COMPOUND INTEREST PROBLEMS

About "Compound interest problems"

Compound interest problems :

We might have already known the stuff "simple interest" in solving interest and investment problems. In this section, we are going to see, how to solve problems which involve compound interest (C.I).

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

The formula given below can be used to find accumulated value in C.I

Compound interest problems - Examples

Example 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

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 x 1.4641

A = $1171.28

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.

Since 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.

Hence, 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 x 0.05 x 2

1200 = P x 0.1

Divide both sides by 0.1.

1200/0.1 = P

12000 = P

Hence, 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 ?

COMPOUND INTEREST PROBLEMS

About "Compound interest problems"

Compound interest problems - Examples

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