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Compound Interest - Aptitude Questions and Answers - RejinpaulPlacement




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50.
If a sum on compound interest becomes three times in 4 years, then with the same interest rate, the sum will become 27 times in:
Answer & Solution
Answer: b) 12 years

Solution:   P[1 + (R/100)]^4 = 3P
=> [1 + (R/100)]^4 = 3 ...... (i)

Let P[1 + (R/100)]^n = 27P
=> [1 + (R/100)]^n = 27 = (3)^3 = {[1 + (R/100)]^4 }^3 [using (i)]
=> [1 + (R/100)]^n = [1 + (R/100)]^12
=> n = 12

Required time = 12 years.
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51.
The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is: [N.I.F.T. 2003]
Answer & Solution
Answer: b) 4

Solution:   P[1 + (20/100)]^n > 2P
=> (6/5)^n > 2

Now, [(6/5) * (6/5) * (6/5) * (6/5)] > 2

So, n = 4 years.
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52.
A man borrows Rs. 2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 years in two equal yearly instalments. How much will each instalment be?
Answer & Solution
Answer: c) Rs. 1352

Solution:   Let the value of each instalment be Rs. x. Then,

(P.W. of Rs. x due 1 year hence) + (P.W. of Rs. x due 2 years hence) = Rs. 2550

=> x/[1 + (4/100)] + x/[1 + (4/100)]^2 = 2550
=> (25x/26) + (625x/676) = 2550
=> 1275x = 2550 * 676
=> x = [(2550 * 676)/1275] = 1352

Value of each instalment = Rs. 1352
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53.
What annual income payment will discharge a debt of Rs. 1025 due in 2 years at the rate of 5% compound interest? [S.S.C. 2000]
Answer & Solution
Answer: b) Rs. 551.25

Solution:   Let each instalment be Rs. x. Then,
x/[1 + (5/100)] + x/[1 + (5/100)]^2 = 1025
=> (20x/21) + (400x/441) = 1025
=> 820x = 1025 * 441
=> x = [(1025 * 441)/820] = 551.25

So, value of each instalment = Rs. 551.25
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54.
A man borrows Rs. 12,500 at 20% compound interest. At the end of every year he pays Rs. 200 as part repayment. How much does he still owe after three such instalments?
Answer & Solution
Answer: d) None of these

Solution:   Balance = Rs. [12500 * {1 + (20/100)}^3] - [2000 * {1 + (20/100)}^2 + 2000 * {1 + (20/100)} + 2000]
= Rs. [12500 * (6/5) * (6/5) * (6/5)] - [2000 * (6/5) * (6/5) + 2000 * (6/5) + 2000]
= Rs. [21600 - (2880 + 2400 + 2000)]
= Rs. 14320
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55.
A sum of money is borrowed and paid back in two annual instalments of Rs. 882 each allowing 5% compound interest. The sum borrowed was: [A.I.M.A.T.S. 2002]
Answer & Solution
Answer: b) Rs. 1640

Solution:   Principal
= (P.W. of Rs. 882 due 1 year hence) + (P.W. of Rs. 882 due 2 years hence)
= 882/[1 + (5/100)] + 882/[1 + (5/100)]^2
= [(882 * 20)/21] + [(882 * 400)/441]
= Rs. 1640
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DATA SUFFICIENCY TYPE QUESTIONS
56.
The difference between the compound interest and the simple interest earned on a sum of money at the end of 4 years is Rs. 256.40. To find out the sum, which of the following informations given in the statements P and Q is/are necessary?
P : Amount of simple interest accrued after 4 years.
Q : Rate of interest per annum.
Answer & Solution
Answer: b) Only Q is necessary

Solution:   To find the sum, difference between C.I. and S.I, the time and the rate of interest are needed.
Correct answer is Q.

Correct answer is (b)
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