**Compound Interest Word Problems Worksheet :**

Worksheet given in this section is much useful to the students who would like to practice solving word problems on compound interest.

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

**Problem 1 :**

Mr. George invests $800 in an account which pays 20% compound interest per year. If interest is compounded half yearly, find the the accumulated value and compound interest after 2 years.

**Problem 2 :**

A person invests certain amount in compound interest scheme. If the money invested is doubled itself in 3 years, how long will it take for the money to become 4 times of itself ?

**Problem 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 ?

**Problem 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 ?

**Problem 5 :**

There is 50% increase in an amount deposited in simple interest for 5 years. What will be the compound interest of $10,000 after 3 years at the same rate ?

**Problem 1 :**

Mr. George invests $800 in an account which pays 20% compound interest per year. If interest is compounded half yearly, find the the accumulated value and compound interest after 2 years.

**Solution :**

The formula to find accumulated value in compound interest is

**Given :** P = 800, r = 20% or 0.2, n = 2 and t = 2.

Then, we have

r /n = 0.2 / 2 = 0.1

nt = 2 ⋅ 2 = 4

Finding accumulated value :

A = 800(1 + 0.1)⁴

A = 800(1.1)⁴

A = 800 x 1.4641

A = $1171.28

Finding compound interest :

C. I = Accumulated value - Principal

C. I = 1171.28 - 800

C. I = 371.28

Hence, the accumulated value is $1171.28 and the compound interest is $371.28.

Let us look at the next problem on "Compound interest word problems worksheet".

**Problem 2 :**

A person invests certain amount in compound interest scheme. If the money invested is doubled itself in 3 years, how long will it take for the money to become 4 times of itself ?

**Solution :**

Let "P" be the amount invested initially.

It is given that the money is doubled in 3 years.

That is, P becomes 2P in 3 years.

Because the investment is in compound interest, the principal in the 4th year will be 2P.

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

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

Hence, the amount deposited will become 4 times of itself in 6 years.

Let us look at the next problem on "Compound interest word problems worksheet".

**Problem 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 :**

**Given :** Simple interest for two years is $1200.

So, simple interest for one year is $600.

**Given :** Compound interest for two years is $1230.

**Fact : **When it is compounded annually, interest earned in both compound interest and simple interest for one year on the same principal would be same.

So, compound interest for 1st year is $600 and for 2nd year is $630.

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

Because, $600 interest earned in the first year earned this additional $30 interest in the second year.

Because it is compounded annually, it can be considered as simple interest for one year.

That is,

Principal = 600, Interest = 30

The formula to find simple interest is

I = P ⋅ r ⋅ t

Plug I = 30, P = 600 and t = 1.

30 = 600 ⋅ r ⋅ 1

30 = 600r

Divide both sides by 600.

30 / 600 = r

0.05 = r

5% = r

**Given : **The rate of interest is same for both compound interest.

So, the rate of interest in simple interest is 5%.

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

I = P ⋅ r ⋅ t

Plug I = 1200, r = 5% or 0.05 and t = 2.

1200 = P ⋅ 0.05 ⋅ 2

1200 = P ⋅ 0.1

Divide both sides by 0.1

1200 / 0.1 = P

12000 = P

Hence, the principal is $12,000.

Let us look at the next problem on "Compound interest word problems worksheet".

**Problem 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

To find the accumulated value for the first year,

Plug P = 15000, r = 12 % or 0.12, n = 1 in the above formula

A = 15000(1 + 0.12)^{1}

A = 15000(1.12)

A = 16800

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

So, the balance to be repaid is

= 16800 - 7000

= 9800

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

Now we need to calculate the accumulated value for the principal 9800 in one year.

That is, the amount to be paid at the end of 2nd year to completely discharge the loan.

A = 9800(1 + 0.12)^{1}

A = 9800 ⋅ 1.12

A = 10976

Hence, he has to pay $10,976 at the end of the second year to completely discharge the loan.

Let us look at the next problem on "Compound interest word problems worksheet".

**Problem 5 :**

There is 50% increase in an amount deposited in simple interest for 5 years. What will be the compound interest of $10,000 after 3 years at the same rate ?

**Solution :**

**Given :** There is 50% increase in an amount in 5 years at simple interest.

Then , the increase in 1 year is

= 50% / 5

= 10%

The amount deposited in simple interest is increased by 10% in 1 year.

So, the rate of interest in simple interest is 10%.

**Given :** The rate of interest is same for both compound interest and simple interest

Because the rate of interest is same for both compound interest and simple interest, we can use the rate of interest 10% in compound interest.

To know compound interest for 3 years, plug

P = 12000, r = 0.1, n = 1 and t = 3

in the formula of compound interest.

So, we have

A = 10000(1 + 0.1)^{3}

A = 10000(1.1)^{3}

A = 10000 ⋅ 1.331

A = 13,310

Compound interest for 3 years :

C.I = A - P

C.I = 13,310 - 10000

C.I = 3,310

Hence the compound interest after 3 years at the same rate of interest is $3310.

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