Compound Interest | Aptitude Questions and Answers | Aptitude Free Online-Tests | LucentBlackBoard

01.

The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the rate of interest per annum?

A.	8	B.	10
C.	12	D.	Cannot be determined
E.	None of these

Answer and Explanation

Answer: 8

Explanation:

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02.

The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is:

A.	6.06%	B.	6.07%
C.	6.08%	D.	6.09%

Answer and Explanation

Answer: 6.09%

Explanation:

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03.

If the difference ^betweensimple interest and compound interest on a certain for 2 years at 5% is Rs. 250, then find the sum

A.	Rs. 99999	B.	Rs. 100000
C.	Rs. 100001	D.	Rs. 989898

Answer and Explanation

Answer: Rs. 100000

Explanation:
(CI – SI)for 2 years = P(R / 100)²
= 250 = P(5/100)²
P = (250×100×100) / 25 = Rs. 100000

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04.

In what time will Rs. 1000 amount to Rs. 1331 at 10% per annum, under compound interest?

A.	2.8 years	B.	3.2 years
C.	2.5 years	D.	3 years

Answer and Explanation

Answer: 3 years

Explanation:
Amount after n years = P( 1 + R/100)n
∴ 1331 = 1000(1 + 10/100)n
= (1331 / 1000) = (110 / 100)n
= (11/10)3 = (11 / 10)n
= 3 years

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05.

If the composed interest on a certain sum for 2 years at 12.5% per annum is Rs. 170 then find the sum.

A.	Rs. 540	B.	Rs. 1360
C.	Rs. 640	D.	Rs. 620

Answer and Explanation

Answer: Rs. 640

Explanation:
Compound interest = A – P = P (1 + R/100)2 – P
= P[ ( 1 + 12.5/100)2 – 1]
∴ 170 – P[(1.125)2 -1)] = P(1.125 + 1)(1.125 – 1)
∴ P = 170/2.125 × 0.125 = Rs. 640

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06.

The value of a refrigerator that was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs. 9680, how much was it purchased

A.	Rs. 12145	B.	Rs. 12142
C.	Rs. 12700	D.	Rs. 12500

Answer and Explanation

Answer: Rs. 12500

Explanation:
Let the cost price if the refrigerator be Rs. V
The present value of the refrigerator = V( 1- 12/100)2 = V(22/25)2
∴ 9680 = V×(22/25) ×(22/25)
V= Rs. 12500

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07.

Find the compound interest on Rs. 16000 at 10% per annum at the end of 1 ½ years, if the interest is calculated half yearly.

A.	Rs. 2460	B.	Rs. 2522
C.	Rs. 5296	D.	Rs. 2722

Answer and Explanation

Answer: Rs. 2522

Explanation:
n = 1 ½ years = 3 half years
R = 10% per annum = 5% per half year
Amount = 16000 (1 + 5/100)3 = 16000 × (105/100) × (105/100) × (105/100) = Rs. 18522
Compound interest = 18522 – 16000 = Rs. 2522

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08.

A person borrowed Rs. 10000 at 10% rate of compound interest. He repaid certain amount at the end of the first year. Then he paid Rs. 5500 at the end of the second year thereby discharging the loan completely. What amount did he repay at the end of the first year?

A.	Rs.5500	B.	Rs. 6600
C.	Rs. 5750	D.	Rs. 6000

Answer and Explanation

Answer: Rs. 6000

Explanation:
The amount of Rs. 10000 borrowed at 19% rate will become 11000 at the end of the first year. At this point a certain amount was repaid and let the amount be repaid Rs. A. Principal for 2nd year = Rs. (1100 – A). At 10% rate of interest, this will amount to (1100 – A) × 1.1 at the end of 2nd year. But by repaying Rs. 5500, the loan has been discharged completely.
∴ (1100 – A) × 1.1 = 5500
∴ 1100 – A = 55000/11 = 5000
∴ A = Rs. 6000
∴ The amount repaid at the end of 1st year = Rs. 6000

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09.

Find the compound interest on Rs. 2000 at 5% per annum payable half yearly for 1 ½ years?

A.	Rs.151.80	B.	Rs. 153.78
C.	Rs. 315.25	D.	Rs. 75.50

Answer and Explanation

Answer: Rs. 153.78

Explanation:
n = 1 ½ years = 3 half years
R= 5% per annum = 2.5% per half year
A = 2000 ( 1 + 2.5/100)3 = 2153.78
∴ CI= A – P = 2153.78 - 2000 = Rs. 153.78

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10.

The bacteria in a culture grow by 5% in the first hour, get destroyed by 5% in the second hour and again grow b 5% in the third hour. If the count of the bacteria at the end of the third hour is 8.379 * 108 , find the original count of bacteria in the sample.

A.	8*106	B.	8*104
C.	8*107	D.	8*108

Answer and Explanation

Answer: 8*108

Explanation:
Let the original count be x. After the 1st hour , the count is x(1+5/100) = 1.05x. After the 2nd hour , the count is 1.05x(1 – 5/100) = 1.05(0.95)x. After the 3rd hour, the count is [1.05(0.95)x]( 1 + 5/100) = (1.05)(0.95) (1.05)x. Therefore, x= 8*108

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11.

138.009 + 341.981 – 146.305 = 123.6 + ?

A.	120.085	B.	120.85
C.	220.085	D.	None of these

Answer and Explanation

Answer: None of these

Explanation:
Let 138.009 + 341.981 – 146.305 = 123.6 + x
Then, x = (138.009 + 341.981) – (146.305 + 123.6) = 479.99 – 269.905 = 210.085

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