10
40
is not equal to
14
42
. Therefore, x and y are not proportional.
iii
Clearly,
x
y
=
5
15
=
7
21
=
9
27
=
25
75
=
1
3
, while
15
60
=
18
72
=
1
4
i. e. ,
5
15
=
7
21
=
9
27
=
25
75
is not equal to
15
60
and
18
72
. Therefore, x and y are not proportional.
Q u e s t i o n : 2
If x and y are directly proportional, find the values of x
1
, x
2
and y
1
in the table given below:
x 3 x
1
x
2 10
y 72 120 192 y
1
S o l u t i o n :
Since x and y are directly propotional, we have:
3
72
=
x 1
120
=
x 2
192
=
10
y 1
Now,
3
72
=
x 1
120
? x
1
=
120×3
72
= 5
And,
3
72
=
x 2
192
? x
2
=
3 × 192
72
= 8
And,
3
72
=
10
y 1
? y
1
=
72×10
3
= 240
Therefore, x
1
= 5, x
2
= 8 and y
1
= 240
Q u e s t i o n : 3
A truck covers a distance of 510 km in 34 litres of diesel. How much distance would it cover in 20 litres of diesel?
S o l u t i o n :
Let the required distance be x km. Then, we have:
Quantity of diesel inlitres

34 20
Distance inkm 510 x
Clearly, the less the quantity of diesel consumed, the less is the distance covered.
So, this is a case of direct proportion.
Now,
34
510
=
20
x
?
1
15
=
20
x
? x ×1 = 20 ×15 = 300
Therefore, the required distance is 300 km.
Q u e s t i o n : 4
A taxi charges a fare of Rs 2550 for a journey of 150 km. How much would it charge for a journey of 124 km?
S o l u t i o n :
Let the charge for a journey of 124 km be x.
Pricein 2550 x
Distanceinkm 150 124
More is the distance travelled, more will be the price.
So, it is a case of direct proportion.
?
2550
150
=
x
124
? x =
2550×124
150
= 2108
Thus, the taxi charges 2,108 for the distance of 124 km.
Q u e s t i o n : 5
A loaded truck covers 16 km in 25 minutes. At the same speed, how far can it travel in 5 hours?
S o l u t i o n :
Let the required distance be x km. Then, we have:
1 h = 60 mini. e. , 5 h = 5 ×60 = 300 min
( )
.
Distance inkm

16 x
Time inmin 25 300
Clearly, the more the time taken, the more will be the distance covered.
So, this is a case of direct proportion.
Now,
16
25
=
x
300
? x =
16×300
25
? x = 192
Therefore, the required distance is 192 km.
Q u e s t i o n : 6
If 18 dolls cost Rs 630, how many dolls can be bought for Rs 455?
S o l u t i o n :
Let the required number of dolls be x. Then, we have:

No of dolls 18 x
Cost of dolls
inrupees
630 455
Clearly, the less the amount of money, the less will be the number of dolls bought.
So, this is a case of direct proportion.
Now,
18
630
=
x
455
?
1
35
=
x
455
? x =
455
35
? x = 13
Therefore, 13 dolls can be bought for Rs 455.
Q u e s t i o n : 7
If 9 kg of sugar costs  238.50, how much sugar can be bought for  371?
S o l u t i o n :
Let the quantity of sugar bought for 371 be x kg.
Quantity inkg 9 x
Pricein 238.50 371
The price increases as the quantity increases. Thus, this is a case of direct proportion.
?
9
238.50
=
x
371
? x =
9×371
238.50
= 14
Thus, the quantity of sugar bought for 371 is 14 kg.
Q u e s t i o n : 8
The cost of 15 metres of a cloth is Rs 981. What length of this cloth can be purchased for Rs 1308?
S o l u t i o n :
Let the length of cloth be x m. Then, we have:
Length of cloth inmetres 15 x
Cost of cloth inrupees 981 1308
Clearly, more length of cloth can be bought by more amount of money.
So, this is a case of direct proportion.
Now,
15
981
=
x
1308
? x =
15×1308
981
? x = 20
Therefore, 20 m of cloth can be bought for Rs 1,308.
Q u e s t i o n : 9
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 15m high. If the length of the ship is 35 metres, how long is the model ship?
S o l u t i o n :
Let x m be the length of the model of the ship. Then, we have:
1 m = 100 cmTherefore, 15 m = 1500 cm35 m = 3500 cm

Length of the mast incm Length of the  ship incm
Actual ship 1500 3500
Model of the
ship
9 x
Clearly, if the length of the actual ship is more, then the length of the model ship will also be more.
So, this is a case of direct proportion.
Now,
1500
9
=
3500
x
? x =
3500×9
1500
? x = 21 cm
Therefore, the length of the model of the ship is 21 cm.
Q u e s t i o n : 1 0
In 8 days, the earth picks up (6.4 × 10
7
) kg of dust from the atmosphere. How much dust will it pick up in 15 days?
S o l u t i o n :
Let x kg be the required amount of dust. Then, we have:

No. of days 8 15
Dust inkg 6. 4 ×10
7
x
( )
Clearly, more amount of dust will be collected in more number of days.
So, this is a case of direct proportion.
Now,
8
6.4×10
7
=
15
x
? x =
15×6.4×10
7
8
? x = 12 ×10
7
Therefore, 12,00,00,000 kg of dust will be picked up in 15 days.
Q u e s t i o n : 1 1
A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 1 hour 12 minutes?
S o l u t i o n :
Let x km be the required distance. Then, we have:
1 h = 60 mini. e. , 1h 12 min = (60 +12) min = 72 min

Distance covered inkm 50 x
Time inmin 60 72
Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.
Now,
50
60
=
x
72
? x =
50×72
60
? x = 60
Therefore, the distance travelled by the car in 1 h 12 min is 60 km.
Q u e s t i o n : 1 2
Ravi walks at the uniform rate of 5 km/hr. What distance would he cover in 2 hours 24 minutes?
S o l u t i o n :
Let x km be the required distance covered by Ravi in 2 h 24 min.
Then, we have:
1 h = 60mini. e. , 2 h 24 min = (120 +24) min = 144 min

Distance covered
inkm
5 x
Time inmin 60 144
Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.
Now,
5
60
=
x
144
? x =
5×144
60
? x = 12
Therefore, the distance covered by Ravi in 2 h 24 min is 12 km.
Q u e s t i o n : 1 3
If the thickness of a pile of 12 cardboards is 65 mm, find the thickness of a pile of 312 such cardboards.
S o l u t i o n :
Let x mm be the required thickness. Then, we have:

Thickness of cardboard inmm 65 x
No. of cardboards 12 312
Clearly, when the number of cardboard is more, the thickness will also be more.
So, it is a case of direct proportion.
Now,
65
12
=
x
312
? x =
65×312
12
? x = 1690
Therefore, the thickness of the pile of 312 cardboards is 1690 mm.
Q u e s t i o n : 1 4
11 men can dig 6
3
4
-metre-long trench in one day. How many men should be employed for digging 27-metre-long trench of the same type in one day?
S o l u t i o n :
Let x be the required number of men.
Now, 6
3
4
m =
27
4
m
Then, we have:
Number of men 11 x
Length of trench inmetres
27
4
27
Clearly, the longer the trench, the greater will be the number of men required.
So, it is a case of direct proportion.
Now,
11
27
4
=
x
27
?
11×4
27
=
x
27
? x = 44
Therefore, 44 men should be employed to dig a trench of length 27 m.
Q u e s t i o n : 1 5
Reenu types 540 words during half an hour. How many words would she type in 8 minutes?
S o l u t i o n :
Let Reenu type x words in 8 minutes.

No. of words 540 x
Time taken inmin 30 8
Clearly, less number of words will be typed in less time.
So, it is a case of direct proportion.
Now,
540
30
=
x
8
? x =
540×8
30
? x = 144
Therefore, Reenu will type 144 words in 8 minutes.
Q u e s t i o n : 1 6
Observe the tables given below and in each case find whether x and y are inversely proportional:
i
x 6 10 14 16
y 9 15 21 24
ii
x 5 9 15 3 45
y 18 10 6 30 2
iii
x 9 3 6 36
y 4 12 9 1
S o l u t i o n :
i
Clearly, 6 ×9 ? 10 ×15 ? 14 ×21 ? 16 ×24Therefore, x and y are not inversely proportional.
ii
Clearly, 5 ×18 = 9 ×10 = 15 ×6 = 3 ×30 = 45 ×2 = 90 = (consant)Therefore, x and y are inversely proportional.
iii
Clearly, 9 ×4 = 3 ×12 = 36 ×1 = 36, while 6 ×9 = 54i. e. , 9 ×4 = 3 ×12 = 36 ×1 ? 6 ×9Therefore, x and y are not inversely proportional.
Q u e s t i o n : 1 7
If x and y are inversely proportional, find the values of x
1
, x
2
, y
1
and y
2
in the table given below:
x 8 x
1 16 x
2 80
y y
1 4 5 2 y
2
S o l u t i o n :
Since x and y are inversely proportional, xy must be a constant.
Therefore, 8 ×y
1
= x
1
×4 = 16 ×5 = x
2
×2 = 80 ×y
2
Now, 16 ×5 = 8 ×y
1
?
80
8
= y
1
? y
1
= 1016 ×5 = x
1
×4 ?
80
4
= x
1
? x
1
= 2016 ×5 = x
2
×2 ?
80
2
= x
2
? x
2
= 4016 ×5 = 80 ×y
2
?
80
80
= y
2
? y
2
=
Q u e s t i o n : 1 8
If 35 men can reap a field in 8 days, in how many days can 20 men reap the same field?
S o l u t i o n :
Let x be the required number of days. Then, we have:

No. of days 8 x
No. of men 35 20
Clearly, less men will take more days to reap the field.
So, it is a case of inverse proportion.
Now, 8 × 35 = x × 20 ?
8 × 35
20
= x ? 14 = x
Therefore, 20 men can reap the same field in 14 days.
Q u e s t i o n : 1 9
12 men can dig a pond in 8 days. How many men can dig it in 6 days?
S o l u t i o n :
Let x be the required number of men. Then, we have:

No. of days 8 6
No. of men 12 x
Clearly, more men will require less number of days to dig the pond.
So, it is a case of inverse proportion.
Now, 8 × 12 = 6 × x ? x =
8 × 12
6
? x = 16
Therefore, 16 men can dig the pond in 6 days.
Page 5

Q u e s t i o n : 1
Observe the tables given below and in each one find whether x and y are proportional:
i
x 3 5 8 11 26
y 9 15 24 33 78
ii
x 2.5 4 7.5 10 14
y 10 16 30 40 42
iii
x 5 7 9 15 18 25
y 15 21 27 60 72 75
S o l u t i o n :
i
Clearly,
x
y
=
3
9
=
5
15
=
8
24
=
11
33
=
26
78
=
1
3
constant Therefore, x and y are proportional.
ii
Clearly,
x
y
=
2.5
10
=
4
16
=
7.5
30
=
10
40
=
1
4
, while
14
42
=
1
3
i. e. ,
2.5
10
=
4
16
=
7.5
30
=
10
40
is not equal to
14
42
. Therefore, x and y are not proportional.
iii
Clearly,
x
y
=
5
15
=
7
21
=
9
27
=
25
75
=
1
3
, while
15
60
=
18
72
=
1
4
i. e. ,
5
15
=
7
21
=
9
27
=
25
75
is not equal to
15
60
and
18
72
. Therefore, x and y are not proportional.
Q u e s t i o n : 2
If x and y are directly proportional, find the values of x
1
, x
2
and y
1
in the table given below:
x 3 x
1
x
2 10
y 72 120 192 y
1
S o l u t i o n :
Since x and y are directly propotional, we have:
3
72
=
x 1
120
=
x 2
192
=
10
y 1
Now,
3
72
=
x 1
120
? x
1
=
120×3
72
= 5
And,
3
72
=
x 2
192
? x
2
=
3 × 192
72
= 8
And,
3
72
=
10
y 1
? y
1
=
72×10
3
= 240
Therefore, x
1
= 5, x
2
= 8 and y
1
= 240
Q u e s t i o n : 3
A truck covers a distance of 510 km in 34 litres of diesel. How much distance would it cover in 20 litres of diesel?
S o l u t i o n :
Let the required distance be x km. Then, we have:
Quantity of diesel inlitres

34 20
Distance inkm 510 x
Clearly, the less the quantity of diesel consumed, the less is the distance covered.
So, this is a case of direct proportion.
Now,
34
510
=
20
x
?
1
15
=
20
x
? x ×1 = 20 ×15 = 300
Therefore, the required distance is 300 km.
Q u e s t i o n : 4
A taxi charges a fare of Rs 2550 for a journey of 150 km. How much would it charge for a journey of 124 km?
S o l u t i o n :
Let the charge for a journey of 124 km be x.
Pricein 2550 x
Distanceinkm 150 124
More is the distance travelled, more will be the price.
So, it is a case of direct proportion.
?
2550
150
=
x
124
? x =
2550×124
150
= 2108
Thus, the taxi charges 2,108 for the distance of 124 km.
Q u e s t i o n : 5
A loaded truck covers 16 km in 25 minutes. At the same speed, how far can it travel in 5 hours?
S o l u t i o n :
Let the required distance be x km. Then, we have:
1 h = 60 mini. e. , 5 h = 5 ×60 = 300 min
( )
.
Distance inkm

16 x
Time inmin 25 300
Clearly, the more the time taken, the more will be the distance covered.
So, this is a case of direct proportion.
Now,
16
25
=
x
300
? x =
16×300
25
? x = 192
Therefore, the required distance is 192 km.
Q u e s t i o n : 6
If 18 dolls cost Rs 630, how many dolls can be bought for Rs 455?
S o l u t i o n :
Let the required number of dolls be x. Then, we have:

No of dolls 18 x
Cost of dolls
inrupees
630 455
Clearly, the less the amount of money, the less will be the number of dolls bought.
So, this is a case of direct proportion.
Now,
18
630
=
x
455
?
1
35
=
x
455
? x =
455
35
? x = 13
Therefore, 13 dolls can be bought for Rs 455.
Q u e s t i o n : 7
If 9 kg of sugar costs  238.50, how much sugar can be bought for  371?
S o l u t i o n :
Let the quantity of sugar bought for 371 be x kg.
Quantity inkg 9 x
Pricein 238.50 371
The price increases as the quantity increases. Thus, this is a case of direct proportion.
?
9
238.50
=
x
371
? x =
9×371
238.50
= 14
Thus, the quantity of sugar bought for 371 is 14 kg.
Q u e s t i o n : 8
The cost of 15 metres of a cloth is Rs 981. What length of this cloth can be purchased for Rs 1308?
S o l u t i o n :
Let the length of cloth be x m. Then, we have:
Length of cloth inmetres 15 x
Cost of cloth inrupees 981 1308
Clearly, more length of cloth can be bought by more amount of money.
So, this is a case of direct proportion.
Now,
15
981
=
x
1308
? x =
15×1308
981
? x = 20
Therefore, 20 m of cloth can be bought for Rs 1,308.
Q u e s t i o n : 9
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 15m high. If the length of the ship is 35 metres, how long is the model ship?
S o l u t i o n :
Let x m be the length of the model of the ship. Then, we have:
1 m = 100 cmTherefore, 15 m = 1500 cm35 m = 3500 cm

Length of the mast incm Length of the  ship incm
Actual ship 1500 3500
Model of the
ship
9 x
Clearly, if the length of the actual ship is more, then the length of the model ship will also be more.
So, this is a case of direct proportion.
Now,
1500
9
=
3500
x
? x =
3500×9
1500
? x = 21 cm
Therefore, the length of the model of the ship is 21 cm.
Q u e s t i o n : 1 0
In 8 days, the earth picks up (6.4 × 10
7
) kg of dust from the atmosphere. How much dust will it pick up in 15 days?
S o l u t i o n :
Let x kg be the required amount of dust. Then, we have:

No. of days 8 15
Dust inkg 6. 4 ×10
7
x
( )
Clearly, more amount of dust will be collected in more number of days.
So, this is a case of direct proportion.
Now,
8
6.4×10
7
=
15
x
? x =
15×6.4×10
7
8
? x = 12 ×10
7
Therefore, 12,00,00,000 kg of dust will be picked up in 15 days.
Q u e s t i o n : 1 1
A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 1 hour 12 minutes?
S o l u t i o n :
Let x km be the required distance. Then, we have:
1 h = 60 mini. e. , 1h 12 min = (60 +12) min = 72 min

Distance covered inkm 50 x
Time inmin 60 72
Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.
Now,
50
60
=
x
72
? x =
50×72
60
? x = 60
Therefore, the distance travelled by the car in 1 h 12 min is 60 km.
Q u e s t i o n : 1 2
Ravi walks at the uniform rate of 5 km/hr. What distance would he cover in 2 hours 24 minutes?
S o l u t i o n :
Let x km be the required distance covered by Ravi in 2 h 24 min.
Then, we have:
1 h = 60mini. e. , 2 h 24 min = (120 +24) min = 144 min

Distance covered
inkm
5 x
Time inmin 60 144
Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.
Now,
5
60
=
x
144
? x =
5×144
60
? x = 12
Therefore, the distance covered by Ravi in 2 h 24 min is 12 km.
Q u e s t i o n : 1 3
If the thickness of a pile of 12 cardboards is 65 mm, find the thickness of a pile of 312 such cardboards.
S o l u t i o n :
Let x mm be the required thickness. Then, we have:

Thickness of cardboard inmm 65 x
No. of cardboards 12 312
Clearly, when the number of cardboard is more, the thickness will also be more.
So, it is a case of direct proportion.
Now,
65
12
=
x
312
? x =
65×312
12
? x = 1690
Therefore, the thickness of the pile of 312 cardboards is 1690 mm.
Q u e s t i o n : 1 4
11 men can dig 6
3
4
-metre-long trench in one day. How many men should be employed for digging 27-metre-long trench of the same type in one day?
S o l u t i o n :
Let x be the required number of men.
Now, 6
3
4
m =
27
4
m
Then, we have:
Number of men 11 x
Length of trench inmetres
27
4
27
Clearly, the longer the trench, the greater will be the number of men required.
So, it is a case of direct proportion.
Now,
11
27
4
=
x
27
?
11×4
27
=
x
27
? x = 44
Therefore, 44 men should be employed to dig a trench of length 27 m.
Q u e s t i o n : 1 5
Reenu types 540 words during half an hour. How many words would she type in 8 minutes?
S o l u t i o n :
Let Reenu type x words in 8 minutes.

No. of words 540 x
Time taken inmin 30 8
Clearly, less number of words will be typed in less time.
So, it is a case of direct proportion.
Now,
540
30
=
x
8
? x =
540×8
30
? x = 144
Therefore, Reenu will type 144 words in 8 minutes.
Q u e s t i o n : 1 6
Observe the tables given below and in each case find whether x and y are inversely proportional:
i
x 6 10 14 16
y 9 15 21 24
ii
x 5 9 15 3 45
y 18 10 6 30 2
iii
x 9 3 6 36
y 4 12 9 1
S o l u t i o n :
i
Clearly, 6 ×9 ? 10 ×15 ? 14 ×21 ? 16 ×24Therefore, x and y are not inversely proportional.
ii
Clearly, 5 ×18 = 9 ×10 = 15 ×6 = 3 ×30 = 45 ×2 = 90 = (consant)Therefore, x and y are inversely proportional.
iii
Clearly, 9 ×4 = 3 ×12 = 36 ×1 = 36, while 6 ×9 = 54i. e. , 9 ×4 = 3 ×12 = 36 ×1 ? 6 ×9Therefore, x and y are not inversely proportional.
Q u e s t i o n : 1 7
If x and y are inversely proportional, find the values of x
1
, x
2
, y
1
and y
2
in the table given below:
x 8 x
1 16 x
2 80
y y
1 4 5 2 y
2
S o l u t i o n :
Since x and y are inversely proportional, xy must be a constant.
Therefore, 8 ×y
1
= x
1
×4 = 16 ×5 = x
2
×2 = 80 ×y
2
Now, 16 ×5 = 8 ×y
1
?
80
8
= y
1
? y
1
= 1016 ×5 = x
1
×4 ?
80
4
= x
1
? x
1
= 2016 ×5 = x
2
×2 ?
80
2
= x
2
? x
2
= 4016 ×5 = 80 ×y
2
?
80
80
= y
2
? y
2
=
Q u e s t i o n : 1 8
If 35 men can reap a field in 8 days, in how many days can 20 men reap the same field?
S o l u t i o n :
Let x be the required number of days. Then, we have:

No. of days 8 x
No. of men 35 20
Clearly, less men will take more days to reap the field.
So, it is a case of inverse proportion.
Now, 8 × 35 = x × 20 ?
8 × 35
20
= x ? 14 = x
Therefore, 20 men can reap the same field in 14 days.
Q u e s t i o n : 1 9
12 men can dig a pond in 8 days. How many men can dig it in 6 days?
S o l u t i o n :
Let x be the required number of men. Then, we have:

No. of days 8 6
No. of men 12 x
Clearly, more men will require less number of days to dig the pond.
So, it is a case of inverse proportion.
Now, 8 × 12 = 6 × x ? x =
8 × 12
6
? x = 16
Therefore, 16 men can dig the pond in 6 days.
Q u e s t i o n : 2 0
6 cows can graze a field in 28 days. How long would 14 cows take to graze the same field?
S o l u t i o n :
Let x be the number of days. Then, we have:

No. of days 28 x
No. of cows 6 14
Clearly, more number of cows will take less number of days to graze the field.
So, it is a case of inverse proportion.
Now, 28 × 6 = x × 14 ? x =
28 × 6
14
? x = 12
Therefore, 14 cows will take 12 days to graze the field.
Q u e s t i o n : 2 1
A car takes 5 hours to reach a destination by travelling at the speed of 60 km/hr. How long will it take when the car travels at the speed of 75 km/hr?
S o l u t i o n :
Let x h be the required time taken. Then, we have:

Speed inkm/h 60 75
Time inh 5 x
Clearly, the higher the speed, the lesser will be the the time taken.
So, it is a case of inverse proportion.
Now, 60 ×5 = 75 ×x ? x =
60×5
75
? x = 4
Therefore, the car will reach its destination in 4 h if it travels at a speed of 75 km/h.
Q u e s t i o n : 2 2
A factory requires 42 machines to produce a given number of articles in 56 days. How many machines would be required to produce the same number of articles in 48 days?
S o l u t i o n :
Let x be the number of machines required to produce same number of articles in 48.
Then, we have:

No. of machines 42 x
No. of days 56 48
Clearly, less number of days will require more number of machines.
So, it is a case of inverse proportion.
Now, 42 ×56 = x ×48 ? x =
42×56
48
? x = 49
Therefore, 49 machines would be required to produce the same number of articles in 48 days.
Q u e s t i o n : 2 3
7 teps of the same size fill a tank in 1 hour 36 minutes. How long will 8 taps of the same size take to fill the tank?
S o l u t i o n :
Let x be the required number of taps. Then, we have:
1 h = 60 min
i.e., 1 h 36 min = 60 +36
min = 96 min

No. of taps 7 8
Time inmin 96 x
Clearly, more number of taps will require less time to fill the tank.
So, it is a case of inverse proportion.
Now, 7 ×96 = 8 ×x ? x =
7×96
8
? x = 84
Therefore, 8 taps of the same size will take 84 min or 1 h 24 min to fill the tank.
Q u e s t i o n : 2 4
8 taps of the same size fill a tank in 27 minutes. If two taps go out of order, how long would the remaining taps take to fill the tank?
S o l u t i o n :
Let x min be the required number of time. Then, we have:
No. of taps 8 6
Time inmin 27 x
Clearly, less number of taps will take more time to fill the tank .
So, it is a case of inverse proportion.
Now, 8 ×27 = 6 ×x ? x =
8×27
6
? x = 36
Therefore, it will take 36 min to fill the tank.
Q u e s t i o n : 2 5
A farmer has enough food to feed 28 animals in his cattle for 9 days. How long would the food last, if there were 8 more animals in his cattle?
S o l u t i o n :
Let x be the required number of days. Then, we have:
```

## Mathematics (Maths) Class 8

187 videos|321 docs|48 tests
 Use Code STAYHOME200 and get INR 200 additional OFF

## Mathematics (Maths) Class 8

187 videos|321 docs|48 tests

Track your progress, build streaks, highlight & save important lessons and more!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;