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%

**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**