RS Aggarwal Solutions: Exercise 9A - Percentage | Mathematics (Maths) Class 8 PDF Download

 Page 1


Question:1
Express each of the following as a fraction:
i
48%
ii
220%
iii
2.5%
Solution:
(i) 48 % =
48
100
=
12
25
(ii) 220 % =
220
100
=
11
5
(iii) 2. 5 % =
2.5
100
=
25
1000
=
1
40
Question:2
Express each of the following as a decimal:
i
6%
ii
72%
iii
125%
Solution:
(i) 6 % =
6
100
= 0. 06
(ii) 72 % =
72
100
= 0. 72
(iii) 125 % =
125
100
= 1. 25
Question:3
Express each of the following as a percentage:
i
9
25
ii
3
125
iii
12
5
Solution:
(i) 
9
25
=
9
25
×100 % = 9 ×4 % = 36 %
(ii) 
3
125
=
3
125
×100 % = 2. 4 %
(iii) 
12
5
=
12
5
×100 % = 240 %
Question:4
Convert the ratio 4 : 5 to percentage.
Solution:
4 : 5 =
4
5
=  
4
5
×100 % = 80 %
Question:5
Express 125% as a ratio.
Solution:
125 % =
125
100
=
5
4
 = 5: 4
Question:6
Which is largest in 6
2
3
%,
3
20
and 0.14?
Solution:
?       We have: 6
2
3
% =
20
3
%             =
20
3
×
1
100
             =
1
15
             = 0. 06  
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Page 2


Question:1
Express each of the following as a fraction:
i
48%
ii
220%
iii
2.5%
Solution:
(i) 48 % =
48
100
=
12
25
(ii) 220 % =
220
100
=
11
5
(iii) 2. 5 % =
2.5
100
=
25
1000
=
1
40
Question:2
Express each of the following as a decimal:
i
6%
ii
72%
iii
125%
Solution:
(i) 6 % =
6
100
= 0. 06
(ii) 72 % =
72
100
= 0. 72
(iii) 125 % =
125
100
= 1. 25
Question:3
Express each of the following as a percentage:
i
9
25
ii
3
125
iii
12
5
Solution:
(i) 
9
25
=
9
25
×100 % = 9 ×4 % = 36 %
(ii) 
3
125
=
3
125
×100 % = 2. 4 %
(iii) 
12
5
=
12
5
×100 % = 240 %
Question:4
Convert the ratio 4 : 5 to percentage.
Solution:
4 : 5 =
4
5
=  
4
5
×100 % = 80 %
Question:5
Express 125% as a ratio.
Solution:
125 % =
125
100
=
5
4
 = 5: 4
Question:6
Which is largest in 6
2
3
%,
3
20
and 0.14?
Solution:
?       We have: 6
2
3
% =
20
3
%             =
20
3
×
1
100
             =
1
15
             = 0. 06  
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      Also, 
3
20
= 0. 15
      The third number is 0. 14. Clearly, 0. 15 is the largest. Hence, 
3
20
 is the largest.
Question:7
i
What per cent of 150 is 96?
ii
What per cent of 5 kg is 200 g?
iii
What per cent of 2 litres is 250 mL?
Solution:
(i) Required percentage =
96
150
×100 % = 64 %
                                    
ii Required percentage =
200
5×1000
×100 % = 4 %
                                  
(iii) Required percentage =
250
2×1000
×100 % = 12. 5 %
                                    
Question:8
Find 4
1
2
%
of Rs 3600.
Solution:
4
1
2
% =
9
2×100
? 
9
200 
 of Rs 3600 =
9
200
×3600 = Rs 162
Question:9
If 16% of a number is 72, find the number.
Solution:
Let the number be x. 16 % of x is 72. ?
16
100
×x = 72 ? 16x = 72 ×100 ? 16x = 7200 ? x =
7200
16
= 450 ? The required number is 450.
Question:10
A man saves 18% of his monthly income. If he saves Rs 1890 per month, what is his monthly income?
Solution:
Let Rs x be his monthly income. His savings = 18 % of Rs x
                = Rs x ×
18
100
= Rs 
9x
50
Now, 
9x
50
= 1890 ? x = Rs 1890 ×
50
9
? x = Rs 10500 ? His monthly income is Rs. 10500.
Question:11
A football team wins 7 games, which is 35% of the total games played. How many games were played in all?
Solution:
Let x be the total number of games played. Percentage of games won = 35 % of x
                             = x ×
35
100
=
35x
100
Now, 
35x
100
= 7      ? x = 7 ×
100
35
      ? x = 20 ? The total number games played is 20.
Question:12
Amit was given an increment of 20% on his salary. If his new salary is Rs 15300, what was his salary before the increment?
Solution:
Let Rs x be Amit's old salary. His salary after increment will be Rs x +
20
100
x According to the question, we have: ? x +
20
100
x = 15300 ?
100x+20x
100
= 15300       (LCM = 100) ?
120x
100
= 15300 ?
Question:13
Sonal attended her school on 204 days in a full year. If her attendance is 85%, find the number of days on which the school was opened.
Solution:
Let x be the number of days the school was opened. Number of days Sonal attended school = 204 daysPercentage of her attendance = 85 % of x
                                = x ×
85
100
=
85x
100
Now, 
85x
100
= 204      ? x = 204 ×
100
85
      ? x = 240 ? The school was opened for 240 day.
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Page 3


Question:1
Express each of the following as a fraction:
i
48%
ii
220%
iii
2.5%
Solution:
(i) 48 % =
48
100
=
12
25
(ii) 220 % =
220
100
=
11
5
(iii) 2. 5 % =
2.5
100
=
25
1000
=
1
40
Question:2
Express each of the following as a decimal:
i
6%
ii
72%
iii
125%
Solution:
(i) 6 % =
6
100
= 0. 06
(ii) 72 % =
72
100
= 0. 72
(iii) 125 % =
125
100
= 1. 25
Question:3
Express each of the following as a percentage:
i
9
25
ii
3
125
iii
12
5
Solution:
(i) 
9
25
=
9
25
×100 % = 9 ×4 % = 36 %
(ii) 
3
125
=
3
125
×100 % = 2. 4 %
(iii) 
12
5
=
12
5
×100 % = 240 %
Question:4
Convert the ratio 4 : 5 to percentage.
Solution:
4 : 5 =
4
5
=  
4
5
×100 % = 80 %
Question:5
Express 125% as a ratio.
Solution:
125 % =
125
100
=
5
4
 = 5: 4
Question:6
Which is largest in 6
2
3
%,
3
20
and 0.14?
Solution:
?       We have: 6
2
3
% =
20
3
%             =
20
3
×
1
100
             =
1
15
             = 0. 06  
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      Also, 
3
20
= 0. 15
      The third number is 0. 14. Clearly, 0. 15 is the largest. Hence, 
3
20
 is the largest.
Question:7
i
What per cent of 150 is 96?
ii
What per cent of 5 kg is 200 g?
iii
What per cent of 2 litres is 250 mL?
Solution:
(i) Required percentage =
96
150
×100 % = 64 %
                                    
ii Required percentage =
200
5×1000
×100 % = 4 %
                                  
(iii) Required percentage =
250
2×1000
×100 % = 12. 5 %
                                    
Question:8
Find 4
1
2
%
of Rs 3600.
Solution:
4
1
2
% =
9
2×100
? 
9
200 
 of Rs 3600 =
9
200
×3600 = Rs 162
Question:9
If 16% of a number is 72, find the number.
Solution:
Let the number be x. 16 % of x is 72. ?
16
100
×x = 72 ? 16x = 72 ×100 ? 16x = 7200 ? x =
7200
16
= 450 ? The required number is 450.
Question:10
A man saves 18% of his monthly income. If he saves Rs 1890 per month, what is his monthly income?
Solution:
Let Rs x be his monthly income. His savings = 18 % of Rs x
                = Rs x ×
18
100
= Rs 
9x
50
Now, 
9x
50
= 1890 ? x = Rs 1890 ×
50
9
? x = Rs 10500 ? His monthly income is Rs. 10500.
Question:11
A football team wins 7 games, which is 35% of the total games played. How many games were played in all?
Solution:
Let x be the total number of games played. Percentage of games won = 35 % of x
                             = x ×
35
100
=
35x
100
Now, 
35x
100
= 7      ? x = 7 ×
100
35
      ? x = 20 ? The total number games played is 20.
Question:12
Amit was given an increment of 20% on his salary. If his new salary is Rs 15300, what was his salary before the increment?
Solution:
Let Rs x be Amit's old salary. His salary after increment will be Rs x +
20
100
x According to the question, we have: ? x +
20
100
x = 15300 ?
100x+20x
100
= 15300       (LCM = 100) ?
120x
100
= 15300 ?
Question:13
Sonal attended her school on 204 days in a full year. If her attendance is 85%, find the number of days on which the school was opened.
Solution:
Let x be the number of days the school was opened. Number of days Sonal attended school = 204 daysPercentage of her attendance = 85 % of x
                                = x ×
85
100
=
85x
100
Now, 
85x
100
= 204      ? x = 204 ×
100
85
      ? x = 240 ? The school was opened for 240 day.
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Question:14
A's income is 20% less than that of B. By what per cent is B's income more than A's?
Solution:
Let B's income be Rs 100Then, A's income = Rs 80Therefore, B's income is more than ?A's income by = 
(100-80)
80
×100 % = 
20
80
×100 % = 25 %                                   
                                                          = Rs125
? B's income is more than that of A's by (125 -100)%, i. e. , 25 %.
Question:15
The price of petrol goes up by 10%. By how much per cent must a motorist reduce the consumption of petrol so that the expenditure on it remains unchanged?
Solution:
Let the consumption of petrol originally be 1 unit and let its cost be Rs 100. New cost of 1 unit of petrol = Rs 110Now, Rs 110 will yield 1 unit of petrol. i. e. , Rs 100 will yield 
1
110
×100 ,
                    
Now, reduction in consumption = 1 -
10
11
=
1
11
unit
Percentage of reduction =
1
11
×
1
1
×100 %
= 9
1
11
%
? A motorist must reduce the consumption of petrol by 9
1
11
%.
Question:16
The population of a town increases by 8% annually. If the present population is 54000, what was it a year ago?
Solution:
Let x be the population of the town a year ago. Then, present population = 108 % of x
                             = x ×
108
100
=
27x
25
Now, 
27x
25
= 54000      ? x = 54000 ×
25
27
      ? x = 50000
Hence, the population of the town a year ago was 50000.
Question:17
The value of a machine depreciates every year by 20%. If the present value of the machine be Rs 160000, what was its value last year?
Solution:
Let Rs x be the value of the machine last year. Then, present value = 80 % of Rs x
                       = Rs x ×
80
100
= Rs
4x
5
Now, 
4x
5
= 160000      ? x = 160000 ×
5
4
      ? x = 40000 ×5 = 200000
Hence, the value of the machine last year was Rs 2, 00, 000.
Question:18
An alloy contains 40% copper, 32% nickel and rest zinc. Find the mass of zinc in one kg of the alloy.
Solution:
Mass of the alloy = 1 kgPercentage of copper = 40 %Percentage of nickel = 32 %Percentage of zinc = {100 -(40 +32)}%
                     = 28 %
                     
? Mass of zinc in 1 kg of alloy =
28
100
×1 kg
                                     = 0. 28 kg = 0. 28 ×1000 g = 280 g
Question:19
Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in his food daily, find in calories the amount of each of these in his
daily food intake.
Solution:
Amount of protein = 12 % of 2600
                       = 2600 ×
12
100
= 312 cal
Amount of fat = 25 % of 2600
                 = 2600 ×
25
100
= 650 cal
Amount of carbohydrate = 63 % of 2600
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Page 4


Question:1
Express each of the following as a fraction:
i
48%
ii
220%
iii
2.5%
Solution:
(i) 48 % =
48
100
=
12
25
(ii) 220 % =
220
100
=
11
5
(iii) 2. 5 % =
2.5
100
=
25
1000
=
1
40
Question:2
Express each of the following as a decimal:
i
6%
ii
72%
iii
125%
Solution:
(i) 6 % =
6
100
= 0. 06
(ii) 72 % =
72
100
= 0. 72
(iii) 125 % =
125
100
= 1. 25
Question:3
Express each of the following as a percentage:
i
9
25
ii
3
125
iii
12
5
Solution:
(i) 
9
25
=
9
25
×100 % = 9 ×4 % = 36 %
(ii) 
3
125
=
3
125
×100 % = 2. 4 %
(iii) 
12
5
=
12
5
×100 % = 240 %
Question:4
Convert the ratio 4 : 5 to percentage.
Solution:
4 : 5 =
4
5
=  
4
5
×100 % = 80 %
Question:5
Express 125% as a ratio.
Solution:
125 % =
125
100
=
5
4
 = 5: 4
Question:6
Which is largest in 6
2
3
%,
3
20
and 0.14?
Solution:
?       We have: 6
2
3
% =
20
3
%             =
20
3
×
1
100
             =
1
15
             = 0. 06  
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      Also, 
3
20
= 0. 15
      The third number is 0. 14. Clearly, 0. 15 is the largest. Hence, 
3
20
 is the largest.
Question:7
i
What per cent of 150 is 96?
ii
What per cent of 5 kg is 200 g?
iii
What per cent of 2 litres is 250 mL?
Solution:
(i) Required percentage =
96
150
×100 % = 64 %
                                    
ii Required percentage =
200
5×1000
×100 % = 4 %
                                  
(iii) Required percentage =
250
2×1000
×100 % = 12. 5 %
                                    
Question:8
Find 4
1
2
%
of Rs 3600.
Solution:
4
1
2
% =
9
2×100
? 
9
200 
 of Rs 3600 =
9
200
×3600 = Rs 162
Question:9
If 16% of a number is 72, find the number.
Solution:
Let the number be x. 16 % of x is 72. ?
16
100
×x = 72 ? 16x = 72 ×100 ? 16x = 7200 ? x =
7200
16
= 450 ? The required number is 450.
Question:10
A man saves 18% of his monthly income. If he saves Rs 1890 per month, what is his monthly income?
Solution:
Let Rs x be his monthly income. His savings = 18 % of Rs x
                = Rs x ×
18
100
= Rs 
9x
50
Now, 
9x
50
= 1890 ? x = Rs 1890 ×
50
9
? x = Rs 10500 ? His monthly income is Rs. 10500.
Question:11
A football team wins 7 games, which is 35% of the total games played. How many games were played in all?
Solution:
Let x be the total number of games played. Percentage of games won = 35 % of x
                             = x ×
35
100
=
35x
100
Now, 
35x
100
= 7      ? x = 7 ×
100
35
      ? x = 20 ? The total number games played is 20.
Question:12
Amit was given an increment of 20% on his salary. If his new salary is Rs 15300, what was his salary before the increment?
Solution:
Let Rs x be Amit's old salary. His salary after increment will be Rs x +
20
100
x According to the question, we have: ? x +
20
100
x = 15300 ?
100x+20x
100
= 15300       (LCM = 100) ?
120x
100
= 15300 ?
Question:13
Sonal attended her school on 204 days in a full year. If her attendance is 85%, find the number of days on which the school was opened.
Solution:
Let x be the number of days the school was opened. Number of days Sonal attended school = 204 daysPercentage of her attendance = 85 % of x
                                = x ×
85
100
=
85x
100
Now, 
85x
100
= 204      ? x = 204 ×
100
85
      ? x = 240 ? The school was opened for 240 day.
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Question:14
A's income is 20% less than that of B. By what per cent is B's income more than A's?
Solution:
Let B's income be Rs 100Then, A's income = Rs 80Therefore, B's income is more than ?A's income by = 
(100-80)
80
×100 % = 
20
80
×100 % = 25 %                                   
                                                          = Rs125
? B's income is more than that of A's by (125 -100)%, i. e. , 25 %.
Question:15
The price of petrol goes up by 10%. By how much per cent must a motorist reduce the consumption of petrol so that the expenditure on it remains unchanged?
Solution:
Let the consumption of petrol originally be 1 unit and let its cost be Rs 100. New cost of 1 unit of petrol = Rs 110Now, Rs 110 will yield 1 unit of petrol. i. e. , Rs 100 will yield 
1
110
×100 ,
                    
Now, reduction in consumption = 1 -
10
11
=
1
11
unit
Percentage of reduction =
1
11
×
1
1
×100 %
= 9
1
11
%
? A motorist must reduce the consumption of petrol by 9
1
11
%.
Question:16
The population of a town increases by 8% annually. If the present population is 54000, what was it a year ago?
Solution:
Let x be the population of the town a year ago. Then, present population = 108 % of x
                             = x ×
108
100
=
27x
25
Now, 
27x
25
= 54000      ? x = 54000 ×
25
27
      ? x = 50000
Hence, the population of the town a year ago was 50000.
Question:17
The value of a machine depreciates every year by 20%. If the present value of the machine be Rs 160000, what was its value last year?
Solution:
Let Rs x be the value of the machine last year. Then, present value = 80 % of Rs x
                       = Rs x ×
80
100
= Rs
4x
5
Now, 
4x
5
= 160000      ? x = 160000 ×
5
4
      ? x = 40000 ×5 = 200000
Hence, the value of the machine last year was Rs 2, 00, 000.
Question:18
An alloy contains 40% copper, 32% nickel and rest zinc. Find the mass of zinc in one kg of the alloy.
Solution:
Mass of the alloy = 1 kgPercentage of copper = 40 %Percentage of nickel = 32 %Percentage of zinc = {100 -(40 +32)}%
                     = 28 %
                     
? Mass of zinc in 1 kg of alloy =
28
100
×1 kg
                                     = 0. 28 kg = 0. 28 ×1000 g = 280 g
Question:19
Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in his food daily, find in calories the amount of each of these in his
daily food intake.
Solution:
Amount of protein = 12 % of 2600
                       = 2600 ×
12
100
= 312 cal
Amount of fat = 25 % of 2600
                 = 2600 ×
25
100
= 650 cal
Amount of carbohydrate = 63 % of 2600
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                             = 2600 ×
63
100
= 1638 cal
Question:20
Gunpowder contains 75% nitre and 10% sulphur. Find the amount of gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.5 kg sulphur?
Solution:
Let x be the amount of gunpowder. Amount of nitre = 75 %Let x kg be the amount of gunpowder containing 9 kg of nitre. i. e. , (75 % of x) = 9 kg
  ? x ×
75
100
= 9 ?
75x
100
= 9 ? x = 9 ×
100
75
? x = 12 kg
Hence, 12 kg of gunpowder contains 9 kg of nitre. Now, amount of sulphur = 10 %Let x kg be the amount of gunpowder containing 2. 5 kg of sulphur. i. e. , (10 % of x) = 2. 5 kg
  ? x ×
10
100
= 2. 5 ?
10x
100
= 2. 5 ?
x
10
= 2. 5 ? x = (2. 5 ×10) ? x = 25 kg
Hence, 25 kg of gunpowder contains 2. 5 kg of sulphur.
Question:21
Divide Rs 7000 among A, B and C such that A gets 50% of what B gets and B gets 50% of what C gets.
Solution:
Let  Rs x be the amount of money recieved by C. Then, amount of money B gets = (50 % of Rs x)Amount of money A gets = (50 % of B)                                         = (25 % of Rs x)Now, x +
? C gets Rs 4000. Amount of money B gets = (50 % of Rs x)
                                   
= (50 % of Rs 4000) = Rs 4000 ×
50
100
= Rs 2000
Amount of money A gets = (25 % of Rs x)
                                  
= (25 % of Rs 4000) = Rs 4000 ×
25
100
= Rs 1000
Question:22
Find the percentage of pure gold in 22-carat gold, if 24-carat gold is 100% pure.
Solution:
22 carat gold contains 22 parts pure gold out of 24 parts. Also, 24 carat gold is given to be 100 % pure. ? Percentage of pure gold in 22 carat gold =
22
24
×100 % 
                                                   = 91
2
3
%
Hence, 22 carat gold contains 91
2
3
% of pure gold.
Question:23
The salary of an officer is increased by 25%. By what per cent should the new salary be decreased to restore the original salary?
Solution:
Let the original salary be Rs 100
Then, after increment of 25% the salary becomes
= 100 1 +
25
100
 = 100
125
100
 = Rs 125
To restore the original salary, let the new salary be decreased by x%.
Thus, we get
125 1 -
x
100
 = 100 ? 1 -
x
100
 = 
100
125
 = 
4
5
?
x
100
 = 
1
5
? x = 
100
5
 = 20 %
Therefore, the new salary must be reduced by 20% to restore the original salary.
   
     
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