COMPOUND INTEREST PROBLEMS

Compound Interest Problems :

In this section, you will learn how to solve problems on compound interest.

The formulas given below will be useful to solve problems on compound interest. 

A  =  P(1 + r/n)^nt

C.I  =  A - P

A -----> Accumulated value (final value )  

P -----> Principal (initial value of an investment) 

r -----> Annual interest rate (in decimal)

n -----> Number of times interest compounded per year

t -----> Time (in years)

C.I -----> Amount of interest

You might have already known the stuff simple interest in solving interest and investment problems.

The picture shown below clearly explains the difference between simple interest and compound interest.

Important Note : 

When we look at the above picture, it is clear that interest earned in S.I and C.I is same ($100) for the 1st year when interest is compounded annually in C.I. 

Compound Interest Problems

Problem 1 :

$800 is invested in C.I where the rate of interest is 20% per year. If interest is compounded half yearly, what will be the accumulated value and C.I after 2 years ?

Solution :

The formula to find accumulated value in C.I is 

A  =  P(1 + r/n)^nt

A -----> Accumulated value (final value )  

P -----> Principal (initial value of an investment) 

r -----> Annual interest rate (in decimal)

n -----> Number of times interest compounded per year

t -----> Time (in years)

C.I -----> Amount of interest

Here,

P  =  800

r  =   20%   =  0.2

n  =  2

t  =  2

Then, the accumulated value is

A  =  800(1 + 0.1)⁴

A  =  800(1.1)⁴

A  =  800 ⋅ 1.4641

A  =  $1171.28

Compound interest is

C.I  =  A - P 

C.I  =  1171.28 - 800

C.I  =  $371.28

Example 2 :

A sum of money placed at compound interest doubles itself in 3 years. In how many years will it amount to four times itself ?

Solution :

Let P be the amount invested initially. 

From the given information,

P becomes 2P in 3 years

Because the investment is in C.I, the principal in the 4th year will be 2P.

And 2P becomes 4P (it doubles itself) in the next 3 years. 

Therefore, at the end of 6 years accumulated value will be 4P.

So, the amount deposited will amount to 4 times itself in 6 years.

Example 3 : 

The compound interest and simple interest on a certain sum for 2 years is $ 1230 and $ 1200 respectively. The rate of interest is same for both compound interest and simple interest and it is compounded annually. What is the principle ?

Solution :

Simple interest for two years  is 1200 and interest for one year  is 600 

So, C.I for 1st year is 600 and for 2nd year is 630.

(Since it is compounded annually, S.I and C.I for 1st year would be same) 

When we compare the C.I for 1st year and 2nd year, it is clear that the interest earned in 2nd year is 30 more than the first year. 

Because, interest 600 earned in 1st year earned this 30 in 2nd year. 

It can be considered as simple interest for one year.

That is, principal  =  600, interest  =  30 

I  =  Pit

30  =  600i(1)

0.05  =  i

5%  =  i

In the given problem, simple interest earned in two years is 1200. 

I  =  Pit

1200  =  P ⋅ 0.05 ⋅ 2

1200  =  P ⋅  0.1

Divide both sides by 0.1.

1200/0.1  =  P

12000  =  P

So, the principal is $ 12,000.

Example 4 :

Mr. David borrowed $ 15,000 at 12% per year compounded annually. He repaid $ 7000 at the end of 1st year. What amount should he pay at the end of second year to completely discharge the load ?

Solution : 

The formula to find accumulated value in C.I is 

A  =  P(1 + r/n)^nt

A -----> Accumulated value (final value )  

P -----> Principal (initial value of an investment) 

r -----> Annual interest rate (in decimal)

n -----> Number of times interest compounded per year

t -----> Time (in years)

C.I -----> Amount of interest

To find the accumulated value for the first year, 

Substitute P = 15000, i = 0.12, n = 1 in the above formula. 

A  =  15000(1 + 0.12)¹ 

A  =  15000(1.12)

A  =  16800 

Amount paid at the end of 1st year is 7000. 

Balance to be repaid :

=  16800 - 7000 

=  9800 

This 9800 is going to be the principal for the 2^nd year. 

Now we need the accumulated value for the principal 9800 in one year. (That is, at the end of 2nd year) 

A  =  9800(1 + 0.12)¹ 

A  =  9800 ⋅ 1.12

A  =  10976 

So, to completely discharge the loan, at the end of 2nd year, Mr. David has to pay $ 10,976.

Example 5 :

There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate ? 

Solution :

Let the principal in simple interest be $ 100. 

Since there is 60% increase, simple interest = 60 

We already know the formula for S.I.

That is, 

I  =  PRT/100 

Here, I  =  60, P  =  100, t  =  6 

60  =  (100 ⋅ R ⋅ 6) / 100

60  =  6R

Divide both sides by 6.

10  =  R or R  =  10%

Because the rate of interest is same for both S.I and C.I, we can use the rate of interest 10% in C.I.

To know compound interest for 3 years,

Substitute P = 12000, r = 0.1, n = 1 and t = 3 in the formula of C.I.

A  =  12000(1 + 0.1)³ 

A  =  12000(1.331) 

A  =  15972 

C.I  =  A - P

C.I  =  15972 - 12000

C.I  =  3972 

So, the compound interest after 3 years at the same rate of interest is $ 3972.

After having gone through the stuff given above, we hope that the students would have understood, how to solve problems on compound interest. 

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