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.

# Compound Interest

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**

$latex A=10000\left ( 1+\frac{2}{100} \right )^{4}&s=1$

$latex =>10000\left ( \frac{51}{50} \right )^{4}&s=1$

$latex =>10000\left ( 1.08243216 \right )=10824.3216$

Since Interest = 10824.3216 – 10000 = Rs.824.32

Hence option [C] is the right answer.

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,

$latex 1331=1000\left ( 1+\frac{10}{100} \right )^{n}&s=1$

$latex =>\frac{1331}{1000}=\left ( \frac{10+1}{10} \right )^{n}&s=1$

$latex =>\left ( \frac{11}{10} \right )^{n}=\left ( \frac{11}{10} \right )^{3}&s=1$

$latex =>n=3$

Hence option [B] is the right answer.

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

$latex \therefore 1352=x\left ( 1+\frac{4}{100} \right )^{2}&s=1$

$latex =>1352=x\left ( 1+\frac{1}{25} \right )^{2}&s=1$

$latex =>1352=x\left ( \frac{26}{25} \right )^{2}&s=1$

$latex =>x=\frac{1352\times 25\times 25}{26\times 26}&s=2$

$latex =>x=Rs. 1250$

Hence option [A] is the right answer.

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,

$latex 2420=2000\left ( 1+\frac{10}{100} \right )^{t}&s=1$

$latex =>\frac{2420}{2000}=\left ( \frac{11}{10} \right )^{t}&s=1$

$latex =>\frac{121}{100}=\left ( \frac{11}{10} \right )^{t}&s=1$

$latex =>\left ( \frac{11}{10} \right )^{2}=\left ( \frac{11}{10} \right )^{t}&s=1$

∴ t = 2 years

Hence option [C] is the right answer.

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.

$latex \therefore 270.40=P\left ( 1+\frac{4}{100} \right )^{2}&s=1$

$latex =>270.40=P\left ( 1+0.04 \right )^{2}&s=1$

$latex =>P=\frac{270.40}{1.04\times 1.04}=Rs. 250&s=1$

Hence option [B] is the right answer.

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**

$latex Amount = 2000\left ( 1+\frac{4}{100} \right )\left ( 1+\frac{3}{100} \right )&s=1$

$latex =>2000\times 1.04\times 1.03$

$latex =>Rs. 2142.40$

$latex \therefore CI = Rs. (2142.40 – 2000)$

$latex => Rs. 142.40$

Hence option [D] is the right ansewr.

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

$latex A = p \left ( 1+\frac{r}{100} \right )^{n}&s=1$

$latex \therefore \left ( 1 + \frac{r}{100} \right )^{n} = \frac{A}{P}&s=1$

$latex =>\left ( 1+\frac{r}{100} \right )^{3} = \frac{3993}{3000} = \frac{1331}{1000}&s=1$

$latex =>\left ( 1+\frac{r}{100} \right )^{3} = \left ( \frac{11}{10} \right )^{3}&s=1$

$latex =>1+\frac{r}{100} = \frac{11}{10}&s=1$

$latex =>\frac{r}{100} = \frac{11}{10}-1&s=1$

$latex => \frac{r}{100} = \frac{1}{10}&s=1$

$latex => r = \frac{100}{10} = 10\%&s=1$

Hence option [B] is the right ansewr.

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.**

$latex Amount = p\left ( 1+\frac{R}{100} \right )^{t}&s=1$

$latex = 8000\left ( 1+\frac{15}{100} \right )^{2\frac{1}{3}}&s=1$

$latex = 8000\left ( 1+\frac{3}{20} \right )^{2}\left ( 1+\frac{3}{20\times 3} \right )&s=1$

$latex = 8000\times \frac{23}{20}\times\frac{23}{20}\times\frac{21}{20}&s=1$

$latex = Rs. 11109 $

∴ Compound Interest = Rs. (11109 – 8000) = Rs. 3109

Hence option [A] is the right ansewr.

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,

$latex P\left ( 1+\frac{R}{100} \right )^{2} = 9680 ………(1)&s=1$

$latex P\left ( 1+\frac{R}{100} \right )^{3} = 10648 ……(2)&s=1$

On dividing equation (2) by (1) :

$latex 1+\frac{R}{100} = \frac{10648}{9680}&s=2$

$latex => \frac{R}{100} = \frac{10648}{9680} – 1&s=2$

$latex => \frac{R}{100} =\frac{10648 – 9680}{9680}&s=2$

$latex => \frac{R}{100} = \frac{968}{9680} = \frac{1}{10}&s=2$

$latex => R = \frac{1}{10}\times 100 = 10\%&s=1$

Hence option [C] is the right ansewr.

At what rate per cent per annum will Rs. 2304 amount to Rs. 2500 in 2 years at compound interest?

[A]$latex 4\frac{1}{3}\%&s=1$

[B]$latex 4\frac{1}{2}\%&s=1$

[C]$latex 4\frac{1}{6}\%&s=1$

[D]$latex 4\frac{1}{5}\%&s=1$

**$latex \mathbf{4\frac{1}{6}\%}&s=1$**

Let the rate per cent per annum be r. Then,

$latex 2500 = 2304 \left ( 1 + \frac{r}{100} \right )^{2}&s=2$

$latex => \left ( 1 + \frac{r}{100} \right )^{2} = \frac{2500}{2304} = \left ( \frac{50}{48} \right )^{2}&s=2$

$latex => 1+\frac{r}{100} = \frac{50}{48} = \frac{25}{24}&s=2$

$latex => \frac{r}{100} = \frac{25}{24} – 1 = \frac{1}{24}&s=2$

$latex => r = \frac{100}{24} = \frac{25}{6} = 4\frac{1}{6}\%&s=2$

Hence option [C] is the right ansewr.