percentage, profit, loss 7 and discount …©edulabz interna tional icse math class vii 4 question...

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ICSE Math Class VII 1 Question Bank

1. Dhoni scored 110 runs which included 6 boundaries and 6 sixes. What per cent

of his total score did he make by running between the wickets ?

Ans. Total score = 110 runs

Boundaries = 6 × 4 = 24 runs, Sixer = 6 × 6 = 36

∴ Run scored by running between the wickets = 110 – (24 + 36)

= 110 – 60 = 50 runs

% scored by running between the wickets = 50

100%10

× 500 5

% 45 %11 11

= =

2. A property dealer charges a commission of 2% on first Rs 25000 and 1.5% on the

remainder. What commission does he charge for selling a plot of land for

Rs 65000 ?

Ans. Ist commission 2% on Rs 25000 = Rs 2

25000100

× = Rs 500

remaining = Rs (65000 – 25000) = Rs 40000

IInd commission 1.5% on Rs 40000 = Rs 1.5

40000100

× = Rs 15

40000100 10

×

×

Hence, total commission = Rs (500 + 600) = Rs 1100

3. In an examination, the maximum marks are 850. Rajat gets 34% marks and fails by

17 marks. What are passing marks ? What is the minimum percentage for passing

the examination ?

Ans. Maximum marks = 850

Rajat gets = 34% of 850 = 34

850 17 17 289100

× = × =

But he fails by 17 marks

∴ Passing marks = 289 + 17 = 306

Percentage for passing marks = 306

100% 18 2 36%850

× = × =

Hence, minimum percentage for passing the examination is 36%.

7

PERCENTAGE, PROFIT, LOSS

AND DISCOUNT

SIMPLE INTEREST

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4. In an election, two candidates X and Y contested X got 60% of the votes. The total

votes polled were 7000. How many votes did each get ?

Ans. Total number of votes polled = 7000

X got 60% of the votes

∴ X got total votes = 60% of 7000 60

7000 4200100

= × =

∴ Y got total votes = 7000 – 4200 = 2800.

5. A person saves 10% of his salary every month. If his salary is Rs 2,500, find his

expenditure.

Ans. Total salary = Rs 2500

∴ Total savings = 10% of Rs 2500 10

2500 250100

= × =

∴ Total expenditure = Rs 2500 – Rs 250 = Rs 2250.

6. Sonia bought a basket of mangoes containing 250 mangoes 20% of these were

found to be rotten. Of the remaining, 8% got crushed. How many mangoes were in

good condition ?

Ans. Total mangoes = 250

Rotten mangoes = 20% of 250 20

250 50100

= × =

Remaining mangoes = 250 – 50 = 200

Mangoes which were crushed = 8% of 200 8

200 16100

= × =

Thus balance mangoes = 200 – 16 = 184.

Hence, 184 mangoes were in good condition.

7. In a Maths Quiz of 60 questions, Vinay got 90% correct answers and Anu got 80%correct answers. How many correct answers did each give ?

What per cent is Anu’s correct answers to Vinay’s correct answers ?

Ans. Number of total questions = 60

Vinay got correct answers of the questions = 60 90

90% of 60 54100

×= =

Anu got correct answers of the questions = 80% of 60 = 80

60 48100

× =

∴ Percentage of Anu’s correct answer of that of Vinay’s

= 48 800

100 %54 9

× = = 8

88 %9

.

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8. In an examination, the maximum marks are 900. Sanjay gets 33% of the maximum

marks and fails by 45 marks. What is the passing mark ? Also, find the pass percent-

age.

Ans. Maximum marks = 900

Sanjay got 33% of 900 marks = 33

900 297100

× =

Number of marks by which he failed = 45

∴ Pass marks = 297 + 45 = 342

Percentage of pass marks = 342 100

38%900

×=

9. In an exam, Moni gets 38% marks and fails by 13 marks. If the maximum marks

are 650, what are the passing marks ? What is the minimum percentage for passing

the exam ?

Ans. Moni secured marks = 38% of 650 38

650 247.100

= × =

As she fails by 13 marks,

Thus, passing marks = 247 + 13 = 260

% age for passing the exam 260

100 40%650

= × =

Hence, the minimum percentange for passing the exam is 40%.

10. An alloy contains 23% copper, 37% zinc and the rest is nickel. Find the quantity of

nickel in 12 kg of alloy.

Ans. Total weight of an alloy = 12 kg

Weight of copper = 23% of 12 kg 23 69

12 2.76100 25

= × = = kg.

Weight of zinc = 37% of 12 kg 37 111

12 4.44100 25

= × = = kg.

Hence, weight of nickel = Total weight of alloy – (Weight of copper + Weight of

zinc)

= 12 – (2.76 + 4.44) = 12 – 7.2 = 4.8 kg.

11. On increasing the salary of a man by 12%, his salary is increased by 1158. Whatwas his original salary ?

Ans. Let the original salary be Rs x

Increase = 12% of x = 12 3

100 25

xx× =

According to the given condition = 3

115825

x=

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⇒ 1158 25

386 25 96503

x×

= = × =

Hence, his original salary is Rs 9650.

12. On decreasing the price of a car by 6%, its value becomes Rs 155100. What was the

original price of the car ?

Ans. Let the original price of the car be x

Decrease = 6% of x = 6 3

100 50

xx× =

Decreased value = x – 3 50 – 3 47

50 50 50

x x x x= =


15510050

x=

⇒ 155100 50

3300 5047

x×

= = × , x = 165000

Hence, the original price of the car is Rs 165000.

13. The salary of Gopal was increased by 10% and then the increased salary was de-

creased by 10%. Find the net increase or decrease per cent in his original salary.

Ans. Let original salary be Rs 100

∴ Increased salary = Rs 100 + 10 = Rs 110

Rate of decrease = 10%

∴ Decreased salary = 110 (100 10) 110 90

100 100

× − ×= = Rs 99

∴ Decrease = Rs 100 – 99 = Re 1

Hence, per cent decreased = 1 100

1%100

×= .

14. 40% of the population of a town are men and 39% are women. If the number of

children is 12600, find the number of men.

Ans. Let the whole population be x

Number of men = 40% of x = 40

100x , Number of women = 39% of x =

39

100x

Number of children = 12600

According to the given condition = 40 39

12600100 100

x x x

− + =

⇒ 40 39

12600100

x xx

+ − =

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⇒ 79

12600100

x x− = ⇒ 100 79

12600100

x x−=

⇒ 21

12600100

x = ⇒ 12600 100

6000021

x×

= =

∴ Total population = 60000

Number of men 40 40

60000 24000100 100

x= = × =

Hence, number of men are 24000.

15. If the price of a watch is increased by 15%, the increase in the price is Rs. 90. What

was the price of watch earlier ?

Ans. Let the price of watch earlier be x.

Increase = 15% of x = 15 3

100 20

x x=


20x x+ = x + 90

⇒ 20 3

9020

x xx

+= + ⇒

2390

20

xx= +

⇒ 23

9020

x x− = ⇒ 23 20

9020

x x−=

⇒ 3

9020

x = ⇒ 90 20

6003

x×

= =

Hence, the price of the watch earlier is Rs 600.

16. Vivek spends 20% of his salary on house rent, 17% on the education of his chil-

dren and 45% on food and other items. If he saves Rs 2340 per month, what is his

monthly salary ?

Ans. Let the monthly be Rs x

Money spend on house rent = 20% of x = 20

100x

Money spent on education of the children = 17% of x = 17

100x

Money spent on food and other items = 45% of x = 45

100x

Money saved = Rs 2340 per month

As per condition

20 17 45

100 100 100x x x x

− + +

= Rs 2340

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20 17 45

100

x x xx

+ + −

= 2340, 82

2340100

x x− = = 100 82

2340100

x x−=

182340

100x =

2340 100

18x

×= = Rs 13000

Hence, his monthly salary is Rs 13000

17. At present, a tank contains 154 litres of water and it is 35% full. Find its fullcapacity. How many litres of water must be put into it so that it is 80% full ?

Ans. Let the full capacity of tank be x

Now, 35% of x = 154 = 35

154100

x = 154 100

35x

×= = 440 litres.

∴ Full capacity of tank = 440 litres.

80% full = 80% of 440 = 80

440 352100

× = litres.

Now, is it is already 35% full, i.e. 154 litres of water is there it.

Full capacity of tank = 352 – 154 = 198 litres.

Hence, 198 litres of water must put into it so that it is 80% full.

18. In a school out of 300 students, 70% are girls and 30% are boys. If 30 girls leaveand no new boy is admitted, what is the new percentage of girls n the school ?

Ans. Total number of children in a school = 300

Number of boys = 30% of 300 = 30

300 90100

× =

Number of girls = 70% of 300 = 70

300 210100

× =

Now number of girls left = 30

∴ Number of girls after leaving 30 girls = 210 – 30 = 180

and Number of children in the school = 180 + 90 = 270

New percentage of girls = 180 200 2

100 % 66 %270 3 3

× = =

19. Kumar bought a transistor of Rs. 960. He paid 1

12 %2

cash money. The rest he

agreed to pay in 12 equal monthly instalments. How much will he pay each month ?

Ans. Price of transistor = Rs 960. Amount paid in cash = 1

12 %2

of Rs 960

= 25

9602 100

×

×

= Rs 120

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Balance amount = Rs 960 – Rs 120 = Rs 840

Number of instalments = 12

∴ Amount of each instalment = Rs 840 ÷ 12 = Rs 70

Hence, amount of each instalment is Rs 70.

20. Out of 1200 students in a school, 900 are boys and the rest are girls. If 20% of the

boys and 30% of the girls wear spectacles, find

(i) how many students in all, wear spectacles ?

(ii) what per cent of the total number of students wear spectacles ?

Ans. Total number of students = 1200

Number of boys = 900 and number of girls = 1200 – 900 = 300

Number of boys who wear spectacles = 20% of 900 20

900 180100

= × =

Number of girls who wear spectacles = 30% of 300 30

300 90100

= × =

(i) ∴ Total number of students who wear spectacles = 180 + 90 = 270

(ii) Percentage of students who wear spectacles 270 100 2700

225 %120 12

×= = =

21. At an election involving two candidates, 68 votes were declared invalid. The win-

ning candidate secures 52% of the valid votes and wins by 98 votes. Find the total

number of votes polled.

Ans. Let valid votes be 100

Then winning candidate secures votes = 52

And losing candidate will get (100 – 52) i.e., 48 votes.

Difference = 52 – 48 = 4

If difference is 4, then valid votes = 100.

And if difference is 98, then valid votes = 100

98 25 98 24504

× = × =

Invalid votes = 68

Hence, total number of votes polled = 2450 + 68 = 2518.

22. A student secures 90%, 60% and 54% marks in test papers with 100, 150 and 200

respectively as maximum marks. Find his aggregate percentage.

Ans. Marks secured by the student in different subject :

90% of 100 = 90 100

90100

×= , 60% of 150 =

60 15090

100

×=

54% of 200 = 54 200

108100

×=

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∴ Total marks = 100 + 150 + 200 = 450

Total marks secured = 90 + 90 + 108 = 288

Hence, aggregate percentage = 288

100 32 2 64%450

× = × =

23 . In an examiation, 5% of the applicants were found ineligible and 85% of the eligible

candidates belonged to the general category. If 4275 candidates belonged to other

categories, then how many candidates applied for the examination ?

Ans. Let total number of candidates applied be 100

Then number of ineligible candidates = 5

Balance number of eligible candidates = 100 – 5 = 95

Candidates belonging to general category = 85% of 95

85 95 17 19 323

100 4 4

× ×= = =

∴ Candidates belonging to other categories 323 380 323 57

954 4 4

−= − = =

If 57

4 are of other categories then total number of candidates = 100. If 4275

belongs to other categories then total number of candidates

100 4275 4

57

× ×= = 100 × 75 × 4 = 30000

24. A fruit-seller bought oranges at the rate of 6 for Rs 8 and sold them at the rate of 8

for Rs 14. Find his gain or loss per cent.

Ans. C. P. of 6 oranges = Rs 8, C. P. of 1 oranges = Rs 8 4

Rs6 3

=

S. P. of 1 orange = Rs 14 7

Rs8 4

=

Gain = S. P. – C.P. = Rs 7 4 21 16 5

Rs4 3 12 12

− − = =

∴ Gain % =

5Gain 12100% 100%

4C. P.

3

× = ×

500 125

% % 31.25 %16 4

= = =

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25. Lemons are bought at the rate of 4 for Rs 3. At what rate must they be sold to gain

20%.

Ans. C. P. of 4 lemons = Rs 3, C. P. of 1 lemons = Rs 3

4

Gain = 20%

S. P. of 1 lemon = Rs 100 gain %

C. P.100

+×

= Rs 100 20 3 120 3 9

Rs Re100 4 100 4 10

+× = × =

Hence, S. P. of 1 lemon = Re 9

10(for the gain of 20%)

26. The selling price of 12 pens in equal to the cost price of 14 pens. Find the gain per

cent.

Ans. Let C. P. of 1 pens be Re 1

C. P. of 12 pens = Rs 12

∴ S. P. of 12 pens = C. P. of 14 pens = Rs 14

Gain = S. P. – C. P. = Rs (14 – 12) = Rs 2

Gain % = Gain 2

100 % 100 %C.P. 12

× = × 50 2

% 16 %3 3

= =

27. The cost price of 12 oranges is equal to the selling price of 15 oranges. Find the

loss per cent.

Ans. Let C. P. of each orange = Re 1

C. P. of 15 oranges = Rs 15

S. P. of 15 oranges = C. P. of 12 oranges = Rs 12

Loss = C. P. – S. P. = Rs (15 – 12) = Rs 3

∴ Loss % = Loss 3

100% 100% 20%C.P. 15

× = × =

28. Oranges are bought at 5 for Rs 10 and sold at 6 for Rs 15. Find profit or loss as per

cent.

Ans. L. C. M. of 5 and 6 = 30

Let 30 oranges are bought

∴ C. P. of 30 oranges = 30 10

5

×

= Rs 60 and S. P. of 30 oranges = 30 15

6

×

= Rs 75

Gain = S. P. – C. P. = Rs 75 – Rs 60 = Rs 15

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∴ Gain% = gain 100 15 100

25%C.P. 60

× ×= = .

29. A certain number of articles are bought at 3 for Rs. 150 and all of them are sold at 4

for Rs. 180. Find the loss or gain as per cent.

Ans. L. C. M. of 3 and 4 = 12

∴ C. P. of 12 articles = Rs 150 12

3

× = Rs 600

and S. P. of 12 articles = Rs 180 12

4

×

= Rs 540

Loss = C. P. – S. P. = Rs 600 – Rs 540 = Rs 60

Loss % = Loss 100 60 100

10 %C. P. 600

× ×= =

30. A vendor bought 120 sweets at 20 p each. In his house, 18 were consumed and he

sold the remaining at 30 p each. Find his profit or loss as per cent.

Ans. Quantity of sweets bought = 120

∴ C. P. of 120 sweets = 120 20

100

×

= Rs 24

Number of sweets consumed = 18, Balance sweets = 120 – 18 = 102

∴ S. P. of 102 sweets = 102 30 3060

100 100

×= = Rs 30.60

Gain = S. P. – C. P. = Rs 30.60 – Rs 24= Rs 6.60

Gain % = gain 100 6.60 100

C. P. 4

× ×=

660 100 5527.5%

100 24 2

×= = =

×

.

31. The cost price of an article is Rs 1,200 and selling price is 5

4 times of its cost

price. Find:

(i) selling price of the article : (ii) profit or loss as per cent.

Ans. Cost price (C. P.) = Rs 1200

(i) S. P. = 5

4of C. P. =

5

4 × 1200 = Rs 1500

So, there is a gain.

(ii) Gain = S. P. – C. P. = Rs 1500 – Rs 1200 = Rs 300

∴ Gain % = gain 100 300 100

25%C. P. 1200

× ×= = .

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32. Mr. Mehta purchased a dozen famous paintings at Rs 20000 per piece. He spent Rs

25000 on advertisement. He sold 10 pieces at Rs 25000 each and the remaining two

at cost price only. Find his profit or loss and calculate it as a percentage.

Ans. Cost price of 1 painting = Rs 20000

Cost price of 12 paintings = Rs 20000 × 12 = Rs 240000

Advertisement cost = Rs 25000

C. P. = Rs 240000 + Rs 25000 = Rs 265000

Selling price of 1 piece = Rs 25000

Selling price of 10 pieces = Rs 25000 × 10 = Rs 250000

Selling price of remaining 2 pieces = Rs 20000 × 2 = Rs 40000

Total selling price 250000 + Rs 40000 = Rs 290000

Profit = S. P. – C. P. = Rs 290000 – Rs 265000 = Rs 25000

Profit % = Profit

100C. P.

× 25000 500 23

100 9 %265000 53 53

= × = = .

33. A fruit seller bought mangoes at Rs 22.50 per dozen and sold them at a loss of 8%.

How much will a customer pay for 40 mangoes?

Ans. C. P. of 1 dozen or 12 mangoes = Rs 22.50. Loss = 8%

∴ S. P. of 1 dozen or 12 mangoes = C. P. (100 loss %)

100

× −

= Rs 22.50 (100 8) 2250 92

100 100 100

− ×=

× = Rs

207

10 = Rs 20.70

S. P. of 40 mangoes = 20.70

4012

× 2070 40

Rs100 12

×=

×

= Rs 69.

34. By selling two transistors for Rs 600 each, a shopkeeper gains 20 per cent on one

transistor and loses 20 per cent on the other.

Find :

(i) C. P. of each transistor (ii) total C. P. and total S. P. of both the transistors

(iii) profit or loss per cent on the whole.

Ans. S. P. of first transistor = Rs 600. Gain = 20%

(i) ∴ C. P. = S. P 600 100

100 gain % 100 20

. 100 ×=

+ +

×

600 100

120

×= = Rs 500.

S. P. of the second transistor = Rs 600 Loss = 20%

∴ C. P. of the other transistor S. P. 100 600 100

Rs100 loss % 100 20

× ×= =

− −

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600 100

Rs80

×= = Rs 750

∴ C. P. of the two transistors are Rs 500 and Rs 750.

(ii) Total C. P. of both the transistor = Rs 500 + Rs 750 = Rs 1250

and total S. P. of both the transistors = Rs 600 + Rs 600 = Rs 1200

(iii) Total loss = C. P. – S. P. = 1250 – 1200 = Rs 50

∴ Loss % = loss 100 50 100

C. P. 1250

× ×= = 4%.

35. Ravi bought one almirah for Rs 4800 and the other for Rs 3640. He sold the first

almirah at a gain of 1

13 %3

and the other at a loss of 15%. How much did he gain

or lose in the whole deal ?

Ans. C. P. of first almirah = Rs 4800

If the gain = 1

13 %3

on it,

Gain = 1

13 %3

of 4800 40 4800

3 100

×=

×

= Rs 640

∴ S. P. = C. P. + Gain = Rs 4800 + Rs 640 = Rs 5440

C. P. of second almirah = Rs 3640

If it is sold at a loss of 15%, then loss 15% of 3640 15 3640

100

×= = Rs 546

∴ S. P. = C. P. – Loss = Rs 3640 – Rs 546 = Rs 3094

Total C. P = Rs 4800 + Rs 3640 = Rs 8440

Total S. P. = Rs 5440 + Rs 3094 = Rs 8534

∴ Gain = S. P. – C. P. = Rs 8534 – Rs 8440 = Rs 94.

36. Anurag bought a certain number of apples at Rs 36 a score and sold them at a

profit of 30%. Find the selling price per apple.

Ans. C. P. of 20 apples = Rs 36

If the profit is 30% on it, profit = 30% of 36 = 30 54

36 Rs100 5

× = = Rs 10.8

∴ S. P. of 20 apples = C. P. + Profit

= Rs 36 + Rs 10.8 = Rs 46.80

S. P. of 1 apple = Rs 46.80

20 = Rs 2.34.

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37. On selling a bed for Rs 6075, a carpenter loses 10%. For what amount should he sell

it to gain 4% ?

Ans. S. P. of a bed = Rs 6075, loss = 10%

C. P. of a bed = Rs 100

S. P.100 loss %

×

−

= Rs 100 100

6075 Rs 6075100 10 90

× = ×

− = 10 × 675 = 6750

∴ S. P. of a bed = Rs 100 gain %

C. P.100

+ ×

= Rs 100 4 104

6750 Rs 6750100 100

+ × = ×

=

266750

25

×

= Rs 7020

He should sell it for Rs 7020 to get a gain of 4%.

38. On selling an almirah for Rs 4356, a man gains 10%. What per cent does he gain on

selling the same for Rs 4158 ?

Ans. First S. P. of almirah = Rs 4356 Gain = 10%

∴ C. P. = S. P. 100 4356 100

Rs(100 gain %) 100 10

× ×=

+ +

= 4356 100

110

×

= Rs 396 × 10 = Rs 3960

Second time S. P. = Rs 4158, Gain = S. P. – C. P. = Rs (4158 – 3960) = Rs 198

∴ Gain % = Total gain 100 198 100 10

%C. P 3960 2.

× ×= = = 5%

39. Kamal sold two scooters for Rs 19800 each, gaining 10% on one and losing 10% on

the other. Find his gain or loss per cent on the whole, transaction.

Ans. S. P. of Ist scooter = Rs 19800, gain = 10%

C. P. of Ist scooter = Rs 100

19800100 10

×

+

10019800

110

= ×

= Rs 10 × 1800

= Rs 18000

S. P. of IInd scooter = Rs 19800. Loss = 10%

C. P. of IInd scooter = Rs 100

19800100 10

×

− = Rs

10019800

90×

= Rs 10 × 2200 = Rs 22000

∴ Total C. P. of two scooters = Rs (18000 + 22000) = Rs 40000

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Total S. P. of two scooters = Rs (19800 + 19800) = Rs 39600

∴ Loss = Rs (40000 – 39600) = Rs 400

Hence, loss % = 400

100%40000

× = 1%.

40. A grocer bought one quintal of rice at Rs 24 per kg. He sold 75 kg of it at the rate of

Rs 26 per kg. At what rate per kg must he sell the remainder to gain 15% on the

whole ?

Ans. Total rice purchased = 1 quintal = 100 kg

∴ Cost price of 100 kg rice = Rs (24 × 100) = Rs 2400

Selling price of 75 kg at the rate of Rs 26 per kg = 75 × Rs 26 = 1950

Required gain = 15%

∴ Selling price of total rice = C. P. (100 gain %)

100

+

= 2400 (100 15)

100

+ = Rs (24 × 115) = Rs 2760

∴ S. P. of balance rice (100 – 75) = 25 kg = Rs (2760 – 1950) = Rs 810

Hence, S. P. of 1 kg = Rs 810

25 = Rs 32.40.

41. Bata Shoe company declared a discount of 8% on each of its varieties. How much a

customer has to pay for a pair of shoes marked at Rs. 568.75 ?

Ans. Marked price of a pair of shoes = Rs 568.75

Discount = 8% or Rs 568.75 = Rs 8 4550.00

568.75 Rs100 100

× =

= Rs 45.50

Hence, S. P. of a pair of shoes = Rs (568.75 – 45.50) = Rs 523.25.

42. The printed price of a book is Rs 165. The publisher allows a discount of 16% on it.

For how much price, is the book available ?

Ans. Marked price of the book = Rs 165, discount = 16%

Discount = 16% of Rs 165 = Rs 16 16 33 4 33

165 Rs100 20 5

× × × = =

132

Rs5

= = Rs 26.40

43. A scooter is marked at Rs 25000. The dealer gives 4% discount on first Rs. 10000

and 2% discount on the remaining Rs 15000. Find (i) the amount of discount

(ii) the price charged by the dealer.

Ans. Marked price of scooter = Rs 25000

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Rate of discount on first Rs 10000 = 4%

∴ Amount of discount = Rs 4

10000100

×

= Rs 400

Rate of discount on next Rs 15000 = 2%

∴ Amount of discount = Rs 15000 2

100

×

= Rs 300

(i) Thus, total discount = Rs (400 + 300) = Rs 700

(ii) Thus, the price charged by the dealer = Rs 25000 – Rs 700 = Rs 24300.

44. A carpenter allows 15% discount on his goods. Find the marked price of a chair

which is sold by him for Rs. 680.

Ans. Let marked price of the chair be Rs x Discount = 15%

Discount = 15%of Rs x = Rs 15 3

Rs100 20

xx× =

∴ S. P. of the chair = Rs 3

20

xx

−

= Rs 20 3 17

Rs20 20

x x x−=


20

x = 680

⇒ 680 20

17x

×= = 40 × 20

⇒ x = 800.

∴ Marked price of the chair Rs. 800

45. A shopkeeper allows 20% discount on the marked price of his articles. Find the

marked price of an article for which he charges Rs. 740.

Ans. Let marked price of the article be Rs x Discount = 20%

Discount = 20% of Rs x = Rs 20

100x

×

= Rs

5

x

∴ S. P. of the articles = Rs 5 4

Rs Rs5 5 5

x x x xx

− − = =


5

x= 740

⇒ 740 5

4x

×=

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⇒ x = 925

Hence, marked price of the article is Rs 925.

46. A shopkeeper marks his goods 25% above cost price and then allows a discount of

12% on the marked price. Find his gain per cent.

Ans. Let C. P. of the goods be Rs 100

Then marked price = Rs 100 (100 25)

100

× +

= Rs 100 125

100

×

= Rs 125

Rate of discount = 12%

∴ Selling price = Rs S. P× (100 rate of discount)

100

−

= Rs 125 (100 12)

100

−

= Rs 125 88

Rs (5 22)100

× = ×

= Rs 110

∴ Gain = S. P. – C. P. = Rs (110 – 100) = Rs 10

Gain per cent = Total gain 100

%C. P.

×

= 10 100

% 10%100

× =

47. A dealer allows a discount of 10% and still gains 5%. What per cent above cost price

must he must mark his goods ?

Ans. Let cost price of goods be Rs 100

∴ Selling price of goods be Rs (100 + 5) = Rs 105


Thus, marked price = S. P. 100

(100 Rate of discount)

×

−

= Rs 105 100 105 100 350

Rs Rs100 10 90 3

× × = =

−

∴ Amount marked above the C. P.

Rs 350

3 – Rs 100 = Rs

350 300 50Rs

3 3

− =

∴ Per cent rate = Total gain 100

%C. P.

×

= 50 100 50 2

% 16 %3 100 3 3

× = =

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48. A dealer buys a T.V. set for Rs. 2500. He marks it at Rs. 3,200 and then gives a

discount of 10% on it. Find:

(i) the selling price of the T.V. set (ii) the profit per cent made by the dealer.

Ans. C.P. of a T.V. set = Rs 2500

M.P. = Rs 3200


∴ Total discount = 10

Rs 3200 Rs 320100

× =

(i) Thus, selling price of T.V. Set = Rs 3200 – Rs 320 = Rs 2880

(ii) Gain = S.P. – C.P. = Rs 2880 – Rs 2500 = Rs 380

∴ Gain % = gain 100 380 100

% %C.P. 2500

× ×= =

76 1% 15 %

5 5= = 15.2 %

Thus profit per cent made by dealer is 15.2 %

49. A dealer allows 8% discount on the marked price of an article. Find the marked

price if the selling price is Rs 621.

Ans. Let Marked price of an article be Rs 100

Discount = 8% of Rs 100

Rs 8

100100

×

= Rs 8

∴ S.P. = M.P. – Discount = Rs 100 – Rs 8 = Rs 92

If S.P. is Rs 92, then M.P. = Rs 100

If S.P. is Re 1, then M.P. = 100

92

If S.P. is Rs 621, then M.P. = 100

Rs 62192

×

= 675 = 675

50. A dealer marks his goods 25% above the cost price and gives a discount of 8%.

Find his profit percentage.

Ans. Let the cost price be Rs. 100

∴ Marked price = Cost price + 25% of C.P. = 25

100 100100

+ × = Rs 125

Amount of discount = 8% of Rs 125

= 8

125 Rs 10100

× =

∴ Selling Price (S.P.) = M.P. – Discount = Rs 125 – Rs 10 = Rs 115

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∴ Profit = S.P. – C.P. = Rs 115 – Rs 100 = Rs 15

Profit % age = Profit

100%C.P.

×

= 15

100% 15%100

× =

51. Raman took a loan of Rs 16500 from a bank on 3rd January, 2006 at 101

%2

per

annum and paid it back on 17th March, 2006. Find the total amount paid by Raman.

Ans. Loan taken by Raman Principal (P) = Rs 16500, Rate (R) = 101

%2

p.a. Time (T) =

3rd Jan, 2006 to 17th March, 2006

= 28 + 28 + 17 = 73 days = 73

365 years =

1

5 year

∴ Simple interest (S.I) = 100

P R T× × = Rs.

16500 21 1 33 21

100 2 5 2

× × ×=

× ×= Rs.

693

2

= Rs. 346.50

∴ Total amount paid by = P + S.I. = Rs 16500 + Rs 346.50 = Rs 16846.50

Hence, total amount paid by Raman Rs 16846.50

52. In what time will Rs 6880 amount to Rs 7224, if simple interest is calculated at

17 %

2 per annum ?

Ans. Amount (A) = Rs 7224, Principal (P) = Rs 6880

∴ Simple Interest = A – P = Rs 7224 – Rs 6880 = Rs 344

Rate (R) = 1 15

7 % %2 2

= p.a.

∴ Time = . 100 344 100 2

6880 15

S I

P R

× × ×=

× ×

= 2

3 years =

212

3× months = 8 months

53. In what time will Rs. 1860 amount to Rs. 2278.50, if simple interest is calculated at

9% per annum?

Ans. Amount (A) = Rs 2778.5

Principal (P) = Rs 1860.00

∴ S.I = A –P = Rs 2278.50 – Rs 1860.00

= Rs 41850

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Rate = 9% p.a

∴ Time (T) = . 100S I

P R

×

×

= 418.50 100

1860 9

×

×

= 41850 100 465

2.5100 1860 9 186

×= =

× ×

years = 5

210

years = 1

22

years.

54. At what rate per cent per annum will Rs. 6360 yield an interest of Rs. 1378 in

12

2 years ?

Ans. Principal (P) = Rs 6360, simple interest (SI) = 1378, Time (T) = 1

22

years = 5

2 years

∴ Rate = . 100 1378 100 2

6360 5

S I

P T

× × ×=

× ×

= 1378 26 2

8 %159 3 3

= = p.a.

55. At what rate per cent per annum will Rs. 1650 amounts to Rs. 2046 in 3 years ?

Ans. Amount (A) = Rs 2046

Principal (P) = Rs 1650

∴ S. I. = A – P = Rs (2046 – 1650) = Rs 396 = Time (T) = 3 years

∴ Rate = . 100 396 100

1650 3

S I

P T

× ×=

× ×

= 1320

8%165

= p.a.

56. The interest on a sum of money at the end of 1

42

years is 3

5 of the sum. Find the rate

per cent per annum.

Ans. Let sum (P) = Rs 100.

∴ S. I = 3

5 of the sum =

3

5 × 100 = Rs 60. Time (T) =

14

2 years =

9

2 years

∴ Rate = . 100 60 100 2

%100 9

S I

P T

× × ×=

× ×

= 40 1

% 13 %3 3

= p.a.

Hence, rate per cent in 1

133

p.a.

57. The simple interest on a certain sum for 3 years at 10% per annum is Rs 829.50.

Find the sum.

Ans. S. I. = Rs 829.50, Rate (R) = 10% p.a, Time (T) = 3 years

∴ Principal = . 100 829.50 100

10 3

S I

R T

× ×=

× × =

82950 100 8295

100 10 3 3

×=

× × = Rs 2765

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58. Karim borrowed Rs. 8000 from a money lender at an interest of 2% per month.

How much simple interest he has to pay for 4 months ?

Ans. Principal = Rs 8000, R = 2% per month, T = 4 months

Simple Interest = 100

P R T× × =

8000 2 4

100

× × = Rs 640

Hence, Karim will pay simple interest Rs 640 for, 4 month.

59. Calculate the simple interest on Rs. 16000 at the rate of 1.2% per month for one

year 3 months.

Ans. P = 16000, R = 1.2% per month, T = 1 year 3 months = 15 months


P R T× × =

16000 1.2 15

100

× ×

= 160 12 15

10

× ×

= Rs 2880

Hence, simple interest for 15 months is Rs 2880

60. Find the simple interest and the amount of Rs. 1250 from June 27 to September 8

same year at 10% per annum.

Ans. Here, P = Rs 1250 R = 10% p.a.

T = 3 June + 31 July + 31 August + 8 September

= 73 days = 73

365 year


P R T× ×=

1250 10 73

100 365

× ×

× = Rs 25

Amount = Principal + Interest = Rs (1250 + 25) = Rs 1275.

61. Find the amount of Rs. 2400 from 18th February 2000 to 13th July 2000 at 1

12 %2

per annum simple interest.

Ans. Here, P = Rs. 2400

R = 1

12 %2

p.a. = 25

%2

p.a.

T = 11 February + 31 March + 30 April + 31 May + 30 June + 13 July

= 146 days = 146

365year


P R T× ×

= 2400 25 146

100 365 2

× ×

× × = Rs 120

Amount = Principal + Interest = Rs (2400 + 120) = Rs 2520.

62. A man invested Rs 10000 at the rate of simple interest of 8 paise per rupee per year.

What amount will he get after 15 months ?

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P = 10000

Ans. R = 8 paise per rupee per year

= 8 rupees per hundred per year

= 8% per annum

T = 15 months = 15 5

12 4= year

Simple Interest = 10000 8 5

100 100 4

P R T× × × ×=

× = Rs 1000

Amount = Principal + Interest

= Rs (10000 + 1000) = Rs 11000

63. Raja borrows Rs 8000 from a money lender for his daughter’s marriage and pays

back Rs11200 after 8 months. Find the rate of simple interest charged per annum.

Ans. Here, P = Rs 8000, A = Rs 11200

∴ Interest = A – P = Rs (11200 – 8000) = Rs 3200

T = 8 months = 8 2

12 3= years

∴ R = 1 100

P T

×

× =

3200 100 3

8000 2

× ×

× = 60% p.a.

64. Rs 14000 is invested at 4% per annum simple interest. How long will it take for the

amount to reach Rs. 16240 ?

Ans. Here, P = Rs14000, A = Rs 16240

S. I = A – P = Rs (16240 – 14000) = Rs 2240, R = 4% p.a

T = . . 100 2240 100

1400 4

S I

P R

× ×=

× × = 4 years.

65. A man borrowed Rs. 3000 on 1st April and returned Rs. 3144 on 13th June same

year. Find the rate of simple interest charged.

Ans. Here, P = Rs 3000, A = Rs 3144

∴ S.I. = A – P

∴ = Rs (3144 – 3000) = Rs 144

T = 30 April + 31 May + 12 June = 73 days

= 73

365years.

R = . . 100 144 100 365

3000 73

S I

P T

× × ×=

× × = 24% p.a.

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66. In what time will a sum of money double itself at 8% p.a.?

Ans. Let the principal (P) = Rs 100

∴ Amount (A) = Rs 100 × 2 = Rs 200

∴ S. I. = A – P = Rs 200 – Rs. 100 = Rs 100

Rate (R) = 8% p.a.

∴ Time = . . 100 100 100

100 8

S I

P R

× ×=

× ×

= 25 1

122 2

= years

67. In how many years will be Rs 870 amount to Rs 1,044, the rate of interest being

12 %

2p.a.?

Ans. Principal (P) = Rs 870, Amount (A) = Rs 1044

∴ S. I. = P – A = Rs 1044 – Rs 870 = Rs 174

Rate (R) = 1 5

2 %2 2

= p.a.

∴ Time = . . 100 174 100 2

870 5

S I

P R

× × ×=

× × = 8 years.

68. Find the sum which will amount to Rs.700 in 5 years at 8% rate p.a.

Ans. Amount = Rs 700, Rate (R) = 8% p.a., Time (T) = 5 years

Let principal (P) = Rs 100 then S. I = 100 8 5

100 100

P R T× × × ×= = Rs 40

∴ Amount (A) = P + S. I = Rs 100 + 40 = Rs 140

If amount is Rs 700, then principal

= Rs 100 700

140

×= Rs 500

69. In 4 years, Rs 6,00 amount to Rs 8,000.In what time will Rs 525 amount to Rs 700

at the same rate ?

Ans. In first case, Principal (P) = Rs 6,000, Amount (A) = Rs 8,000

∴ S.I = A – P = Rs 8,000 – Rs 6,000 = Rs 2000

Time (T) = 4 years

∴ R = . . 100 2000 100

6000 4

S I

P T

× ×=

× × =

25 1% 8 %

3 3= p.a.

In second case, Principal (P) = Rs 525, Amount (A) = Rs 700

∴ S. I. = A – P = Rs 700 – Rs. 525= Rs 175,

© EDULABZ

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Rate (R) = 25

%3

of p.a.

∴ Time = . . 100 175 100 3

525 25

S I Rs

P R

× × ×=

× ×

= 4 years

70. The interest on a sum of money at the end of 1

22

years is 4

5of the sum. What is the

rate per cent ?

Ans. Let the sum (P) = Rs 100

∴ S.I. = Rs 100 × 4

5 = Rs 80

Period (T) = 1 5

22 2

= years.

∴ Rate = . . 100 80 100 2

100 5

S I

P T

× × ×=

× ×

= 32% p.a

71. What sum of money lent out at 5% for 3 years will produce the same interest as

Rs. 900 lent out at 4% for 5 years ?

Ans. In second case, Principal (P) = Rs 900, Rate (R) = 4%, Time (T) = 5 years

∴ S.I = 900 4 5

100 100

P R T× × × ×= = Rs 180

In first case, S.I = Rs 180

Rate = 5%, Time = 3 years

∴ Sum = . 100 180 100

5 3

S I

R T

× ×=

× ×

= Rs. 1200

72. A sum amount to Rs 2,652 in 6 years at 5% p.a. simple interest. Find :

(i) the sum

(ii) the time in which the same sum will double itself at the same rate of interest.

Ans. (i) In first case, Let Principal (P) = Rs 100, Rate (R) = 5% p.a. Time (T) = 6 yrs

∴ S. I. = 100 5 6

100 100

P R T× × × ×= = Rs 30 and,

Amount = Rs 100 + Rs 30 = Rs 130

If amount is Rs 130, then principal

= Rs 100 and if amount is Rs 2652, then

Principal = 100 2,652

130

×

= Rs 2040

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In second case, Let sum (P) = Rs 100, Amount (A) = Rs. 100 × 2 = Rs 200

S. I. = A – P = Rs. 200 – 100 = Rs 100

Rate = 5%p.a.

Time = . . 100 100 100

100 5

S I

P R

× ×=

× ×

= 20 years

73. What sum of money will amount to Rs. 6860 in 2 years 6 months at 9% p.a. simple

interest ?

Ans. Here, Amount A = Rs 6860, Rate of Interest, R = 9% p.a

T = 2 years, 6 months

= 6

212

+ years = 1

22

years = 5

2 years

Let sum, P = Rs 100

S. I. = 100 9 5

100 100 2

P R T× × × ×=

×

= 45

2 = Rs 22.5

∴ Amount = P + I

= Rs 100 + 22.5 = Rs 122.5.

If amount is Rs 122.5, then

Sum P = Rs 100

If amount is Re 1, then sum P = 100

122.5

If amount is Rs 6860, then sum

P = 100

6860122.5

× = Rs 5600.

74. What sum of money lent out at simple interest will amount to Rs. 1.13 lakh in

13

4 years at 4% at 4% p.a. ?

Ans. Here, Amount A = Rs. 1.13 lakh = Rs . 113000

Time, T = 1

34

years

Rate of Interest, R = 4% p.a

Let sum, P = Rs 100

∴ S. I. = 100

P R T× ×

= 100 4 13

100 4

× ×

×

= Rs 13

∴ Amount = P + I = Rs (100 + 13) = Rs 113

If amount is Rs 113, then the sum, P = Rs 100

© EDULABZ

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If amount is Re 1, then the sum, P = 100

113

If amount is Rs. 113000, then the sum

P = 100

113000113

× = Rs 100000

= Rs 1 lakh

75. What sum of money lent out at simple interest at 12% per annum for 1

32

years will

produce the same interest as Rs. 6000 lent at 10.5% per annum in 3 years ?

Ans. In second case, Here, P = Rs 6000, R = 10.5% p.a

T = 3 years

S. I. = 100

P R T× × =

6000 105 3

100 10

× ×

× = Rs 1890.

Now, In first case,

S. I. = Rs 1890, R = 12% p.a. T = 1

32

years = 7

2 years

∴ P = . 100S I

R T

×

× =

1890 100 2

12 7

× ×

× = Rs 4500

76. A sum of money invested for 2 years at 6.5% per annum simple interest amounted to

Rs. 9040. What will it amount to in 3 years 3 months at 9% per annum simple

interest ?

Ans. Here, in first case,

T = 2 years, R = 6.5% p.a.

A = Rs 9040

Let P be Rs 100

∴ S.I. = 100 6.5 2

100 100

P R T× × × ×=

= 65 2

10

×

= Rs. 13

Amount = Principal + Interest = Rs (100 + 13) = Rs 113

If Amount is Rs 113, then principal P = Rs 100

If amount is Re1, then principal P = 100

113

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If amount is Rs 9040, then principal P = 100

9040113

× = 904000

113 = Rs 8000

Now, in second case,

P = Rs 8000, T = 3 years 3 months = 1

34

years

R = 9% p.a.

∴ S. I = 100

P R T× ×

= 8000 9 13

100 4

× ×

×= Rs. 2340

∴ Amount = Principal + S. I. = Rs (8000 + 2340) = Rs 10340

77. If Rs. 4000 amounts to Rs. 4500 in 2 years, find how much would it amount in

5 years at the same rate of interest.

Ans. Here, in first case

P = Rs 4000, A = Rs 4500

∴ S. I. = A – P =

Rs (4500 – 4000) = Rs 500, T = 2 years

R = . . 100 500 100

4000 2

S I

P T

× ×=

× × =

25%

4p.a. =

16 %

4p.a.

Now, in second case, R = 1

6 %4

p.a, P = Rs 4000, T = 5 years

S. I. = 100

P R T× × =

4000 25 5

100 4

× ×

× = Rs 1250

∴ Amount = Principal + Interest

= Rs (4000 + 1250) = Rs 5250

78. To purchase a cow, farmer borrowed some money from a bank. He cleared his debt

by paying Rs.4720 at the end of one and a half year. If the bank charges interest at the

rate of 12% per annum, find cost of the cow.

Ans. The cost of the cow is equal to sum of money = P. Let it be Rs 100

T = 1

12

years, R = 12% p.a. S. I.= 100

P R T× ×

= 100 12 3

100 2

× ×

× = Rs 18

∴ Amount = Principal + Interest = Rs 100 + Rs 18 = Rs 118

By Amount = Rs 4720 (Given)

If amount is Rs 118, then principal P = Rs 100

© EDULABZ

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If amount is Re 1, then principal P = 100

118

If amount is Rs 4720, then principal = 100

4720118

× = 236000

59 = Rs 4000

Hence, the cost of the cow is Rs 4000

79. A sum of money invested for 4 years at 9.5% per annum simple interest amounts to

Rs 12144. What will it amount to in 2 years 6 months at 6% per annum simple

interest ?

Ans. Here, in first case,

T = 4 years, R = 9.5% p.a, A = Rs 12144

Let principal, P = Rs 100

∴ S. I. = 100

P R T× ×

= 100 9.5 4

100

× ×

= Rs 38

∴ Amount = Principal + Interest = Rs 100 + Rs. 38 = Rs 38

If amount is Rs 138, the principal,

P = Rs 100

If amount is Re 1, then principal, P = 100

138

If amount is Rs 12144, then principal P = 100

12144138

× = 607200

69= 8800

Now, in second case,

P = Rs. 8800, R = 6% p.a., T = 1

22

years. S.I.= 100

P R T× ×

= 8800 6 5

100 2

× ×

× = Rs 1320

∴ Amount = Principal + Interest = Rs (8800 + 1320) = Rs 10120.

80. At what rate per cent per annum simple interest will a sum treble itself in 16 years ?

Ans. Let the sum (P) = Rs 100

∴ Amount (A) = Rs 100 × 3 = Rs 300

Time (T) = 16 years

∴ Rate = . 100 200 100

100 16

S I

P T

× ×=

× × =

25 1% 12

2 2= p.a.

percentage, profit, loss 7 and discount …©edulabz interna tional icse math class vii 4 question...

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