# COMPOUND INTEREST EXAMPLES

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

 (A) 5 years                   (B) 6 years (C) 7 years                   (D) 8 years

jQuery UI Dialog functionality
Let "P" be the amount invested initially.

From the given information, P becomes 2P in 3 years.
Since the investment is in compound interest, in the 4th year the principal 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.

2. A man borrows a certain sum of money and pays back in two years in two installments. If compound interest is reckoned at 5% per year compounded annually and he pays back annually \$ 441, what sum did he borrow?

 (A) \$ 820                   (B) \$ 825 (C) \$ 830                   (D) \$ 835

jQuery UI Dialog functionality
Let "x" be the amount borrowed which becomes \$ 441 in one year.
Let "y" be the amount borrowed which becomes \$ 441 in two years.

We know the formula in compound interest. That is
A = P(1+i)n ===> P = A/(1+i)n

P = x,A = 441,i = 0.05/1 = 0.05 (Compounded annually, So divided by 1)
Therefore, x = 441/(1+0.05)1= 441/1.05 = 420
x = 420

P = y,A = 441,i = 0.05/1 = 0.05 (Compounded annually, So divided by 1)
Therefore, y = 441/(1+0.05)2= 441/1.1025 = 400
y = 400

Total = x+y = 420+400 = 820

Hence, the total money borrowed is \$ 820

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?

 (A) \$ 11,000              (B) \$ 12,000 (C) \$ 13,000              (D) \$ 14,000

jQuery UI Dialog functionality
Simple interest for two years = 1200 and interest for one year = 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 if it were simple interest for one year. That is principle = 600, interest = 30
I = Pit ===> 30 = 600i(1) ===> i = 0.05 ===>rate of interest = 5%

In the given problem, simple interest earned in two years = 1200.
I = Pit ===> 1200 = PX0.05X2 ===> P = 1200/0.1 = 12000

Hence, the principal is \$ 12,000

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?

 (A) \$ 10676                   (B) \$ 10776 (C) \$ 10876                   (D) \$ 10976

jQuery UI Dialog functionality
The formula for accumulated value in C.I is A = P(1+i)n
To find the accumulated value for the first year, plug P = 15000, i = 0.12, n = 1 in the above formula

A = 15000(1+0.12)1 = 15000(1.12) = 16800
Amount paid at the end of 1st year = 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 = 9800X1.12 = 10976

Hence to completely discharge the loan, at the end of 2nd year, he has to pay \$ 10,976

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

 (A) \$ 3672                   (B) \$ 3772 (C) \$ 3872                   (D) \$ 3972

jQuery UI Accordion - Default functionality
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 = Pit
Here I = 60, P = 100, t = 6
60 = 100i(6) ===> i = 60/600 = 0.1 = 10%
Therefore, rate of interest is 10% or i = 0.1

In C.I, formula for accumulated value, A = P(1+i)n
Here P =12000,i = 0.1/1 = 0.1,n = 3X1 = 3 (compounded annually)
A = 12000(1+0.1)3 = 12000(1.331)
A = 15972
C.I = A - P = 15972 - 12000 = 3972

Hence the compound interest after 3 years at the same rate of interest is \$ 3972.

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