# Compound Interest Question-Quantitative Aptitude

This is another post related to compound interest problems. All the problems are fully solved step by step for your convenience so that there is no room of doubt left among the students.

I think this chapter is the one where you should invest some time because apart from its examination significance, this chapter is also helpful in your practical life.
When ever you borrow money or take a loan for your home, car or personal need, it involves the application of compound interest. The monthly installment, interest amount or EMI is calculated using the compound interest formula.
So, if you want to make yourself practically aware invest some time in this subject.

## Compound Interest Questions for Competition Exams

### (01) On a certain principal, the compound interest compounded annually for second year at 10% per annum is Rs. 132. The principal is?

2nd year compound interest (C.I) = Rs. 132
Rate of interest (R) = 10%
Time (T) = 2 years
We need to find Principal (P)=?

Interest in 1st year = P * R * T/100
Interest in 1st Year = P * 10 * 1/100 => P/10

Two year Compound Interest – interest in 1st year = Compound interest for 2nd year

=>P[{ (1+\frac { R }{ 100 } ) }^{ 2 }-1]-\frac { P }{ 10 } =132\\\ \\ =>P[{ (1+\frac { 10 }{ 100 } ) }^{ 2 }-1]-\frac { P }{ 10 } =132\\\ \\ =>P[{ (\frac { 10+1 }{ 10 } ) }^{ 2 }-1]-\frac { P }{ 10 } =132\\\ \\ =>P[{ (\frac { 11 }{ 10 } ) }^{ 2 }-1]-\frac { P }{ 10 } =132\\\ \\ =>P[(\frac { 121 }{ 100 } )-1]-\frac { P }{ 10 } =132\\\ \\ =>P(\frac { 121-100 }{ 100 } )-\frac { P }{ 10 } =132\\\ \\ =>P(\frac { 21 }{ 100 } )-\frac { P }{ 10 } =132\\\ \\ =>\frac { 21P }{ 100 } -\frac { P }{ 10 } =132\\\ \\ =>\frac { 21P-10P }{ 100 } =132\\\ \\ =>\frac { 11P }{ 100 } =132\\\ \\ =>11P=132\times 100\\\ \\ =>P=₹1200

Hence the initial principal is Rs. 1200

### (02) Find the amount on principal Rs. 25,000 in 2 years at annual compound interest if the rates for successive years be 4 and 5 percent respectively.

Given:
Principal (P) = Rs 25,000
Time = 2 years
Rate (R1) =4%
Rate (R2) = 5%

Amount=P(1+\frac { { R }<em>{ 1 } }{ 100 } )(1+\frac { { R }</em>{ 2 } }{ 100 } )\\\ \\ Amount=25000(1+\frac { 4 }{ 100 } )(1+\frac { 5 }{ 100 } )\\\ \\ Amount=25000(\frac { 100+4 }{ 100 } )(\frac { 100+5 }{ 100 } )\\\ \\ Amount=25000(\frac { 104 }{ 100 } )(\frac { 105 }{ 100 } )\\\ \\ Amount=25000(\frac { 26 }{ 25 } )(\frac { 21 }{ 20 } )\\\ \\ Amount=50\times 26\times 21\\\ \\ Amount=₹27300

Hence the final amount is Rs 27,300

### (03) A certain sum of money, the simple interest for 2 years is Rs. 350 at the rate of 4% per annum. If it was invested at compound interest at same rate and same duration as before, how much more interest would be earned?

Simple interest = Rs. 350
Time (n) = 2 years
Rate = 4%

Simple Interest (S.I) = (P * R * T)/100
==> 350 = P * 4 * 2/100
On solving we get;
==> P = Rs 4375

Now calculating the compound interest
Compound\quad Interest(C.I)=P[(1+\frac { R }{ 100 } )^{ n }-1]\\\ \\ C.I=P[{ (1+\frac { 4 }{ 100 } ) }^{ 2 }-1]\\\ \\ C.I=4375[(1+\frac { 1 }{ 25 } )^{ 2 }-1]\\\ \\ C.I=4375[(\frac { 25+1 }{ 25 } )^{ 2 }-1]\\\ \\ C.I=4375[(\frac { 26 }{ 25 } )^{ 2 }-1]\\\ \\ C.I=4375(\frac { 676 }{ 625 } -1)\\\ \\ C.I=4375(\frac { 676-625 }{ 625 } )\\\ \\ C.I=4375(\frac { 51 }{ 625 } )\\\ \\ C.I=7\times 51\\\ \\ C.I=₹357

Using compound interest, the interest earned = Rs 357
Using Simple Interest, interest earned= Rs. 350

The difference in earnings => 357-350 => Rs. 7

### (04) The compound interest on a sum of Rs. 5000 at 8 per annum for 9 months when the interest is compounded quarterly is?

Rate (R) = 8/4 = 2% per quarter
Time (n) = 3 quarter
Principal (P) = Rs. 5000

We need to find compound interest

C.I=P[(1+\frac { R }{ 100 } )^{ n }-1]\\\ \\ C.I=5000[{ (1+\frac { 2 }{ 100 } ) }^{ 3 }-1]\\\ \\ C.I=5000[(\frac { 100+2 }{ 100 } )^{ 3 }-1]\\\ \\ C.I=5000[(\frac { 102 }{ 100 } )^{ 3 }-1]\\\ \\ C.I=5000[(1.02)^{ 3 }-1]\\\ \\ C.I=5000(1.061208-1)\\\ \\ C.I=5000\times 0.061208\\\ \\ C.I=₹306.04

The compound interest is Rs 306.4