9 PERCENTAGE AND ITS APPLICATION - Testlabz.com

1Class - VI Mathematics Question Bank

1.(A) Express each of the following statements in the percentage form :

(i) 21 eggs out of 30 are good (ii) 47 students out of 50 are present

(iii) Rs 134 out of Rs 200 is spent.

Ans. (i) 21 eggs out of 30 are good = 21

100 7 10 70%30

× = × =

(ii) 47 students out of 50 are present = 47

100 47 2 94%50

× = × =

(iii) Rs 134 out of Rs 200 is spent = 134 134

100 67%200 2

× = =

(B) Express the following fractions as percents : (i)65

80(ii)

7

5

Ans. (i)65 65 5 325 1

100 81 %80 4 4 4

×× = = = or 81.25%

(ii)7

100 7 20 140%5

× = × =

(C) Express as percents : (i) 0.09 (ii) 0.032 (iii) 0.088 (iv) 0.90

Ans. (i)9

0.09 100 9%100

= × = (ii) 32

0.032 100 3.2%1000

= × =

(iii)88

0.088 100 8.8%1000

= × = (iv) 90

0.90 100 90%100

= × =

(D) Convert into fractions in lowest terms :

(i)1

12 %2

(ii)2

6 %3

(iii)1

37 %2

(iv)2

16 %3

Ans. (i)1 25 25 1

12 % %2 2 2 100 8

= = =×

(ii)2 20 0 1

6 % %3 3 3 100 15

2= = =

×

(iii)1 75 75 3

37 % %2 2 2 100 8

= = =×

(iv)2 50 50 1

16 % %3 3 3 100 6

= = =×

(E) Express as decimal fractions: (i) 108% (ii) 29.2%

Ans. (i)108 54

108% 1.08100 50

= = = (ii)292 292

29.2% 0.29210 100 1000

= = =×

9PERCENTAGE AND ITS

APPLICATION

Class - VI Mathematics Question Bank2

2. (i) What percent of 95 is 19 ?

Ans. Required percentage = 19 100

10095 5

× = = 20%

(ii) What percent of 65 is 117 ?

Ans. Required percentage = 117

100 9 20 180%65

× = × =

(iii) What percent of 42 is 29.4 ?

Ans. Required percentage = 294

100 7 10 70%10 42

× = × =×

3. (i) What percent is 62 P of Re 1 ?

Ans. Required percentage = 62 62

100 100 62%Re1 100

p

p× = × = (∵ 1 Re = 100p)

(ii) What is 16 mm of 1 cm ?

Ans. Required percentage = 16mm

1001cm

× = 16mm

100 160%10 1mm

× =×

(iii) What percent is 8 cm of 1 m ?

Ans. Required percentage = 8cm

1001m

× = 8cm

100 8%100m

× = (1 m = 100 cm)

4 . (i) What percent is 189 m of 1.05 km ?

Ans. Required percentage = 189m

1001.05km

× = 189

1001.05 100 m

××

= 1890

18%105

=

(ii) What percent is 175 gm of 2.5 kg ?

Ans. Required percentage = 175gm

1002.5kg

× = 175gm

1002.5 100 gm

××

= 7%

(iii) What percent is 2 minute 24 seconds of 1 hour ?

Ans. Required percentage = 2minute24seconds

1001hour

×

= (2 60 24)seconds

10060 60seconds

× +×

× =

144100 4%

60 60× =

×

(iv) What per cent is 3

4 litre of

14

2 litres ?

Ans. Required percentage =

3litre

4 1001

4 litres2

× =

3100

49

2

×

= 50 2

% 16 %3 3

=


5. What is the number whose 8% is 18 ?

Ans. Let the number be ‘x’. As per condition,

8% of x = 18 ⇒ 8

18100

x× = ⇒ 18 100

9 258

x×

= = × = 225

Thus, the number is 225.

6. What is the number whose 6% is 8.1 ?

Ans. Let the number be ‘x’

∴ 6% of x = 8.1 ⇒ 6

8.1100

x× = ⇒ 8.1 100

6x

×=

81 100 810

10 6 6

×= =

× = 135.

Thus, the number is 135,

7. What is the amount whose 12% is Rs 21 ?

Ans. Let the amount be ‘x’. As per condition, 12% of x = Rs 21

⇒12

21100

x× = ⇒ 21 100

7 25 17512

x×

= = × = .

Thus, the amount = Rs 175.

8. What is the length whose 16% is 36 cm ?

Ans. Let the length be ‘x’, As per condition,

16% of x = 36 cm ⇒ 16

36100

x× =

⇒36 100

9 25 22516

x×

= = × =

Thus, the length = 225 cm = 225

100 m = 2.25 m

9. What is the number whose 12.5% is 3.5 ?

Ans. Let the number be ‘x’.

As per condition, 12.5% of x = 3.5 ⇒ 12.5

3.5100

x× =

⇒ 3.5 100

12.5x

×= ⇒

35 1007 4 28

125

×= = × =

Thus, the number is 28.

10. A class contains 25 children, out of which 6 are girls. What percentage of the class

are boys ?

Ans. Total number of students = 25. Number of boys = (25 – 6) = 19

Thus, percentage of boys = 19

10025

× = 76%


11. A girls gets 65 marks out of 80. What percent of marks did she get ?

Ans. Total marks = 80. Marks obtained by the girl = 65

∴ Required percentage = 65

10080

× = 325

81.25%4

=

12. In a class test, the marks were awarded out of 80. Kamal obtained 65% marks. How

many marks did he get ?

Ans. Total marks = 80. Kamal obtained = 65% of 80 = 65 520

80 52100 10

× = =

13. In an examination, Renu obtained 480 marks out of 750. What percentage of marks

did she get ?

Ans. Total marks = 750. Renu obtained marks = 480

Thus, percentage of marks she get = 480 48 4

10 64%750 3

×× = =

14. In her annual examination Geeta scored 39 marks out of 60 in English and 51 out

of 75 in Hindi. In which subject her performance was better ?

Ans. Percentage of marks in English = 39

10060

× = 39 5

65%3

×=

Thus, percentage of marks in Hindi = 51 51 4

100 68%75 3

×× = =

∴ In Hindi Geeta’s performance is better.

15. In an examination, 640 students appeared. Of them, 544 passed and the rest failed.

What percentage of students failed ?

Ans. Total sutdents = 640, Passed students = 544, ∴ Failed students = 640 – 544 = 96

∴ Percentage of failed students = 96

100640

× = 6 100

%40

× = 15%

16. In a plot of ground of area 6000 sq. mm only 4500sq. m is allowed for construc-

tion. What percent is to be left without construction?

Ans. Total area of the ground = 6000 sq. m.

Area allowed for construction = 4500 sq. m.

Area left without construction = 6,000 sq. m – 4500 sq. m = 1500 sq.m

Thus, percentage of the ground without construction = 1500

100 25%6000

× =

17. The population of a small locality was 4000 in 1979 and became 4500 in 1981. By

what percent did the population increase ?

Ans. Production in year 1979 = 4000. Production in year 1981 = 4500


Increase in population = (4500 – 4000) = 500

Thus, percentage of increase in population = 500

100 12.5%4000

× =

18. The price of a scooter was Rs 8000 in 1975. It reduced to Rs 6000 in 1980. By

what percent did the price of the scooter reduce ?

Ans. Original cost of scooter = Rs 8000

Reduced cost of scooter = Rs 6000

Reduction in price of scooter = Rs 8,000 – Rs 6,000 = Rs 2,000

Thus, percentage of reduction = 2000

1008000

× = 25%

19. In an examination 1360 students appeared. Of them, 85% passed. How many

students failed ?

Ans. Total students = 1360. Number of students passed = 85% of 1360 =

851360 1156

100× =

Thus, failed students = 1360 – 1156 = 204.

20. A tin contained 25 litres of oil. Due to leaksage 2 litres of oil was lost. What

percentage of oil is still there in the tin ?

Ans. Total oil in the tin = 25 litres oil leak out = 2 litres. ∴ Oil left in the tin = (25 – 2)

litres = 23 litres.

Thus, percentage of oil left in tin = 23

100 23 4 92%25

× = × =

21. An alloy consists of copper and zinc. It contains 64% copper and rest is zinc. Find

the quantity of copper and zinc in 425 g of this alloy ?

Ans. Total alloy = 425 gm. Quantity of copper in alloy = 64% of 425 gm

= 64

425100

× gm = 64 17

2724

×= gm

Thus, quantity of zinc in alloy = (425 – 272) gm = 153 gm.

22. Mr Mehat had Rs. 25600. Out of this money, he gave 20% to his daughter, 35% to

his son. 30% to his wife and the rest he donated to a Charitable trust. How much

did he donate to the Charitable trust ?

Ans. Total money = Rs 25600. Daughter’s share = 20% of Rs 25600

= Rs 20

25600100

× = Rs 5120.

Son’s share = 35% of Rs 25600 = Rs 35

25600100

× = Rs 8960


Wife’s share = 30% of Rs 25600 = Rs 30

25600100

× = Rs 7680

Amount he donated to a Charitable Trust = Rs [25600 – (5120 + 8960 + 7680)]

= Rs [25600 – 21760] = Rs 3840.

23. Mr. Vasudevan was a clerk in bank. His salary was Rs 8250 per month. Now, he has

been promoted as an accountant. His salary is now Rs 9735. Find the percentage

increase in his salary.

Ans. Total salary = Rs 8250 per month. After promontion total salary = Rs 9735 per

month.

Increase in salary = Rs (9735 – 8250) = Rs 1485

Thus, percentage of increased in salary = 1485 99

100 10 18%8250 55

× = × =

24. A tank holds 126 litres of water. At present, it is 35% full. How many litres of

water must be put into it so that it is 60% full ?

Ans. Let total capacity of tank = x litres

At present water in tank = 35% of x litres = 35

126100

x× =

⇒126 100 126 20

35 7x

× ×= = litres = 360 litres

Thus, total capacity of tank = 360 litres 60% of water in tank = 60% of 360

= 60

360100

× litres = 216 litres

Hence required water to be put into tank = (216 – 126) litres = 90 litres

25. On a particular day, 96% students were present in a school. If the number of

obsentees on that day was 154, find the total strength of school.

Ans. Student present in school = 96%. Present age of absentees percentage = (100 –

96)% = 4%.

Let total strength of school be x, then

= 4% of x = 154

⇒4

154100

x× = ⇒ 154 100

4x

×= ⇒ x = 154 × 25 = 3850

Thus strength of school = 3850.

26. In an examination, a student has to secure 48% marks to pass. If Rahul gets 247

marks and fails by 5 marks, what are the maximum marks ?

Ans. Let the maximum mark be x. Passing marks = 48% of x = 12

25

x


As per condition, 12

247 525

x= + ⇒

12252

25

x=

252 25

12x

×= = 21 × 25 = 525

Thus, maximum marks = 252

27. The value of a car depreciates (decreases) by 10% every year. If its present value

is Rs 18,260, what will be its value after one year ?

Ans. Present value of car = Rs 185260

Depreciates every year = 10% of Rs 185260

= Rs 10

185260100

× = Rs 18526.

∴ Value of car after one year = Rs (185260 – 18526) = Rs 166734.

28. The price of a radio was Rs 675. Now, its price has been increased by 6%. What is

the increased price of the radio ?

Ans. Price of radio = Rs 675. Increase in price = 6% of Rs 675 = Rs 6

675100

×

= Rs 3

272

× = Rs 81

2 = Rs 40.50

∴ Increased price of radio = Rs (675 + 40. 50) = Rs 715.50

29. The price of a cooler during last summer was quoted at Rs 2750. In off season, the

price has been reduced by 14%. What is the reduced price of the cooler ?

Ans. Price of cooler = Rs 2750

Reduction in price = 14% of Rs 2750 = Rs 14

2750100

× = Rs 7

2755

× = Rs 385

∴ Reduced price of cooler = Rs (2750 – 385) = Rs 2365.

30. An election was contested by two candidates A and B. In all there were 75600

voters. 80% of votes were polled. If A got 60% of total votes polled, how many

votes did B get ?

Ans. Total voters = 75600.

Votes polled = 80% of 75600 = 80

75600 80 756 60480100

× = × =

Number of votes A got = 60% of 60480 = 60

60480 6 6048 36288100

× = × =

Number of votes B got = 60480 – 36288 = 24192.

31. Rate of basmati rice last year was Rs 40 a kg. This year they are costler by 20%.

What is the price this year ?


Ans. Rate of basmata rice = Rs 40 kg. Increase in rate = 20%

New price = 20

40 40 40 8100

+ × = +

= Rs 48 per kg.

32. A tin contains 24 litres of milk. Due to leakage, 720 ml is lost. What percent of

milk is still present in the tin ?

Ans. Total milk = 24 litre = 24 × 1000 ml. = 24000 ml. Lost milk = 720 ml.

Milk left in the tin = 24000 ml – 720 ml = 23280 ml.

Percentage of milk 23280

10024000

= ×2328

97%24

= =

33. During October, a shop sold sarees worth Rs 1.40 lakhs. In the festival month of

November, it sold sarees worth Rs 1.68 lakhs. Find the percentage increase in the

sale.

Ans. Sale in Oct. = Rs 1.40 lakh. Sale in Nov. = Rs 1.68 lakh.

Increase in sale = 0.28 lakh

Percentage of increase in sale = 0.28

1001.40

× = 28 1

100 100 20%140 5

× = × = .

34. A person having a monthly salary of Rs 3600 earns 15% raise.Find his new monthly

salary.

Ans. Original salary = Rs 3600. Increased salary

= 3600 + 15

3600100

×

= 3600 + 540 = Rs 4140.

35. In a student election, Rahul got 66% of the votes polled. If he got 363 votes, find

the total number of votes polled.

Ans. Let the total votes be x. Number of votes Rahul got = 363

66% of x = 363 ⇒ 100 33 100

36366 6

x×

= × = = 550 votes.

36. The population of a village has decreased by 6%. If the original population was

3650, find the population after decrease.

Ans. Original population = 3650 Percentage of decrease in population = 6%

= 6

3650 – 3650100

×

= 3650 – 219 = 3431.

37. Rajesh scored 555 marks out of a total of 600 marks and Ajay scored 473 marks

out of a total of 550 marks. Whose performance is better ?

Ans. ∴ Percentage of Rajesh’s mark = 555

100 92.5%600

× =


Percentage of Ajay’s marks = 473

100550

× = 473 2 946

11 11

×= = 86%

Thus, Rajesh has better performance.

38. 43% of the students in a school are girls. If the number of boys is 1482, find :

(i) the total strength of the school. (ii) number of girls in the school.

Ans. (i) Let total student be x-percentage of girls = 43%

Percentage of boys = 100 – 43 = 57%

Number of boys = 1482 ⇒ 57% of x = 1482

⇒ 100

1482 260057

x = × =

Number of girls = (2600 – 1482) = 1118

39. (i) After an increase of 20% ; a number becomes 540. Find the original number.

(ii) After a decrease of 35% ; number becomes 520. Find the original number.

Ans. (i) Let the number be = x. 20% increase = 20 20

100 100

xx× =

As per condtion, 20

540100

xx + = ⇒

100 20540

100

x x+=

120 x = 540 × 100 ⇒ 540 100

120x

×= = 90 × 5 = 450

Thus, original number = 450

(ii) Let the number be x. 35% decrease 35 35

100 100

xx= × =

As per condition, 35

– 520100

xx = ⇒

100 – 35520

100

x x=

65520

100

x= ⇒ 65x = 520 × 100 ⇒ x

520 100

65

×= = 20 × 40 = 800

Thus, original number = 800.

40. A boy got 60 out of 80 in Hindi, 75 out of 90 in English and 35 out of 40 in

Arithmetic. In which subject his percentage of marks is best ?

Ans. In Hindi marks precentage of his obtained by the boy = 60

10080

× = 75%

∴ Percentage of marks obtained by boy in English = 75

10090

× = 500

6 = 83.33%


Percentage of marks obtained by boy in Arithmetic = 35

10040

×1

87 87.5%2

= =

On comparing all percentage of marks in different subjects

The boy scored highest marks in Arithmetics.

41. An agent gets commission at the rate of 1

122

% on the business done by him. In a

particular week, he does a business of Rs 50,000, find his commmission.

Ans. Total business done by agent = Rs 50,000. Rate of commission = 1 25

12 % %2 2

=

∴ Commission of agrent = Rs 50,000 × 25

2 100×= 250 × 25 = Rs 6250.

42. Out of a salary of Rs 4,500, I keep 1/3 as savings. Out of the remaining money, I

spend 50% on food and 20% on house rent. How much do I spend on food and

house rent ?

Ans. Total salary = Rs 4,500. Saving = 1

3 of 4,500 = Rs 1,500

Balance salary = Rs 4,500 – Rs 1,500 = Rs 3,000

Amount spending on food = 50% of Rs 3,000 = 50

100× Rs 3,000 = Rs 1,500

Amount spend on house rent = 20% of Rs 3,000 = 20

100× Rs 3,000 = Rs 600

Thus, total amount spend on food and house rent = Rs (1,500 + 600) = Rs 2,100.

43. 20% of length of a flagpole is painted green, 45% is painted yellow and the

remaining red. If the length of the pole is 18 m, what length of it is painted red ?

Ans. Total length of pole = 18 m. Green part = 20%. Yellow part = 45%. Red part = ?

Length of pole painted green painted = 20

18100

× = 18

3.6m5

=

Length of pole painted yellow = 45

18100

× = 9

18 8.1m20

× =

Hence, length of pole painted red = 18 – (3.6 + 8.1) = 18 – 11.7 = 6.3 m.

44. A grocer bought 250 eggs at 5 for Rs 8 and sold them at 2 for Rs 5. Find :

(i) the cost price of one egg ; (ii) the profit or loss on selling one egg ;

(iii) his total profits or loss on selling all the eggs.

Ans. (i) The cost price of one egg = Rs 8

5 = Rs 1.60


(ii) The cost price of one egg = Rs 1.60

The selling price of one egg = Rs 2.50

The profit on selling one egg = Rs 2.50 – 1.60 = Rs 0.90

(iii) The cost price of 250 eggs = Rs 250 × 1.60 = Rs 400

The selling price of 250 eggs = Rs 250 × 2.50 = 625

Profit = Rs 625 – 400 – 400 = Rs 225

45. A cloth seller purchased 80 metres of cloth for Rs 3240 and sold it at the rate of Rs

44 per metre. Find his profit or loss.

Ans. The cost price of 80 metres = Rs 3240. The selling price of 80 metres = Rs 80 × 44

= Rs 3520.

Profit = Rs 3550 – 3240 = Rs 280.

46. A fruit seller bougth 300 oranges at 6 for Rs 5 and sold them at 5 for Rs 6. Find :

(i) the cost price of all the oranges; (ii) the selling price of all the oranges;

(iii) his overall profit and profit percentage.

Ans. (i) The cost price of one orange = Rs 5

6.

The cost price of 300 orange = Rs 5

3006

× = Rs 250

(ii) The selling price of one orange = Rs 6

5

The selling price of all the orange = Rs 6

3005

× = Rs 360

(iii) CP of 300 orange = Rs 250. SP of 300 orange = Rs 360

Profit = S . P – C. P = Rs 360 – 250 = Rs 110

Profit percentage = 110

100 44%250

× =

47. A dealer buys a washing machine for Rs 12500 and sells it for Rs 13500. Find his

profit and profit percentage.

Ans. Cost price of machine = Rs 12,500. Selling price of machine = Rs 13,500

Profit = S. P. – C. P. = (13,500 – 12,500) = Rs 1000

Profit % = Profit

100C.P

× = 1000

10012,500

× = 8%

48. A dealer buys a motor cycle for Rs 35500 and sells it for Rs 33370. Find his loss

and loss percentage.

Ans. Cost price of motor cycle = Rs 35,500; Selling price of motor cycle = Rs 33,370

Loss = C. P. – S. P. = 35,500 – 33,370 = Rs 2130

Loss % = loss 2130

100 100 6%C.P 35500

× = × =


49. Mr Basu purchased 20 plastic chairs at Rs 225 per chair. He sold 12 chairs at

Rs 275 per chair and the remaining chairs at Rs 200 per chair. Find his gain or loss

percent.

Ans. Cost price of 1 chair = Rs 225; Cost of 20 chair = 225 × 20 = Rs 45500

S. P. of 12 chairs at the rate of Rs 275 per chair = 12 × 275 = Rs 3300

S. P. of 8 chairs at the rate of Rs 200 per chair = 8 × 200 = Rs 1600

Total S. P. of 20 chairs = Rs 3300 + Rs 1600 = Rs 4900

Profit = S. P. – C. P. = 4900 – 4500 = Rs 400

Profit% = Profit

100C.P.

× = 400 80 8

100 8 %4500 9 9

× = =

50. A shopkeeper buys three articles for Rs 375, Rs 580, and Rs 428. He is able to sell

these articles for Rs 436, Rs 635 and Rs 350 respectively. Find the gain or loss to

the shopkeeper on the whole.

Ans. C. P. of three articles = Rs 375 + Rs 580 + Rs 428 = Rs 1283

S. P of three articles = Rs 436 + Rs 635 + Rs 350 = Rs 1421

As S. P. is more than C. P. hence profit = S P – C P = Rs 1421 – Rs 1383 = Rs 38.

51. Rajesh buys an old sofa-set for Rs 1,250 and spends Rs 350 on its repairs.

(i) Find the total cost price of the sofa to Rajesh.

(ii) Find the profit or the loss made, if Rajesh is able to sell the repaired sofa-set

for Rs1,540.

Ans. (i) Total C. P. of sofa-set = cost price of sofa + money spend on repairs

= Rs 1,250 + Rs 350 = Rs 1600

(ii) S. P. = Rs 1,540. Since S. P. is less than C. P.

Loss = C. P. – S. P. = Rs 1600 – Rs 1540 = Rs 60.

52. A florish bought 240 roses at Rs 9 per dozen. If he sold all of them at Re 1 each,

what profit did he make ? What is his profit percent ?

Ans. C. P. of 1 rose = 9

12 = Rs 0.75. S.P. of 1 rose = Rs 1.00

Since S. P is more than C.P, hence

Profit = S. P. – C. P. 1.00 – Rs 0.75 = Rs 0.25

Profit on sale of 1 rose = Rs 0.25

Profit on sale of 240 rose = 240 × Rs 0.25 = Rs 60.00.

Profit% = Ppofit 0.25

100 100C.P. 0.75

× = × = 100 1

33 %3 3

= .

53. At rate of Rs 10 per dozen, 12 dozen eggs are bought. Two dozen eggs were

broken. They were sold at Rs 15 per dozen. What was the gain or loss percent ?


Ans. C. P. of one dozen of eggs = Rs 10 C.P. of 12 dozen of eggs = Rs 10 × 12 = Rs 120

S .P. of one dozen of eggs = Rs 15

S. P. of 10 dozen (12 dozen – 2 dozen broken) = Rs 15 × 10 = Rs 150

Profit = S. P. – C. P = Rs 150 – Rs 120 = Rs 30

Profit % = Profit

100C.P.

× = 30

100 25%120

× = .

54. A man purchased an iron safe for Rs 5320 and paid Rs 80 on its transportation. He

sold it for Rs 5238. Find his loss per cent.

Ans. C. P. of an iron safe = Rs 5320. Amount spend on transportation = Rs 80

Net C. P. of iron safe = Rs (5320 + 80) = Rs 5400

S . P. of iron safe = Rs 5238. Loss = Rs (5400 – 5238) = Rs 162

∴ Loss percentage = 162

100 3%5400

× =

55. A man bought a car for Rs 80000. He paid 2% of cost price to the broker and spent

Rs 3400 on the repair of car. Then, he sold it for Rs 91800. Find gain percent.

Ans. C. P. of car Rs 80000. Amount paid to broker = 2% of Rs 80000.

= 2

80000100

× =Rs 1600

Amount spent on repair = Rs 3400. Net C. P. of car = Rs (80000 + 1600 + 3400) =

Rs 85000.

S. P. of car = Rs 91800. Gain = Rs (91800 – 85000) = Rs 6800

Gain% = 6800

100 8%85000

× =

56. A grocer buys 108 eggs at Rs 25 per dozen and sells them at Rs 2.50 each. Find his

gain per cent.

Ans. The cost price of one egg = Rs 25

12

The cost price of 108 eggs = Rs 25

10812

× = Rs 25 × 9 = Rs 225

The selling price of 108 eggs = Rs 2.50 × 108 = Rs 270

Gain = Rs 270 – 225 = Rs 45

Gain per cent = 45 45 4

100225 9

×× = = 20%

57. A shopkeeper bought 50 kg of one variety of wheat at the rate of Rs

14.30 per kg and 25 kg of another variety at the rate of Rs 9.80 kg.

He mixed the two varieties and sold the mixture at Rs 14.40 per kg.

Find the profit per cent earned by the trader.


Ans. The cost price of one variety of wheat of 50 kg = Rs 50 × 14.30 = Rs 715

The cost price of another vadriety of wheat of 25 kg = Rs 25 × 9.80 = Rs 245

Total cost of mixture of wheat = Rs (715 + 245) = Rs 960

The selling price of mixture of wheat = Rs 75 × 14.40 = Rs 1080

Profit = Rs (1080 – 960) = Rs 120

Profit precent = 120 100 1

100 12 %960 8 2

× = =

58. A chair is sold for Rs 437 at a loss of 8%. At what price must it be sold to gain 8% ?

Ans. S.P. of the chair = Rs 437, Loss = 8%

∴ C.P. of the chair = Rs100

437100 – 8

× = Rs100 25

437 437 25 1992 23

× = × = × =Rs 475

Now, gain = 8%. ∴ S. P. of chair = Rs 100 8

475100

+× = Rs

108475

100×

= 27 × 19 = Rs 513.

59. A radio is sold for Rs 1071 at a gain of 5% At what price must it be sold to gain 10%

?

Ans. S. P. of radio = Rs 1071, Gain = 5%

C. P. of radio = Rs 100

1071100 5

×+

= Rs 100

1071 20 51105

× = × = Rs 1020

Now, gain = 10%

S. P of the radio = Rs 100 10

1020100

+×

= Rs 110

1020100

× = Rs 11 × 102 = Rs 1122

60. A fan is sold for Rs 1296 at a loss of 4%. At what price must it be sold it be sold

to lose 2% only ?

Ans. S. P. fan = Rs 1296, Loss = 4% ∴ C. P. of fan = Rs 100

1296100 – 4

×

= Rs 100

129696

× = Rs 1350

Now, loss = 2%

S. P. of fan = Rs 100 – 2

1350100

× = Rs 98

1350100

×

= 98

135 49 2710

× = × = Rs 1323


61. A buys a transistor for Rs 425 and sells it to B at a gain of 16%. At what price did B

buy it ? If B sells it to C at a loss of 10%, what is the price paid by C ?

Ans. C.P. of transistor for A = Rs 425, gain = 6%

S.P. of transistor by A = Rs 100 16 Rs 116

425 425 Rs 493100 100

+× = × =

C. P. of transistor for B = Rs 493, loss = 10%

S. P. of transistor by B = Rs100 –10

493100

× = Rs 90

493100

× = Rs 4437

10 = Rs 443.70.

∴ C. P of transistor for C = Rs 443.70.

62. A dialer bought two table lamps for Rs 750 each. He sold one of the lamps at 10%

gain and the other at 5% loss. Find his gain per cent on the whole transaction.

Ans. The cost price of two table lamps = Rs 750 × 2 = Rs 1500

Selling price ot the one lamp = Rs 10

750 750100

+ × = Rs 750 + 75 = Rs 825

The selling price of the another lamp

= Rs 5

750 – 750100

× = Rs (750 – 37.50) = Rs 712.50

Total selling price = Rs (825 + 712.50) = Rs 1537.50

Gain = Rs (1537.50 – 1500) = Rs 37.50

Gain per cent on the whole transaction = 37.50 3750 1

100 2 %1500 1500 2

× = =

63. A man borrowed Rs 12500 from a bank at the rate of 1

9 %2

per annum simple

interest. Find the amount that he has to pay at the end of 3 years, to clear off thedebt.

Ans. Hence P = Rs 12500. R = y = 1 19

9 % %2 2

=

Time = t = 3 years

∴ S. I = Rs 12500 19 3

100 2

× ×

× = Rs

12500 19 3

3

× × = Rs

7125

2= Rs3562.50

Amount = P + S. I. = Rs (12500 + 3562.50) = Rs 16062.50.

To clear off the debt the man has to pay = Rs 16062.50 at the end of 3years,

64. A man borrowed Rs 4 lakhs from a bank at 18% per annum simple interest. At the

end of 1

22

years, he cleared his debt by paying Rs 4.25 lakhs and his old ear.

Calculate the market value of the old car.


Ans. P = 4 lakh, R = 18% , Time = 1

22

years = 5

2years

S.I = P R T

100

× × =

400000 18 51,80,000

100 2

× ×=

×

Amount after 1

22

= (4,00,000 + 1,80,000) = Rs 5,80,000.

Amount paid = Rs 4,25,000

Value of old car = Rs 5,80,000 – Rs 4, 25,000 = Rs 1,55,000.

65. Raghu borrowed Rs 10400 from a money lender at 1

8 %2

per annum simple inter-

est. After a period of 2 years 9 months, he gave Rs 8531 and a buffalo to clear offthe debt. What is the cost of the buffalo ?

Ans. Here, P = Rs 10400. R = 1 17

8 % %2 2

=

Time = 2 years 9 month = 9

212

+

years =

32

4

+

years =

11

4 years

S.I. = Rs 10400 17 11

100 2 4

× ×

× × = Rs 13 × 17 × 11 = Rs 2431

Amount = P + S. I. = Rs (10400 + 2431) = Rs 12831

Cost of the buffalo = Rs (12831 – 8531) = Rs 4300

66. Find the principal that will yield an interest of Rs 306 in 2 years at 9% per annum ?

Ans. Let Principal be Rs x. R = 9%, T = 2 years

S. I. = 9 2

100 100

P R T x× × × ×= = Rs

9

50

x ⇒

9306

50

x= (given)

∴ ⇒ 306 50

9x

×= = Rs 34 × 50 = 1700

The principal = Rs 1700.

67. On what sum of money the simple interest at 8% per annum would be Rs 936 in

9 months.

Ans. Let principal be Rs x. R = 8%. T = 9 months = 9

12years =

3

4years

S. I. = 8 3

100 100 4

P R T x× × × ×=

× = Rs

3

50

x ⇒

3

50

x = 936 (given)


∴ 936 50

3x

×= = Rs 312 × 50 = 15600

Thus, sum of money = Rs 15600.

68. What sum will amount to Rs 5082 at 7% per annum in 3years ?

Ans. Let principal be x.

R = 7% = 3 years

S. I. = 7 3

100 100

P R T x× × × ×= = Rs

21

100

x

Amount = P + S. I. = x + 21

100

x

100 215082

100

x x+= ⇒

508242 100 4200

121x = = × =

Thus, sum of money = Rs 4200.

69. Rohan borrowed a certain sum from David at1

7 %2

per annum simple interest.

After a period of 2years 8 months, he paid Rs 3024 to David and settled the

accounts. What sum did he borrow ?

Ans. Let the sum be Rs 100.

Then, P = 100. R = 1 15

7 % %2 2

= per annum, T = 2 years 8 months = 2 8

12

= 2 8

23 3

=

∴ S.I. = 100 15 8

Rs.100 100 2 3

P R T× × × × = =

× × Rs 4 × 5 =Rs 20

Thus, Amount = Rs. (100 + 20) = Rs 120

If the amount is Rs 120, then sum = Rs 100

If the amount is Rs 3204, then sum = Rs 100

3024120

×

= Rs 10 × 252 = Rs 2520

Thus, the sum borrowed was Rs 2520.

70. In how many years, the simple interest on Rs 5000 will be Rs 1125 at 1

7 %2

per

annum ?

Ans. P = Rs 5000, R = 1 15

7 %2 2

= S.I. = Rs.1125


Time = S.I. 100

P R

×

× =

1125 100 2 75 23

5000 15 50

× × ×= =

× years.

71. At what rate per cent per annum will Rs 1640 amount to Rs 1968 in 1

22

years ?

Ans. Here, P = Rs 1640, A = Rs 1968, T = 1

22

years 5

2 years

S. I = Rs (1968 – 1640) = Rs 328

Let the rate be R% per annum. R = 100 . 100 328 2

%1640 5

S I

P T

× × × =

× × = 8%

Hence, the rate of interest is 8% per annum.

72. In how many years would a sum of money double itself at 8% per annum ?

Ans. Let sum of money (P) = Rs x, Amount = Rs 2 x (given)

Rate = 8% per annum, S.I = Rs (2 x – x) = Rs x

T = S.I. 100 100 25

P R 8 2

x

x

× ×= =

× × ∴ Time =

112

2 years.

73. At what rate percent per annum will a sum of money be five times of itself in 10

years ?

Ans. Let the principal be Rs 100.

Amount = 5 time of principal = 5 × 100 = Rs 500

∴ Interest = Rs 500 – Rs 100 = Rs 400. Time = 10

∴ R = 100 400 100

40%100 10

I

P T

× ×= =

× ×

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