**Compound Interest Examples :**

In this section, we are going to see some examples on compound interest.

The formula given below can be used to find accumulated value in compound interest.

**A = P(1+i)ⁿ**

A = Accumulated value (final value of an investment)

P = Principal (initial value of an investment)

**How to find the value of "i" : **

To find the value of "i", first convert the rate of interest in to decimal form. Then divide the decimal value by the number of conversion periods per year.

For example,in an investment rate of interest is 15% and compounded quarterly. First we write 15% as decimal form.That is 0.15. Now we have to divide this value by 4 (because compounded quarterly).

Finally, i = 0.15/4 = 0.0375

**How to find the value of "n" : **

n = (no. of years) X (no. of conversion periods per year)

For example, in an investment no.of years = 2 and compounded quarterly.The value of n = 2x4 = 8.

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 simple interest and compound interest is same ($100) for the 1st year when interest is compounded annually in compound interest.

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 ?

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.

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 ?

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

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)

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

Therefore, x = 441/(1+0.05)

x = 420

P = y,A = 441,i = 0.05/1 = 0.05 (Compounded annually, So divided by 1)

Therefore, y = 441/(1+0.05)

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 ?

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

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 ?

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

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)

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)

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 ?

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.

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)

Here P =12000,i = 0.1/1 = 0.1,n = 3X1 = 3 (compounded annually)

A = 12000(1+0.1)

A = 15972

C.I = A - P = 15972 - 12000 = 3972

Hence 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 compound interest.

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

HTML Comment Box is loading comments...

You can also visit our following web pages on different stuff in math.

**WORD PROBLEMS**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Trigonometry word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**