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# Compound Interest Aptitude Questions

Quantitative Aptitude Questions and Answers section on “Compound Interest” with solution and explanation for competitive examinations such as CAT, MBA, SSC, Bank PO, Bank Clerical and other examinations.

1.

The compound interest on Rs. 10000 in 2 years at 4% per annum, the interest being compounded half-yearly, is:
[A]Rs. 912.86
[B]Rs. 828.82
[C]Rs. 824.32
[D]Rs. 636.80

Rs. 824.32
$A=10000\left ( 1+\frac{2}{100} \right )^{4}$
$=>10000\left ( \frac{51}{50} \right )^{4}$
$=>10000\left ( 1.08243216 \right )=10824.3216$
Since Interest = 10824.3216 – 10000 = Rs.824.32
Hence option [C] is the right answer.

2.

In what time will Rs. 1000 becomes Rs. 1331 at 10% per annum compounded annually?
[A]2 years
[B]3 years
[C]3.5 years
[D]2.5 years

3 years
Let the required time be n years, Then,
$1331=1000\left ( 1+\frac{10}{100} \right )^{n}$
$=>\frac{1331}{1000}=\left ( \frac{10+1}{10} \right )^{n}$
$=>\left ( \frac{11}{10} \right )^{n}=\left ( \frac{11}{10} \right )^{3}$
$=>n=3$
Hence option [B] is the right answer.

3.

A sum becomes Rs. 1352 in 2 years at 4% per annum compound interest. The sum is :
[A]Rs. 1250
[B]Rs. 1270
[C]Rs. 1245
[D]Rs. 1225

Rs. 1250
Let the sum be Rs. x
$\therefore 1352=x\left ( 1+\frac{4}{100} \right )^{2}$
$=>1352=x\left ( 1+\frac{1}{25} \right )^{2}$
$=>1352=x\left ( \frac{26}{25} \right )^{2}$
$=>x=\frac{1352\times 25\times 25}{26\times 26}$
$=>x=Rs. 1250$
Hence option [A] is the right answer.

4.

In how may years will Rs. 2000 amounts to Rs. 2420 at 10% per annum compound interest?
[A]2.5 years
[B]1.5 years
[C]2 years
[D]3 years

2 years
According to question,
$2420=2000\left ( 1+\frac{10}{100} \right )^{t}$
$=>\frac{2420}{2000}=\left ( \frac{11}{10} \right )^{t}$
$=>\frac{121}{100}=\left ( \frac{11}{10} \right )^{t}$
$=>\left ( \frac{11}{10} \right )^{2}=\left ( \frac{11}{10} \right )^{t}$
∴ t = 2 years
Hence option [C] is the right answer.

5.

The principal, which will amount to Rs. 270.40 in 2 years at the rate of 4% per annum compound interest, is
[A]Rs. 220
[B]Rs. 250
[C]Rs. 225
[D]Rs. 200

Rs. 250
Let the principal be Rs.P.
$\therefore 270.40=P\left ( 1+\frac{4}{100} \right )^{2}$
$=>270.40=P\left ( 1+0.04 \right )^{2}$
$=>P=\frac{270.40}{1.04\times 1.04}=Rs. 250$
Hence option [B] is the right answer.

6.

The compound interest on Rs. 2000 in 2 years if the rate of interest is 4% per annum for the first year and 3% per annum for the second year, will be
[A]Rs. 143.40
[B]Rs. 140.40
[C]Rs. 141.40
[D]Rs. 142.40

Rs. 142.40
$Amount = 2000\left ( 1+\frac{4}{100} \right )\left ( 1+\frac{3}{100} \right )$
$=>2000\times 1.04\times 1.03$
$=>Rs. 2142.40$
$\therefore CI = Rs. (2142.40 - 2000)$
$=> Rs. 142.40$
Hence option [D] is the right ansewr.

7.

At what percent per annum will Rs. 3000/- amounts to Rs. 3993/- in 3 years if the interest is compounded annually?
[A]13%
[B]10%
[C]9%
[D]11%

10%
P = Rs. 3000 , A = Rs. 3993 , n = 3 years
$A = p \left ( 1+\frac{r}{100} \right )^{n}$
$\therefore \left ( 1 + \frac{r}{100} \right )^{n} = \frac{A}{P}$
$=>\left ( 1+\frac{r}{100} \right )^{3} = \frac{3993}{3000} = \frac{1331}{1000}$
$=>\left ( 1+\frac{r}{100} \right )^{3} = \left ( \frac{11}{10} \right )^{3}$
$=>1+\frac{r}{100} = \frac{11}{10}$
$=>\frac{r}{100} = \frac{11}{10}-1$
$=> \frac{r}{100} = \frac{1}{10}$
$=> r = \frac{100}{10} = 10\%$
Hence option [B] is the right ansewr.

8.

The compound interest on Rs. 8,000 at 15% per annum for 2 years 4 months, compounded annually is :
[A]Rs. 3109
[B]Rs. 3091
[C]Rs. 2980
[D]Rs. 3100

3109 Rs.
$Amount = p\left ( 1+\frac{R}{100} \right )^{t}$
$= 8000\left ( 1+\frac{15}{100} \right )^{2\frac{1}{3}}$
$= 8000\left ( 1+\frac{3}{20} \right )^{2}\left ( 1+\frac{3}{20\times 3} \right )$
$= 8000\times \frac{23}{20}\times\frac{23}{20}\times\frac{21}{20}$
$= Rs. 11109$
∴ Compound Interest = Rs. (11109 – 8000) = Rs. 3109
Hence option [A] is the right ansewr.

9.

A sum of money on compound interest amounts to Rs. 10648 in 3 years and Rs. 9680 in 2 years. The rate of interest per annum is :
[A]20%
[B]5%
[C]10%
[D]15%

10%
Let the sum be Rs. P and rate of interest be R% per annum. Then,
$P\left ( 1+\frac{R}{100} \right )^{2} = 9680 .........(1)$
$P\left ( 1+\frac{R}{100} \right )^{3} = 10648 ......(2)$
On dividing equation (2) by (1) :
$1+\frac{R}{100} = \frac{10648}{9680}$
$=> \frac{R}{100} = \frac{10648}{9680} - 1$
$=> \frac{R}{100} =\frac{10648 - 9680}{9680}$
$=> \frac{R}{100} = \frac{968}{9680} = \frac{1}{10}$
$=> R = \frac{1}{10}\times 100 = 10\%$
Hence option [C] is the right ansewr.

10.

At what rate per cent per annum will Rs. 2304 amount to Rs. 2500 in 2 years at compound interest?
[A]$4\frac{1}{3}\%$
[B]$4\frac{1}{2}\%$
[C]$4\frac{1}{6}\%$
[D]$4\frac{1}{5}\%$

$\mathbf{4\frac{1}{6}\%}$
Let the rate per cent per annum be r. Then,
$2500 = 2304 \left ( 1 + \frac{r}{100} \right )^{2}$
$=> \left ( 1 + \frac{r}{100} \right )^{2} = \frac{2500}{2304} = \left ( \frac{50}{48} \right )^{2}$
$=> 1+\frac{r}{100} = \frac{50}{48} = \frac{25}{24}$
$=> \frac{r}{100} = \frac{25}{24} - 1 = \frac{1}{24}$
$=> r = \frac{100}{24} = \frac{25}{6} = 4\frac{1}{6}\%$
Hence option [C] is the right ansewr.