# Simple and Compound Interests

A little bit of heavy duty calculation?

## Compound Interests

Q.16: On a certain sum of money, compound interest earned at the end of three years = Rs. 1456. Compound interest at the end of two years is Rs. 880. Compute the principal invested.
1. Rs. 2,400
2. Rs. 2,800
3. Rs. 2,000
4. Rs. 1,600

Choice C. Rs. 2,500

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## Detailed Solution

Let principal = P, rate of interest = r%

CI earned at the end of three years = P(1 + r)3 - P = 1456
=> P(3r2 + 3r + r3) = 1456

CI earned at the end of two years = P(1 + r)2 - P = 880
=> P(r2 + 2r) = 880

Dividing one by the other we get: ${\frac{ 3r^{2} + 3r + r^{3} }{ r^{2}+2r }= \frac{1456}{880}}$

We can cancel ‘r’ and solve the resulting quadratic. However, let us see if we can spot something in the numbers.

${\frac{ 1456 }{ 880 }= \frac{728}{440}}$

Remember that the amounts at the end of three years and two years are linked to (1 + r)3and (1 + r)2 respectively.

Now observe that 1728 = 123 and 144 is 122.

So,
${\frac{728}{440} = \frac{1728 - 1000}{1440 - 1000} = \frac{1.728 - 1}{1.44 - 1}}$

=>r = 20%

CI at the end of 2 years = (1.22 – 1)P => 880 = 0.44P => P = Rs. 2000.

Alternatively:

(r2 + 3r + 3) x 440 = (r + 2) x 728
(r2 + 3r + 3) x 55 = (r + 2) x 91
55r2 + 165r + 165 = 91r + 182
55r2 + 74r -17 = 0
55r2 + 85r - 11r - 17 = 0
5r (11r + 17) -1 (11r + 17) = 0

r = 0.2 or a negative number. Or, r has to be 20%.

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