# Important Compound Interest questions

In this post we will discuss important compound interest questions which are mostly related to payment of loan installments. These loan installment questions basically involved calculation of present value of future installments. The concept is very simple and we have explained in step by step manner.

All these questions have been previously asked in competition exams like GRE, GMAT, Math Olympiad, SSC, SBI, NDA, CAT, IBPS, NABARD etc. so make sure you understand and practice each of the questions given below.

## Compound Interest Solved Problems

### (01) A man borrows Rs. 21,000 at 10% compound interest. How much he has to pay annually at the end of each year, to settle his loan in 2 years

Let the installment be x

First\quad Installment=\frac { x }{ 1+\frac { 10 }{ 100 } }\\ \\ =\frac { x }{ \frac { 100+10 }{ 100 } } =\frac { 100x }{ 110 } =\frac { 10x }{ 11 } \\\ \\ Second\quad installment=\frac { x }{ (1+\frac { 10 }{ 100 } )^{ 2 } }\\ \\ =\frac { x }{ { (\frac { 100+10 }{ 100 } ) }^{ 2 } } =\frac { x }{ { (\frac { 110 }{ 100 } ) }^{ 2 } } =\frac { x }{ \frac { 121 }{ 100 } } =\frac { 100x }{ 121 }

First installment + Second installment = Principal Amount

=>\frac { 10x }{ 11 } +\frac { 100x }{ 121 } =21000\\\ \\ =>\frac { 110x+100x }{ 121 } =21000\\\ \\ =>210x=21000\times 121\\\ \\ =>x=\frac { 21000\times 121 }{ 210 } =100\times 121\\\ \\ =>x=\quad ₹12100

Hence each installments are Rs. 12,100

### (02) A sum of Rs. 3200 invested at 10% per annum compounded quarterly amount to Rs. 3362. Compute the time period?

Given:
Amount = Rs 3362
Principal : Rs 3200
Rate of interest = 10% per annum
Time (n)= 4n (quarterly)

A=P{ (1+\frac { R }{ 100 } ) }^{ n }\\\ \\ Interest\quad is\quad compounded\quad quarterly\\\ \\ 3362=3200{ (1+\frac { 10 }{ 400 } ) }^{ 4n }\\\ \\ \frac { 3362 }{ 3200 } ={ (\frac { 400+10 }{ 400 } ) }^{ 4n }\\\ \\ \frac { 1681 }{ 1600 } ={ (\frac { 410 }{ 400 } ) }^{ 4n }\\\ \\ { (\frac { 41 }{ 40 } ) }^{ 2 }={ (\frac { 41 }{ 40 } ) }^{ 4n }\\\ \\ 4n=2\\\ \\ n=\frac { 2 }{ 4 } =\frac { 1 }{ 2 } years.

Hence the required time period is 1/2 years

### (03) The compound interest in a certain sum of money at a certain rate per annum for two years is Rs. 2050 and the simple interest on the same amount of money at the same rate for 3 years is Rs. 3000. Then the sum of money is?

Given:
S.I = Rs. 3000, Time( T ) = 3 years
C.I = Rs. 2050, Time (n) = 2 years

we know that S.I = P * R * T/100
==> 3000 = PR * 3/100
==> P R= 100000 —- eq(1)

Compound\quad Interest(C.I)=P{ (1+\frac { R }{ 100 } ) }^{ n }-1\\\ \\ 2050=\frac { 100000 }{ R } { (\frac { 100+R }{ 100 } ) }^{ 2 }-1\\\ \\ 2050=\frac { 100000 }{ R } (\frac { 10000+{ R }^{ 2 }+200R }{ 10000 } )-1\\\ \\ 2050=\frac { 100000 }{ R } (\frac { 10000+{ R }^{ 2 }+200R-10000 }{ 10000 } )\\\ \\ 2050=\frac { 10 }{ R } ({ R }^{ 2 }+200R)\\\ \\ 2050=10({ R }+200)\\\ \\ \frac { 2050 }{ 10 } ={ R }+200\\\ \\ 205=R+200\\\ \\ R=205-200\\\ \\ R=5

From eq (1), we know that
==> PR = 100000
==> P * 5 = 100000
==> P = Rs 20,000

Hence the sum of money is Rs. 20,000

### (04) A sum of Rs. 210 is taken as a loan. This is to be paid back in two equal installments. If the rate of interest is 10% compounded annually, then find the value of each installment?

Let the installment be x

First\quad Installment=\frac { x }{ 1+\frac { 10 }{ 100 } } \\\ \\ =>\frac { x }{ \frac { 100+10 }{ 100 } } \\\ \\ =>\quad \frac { 100x }{ 110 } =>\quad \frac { 10x }{ 11 } \\\ \\ Second\quad Installment=\frac { x }{ (1+\frac { 10 }{ 100 } )^{ 2 } } \\\ \\ =>\frac { x }{ { (\frac { 100+10 }{ 100 } ) }^{ 2 } } \\\ \\ =>\frac { x }{ { (\frac { 110 }{ 100 } ) }^{ 2 } } =\frac { 100x }{ 121 }

First installment + Second installment = Principal Amount

=>\frac { 10x }{ 11 } +\frac { 100x }{ 121 } =210\\\ \\ =>\frac { 110x+100x }{ 121 } =210\\\ \\ =>210x=210\times 121

Hence x = 121

Hence the value of each installment is Rs. 121

### (05) A sum of money is paid back in two annual installments of rs 17,640 each, allowing 5% compound interest compounded annually. The sum borrowed was?

Sum borrowed = Present worth of 17,640 due 1st year + Present worth of 17,640 due 2nd year

=> [\frac { P }{ (1+\frac { R }{ 100 } ) } +\frac { P }{ (1+\frac { R }{ 100 } )^{ 2 } } ]\

Sum borrowed => [\frac { 17640 }{ (1+\frac { 5 }{ 100 } ) } +\frac { 17640 }{ (1+\frac { 5 }{ 100 } )^{ 2 } } ]

Sum\quad borrowed=[\frac { 17640 }{ (\frac { 105 }{ 100 } ) } +\frac { 17640 }{ (\frac { 105 }{ 100 } )^{ 2 } } ]\\\ \\ \ Sum\quad borrowed=[\frac { 17640 }{ (\frac { 21 }{ 20 } ) } +\frac { 17640 }{ (\frac { 21 }{ 20 } )^{ 2 } } ]

By doing all the calculations we get

Sum borrowed => 16800 + 16000 ==>32,800

Hence the initial sum borrowed is Rs. 32,800