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Mathematical Ability Test [Topic: Compound Interest] – MAA7

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Test MATHEMATICS

Test Code MAA7

Topic COMPOUND INTEREST

No. of Questions 20

Maximum Marks 20

Competitive score 16

Time allowed 18 minutes

Answers with explanations have been provided for self-evaluation. Do send us your score after completing the test.

1. A borrower returns Rs 14,641 after 3 years from the date of borrowing. If the rate

of interest is 10% per annum and interest compounded annually, what is the amount borrowed?

a. Rs 10000 b. Rs 11000 c. Rs 12500 d. Rs 11200

2. What is the compound interest on Rs 20,000 for 3 years if rate of interest is 5% for 1st year, 10% for 2nd year and 20% for 3rd year.

a. Rs 7750 b. Rs 7720 c. Rs 7790 d. Rs 7620

3. Find the CI for Rs 16,000 at 5% for 2 years.

a. Rs 2512 b. Rs 2222 c. Rs 2533 d. Rs 2522



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4. If CI for a certain sum for 2 years at 10% p.a is Rs 693, find out the sum lent out.

a. Rs 3300 b. Rs 3500 c. Rs 3100 d. Rs 3870

5. If CI for a certain sum at 10% per annum for the 2nd year is Rs 2860, what is the

principal? a. Rs 26,000 b. Rs 28,000 c. Rs 25,000 d. Rs 24,000

6. A certain sum of money at a certain rate of CI doubles in 6 years. In how many

years, will it become 16 times? a. 26 years b. 24 years c. 27 years d. 21 years

7. A certain sum on compound interest becomes Rs 4000 in 3 years and Rs 8000 in 6

years. What will be the amount after 9 years? a. Rs 16000 b. Rs 18000 c. Rs 14000 d. Rs 16500



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8. If a CI on a certain sum of money at 10% per annum is Rs 2,000 what is the CI for the succeeding year when interest is computed annually.

a. Rs 2300 b. Rs 2200 c. Rs 2100 d. Rs 2000

9. An amount is borrowed at 15% CI for 2 years. What is the percentage of increase

in the amount at the end of 2 years? a. 32.25% b. 34.35 c. 31.25% d. 33.33%

10. The intake of skilled laborers in a small industry increase at 10% per annum. If

the present strength is 1000, after how many years, the strength would increase to 1331?

a. 3 years b. 3.5 years c. 4 years d. 6 years

11. If Rs 8000 increases to Rs 9261 in 3 years at CI, what is the rate of interest per annum?

a. 5% b. 6% c. 4.5% d. 4%



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12. At what rate percentage of CI, a sum would become 16 times of itself in 4 years? a. 100% b. 122% c. 121% d. 110%

13. A machine costs Rs 10,000 now, and its annual depreciation is 25%. What would

be its value after 2 years? a. Rs 6625 b. Rs 5625 c. Rs 6750 d. Rs 6650

14. A scooter depreciates at 25% per annum. If the present value is Rs 16,200, find

out the cost price 2 years ago. a. Rs 28,800 b. Rs 26,700 c. Rs 28,600 d. Rs 25,000

15. CI on a certain sum for 2 consecutive years is Rs 150 and Rs 155 respectively.

What is the rate of interest? a. 32/3 % b. 31/3 % c. 3% d. none of the above



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16. Raja borrows a certain sum of money and returns in 2 equal annual instalments. If he pays back Rs 441annually at the rate of interest of 5% per annum, what is the sum borrowed?

a. Rs 940 b. Rs 930 c. Rs 820 d. Rs 850

17. The difference between CI and SI on Rs 2,500 for 2 years at 4% per annum is?

a. Rs 4 b. Rs 6 c. Rs 11 d. None of the above

18. What is the effective annual rate of interest corresponding to a rate of 6% per

annum payable half yearly? a. 6.19% b. 5.09% c. 6.29% d. 6.09%

19. If the difference between CI and SI on a certain sum of money for 2 years at 4% per annum is Rs 40, find the sum.

a. Rs 30,000 b. Rs 25,000 c. Rs 28,000 d. None of the above

20. A sum of money invested at CI amounts to Rs 2,400 in 3 years and Rs 2,640 in 4 years. What is the rate of interest?

a. 10% b. 12% c. 9% d. none of the above

ANSWERS ON NEXT PAGE



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S.No Answer Explanation/Hints Difficulty level

1 b Principal = A = 14641 -------------- --------- ( 1 + R/100)T (11/10)3

= 14641 x 10 x 10 x 10 = 11000 11 x 11 x 11

MODERATE

2 b Amount after 3 years = P X 100+R1 X 100+R2 X 100+R3 100 100 100 = 20000 X 21/20 X 11/10 X 6/5 = 27720. Hence CI = 27720 – 20000 = Rs 7720,

DIFFICULT

3 d 1 + r/100 = 1+ 5/100 = 21/20 Amount = 16000 X 21/20 X 21/20 X 21/20 = 18522 CI = 18522 – 16000 = Rs 2522

SIMPLE

4 a Let the principal be Rs 100 Amount = 100 X (11/10)2 = 121. Hence CI = 21. If principal is 100, CI is 21. If principal is 693, CI = 693 X 100/21 = Rs 3300

SIMPLE

5 a Let principal be Rs 100. Amount after 1st year = Rs 110. Amount after 2nd year = 100 X (11/10)2 = 121. Hence CI for 2nd year = 11. If CI for 2nd year is Rs 11, principal =100. If CI for 2nd year is Rs 2860, principal = 2860 X 100/11 = Rs 26,000, option ‘a’

MODERATE



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S.No Answer Explanation/Hints Difficulty

level

6 b The amount becomes 21 times in 6 years=> It will become 24 times in 4 X 6 = 24 years, option ‘b’ Rule: If a sum becomes ‘x’ times in ‘n’ years, then it will

become xa times in ‘an’ years.

DIFFICULT

7 a In 3 years the amount doubles to Rs 8000. In the next 3 years, the amount doubles to 8000 X 2 = Rs 16,000. Hence, after 9 years the amount will be Rs 16,000 option’a’

MODERATE

8 b CI for succeeding year = 100 +r% times of CI of previous year. = 2000 (100 + 10%) = 2000 X 11/10 = Rs 2,200

MODERATE

9 a Let the amount be 100. Amount after 1st year = 115. Amount after 2nd year = 115 + 15% = 132.25 => hence percentage increase = 32.25

MODERATE

10 a Amount/Principal = (1 + R/100)time 1331 = (11/10)3 1000 Hence time = 3 years

SIMPLE

11 a (1+ r/100)T = Amount/Principal = (1+ r/100)3 = 9261/8000 = (21/20)3 Hence, 1+ r/100 = 21/20 => r = 1/20 = 5%, option ‘a’

MODERATE



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level

12 a Let P be Re1 => amount = 16. Hence, (1+r/100)4 = 24 => 1+r/100 = 2 r = 100%

MODERATE

13 b Value after 2 years = 10000 X (1-25/100)2 = 10000 X ¾ X ¾ = Rs 5625, option ‘b’ Rule: If present value is P, annual depreciation is r%, value after ‘n’

years = P X (1-r/100)n and ‘n’ years before = P / (1-r/100)n

DIFFICULT

14 a Value two years ago = 16200/(1-25/100)2 = Rs 28,800 DIFFICULT

15 b Difference in CI for 2 consecutive years = Rs 5, which is equal to SI on Rs 150 for 1 year. Hence rate of interest = 100/150 X 5 = 31/3 %, option ‘b’

SIMPLE

16 c 1 + r = 1 + 5/100 = 21/20. Principal of 1st installment = 20/21 X 441= 420 Principal of 2nd installment = 441 X 20/21 X 20/21 = 400. Hence sum borrowed = Rs 820

MODERATE

17 a CI – SI = P X (R/100)2 = 2500 X 4/100 X 4/100 = Rs 4

MODERATE



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level

18 d Let P be 100. Amount when compounded half yearly = 100 X (1+3/100)2 = 106.09, hence effective rate of interest = 6.09%

SIMPLE

19 b CI-SI = P (R/100)2 => 40 = P X 1/25 X 1/25 => P = 25,000

MODERATE

20 a SI for 1 year = 2640 – 2400 = 240. P= 2400. Rate = 240 X 100/2400 X 1 = 10%

SIMPLE

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