COMPOUND INTEREST EXAMPLE PROBLEMS

Problem 1 :

a) What will an investment of $3000 at 10% p.a. Compound interest amount to after 3 years?

b) What part of this is interest?

Solution :

P= $3000, R = 10% ==> 0.10 and T = 3 years

a)  Amount A = P(1 + R/100)^t

= 3000(1 + 0.1)³

= 3000(1.1)³

= 3000(1.331)

= $3993

b) Compound interest = Amount – Principal

= 3993 – 3000

= $993

So, interest is $993.

Problem 2 :

How much compound interest is earned by investing $20,000 for 4 years at 12% p.a.?

Solution :

P= $20000, R = 12% and T = 4 years

Amount A = P(1 + R/100)^t

= 20000(1 + 0.12)⁴

= 20000(1.12)⁴

= 20000(1.57351936)

= 31470.39

Compound interest = Amount – Principal

= 31470.39 – 20000

= 11470.39

So, compound interest is $11470.39.

Problem 3 :

$5000 is invested for 2 years at 10% p.a. What will this investment amount to if the interest is calculated as :

a) Simple interest

b) Compound interest?  

Solution :

P= $5000, R = 10% ==> 0.10 and T = 2 year

a)  Simple interest I = (P × T × R)/100

= (5000 × 2 × 10)/100

= 100000/100

= 1000

So, simple interest is $1000.

Total amount = P + I

= 5000 + 1000

= 6000

So, total amount is £6000.

b) Compound interest = P(1 + R/100)^t

= 5000(1 + 0.1)²

= 5000(1.1)²

= 5000(1.21)

= $6050

So, compound interest is $6050.

Problem 4 :

a) What will an investment of $30,000 at 10% p.a. Compound interest amount to after 4 years?

b) What part of this is interest?

Solution :

P= $30000, R = 10% and T = 4 years

a) Amount A = P(1 + R/100)^t

= 30000(1 + 0.1)⁴

= 30000(1.1)⁴

= 30000(1.4641)

= $43923

b) Compound interest = Amount – Principal

= 43923 – 30000

= $13923

So, interest is $13923.

Problem 5 :

How much compound interest is earned by investing €80,000 at 9% p.a. over a 3 year period?

Solution :

Principal amount P= $80000

Rate R = 9%

Time T= 3 years

Amount A = P(1 + R/100)^t

= 80000(1 + 0.09)³

= 80000(1.09)³

= 80000(1.295029)

= 103602.32

Compound interest = Amount – Principal

= 103602.32 – 80000

= 23602.32

So, compound interest is $23602.32.

Problem 6 :

$6000 is invested for 2 years at 15% p.a. What will this investment amount to if the interest is calculated as :

a) Simple interest

b) Compound interest?  

Solution :

P= $6000, R = 15% ==> 0.15 and T = 2 years

a) Simple interest = (P × T × R)/100

= 6000 × 2 × 0.15

 = 1800

So, simple interest is $1800.

Total amount = P + I

= 1800 + 6000

= 7800

So, total amount is $7800.

b)  Compound interest = P(1 + R/100)^t

= 6000(1 + 0.15)²

= 6000(1.15)²

= 6000(1.3225)

= 7935

So, compound interest is $7935.

Problem 7 :

The difference between the compound interest and simple interest on a certain sum of money for 3 years at the rate of 4% per annum is $76, then what is the principal ?

a) $16725     b) $12925        c) $15625      d)  $18825

Solution :

Let P be the principal investing.

Simple Interest = PTR/100

= (P ⋅ 3 ⋅ 0.04)

= 0.12P -----(1)

Compound interest = P(1 + R/100)^t - P

= P[(1 + R/100)^t  - 1]

= P[(1 + 0.04)³ - 1]

= P[(1.04)³ - 1]

= P[1.1248 - 1]

= P(0.1248)-----(2)

(2) - (1)

P(0.1248) - 0.12P = 76

0.0048P = 76

P = 76/0.0048

P = 15833.3

So, option d is correct.

Problem 8 :

When a tennis ball is dropped, it bounces and then rises. The ball rises to 60% of the height from which it is dropped. The ball is dropped from a height of 2 metres.

(a) Calculate the height of the rise after the first bounce.

b) Calculate the height of the rise after the second bounce. The ball carries on bouncing, each time rising to 60% of the last rise.

(c) For how many bounces does it rise to a height greater than 20 cm? Show your working

Solution :

a) Let H be the height of the ball when it drops.

Height of the ball after it bounce first,

= 2 x 1 x (60/100)

= 2 x 0.6

= 1.2 

Height of the rise after the first bounce is 1.2 m.

b) Height of the ball after the second rise.

= 1.2 x 1 x 0.6

= 0.72

So, the height of the ball after the second rise is 0.72 m.

c) 

So, 5 bounces.

Problem 9 :

A $1000 investment is made in a trust fund that pays 12% a compounded monthly. How long will it take the investment to grow to $5000?

Solution :

P = 1000, rate of interest = 12%

= P(1 + R/12)^12t

= 1000(1 + 0.12/12)^12t

= 1000(1 + 0.01)^12t

5000 = 1000(1.01)^12t

5 = (1.01)^12t

log 1.01/log 5 = 12t

0.0043/0.6989

0.006 = 12t

t = 0.006/12

t = 5.12

Approximately 5 years.