8
PERCENTAGE - PROFIT
AND LOSS
1. 36% of the students in a school are girls. If the number of boys is
1440, find the total strength of the school.
Ans.
Let total number of students in the school be x
Total number of boys in school = 1440
36
x
Number of girls in school = 36% of x =
100
36 x 64 x
= 1440
∴ Number of boys in school = x −
=
100 100
1440 × 100
Thus,
x=
= 2250
64
Hence, number of total students is 2250
2. Reeta saves 18% of her monthly salary. If she spends Rs 10250 per
month, what is her monthly salary?
Ans. Monthly saving of Reeta = 18%
and monthly expenditure of Reeta = Rs 10250
Let monthly income of Reeta be Rs x
18
x
then, saving of Reeta = 18% of x =
100
18
82
x=
x
and expenditure of Reeta = x −
100
100
10250 × 100
82
x = 10250 ⇒ x =
Thus
82
100
x = 12500
⇒
Hence, monthly income of Reeta is Rs 12500.
3. In an examination, a student has to secure 40% marks to pass. Rohit
gets 178 marks and fails by 32 marks. What is the maximum marks?
Math
VIIIgets marks = 178
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Ans.Class
Rohit
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Marks by which Rohit fails = 32
Thus pass marks for Rohit = 178 + 32 = 210
But pass marks is 40%
∴ 40% of total marks = 210
210 × 100
= 525
Total marks =
40
Hence, 525 is the maximum marks.
4. In a straight contest, the loser polled 42% votes and lost by 14400
votes. Find the total number of votes polled. If the total number of
eligible voters was 1 lac, find what percentage of voters did not vote.
Ans. Since the losing candidate secured 42% of the votes polled
Thus winning candidate secures
= (100 – 42)% of the votes polled = 58% of the votes polled
Difference of votes = 58% – 42% = 16% of the votes polled
Thus 16% of the votes polled = 14400
16
of the votes polled = 14400
⇒
100
100
14400
×
votes
polled
=
= 900 × 100 = 90000
⇒
16
Total number of eligible voters = 100000
(Given)
Number of voter who did not vote = eligible voters – votes polled
= 100000 – 90000 = 10000
Thus, percentage of the number of voters who did not vote
 10000

10000
×
100
% = 10%
=
% =
1000
 100000

Hence, 10% of voters did not vote.
5. Out of 8000 candidates, 60% were boys. If 80% of the boys and
90% of the girls passed the exam, find the number of candidates
who failed.
Ans. Total number of candidates = 8000
Number of boys candidates = 60% of 8000
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60
× 8000 = 60 × 80 =4800
100
Number of girls candidates = 8000 – 4800 = 3200
Number of passed boys = 80% of number of boys
=
80
× 4800 = 80 × 48 = 3840
100
Number of passed girls = 90% of number of girls
=
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90
× 3200 = 90 × 32 = 2880
=
100
Thus number of passed candidates = 3840 + 2880 = 6720
and number of failed candidates = 8000 – 6720 = 1280
Hence, 1280 candidates are failed.
6. Out of 120 employees in a company, 80% are males. Eighty male
employees and three-fourth of the female employees are married. In
aggregate, what percent of company’s employees are married?
Ans. Total number of employees in the company = 120
Number of males employees in the company = 80% of 120
80
× 120 = 96
100
Number of female employees in the company = 120 – 96 = 24
Number of married employees in the company
=
3

= 80 +  of 24  = 80 + 18 = 98
4

Percentage of married employees in the company
98
98
× 100% = × 10%
=
120
12
98
49
245
2
× 5% =
× 5% =
= 81 %
=
6
3
3
3
Math Class VIII
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2
Hence, 81 % of company’s employees are married.
3
7.(i) A number 3.625 is wrongly read as 3.265; find the percentage error.
(ii) A number 5.78 × 103 is wrongly written as 5.87 × 103 ; find the
percentage error.
Ans. (i) Correct number = 3.625
Number wrongly read = 3.265
Then error = (3.625 – 3.265) = 0.360
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0.360
×100%
3.625
360
36000
× 100% =
% = 9.93%
=
3625
3625
Hence, percentage error is 9.93%.
(ii) Correct number = 5.78 × 103
Number wrongly written as = 5.87 × 103
Thus, error = 5.87 × 103 – 5.78 × 103 = 0.09 × 103
Thus, percentage error =
0.09 × 103
0.09
100%
×
× 100%
=
5.78
5.78 × 103
9
900
=
× 100% =
% = 1.56%
578
578
Hence, percentage error is 1.56%.
8. Bhola Ram gave 35% of his money to his elder son and 40% of the
remainder to the younger son. Now, he is left with Rs 11700. How
much money had he?
Ans. Let Bhola Ram had Rs x.
Thus percentage error =
Amount given to elder son = 35% of x =
35 x
100
35 x  65 x

x
−
Remaining amount = Rs 
=
100

 100
Math Class VIII
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65 x 40
26
×
=
x
100 100 100
 65 x 26  39
x =
x
−
Amount left with him = Rs 
 100 100  100
39
11700 × 100
Thus,
x = 11700 ⇒ x =
100
39
⇒ x = 30000
Hence, total amount Bhola Ram had = Rs 30000.
9. 5% of the population of a town were killed in an earthquake and 8%
of the remainder left the town. If the population of the town now is
43700, what was its population at the beginning?
Ans. Let population of the town at the beginning be x%
People killed in an earthquake = 5% of the population
Amount given to his younger son =
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Thus, total people killed = x ×
5
5x
=
100 100
5 x 95 x
=
100 100
95 x 8
76
×
=
x
Those people who left the town =
100 100 1000
95 x
76
−
x
Hence, remaining population of the town =
100 1000
950 x – 76 x 874
=
x
=
1000
1000
43700 × 1000
874
x
=
= 50000
⇒
∴
x = 43700
874
1000
Hence, population of the town at the beginning was 50000.
10. A and B are two towers. The height of tower A is 20% less than that
of B. How much per cent is B’s height more than that of A?
Ans. Let height of tower B = 100 m
Since, height of tower A is 20% less than B
Then remaining population of the town = x −
Math Class VIII
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20 

∴ Height of tower A = 100 −  100 ×
 = 100 – 20m = 80m
100


Thus, height of B which is more than A’s = (100 – 80)m = 20m
20 × 100
= 25%
Percent of B’s height more than that of A =
80
Hence, B’s height is 25% more than A’s height.
11. In an examination, 30% of the candidates failed in English, 35%
failed in GK and 27% failed in both the subjects. If 310 candidates
passed in both, how many candidates appeared in the examination?
Ans. Let total number of candidates who appeared in examination
= 100
Failed candidates in English = 30% of 100 = 30
Failed candidates in GK = 35% of 100 = 35
Failed candidates in both = 27% of 100 = 27
Thus, total failed candidates = 30 + 35 – 27 = 65 – 27 = 38%
∴ Total passed candidates = (100 – 38) = 62.
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100
× 310 = 500
62
Hence, 500 candidates appeared in the examination..
12. In a combined test in History and Civics; 36% candidates failed in
History; 28% failed in Civics and 12% in both; find :
(i) the percentage of passed condidates
(ii) the total number of candidates appeared, if 208 candidates have
failed.
Ans. Candidates failed only in History = 36% – 12% = 24%
Candidates failed only in Civics = 28% – 12% = 16%
Candidates failed in both subjects = 12%
Total failed candidates = 24% + 16% + 12% = 52%
(i) Thus percentage of passed candidates = 100% – 52% = 48%
Hence total candidates =
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(ii) If failed candidates are 52,
then total appeared candidates = 100
If failed candidate is 1, then total appeared candidates =
100
52
If failed candidates are 208,
100
× 208
52
= 100 × 4 = 400
13. Nandani purchased some parrots. 20% flew away and 5% died. Of
the remaining, 45% were sold. Now 33 parrots remain. How many
parrots had Nandani purchased?
Ans. Let Nandani purchased x parrots.
Then number of parrots flew away = 20% of x
then total appeared candidates
=
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20
1
x
×x = ×x=
100
5
5
5
x
×x=
Number of died parrots = 5% of x =
100
20
x x 
 4x + x 
x
–
+
x
−
Number of remaining parrots =

=


 5 20 
 20 
5x
x 4 x − x 3x
= x− =
=
= x−
20
4
4
4
3x
45 3x
×
Number of parrots sold = 45% of
=
4 100 4
9 3x 27 x
× =
=
20 4
80
Number of those parrots which are not sold
=
=
3 x 27 x 60 x − 27 x 33x
−
=
=
4
80
80
80
As per condition,
Math Class VIII
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33x
33x = 33 × 80
= 33 ⇒
80
33 × 80
x=
x = 80
⇒
⇒
33
Hence, Nandani purchased 80 parrots.
14. A lunch interval of half an hour is 5% of total office hours. Calculate
(i) the total office hours
(ii) the working hours
1
Ans. (i) 5% of total office hours = hour
2
5
1
of total office hours = hour
⇒
100
2
100
hours = 10 hours.
⇒ total office hours =
5× 2
1
(ii) Working hours = 10 hours – hours
2
1

19
1
10
−
hours = 9 hours.
=
 hours =
2
2
2

15. Price of bananas has changed from 5 for a rupee to 4 for a rupee.
Find the percentage increase in the price.
Ans. Original price of 5 bananas = Rs 1
1
Original
price
of
1
banana
=
Rs
∴
5
Changed price of 4 bananas = Rs 1
1
Changed
price
of
1
banana
=
Rs
∴
4
1
1 1
5−4
Increased price = Rs  −  = Rs 
=
Rs

20
 4 5
 20 
1
20
Hence, percentage increased price = 1 × 100%
5
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Math Class VIII
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Question Bank
1 5
× × 100%
20 1
1
= × 100% = 25%
4
16. Two numbers are respectively 25% and 35% less than a third number.
What percent is the second number of the first?
Ans. Let the third number be x
=
Z
B
25 
75
3

x= x
Then, first number =  1 −
x =
100
4
 100 
35 
65
13

x
=
x
x
And, second number =  1 −
=

100
20
100


13
x
13
4
20
x × × 100%
× 100% =
Thus, required percentage =
3
20
3x
x
4
13
13
260
× 100% = × 20% =
%
=
5×3
3
3
2
= 86 %
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17. 75% of the students in a class VIII passed an exam. If 2 more students
had passed the exam, 80% would have been successful. How many
students were there in the class VIII?
Ans. Let the total number of students in the class VIII be x,
Number of passed students in the calss VIII = 75% of x
75
75 x
×x=
=
100
100
As per condition,
75 x
+ 2 = 80% of x
100
75 x
80 x
75 x
80
+2=
⇒
+2=
×x ⇒
100
100
100
100
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Math Class VIII
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Question Bank
80 x 75 x
5x
−
⇒
2=
100 100
100
x
x
=2
⇒
⇒
2=
20
20
⇒
⇒
x = 2 × 20
x = 40
Hence, the total number of students in class VIII were 40.
18. The monthly salary of a school teacher in 2005 was Rs. 12000. It
increased by 10% in 2006 and again by 10% in 2007. What is his
salary in 2007?
Ans. Monthly salary of the school teacher in 2005 = Rs 12000
⇒
2=
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10 

Monthly salary of the school teacher in 2006 = Rs  1 +
 × 12000
100


11
 110 
× 12000
= Rs 
=
Rs
×
1200

10
100


= Rs 11 × 1200 = Rs 13200
Monthly salary of the school teacher in 2007
10 

1
+
=
 Rs 13200
 100 
11
 110 
× 13200
13200
×
=
Rs
= Rs 

10
 100 
= Rs 11 × 1320 = Rs 14520
Hence, monthly salary of the school teacher in 2007 = Rs 14520
19. Dolly’s height increased by 20% last year and by 15% this year. What
is the total percent increase in 2 years?
Ans. Let the original height of Dolly be x
20 
120
120 x

×
x
×
x
=
=
Increased height in last year =  1 +

100
100
 100 
15  120 x

Increased height in this year =  1 +
×
 100  100
Math Class VIII
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Question Bank
115 120 x 23 6 x 138 x
×
=
×
=
100 100 20 5
100
138 x
138 x − 100 x 38 x
−x =
=
Total increases in 2 years =
100
100
100
38 x
Total percentage increases in 2 years = 100 × 100%
x
38 x
× 100% = 38%
=
100 × x
Hence, 38% increase in 2 years.
20. Price of a commodity decreased by 10% last year and increased by
20% this year. Find the percentage change in two years.
Ans. Let the price of commodity be Rs x
When the price of commodity is decreased by 10% in last year
=
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10 

Then price of commodity in last year =  1 −
 × Rs x
100


90
9x
x = Rs
= Rs
10
10
When the price of commodity increased by 20% in this year
9x
20 

Then price of commodity in this year =  1 +
 × Rs 10
 100 
 1  9x
= Rs  1 +  ×
 5  10
6 9x
27
x
= Rs ×
= Rs
5 10
25
Increased of price of commodity in these 2 years
27 x
27 x – 25 x 2 x
–x=
=
=
25
25
25
Hence, percentage increased in these 2 years
Math Class VIII
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Question Bank
2x
= 25 × 100% = 2 x × 100%
x
25 × x
= 2 × 4% = 8%
21. The length and the breadth of a rectangle are 10 cm and 8 cm. If its
length increases by 15% and breadth by 20%, find the percentage
increase in its area.
Ans. Length of rectangle increases by 15% and the breadth increases by
20% (given)
15 

New length of rectangle =  1 +
 of 10 cm
100


115
115
× 10cm =
cm = 11.5 cm
=
100
10
20 

New breadth of rectangle =  1 +
 of 8 cm
 100 
120
6
× 8cm = × 8cm
=
100
5
48
cm = 9.6cm
=
5
Thus, new area of rectangle = (11.5 × 9.6) cm2 = 110.4 cm2
Original area of rectangle = (10 × 8) cm2 = 80 cm2
Thus, increase in area = (110.4 – 80) cm2 = 30.4 cm2
 30.4

 30.4 
× 100  % = 
× 5 %
∴ Percentage increase in area = 
 80

 4

= (7.6 × 5)% = 38%
Hence, the area of the rectangle is increased by 38%.
22. Two numbers are respectively 20 percent and 50 percent more than a
third number. What percent is the second of the first?
Ans. Let the third number be x.
20
100 x + 20 x 120 x
x
+
x
=
First
number
=
=
∴
100
100
100
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Math Class VIII
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Question Bank
50
100 x + 50 x 150 x
x=
=
100
100
100
150 x
150 x 100
100 ×
×
× 100 %
Thus, required percent =
100% =
120 x
100 120 x
100
150 × 100
1500
=
% = 125%
%=
120
12
23. Two numbers are respectively 30 percent and 40 percent less than a
third number. What percent is the second of the first?
Ans. Let the third number be x
Second number = x +
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30 x 100 x − 30 x 70 x 7 x
=
=
=
100
100
100 10
40 x 100 x − 40 x 60 x 6 x
=
=
and second number = x −
=
100
100
100 10
6x
Thus, required percentage = 10 × 100
7x
10
6 x 10
600
5
× × 100 =
= 85 %
=
10 7 x
7
7
5
Hence, 85 percentage of the first is the second.
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24. During 2006, the production of a publication factory decreased by
25%. But, during 2007, it (production) increased by 40% of what it
was at the beginning of 2007. Calculate the resulting change (increase
or decrease) in production during these two years.
Ans. Let at the begining of 2006, production = 100
decrease in production = 25%
Thus, new production = 100 – 25 = 75
In 2007, it is increased by 40%
∴ First number = x −
Math Class VIII
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Question Bank
75 × 40
= 30
100
∴ New production = 75 + 30 = 105
∴ Resulting change in two years (increase) = 105 – 100 = 5
Hence, precentage change in increase in production
5
× 100% = 5%
=
100
25. Motilal bought a certain number of oranges at Rs 60 per score and
sold them at 20% profit. Find the selling price of each orange.
Ans. Cost price of 20 oranges = Rs 60
(∵ 1 score = 20)
60
= Rs 3
∴ Cost price of 1 orange =
20
Profit = 20%
3 × 120 360
=
= Rs 3.60
∴ Selling price of 1 orange =
100
100
26. Nakvi sells two articles for Rs 4,500 each, making 25% profit on
one and 25% loss on the other. Find:
(i) CP of each article.
(ii) total CP of both the articles.
(iii) total SP of both the articles.
(iv) profit % or loss % on the whole.
Ans. (i) Selling price of two articles = Rs 4500 each.
∴ Increase in production =
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100 × 4500 100 × 4500
=
100 + 25
125
4
= × 4500 = 4 × 900 = Rs 3600
5
100 × 4500 100 × 4500
=
Cost price of other =
100 − 25
75
4
= × 4500 = 4 × 1500 = Rs 6000
3
Cost price of one =
Math Class VIII
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Question Bank
(ii) Total cost price of both the articles = Rs 3600 + Rs 6000
= Rs 9600
(iii) Total selling price of both the articles = Rs 2(4500) = Rs 9000
(iv) Loss = Rs (9600 – 9000) = Rs 600
600
100
25
× 100% =
% = % = 6.25%
Loss % =
9600
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27. A wine bottle was sold at a loss 8%. Had it been sold for Rs 56
more, there would have been a gain of 8%. What is the cost price of
the wine bottle?
Ans. Let cost price of the wine bottle be Rs x
8
2x
×x=
Loss = 8% of x =
100
25
2 x 25 x − 2 x 23 x
=
=
∴ Selling price of the wine bottle = x −
25
25
25
8
2x
×x=
Now, gain = 8% of x =
100
25
2x
Thus, selling price of the wine bottle = x +
25
25 x + 2 x 27 x
=
=
25
25
4x
27 x 23 x
27 x − 23 x
= 56,
–
= 56 ⇒
= 56 ⇒
∴
25
25
25
25
56 × 25
, ⇒ x = 350.
⇒ x=
4
Hence, cost price of the wine bottle = Rs 350.
28. The selling price of 18 mangoes is equal to the cost price of 21
mangoes. Find the gain or loss per cent.
Ans. Let cost price of each mangoes = Re 1
Cost price of 18 mangoes = Rs 18
Selling price of 18 mangoes = Cost price of 21 mangoes = Rs 21
Gain = Rs (21 – 18) = Rs 3
3
50
2
× 100% = % = 16 %
Hence,
gain
%
=
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29. The cost price of 12 fans is equal to the selling price of 16 fans. Find
the gain or loss per cent.
Ans. Let selling price of one fan = Rs 100
∴ Selling price of 16 fans = Rs 100 × 16 = Rs 1600
and cost price of 12 fans = Rs 1600
1600
400
= Rs
∴ Cost price of one fan =
12
3
Loss = cost price – selling price
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100
 400 100  400 − 300
−
=
= Rs 
=
Rs

3
1 
3
 3
100
× 100
Loss × 100
3
%=
%
Loss percent =
400
cost price
3
100 × 100 × 3
% = 25%
=
3 × 400
30. A man buys two bats, one for Rs. 340 and the other for Rs 260. He
sells the first bat at a gain of 15% and the second one at a loss of
15%. Find the gain or loss per cent in the whole transaction.
Ans. Cost price of first bat = Rs 340
15
× 340 = Rs 51
100
Selling price of first bat = Rs (340 + 51) = Rs 391
Now, cost price of second bat = Rs 260
15
Loss = 15% of Rs 260 = Rs
× 260 = Rs 39
100
Selling price of second bat = Rs (260 – 39) = Rs 221
∴ Total selling price of two bats = Rs (391 + 221) = Rs 612
Gain = Rs (612 – 600) = Rs 12
12
× 100% = 2%
Hence, gain% =
600
Hence, his gain in whole transaction is 2%.
Gain = 15% of Rs 340 = Rs
Math Class VIII
16
Question Bank
31. Nandlal bought 20 dozen note-books at Rs 48 per dozen. He sold 8
dozens of them at 10% gain and the remaining 12 dozens at 20%
gain. What is his gain percent in the whole ransaction?
Ans. Cost price of 1 dozen note-books = Rs 48
Cost price of 20 dozen note-books = Rs 48 × 20 = Rs 960
Cost price of 8 dozen note-books = Rs 48 × 8 = Rs 384
10
× 384 = Rs38.40
Gain = 10% of Rs 384 = Rs
100
Selling price of 8 dozen note-books = Rs (384 + 38.40) = Rs 422.40
Now, cost price of 12 dozen note-books = Rs 48 × 12 = Rs 576
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20
× 576 = Rs 115.20
100
Thus, selling price of 12 dozen note-books
= Rs (576 + 115.20) = Rs 691.20
Selling price of 20 dozen note-books
= Rs (422.40 + 691.20) = Rs 1113.60
∴ Whole gain = Rs (1113.60 – 960) = Rs 153.60
Gain % = 20% of Rs 576 = Rs
153.60
× 100% = 16%
960
Hence, his gain percent in whole transaction is 16%.
32. A shopkeeper buys a certain number of pens. If the selling price of 5
pens is equal to the cost price of 7 pens, find his profit or loss
percentage.
Ans. Let the cost price of 7 pens be Rs x.
Gain% =
Cost price of 1 pen = Rs
x
7
As per condition,
Selling price of 5 Pens = Rs x
Selling price of 1 Pen = Rs
Math Class VIII
17
x
5
Question Bank
Profit = selling priec – cost price = Rs
x
x
– Rs
5
7
2x
 7 x − 5x 
= Rs 
=
Rs

35
 35 
2x
Profit
× 100% = 35 × 100%
Profit % =
x
cost price
7
2x 7
2
× × 100% = × 100% = 2 × 20% = 40%
=
35 x
5
Hence, profit percentage is 40%.
33. Coffee costing Rs 100 per kg was mixed with chicory costing Rs. 50
per kg in the ratio 5 : 2 for a certain blend. If the mixture was sold at
Rs. 90 per kg. Find the gain or loss per cent.
Ans. Cost price of 1 kg of coffee = Rs 100
Cost price of 5 kg of coffee = Rs 5 × 100 = Rs 500
Cost price of 1 kg of chicory = Rs 50
Cost price of 2 kg of chicory = Rs 2 × 50 = Rs 100
∴ Cost price of (5 + 2) kg of mixture
= Rs (500 + 100) = Rs 600
Now, selling price of 1 kg of mixture = Rs 90
Selling price of 7 kg of mixture = Rs 7 × 90 = Rs 630
Gain = Rs (630 – 600) = Rs 30
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30
× 100% = 5%
600
34. A sells a bicycle to B at a profit of 20% and B sells it to C at a profit
of 5%. If C pays Rs 1890, what did A pay for it?
Ans. C’s cost price for the cycle = Rs 1890
or B’s selling price = Rs 1890
gain = 5%
Hence, gain percent =
Math Class VIII
18
Question Bank
∴ B’s cost price =
cost price × 100
1890 × 100
= Rs
100 + gain%
100 + 5
1890 × 100
= Rs 1800
105
or A’s selling price = Rs 1800
A’s gain = 20%
= Rs
Z
B
1800 × 100
1800 × 100
= Rs
= Rs 1500
100 + 20
120
Hence, A will pay Rs 1500 for bicycle.
35. Roma sold a watch to Monu at 12% gain and Monu had to sell it to
Lavish at a loss 5%. If Lavish paid Rs 1330, how much did Roma
pay for it?
Ans. Let cost price of the watch by ‘Roma’ be Rs x
12
3x
×x=
Profit = 12% of x =
100
25
3 x 25 x + 3x 28 x
=
Selling price of the watch by ‘Roma’ = x +
=
25
25
25
28 x
∴ Cost price of the watch by ‘Monu’ =
25
28 x
5 28 x 7 x
=
×
=
Loss = 5% of
25 100 25 125
28 x 7 x
−
Selling price of the watch by ‘Monu’ =
25 125
140 x − 7 x 133 x
=
=
125
125
133x
∴ Cost price of the watch by ‘Lavish’ =
125
133 x
1330 × 125
= 1330, x =
Thus,
125
133
x = 1250
∴
Hence, cost price of the watch for Roma is Rs 1250.
∴ A’s cost price = Rs
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Question Bank
36. Dhani Ram purchased a cow for Rs 6000 and a buffalo for
Rs 12000. He sold the buffalo at a profit of 15% and the cow at a
loss of 10%. Calculate his overall profit or loss percentage.
Ans. Cost price of cow = Rs 6000
and cost price of buffalo = Rs 12000
Thus, total cost price of cow and buffalo = Rs (6000 + 12000)
= Rs 18000
Dhani Ram sold the cow at the loss of 10%, then selling price of cow
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10 

= 1 −
 of (cost price of cow)
 100 
90
10 

× 6000
6000
×
=
Rs
= Rs  1 −

100
 100 
= Rs 90 × 60 = Rs 5400
Dhani Ram sold the buffalo at the profit of 15%
15 

then, Selling price of buffalo =  1 +
 of (cost price of buffalo)
100


15 

= Rs  1 +
 × 12000
100


115
× 12000
= Rs
100
= Rs 115 × 120 = Rs 13800
∴ Selling price of cow and buffalo = Rs (5400 + 13800)
= Rs 19200
Thus, profit = selling price – cost price
= Rs (19200 – 18000) = Rs 1200
Hence,
Math Class VIII
 Profit

× 100  %
profit% = 
 Cost price

 1200

× 100  %
=
 18000

20
Question Bank
1200
120
20
2
%=
%= %=6 %
180
18
3
3
2
Hence, overall percentage of Dhani Ram is 6 % .
3
37. By selling an almirah for Rs 3920, a shopkeeper would gain 12%. If
it is sold for Rs 4375, find his gain or loss percentage.
Ans. When selling price of almirah = Rs 3920 and gain = 12%
then cost price P = ?
=
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12 

We know that, selling price, SP =  1 +
 of CP
 100 
 100 + 12 
Rs
3920
=

 of CP
⇒
 100 
3920
× 100
CP = Rs
⇒
112
= Rs 35 × 100 = Rs 3500
When cost price of almirah = Rs 3500
and selling price of almirah = Rs 4375
Then, gain = SP – CP = Rs (4375 – 3500) = Rs 875
 gain

× 100  %
Hence, gain% = 
 CP

875
 875

× 100  % =
% = 25%
=
3500
35


38. By selling a bicycle at Rs 1334, a shopkeeper would suffer a loss of
8%. At how much amount should he sell it to make a profit of
1
12 % ?
2
Ans. Selling price of bicycle = Rs 1334
Shopkeeper would suffer a loss % = 8%
Cost price = ?
we know that selling price = (1– loss%) of CP
Math Class VIII
21
Question Bank
8 

−
1
Rs
1334
=

 × CP
⇒
 100 
1334
× 100
⇒ Cost price of bicycle = Rs
92
1334
× 25
⇒ Cost price of bicycle = Rs
23
⇒ Cost price of bicycle CP = Rs 58 × 25 = Rs 1450
Now, CP = Rs 1450
1
25
= %
Profit on bicycle = 12
2% 2
25 

Selling price of bicycle =  1 +
 × CP
2
100
×


25 

⇒ Selling price of bicycle =  1 +
 × (Rs 1450)
2
×
100


 1
⇒ Selling price of bicycle =  1 +  × 1450
 8
9
⇒ Selling price of bicycle = × 1450
8
⇒ Selling price of bicycle = (9 × 181.25) = Rs 1631.25
39. Sohan bought a certain number of note-books for Rs 600. He sold
1
of them at 5 per cent loss. At what price should he sell the
4
remaining note-books so as to gain 10% on the whole?
Ans.
Cost price of note-books = Rs 600
Gain desired on the whole = 10%
Then total selling price of all the note-books
(100 + gain% ) × CP =  100 + 10 
=

 × Rs 600
100
 100 
110
× 600 = Rs 660
= Rs
100
Math Class VIII
22
Question Bank
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1
1
of the books = × Rs 600 = Rs 150
4
4
Loss on these note-books = 5%
Then, selling price of these note-books
(100 − Loss % )
(100 − 5)
× CP =
× Rs 150
=
100
100
95
14250
× 150 = Rs
= Rs
= Rs 142.50
100
100
Thus, required selling price of the remaining note-books
= Rs(660 – 142.50) = Rs 517.50
40. Sanjay sold a bicycle at 5% profit. If the cost had been 30% less and
the selling price Rs 63 less, he would have made a profit of 30%.
What is the cost price of the bicycle?
Ans. Let cost price of the bicycle = Rs 100
When profit = 5% ;
S.P. = Rs (100 + 5) = Rs 105
30


100
−
× 100  = Rs (100 – 30) = Rs 70
New cost price = 
100


Profit on bicycle = 30%
(100 + Profit )
× cost price
Thus selling price =
100
(100 + 30 )
130
× Rs 70 =
× Rs 70
=
100
100
130 × 70
= Rs 91
= Rs
100
Thus, difference of two selling prices = Rs (105 – 91) = Rs 14
If difference is Rs 14 then cost price of the bicycle = Rs 100
100
If difference is Rs 1 then cost price of bicycle = Rs
14
100 × 63
If difference is Rs 63 then cost price of bicycle = Rs
14
=
Rs
50
×
9
=
Rs
450.
Math Class VIII
23
Question Bank
Cost price of
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41. Mona sold a pressure cooker at a loss of 8%. Had she bought it at
10% less and sold for Rs 88 more, she would have gained 20%.
Find the cost price of the pressure cooker.
Ans. Let the cost price of pressure cooker = Rs 100
loss on pressure cooker = 8%
Then, selling price of pressure cooker = Rs (100 – 8) = Rs 92
Again, cost price of pressure cooker = Rs (100 – 10) = Rs 90
gain = 20%
C.P. × (100 + gain%)
Thus, selling price of pressure cooker =
100
90(100 + 20)
= Rs
100
90 × 120
= Rs
= Rs 108
100
∴ Difference in two selling prices (108 – 92) = Rs 16
If difference is Rs 16, then cost price = Rs 100
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100 × 88
= Rs 550
16
Hence, cost price of pressure cooker is Rs 550.
42. The manufacturing price of a T.V. set was Rs 5000. The company
sold it to a distributor at 16% profit. The distributor sold it to a
dealer at 10% profit. The dealer sold it to a customer at 20% profit.
Find the price the customer paid.
Ans. The manufacturing price of T.V. set = Rs 5000
Profit = 16%
16 

Selling price of T.V. set =  1 +
 of Rs 5000
100


 100 + 16 
=
 of Rs 5000
 100 
116
116
× Rs 5000 = Rs
× 5000
=
100
100
= Rs 116 × 50 = Rs 5800
and if difference is Rs 88, then cost price =
Math Class VIII
24
Question Bank
Thus, cost price of T.V. set for distributor = Rs 5800
Now, Profit = 10%
Selling price of T.V. set for distributor
10 

1
+
=
 of Rs 5800
 100 
10 

110
1
+
× 5800
=
 of Rs 5800 = Rs
100
 100 
= Rs 110 × 58 = Rs 6380
Thus, cost price of T.V. set for dealer = Rs 6380
Again profit = 20%
Selling price of T.V. set for dealer
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20 

= 1 +
 of Rs 6380
 100 
 100 + 20 
=
 of Rs 6380
 100 
120
120
× 6380
of Rs 6380 = Rs
=
100
100
= Rs 12 × 638 = Rs 7656
Hence, the price the customer paid = Rs 7656.
43. A sells an article to B at a profit of 20% and B sells it to C at a loss of
6%. If C pays Rs 846, find how much did A pay for it.
Ans. Let the amount A pays for the article be x
when A sells to B at a profit of 20%
20 

Then, cost price for B or the selling price for A =  1 +
 cost
100


price for A.
2
12
12 x

=  1 +  × Rs x = Rs × x = Rs
10
10
 10 
12 x
i.e. cost price of the article for B = Rs
10
Math Class VIII
25
Question Bank
But, B sells to C at a loss of 6%
Then, cost price for C or the selling price for B
12 x
6 

6 

= 1 −
of
cost
price
for
B
=
of
Rs
1
−



10
 100 
 100 
3 
12 x
564 x

47 12 x
=  1 −  of Rs
= Rs
×
= Rs
10
500
50 10
 50 
564
x = Rs 846
But, C pays Rs 846, for an article then,
500
500
250
x = Rs 846 ×
= Rs 846 ×
⇒
564
282
= Rs 3 × 250 = Rs 750
Hence, cost price of an article for A = Rs 750
44. Saurav sold an article at a profit of 12%. Had it been sold for
Rs 16 more, the profit would have been 20%. Find, the selling price
of the article.
Ans. Let the selling price of an article be Rs x,
Then, profit on article = 12%
P 

1
+
Selling
price
=
∴

 of cost price
100


12 

Rs x =  1 +
∴
 of cost price
 100 
 112 
=
 of cost price
 100 
100 x
100
⇒
Cost price = Rs
×x=
112
112
To make 20% profit,
20 

Selling price =  1 +
 of cost price
 100 
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Question Bank
6
 1
=  1 +  of cost price = of cost price
5
 5
100 x
6
 6 100 x 
= Rs  ×
= of Rs

112
5
 5 112 
120
 6 × 20 x 
x
= Rs 
=
Rs

112
112


As per condition,
120
120
x = x + 16
x − x = 16
⇒
112
112
8x
120 x − 112 x
= 16
= 16
⇒
⇒
112
112
16 × 112
= Rs 2 × 112 = Rs 224
⇒ Rs
8
Hence, the selling price of the article is Rs 224.
45. After allowing a discount of 15% a navycut was sold for Rs 578.
Find its marked price.
Ans. Let market price of the navycut be Rs x.
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15 x
3x
× x = Rs
= Rs
100
100
20
3x 

Selling price of the navycut = Rs  x − 
20 

17 x
 20 x − 3 x 
= Rs 
=
Rs

20
 20 
578 × 20
17 x
⇒ x = 680.
∴
= 578 ⇒ x =
17
20
Hence, market price of the navycut is Rs 680.
46. A coat was bought for Rs 435 after getting a discount of 13%. Find
the marked price of the coat.
Ans. Let marked price of the coat be Rs x
Discount = 15% of Rs x = Rs
Math Class VIII
27
Question Bank
Discount = 13% of Rs x
13
13 x
× x = Rs
= Rs
100
100
13x 

x
−
Then, selling price of the coat = Rs 

100 

87 x
 100 x − 13x 
= Rs 
=
Rs

100
100


87 x
435 × 100
= 435 ⇒ x =
⇒ x = 500
100
87
Hence, marked price of the coat is Rs 500.
47. A shopkeeper buys a Tea set for Rs 1200 and marks it 80% above the
cost price. If he gives 15% discount on it, find:
(i) the marked price
(ii) the selling price
(iii) his profit percentage
Ans. (i) Cost price of a tea set = Rs 1200
Marked price of tea set
= Rs1200 + 80% of (Rs 1200)
80


× 1200 
= Rs 1200 +
100


= Rs [1200 + 80 × 12]
= Rs [1200 + 960] = Rs 2160
(ii) Marked price = Rs 2160, discount = 15%
Selling price = ?
d 

Selling price =  1 −
 of marked price
 100 
15 

Selling price =  1 −
 × Rs 2160
100


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Math Class VIII
28
Question Bank
85
× 2160
100
17
× 2160
Selling price = Rs
20
Selling price = Rs 17 × 108 = Rs 1836
(iii) Profit = selling price – cost price
= Rs (1836 – 1200) = Rs 636
 Profit

× 100  %
Profit % = 
 Cost price

Selling price = Rs
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 636

636
× 100  % =
% = 53%
=
12
 1200

48. The cost price of a Reliance mobile is Rs 1600, which is 20% below
the marked price. If the article is sold at a discount of 16%, find:
(i) the marked price
(ii) the selling price
(iii) profit percentage.
Ans. (i) Cost price of Reliance mobile = Rs 1600
Since the cost price of Reliance mobile is 20% below the marked
price.
Let the marked price of Reliance mobile = Rs x
Then, Cost price = Marked price – 20% of Marked price
Rs 1600 = x – 20% of x
⇒
20
×x
Rs 1600 = x −
⇒
100
80 x
Rs 1600 =
⇒
100
100
1600
×
x
=
Rs
⇒
80
= Rs 20 × 100 = Rs 2000
Hence, the marked price of Reliance mobile = Rs 2000
Math Class VIII
29
Question Bank
(ii) Marked price of Reliance mobile = Rs 2000, discount = 16%,
Selling price = ?
16 

Selling price =  1 −
 of marked price
 100 
 100 − 16 
=
 of Rs 2000
100


84
× 2000 = Rs 84 × 20 = 1680
= Rs
100
(iii) Profit = selling price – cost price = Rs (1680 – 1600) = Rs 80
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 Profit

× 100  %
Profit % = 
 Cost price

80
 80

% = 5%.
× 100  % =
=
16
 1600

49. An umbrella was marked 40% above cost price and a discount of
35% was given on its marked price. Find the gain or loss per cent
made by the shopkeeper.
Ans. Let cost price of the umbrella be Rs x.
Marked price of the umbrella = Rs (x + 40% of x)
40
2x 



× x  = Rs  x + 
= Rs  x +
100 
5 


7x
 5x + 2x 
= Rs 
 = Rs 5
 5 
7x
Discount on umbrella = 35% of Rs
5
35
7x
49 x
= Rs
×
= Rs
100
5
100
 7 x 49 x 
–
Thus, selling price of the umbrella = Rs 

5
100 

Hence,
Math Class VIII
30
Question Bank
91x
 140 x – 49 x 
= Rs 
 = Rs
100
100


100 x – 91x
9x
91x 

Loss = Rs  x –
=
Rs
=
Rs

100
100
100 

9x
Hence, Loss% =
× 100% = 9%
100 × x
50. Mr Kabuliwala purchased a washing machine for Rs 7660. After
allowing a discount of 12% on its marked price he sells it at a gain
of 10%. Find the marked price.
Ans. Let marked price of the washing machine be Rs x.
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12
3x
× x = Rs
100
25
Thus, selling price of the washing machine
Discount = 12% of Rs x = Rs
22 x
3x 

 25 x – 3 x 
=
Rs
= Rs  x –  = Rs 

25
25 

 25 
Cost price of the washing machine = Rs 7660
Gain = 10%
Thus, selling price of the washing machine
100 + 10
110
× 7660 = Rs
× 7660 = Rs 8426
100
100
22 x
If selling price Rs
, then marked price = Rs x
25
x × 25
If selling price Re 1, then marked price = Rs
22 x
25
If selling price Rs 8426, then marked price = Rs
× 8426
22
= Rs 9575
51. Find a single discount equivalent to two successive discounts of
40% and 5%.
= Rs
Math Class VIII
31
Question Bank
Ans. Let marked price = Rs 100
Rate of first discount = 40%
and rate of second discount = 5%
marked price × (100 – discount%)
Thus,
selling price =
100
100 × (100 – 40) × (100 – 5)
=
100 × 100
100 × 60 × 95
=
= Rs 57
100 × 100
Thus, total discount = Rs (100 – 57) = Rs 43
Hence, rate of single discount is 43%
52. Find a single discount equivalent to three successive discounts of
20%, 5% and 1%.
Ans. Let marked price = Rs 100
Rate of first discount = 20%
Rate of second discount = 5%
Rate of third discount = 1%
marked price × (100 – discount%)
Selling price =
∴
100
100(100 – 20) × (100 – 5) × (100 – 1)
=
100 × 100 × 100
100 × 80 × 95 × 99 7524
=
=
= Rs 75.24
100 × 100 × 100
100
Thus, total amount of discount = Rs (100 – 75.24) = Rs 24.76
Hence, single discount = 24.76%.
53. If after giving a discount of 10%, a shopkeeper still makes a profit of
12.5%, find how much above the cost price he had marked the prices.
Ans. Let the cost price of the article be Rs x.
Profit is 12.5%
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Math Class VIII
32
Question Bank
 12.5 
Selling price of article =  1 +
 of Rs x
 100 
112.5
 100 + 12.5 
x
of
Rs
x
=
Rs
=

100
 100 
Shopkeeper allows a discount of 10%
10 

Selling price =  1 –
 of marked price
 100 
112.5 x
90
=
of marked price
⇒
100
100
112.5 x
9
=
of marked price
⇒
100
10
10 112.5 x
Marked price =
×
⇒
9
100
1 112.5 x 112.5
x
= ×
=
9
10
90
Thus, the excess value to be marked = marked price – cost price
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112.5 x
 112.5 x

– x
– Rs x = Rs 
= Rs
90
 90

 112.5 x – 90 x 
22.5 x
= Rs 
 = Rs
90
90


Thus, the percentage of excess value to be marked
 excess value

× 100  %
=
 cost price

 22.5

x
 90

 22.5

× 100  %
×
100
%
=
 =
 90

 x



Math Class VIII
33
Question Bank
225
% = 25%
9
Hence, the shopkeeper marks the article above the cost price by 25%.
=
54. If a shopkeeper marks his goods at 20% above the cost price and
then gives 20% discount, find his gain or loss percentage.
Ans. Let the cost price of an article be Rs x.
Since the dealer marks his goods 20% above the cost price.
Marked price = Cost price + 20% of Cost price
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20
of Rs x = Rs
= Rs x +
100
= Rs
20 x 

x
+


100 

120 x
6x
= Rs
100
5
d 

Selling price =  1 –
 of Marked price
100


6x
80
6x
96 x
20 

= 1 –
of
Rs
=
Rs
×
=
Rs

5
100
5
100
 100 
Loss = cost price – selling price
= Rs x – Rs
96 x
= Rs
100
96 x 

x–

100 

4x
x
 100 x – 96 x 
= Rs 
=
Rs
=
Rs

100
25
100


 Loss

× 100  %
Loss% = 
 Cost price

 x

 25

 100 
 x

×
100
× 100  % = 
=
% = 
 % = 4%
x
25
25
×
x








Math Class VIII
34
Question Bank
55. After allowing a discount of 10% on the marked price of an article, a
shopkeeper still makes a profit of 17%. By what percent is the marked
price above the cost price?
Ans. Let the cost price of the article be Rs x.
Profit is 17%
P 

Selling price =  1 +
 of Cost price
100


117
17 

x
= 1 +
of
Rs
x
=
Rs

100
 100 
Since the dealer allows a discount of 10%
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10 

Selling price =  1 –
 of Marked price
100


117 x
90
Rs
=
of Marked price
⇒
100
100
117 x
9
Rs
=
of Marked price
⇒
100
10
117 10
13
x
Marked
price
=
Rs
×
=
Rs
⇒
100
9
10
Thus, the excess value to be marked
= Marked price – Cost price
 13x

13 x
3x
– x  = Rs
= Rs
– Rs x = Rs 
10
10
 10

Thus, the percentage of excess value to be marked
 excess value

× 100  %
=
 cost price

 3x

 10

3x
×
100
=
× 100
% =
x
10
×
x




Math Class VIII
35
Question Bank
 3

=  × 100  % = 30%
 10

Hence, the dealer marks the article above the cost price by 30%.
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B
L
A
A
N
L IO
U AT
D
N
© E ER
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IN
Math Class VIII
36
Question Bank