Compound Interest Questions and Answers

Q: Find the least number of complete years in which a sum of money put out at 25% compound interest will be more than double of itself?

The sum more than doubles once \( (1.25)^{n} > 2 \).\[ (1.25)^{2} = 1.5625, \qquad (1.25)^{3} = 1.953125 \]\[ (1.25)^{4} = 2.4414 \]The third year still falls just short of doubling, so the answer is 4 years.The near miss at year 3 - 1.953, only 2.3% below 2 - is exactly what the question is testing.

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61.

A man gets a simple interest of Rs.500 on a certain principal at the rate of 5% p.a in two years. Find the compound interest the man will get on twice the principal in two years at the same rate.

A.Rs.500
B.Rs.250
C.Rs.5012.50
D.None of these.

Answer: D

Let the principal be Rs.P

S.I at 5% p.a in 8 years on Rs.P = Rs.500

(P)(8)(5)/100 = 500

P = 1250

C.I on Rs.2P i.e., Rs.2500 at 5% p.a in two years

=2500{ [1 + 5/100]² - 1} = 2500{ 21² - 20² /20²}

= 2500/400(441 - 400)

= 25/4(41) = 1025/4 = Rs.256.25

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62.

Rohit invests a sum of money at compound interest which doubles itself in 3 years, find in how many years will it amount to four times itself ?

A.4
B.6
C.10
D.14

Answer: B

Compound growth multiplies, so two doubling periods give a quadrupling.

\[ \left(1 + \frac{R}{100}\right)^{3} = 2 \quad\Longrightarrow\quad \left(1 + \frac{R}{100}\right)^{6} = 2^{2} = 4 \]

So it takes 6 years. The rate itself is never needed.

Under simple interest the answer would instead be 9 years, since there the interest earned each year stays constant.

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63.

Simple interest on a sum at 4% per annum for 2 years is Rs.80. The C.I. on the same sum for the same period is?

A.Rs.81.60
B.Rs.160
C.Rs.1081.60
D.Rs.99

Answer: A

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

Recover the principal from the simple interest.

\[ 80 = \frac{P \times 4 \times 2}{100} = 0.08P \quad\Longrightarrow\quad P = \text{Rs. } 1000 \]

\[ \text{CI} = 1000\left[(1.04)^{2} - 1\right] = 1000(1.0816 - 1) = \text{Rs. } 81.60 \]

The Rs. 1.60 excess is 4% of the first year's Rs. 40 of interest - the whole of what compounding adds over two years.

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64.

A divided Rs. 25,500 between his two grandsons, 13-year-old P and 14-year-old Q, in such a way that each would receive the same amount on turning 25. Until then the money stays in a bank earning 4% compounded annually. How much of it was Q's share?

A.Rs. 12500
B.Rs. 13000
C.Rs. 15000
D.Rs. 11500

Answer: B

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

P is 13 and so waits 12 years; Q is 14 and waits 11. Both end with the same amount:

\[ P_{\text{share}}(1.04)^{12} = Q_{\text{share}}(1.04)^{11} \]

\[ Q_{\text{share}} = P_{\text{share}} \times 1.04 \]

The younger boy's money compounds for an extra year, so he needs less of it. Their shares are therefore in the ratio \( 25 : 26 \):

\[ P_{\text{share}} + 1.04\,P_{\text{share}} = 25500 \quad\Longrightarrow\quad 2.04\,P_{\text{share}} = 25500 \]

\[ P_{\text{share}} = 12{,}500, \qquad Q_{\text{share}} = \text{Rs. } 13{,}000 \]

Checking: \( 12500 \times 1.04 = 13000 \), so after Q's extra year the two balances coincide.

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65.

Albert invested Rs. 8000 in a scheme for 2 years at compound interest rate 5% p.a. How much amount will Albert get on maturity of the fixed deposit?

A.Rs. 8600
B.Rs. 8620
C.Rs. 8800
D.None of these

Answer: D

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

\[ A = 8000 \times (1.05)^{2} = 8000 \times 1.1025 = \text{Rs. } 8820 \]

Rs. 8820 is not among the choices, so the answer is none of these.

The Rs. 8800 on offer is what simple interest would give (Rs. 800 of interest); the extra Rs. 20 is the second year's 5% on the first year's Rs. 400.

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66.

A sum amount to Rs.1344 in two years at simple interest. What will be the compound interest on the same sum at the same rate of interest for the same period?

A.Rs.150
B.Rs.140
C.Rs.130
D.Data insufficient for the answer

Answer: D

Compound interest needs both the principal and the rate, but the question supplies only the final amount.

\[ P + \frac{P \times R \times 2}{100} = 1344 \]

That is one equation in two unknowns, so infinitely many pairs fit. For example:

\( P = 1200 \) at 6% gives 1344, and CI would be Rs. 148.32
\( P = 1120 \) at 10% also gives 1344, and CI would be Rs. 235.20

Since the two produce different answers, the data are insufficient. Any of these questions that fixes only the amount, and not the principal or rate separately, is underdetermined.

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67.

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:

A.Rs. 400
B.Rs. 500
C.Rs. 600
D.Rs. 800

Answer: B

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

Recover the principal from the compound interest.

\[ 525 = P\left[(1.1)^{2} - 1\right] = 0.21P \quad\Longrightarrow\quad P = \text{Rs. } 2500 \]

Double the time is 4 years and half the rate is 5%:

\[ \text{SI} = \frac{2500 \times 5 \times 4}{100} = \text{Rs. } 500 \]

Doubling the time while halving the rate leaves simple interest unchanged, since \( Rn \) is the same - the answer would be identical for 2 years at 10%.

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68.

A town has population of 50,000 in 1988. In one year i.e. by 1989 it increased by 25%. Next year i.e. in 1990, it decreased by 30%. The next year in 1991 there was an increase of 40%. What is the population at end of 1991?

A.60250
B.62250
C.66550
D.61250

Answer: D

Apply each year's change as a multiplier in turn; a decrease multiplies by less than 1.

\[ 50000 \times 1.25 \times 0.70 \times 1.40 \]

\[ 1.25 \times 0.70 = 0.875, \qquad 0.875 \times 1.40 = 1.225 \]

\[ 50000 \times 1.225 = 61{,}250 \]

Adding the percentages (\( +25 - 30 + 40 = +35\% \)) would give 67,500. Successive percentage changes never simply add, because each acts on a different base.

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69.

The simple interest accrued on an amount of Rs.49500 at he end of three is Rs.17820. What would be the compound interest accured on the same amount at teh same rate in the same period? (Round off your answer to two decimal places)

A.Rs.20043.94
B.Rs.18654.81
C.Rs.20264.46
D.Rs.19612.41

Answer: A

Let the rate of interest be R% p.a.

(49500)(3)(R)/100 = 17820

495R = 5940

R = 12

CI on Rs.49500 at 12% p.a. for three years

= 49500{[1 + 12/100]³ - 1} = 69543.936 - 49500 = Rs.20043.94

when rounded off to two decimal places.

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70.

Find the least number of complete years in which a sum of money put out at 25% compound interest will be more than double of itself?

A.6 years
B.1 year
C.2 years
D.4 years

Answer: D

The sum more than doubles once \( (1.25)^{n} > 2 \).

\[ (1.25)^{2} = 1.5625, \qquad (1.25)^{3} = 1.953125 \]

\[ (1.25)^{4} = 2.4414 \]

The third year still falls just short of doubling, so the answer is 4 years.

The near miss at year 3 - 1.953, only 2.3% below 2 - is exactly what the question is testing.

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