# Compound Interest Solved Questions

In this post we will discuss problems of compound interest. There are mainly two forms of money compounding mechanism in quantitative aptitude syllabus of major competition exas:
1. Simple Interest
2. Compound Interest

Compound interest is some what more technical than simple interest but if you understand the basics, none of their problems is difficult to solve.

In this post, we have compiled important problems of compound interest and all of them are solved step by step for your understanding. Our aim is to make you comfortable solving math related questions without any external help.

## Compound Interest Questions

### (01) In what time will Rs.1000 becomes Rs.1331 at 10% per annum compounded annually

Given:
Principal (P) = Rs 1000
Amount (A) = Rs. 1331
Rate (r) = 10% per annum

we have to find time (n)

According to question:

Hence at 3 years time, Rs. 1000 becomes Rs. 1331

### (02) The difference between compound interest and simple interest on a sum of Rs. 1000 at a certain rate of interest for 2 years is Rs. 10. The rate of interest per annum is?

In this type of question we will use following formula for fast calculation

### (04) A sum of money becomes 8 times of itself in 3 years at compound interest. Find the rate of interest per annum

Hence at 100% rate of interest, the money becomes 8 times itself in 3 years

### (06) A sum of money placed at a compound interest doubles itself in a 15 years. In how much years, it would amount to eight times of itself at the same rate of interest

we have to find number of years in which compound interest amount to eight times

we know that A=P(1+\frac { R }{ 100 } )^{ n }

where
A = Amount
P = Principal
R= rate of interest
n = time in years

Let
A = 2x
P = x
n =15

Now putting the values in above formula
2x=x(1+\frac { R }{ 100 } )^{ 15 }

Cancelling x on both sides
2=(1\frac { R }{ 100 } )^{ 15 }\\\ \\ { 2 }^{ \frac { 1 }{ 15 } }=\quad 1+\frac { R }{ 100 } \quad …………..(a)

Now we have to find number of years in which the sum will become 8 times

Let A = 8x and P=x
8x=x(1+\frac { R }{ 100 } )^{ n }\\\ \\ 8=(1+\frac { R }{ 100 } )^{ n }……………..(b)\

Put the value of equation (a) in equation (b)

8=({ 2 }^{ \frac { 1 }{ 15 } })^{ n }\\\ \\ { 2 }^{ 3 }={ 2 }^{ \frac { n }{ 15 } }\

Now comparing both the sides we get
n/15 = 3
n = 45

In 45 years the sum will become 8 times at the same rate of interest

### (07) If the rate of interest be 4% per annum for first year, 5% per annum for second year and 6% per annum for third year, then the compound interest of Rs. 10,000 for 3 years will be?

Given that
Rate for first year (R1) = 4%
Rate for second year (R2)= 5%
Rate for third year (R3) = 6%
Principal = Rs. 10,000

\\ A=10000(1+\frac { 4 }{ 100 } )(1+\frac { 5 }{ 100 } )(1+\frac { 6 }{ 100 } )\\\ \\ A=10000(\frac { 100+4 }{ 100 } )(\frac { 100+5 }{ 100 } )(\frac { 100+6 }{ 100 } )\\\ \\ A=10000(\frac { 104 }{ 100 } )(\frac { 105 }{ 100 } )(\frac { 106 }{ 100 } )\\\ \\ A=10000\times \frac { 26 }{ 25 } \times \frac { 21 }{ 20 } \times \frac { 53 }{ 50 } \

A= 10,000 * 1.04 * 1.05 * 1.06

A= 11575.20

We know that
Compound Interest = Amount – Principal
C.I = 11575.20 – 10,000 => 1575.2 Rs

Hence the compound interest is 1575.2 Rs

### (08) A person deposited a sum of Rs. 6000 in a bank at 5% per annum simple interest. Another person deposited Rs. 5000 at 8 percent per annum compound interest. After 2 years the difference of their interest will be?

First Person
Principal = Rs. 6000
Rate (R1) = 5% per annum
Time (T) = 2 years

Simple Interest Formula = (P * R * T)/100

S.I = (6000 * 5 * 2)/100
S.I = Rs .600

Second Person
Principal = Rs. 5000
Rate (R2) = 8 percent
Time (n) = 2 years

Compound\quad Interest(C.I)=p[(1+\frac { r }{ 100 } )^{ n }-1]\\\ \\ C.I=5000[(1+\frac { 8 }{ 100 } )^{ 2 }-1]\\\ \\ C.I=5000[(\frac { 100+8 }{ 100 } )^{ 2 }-1]\\\ \\ C.I=5000[(\frac { 27 }{ 25 } )^{ 2 }-1]\\\ \\ C.I=5000(\frac { 729 }{ 625 } -1)\\\ \\ C.I=5000(\frac { 729-625 }{ 625 } )\\\ \\ C.I=5000(\frac { 104 }{ 625 } )\\\ \\ C.I=\frac { 5000\times 104 }{ 625 } \\\ \\ C.I=\frac { 520000 }{ 625 }

C.I = Rs 832

Therefore, difference in interest==> 832 – 600 = 232 rs
Hence rs 232 is the right answer

### (09) The compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually is?

Principal (P) = Rs. 8000

Interest for first year = 15% of 8000
==> 8000 * 15/100 ==> 1200 Rs.
So, new principal after 1st year = 8000 + 1200 => 9200 Rs

Interest for second year = 15% of 9200
==> 9200 * 15/100 ==> 1380 Rs

So new principal after second year = 9200 + 1380 = 10580 Rs.

Interest for 4 months (1/3 years) ==> 10580 * 1/3 * 15/100 => Rs. 529

So, interest after 2 years 4 months => 1200+1380+529=> 3109 rs.

### (10) At what rate per annum will Rs 32000 yield a compound interest of rs. 5044 in 9 months interest being compounded quarterly?

Here
Principal (P) = Rs. 32,000
Compound Interest (C.I) = 5044 Rs
Time (n)= 9 months ==> 9/12 year ==>4/3 year

Let Rate of interest be R% per annum
If the interest is compounded quarterly then Rate of interest ==> R/4 % compounded quarterly

C.I=P[(1+\frac { R }{ 100 } )^{ n }-1]\\\ \\ Interest\quad is\quad compounded\quad quarterly\\\ \\ =>5044=32000[(1+\frac { R }{ 400 } )^{ 3 }-1]\\\ \\ =>\frac { 5044 }{ 32000 } =(1+\frac { R }{ 400 } )^{ 3 }-1\\\ \\ =>(1+\frac { R }{ 400 } )^{ 3 }-1=\frac { 1261 }{ 8000 } \\\ \\ =>(1+\frac { R }{ 400 } )^{ 3 }=1+\frac { 1261 }{ 8000 } \\\ \\ =>(1+\frac { R }{ 400 } )^{ 3 }=\frac { 9261 }{ 8000 } =(\frac { 21 }{ 20 } )^{ 3 }\\\ \\ =>1+\frac { R }{ 400 } =\frac { 21 }{ 20 } \\\ \\ =>\frac { R }{ 400 } =\frac { 21 }{ 20 } -1=\frac { 1 }{ 20 } \\\ \\ R=\frac { 400 }{ 20 } =20percent\quad per\quad annum\

Hence the required rate of interest is 20% per annum