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Compound Interest - Aptitude Questions and Answers - RejinpaulPlacement




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29.
The principal that amounts to Rs. 4913 in 3 years at 6   1/4% per annum compound interest compounded annually, is: [S.S.C. 2000]
Answer & Solution
Answer: d) Rs. 4096

Solution:   Principal = Rs. [4913/{1 + [25/(4 * 100)]}^3 ]
= Rs. [4913 * (16/17) * (16/17) * (16/17)]
= Rs. 4096.
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30.
The present worth of Rs. 160 due in 2 years at 4% per annum compound interest is:
Answer & Solution
Answer: c) Rs. 156.25

Solution:   Present worth = Rs. [169/{1 + (4/100)}^2 ]
= Rs. [169 * (25/26) * (25/26)]
= Rs. 156.25
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31.
In how many years will a sum of Rs. 800 at 10% per annum compounded semi-annually become Rs. 926.10? [Section Officers' , 2001]
Answer & Solution
Answer: b) 1    1/2

Solution:   Let the time be n years. Then,

800 * [1 + (5/100)]^(2n) = 925.10
=> [1 + (5/100)]^(2n) = 9261/8000
=> (21/20)^2n = (21/20)^3
=> 2n = 3
=> n = 3/2

n = 1 1/2 years
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32.
If the compound interest on a sum for 2 years at 12   1/2 % per annum is Rs. 510, the simple interest on the same sum at the same rate for the same period of time is: [S.S.C. 2004]
Answer & Solution
Answer: d) Rs. 480

Solution:   Let the sum be Rs. P. Then,

[P{1 + [25/(2 * 100)]}^2 - P] - 510
=> P[(9/8)^2 - 1] = 510
=> P = [(510 * 64)/17] = 1920
Sum = Rs. 1920

So, S.I. = Rs.[(1920 * 25 * 2)/(2 * 100)] = Rs. 480
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33.
The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525. The simple interest on the same sum for double the time at half the rate percent per annum is: [ C.B.I. 1997]
Answer & Solution
Answer: b) Rs. 500

Solution:   Let the sum be Rs. P. Then,

[P{1 + [10/100)]}^2 - P] - 525
=> P[(11/10)^2 - 1] = 525
=> P = [(525 * 100)/21] = 2500
Sum = Rs. 2500

So, S.I. = Rs. [(2500 * 5 * 4)/100] = Rs. 500
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34.
The simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is: [S.S.C. 2003]
Answer & Solution
Answer: c) Rs. 1750

Solution:   C.I = Rs. [4000 * [1 + (10/100)]^2 - 4000]
=Rs. [4000 * (11/10) * (11/10) - 4000]
= Rs. 840

Sum = Rs. [(420 * 100)/(3 * 8)] = Rs. 1750
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35.
There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate? [SIDBI, 2000]
Answer & Solution
Answer: c) Rs. 3972

Solution:   Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.

R = [(100 * 60)/(100 * 6)] = 10% p.a.

Now, P = Rs. 12000, T = 3 years and R = 10% p.a.

C.I = Rs. [12000 * {(1 + (10/100))^3 - 1}]
= Rs. [12000 * (331/1000)]
= Rs. 3972.
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