# Miscellaneous Formulae (Compound Interest)

## Formula 1

If a sum of money P amounts to $A_1$ after n years at CI and the same sum of money amounts to $A_2$ after (n + 1) years at CI, then

r = $\frac{(A_2 – A_1)}{A_1}$ x 100

##### Q. If a sum of money amounts to Rs. 2420 in 2 years and to Rs. 2662 in 3 years, then what was the original sum lent (assume compound interest)?

Explanations :

If a sum of money P amounts to $A_1$ after n years at CI and the same sum of money amounts to $A_2$ after (n + 1) years at CI, then

r = $\frac{(A_2 – A_1)}{A_1}$ x 100 = $\frac{(2662 – 2420)}{2420}$ x 100 = $\frac{242}{2420}$ x 100 = 10%

So, P amounts to Rs. 2420 in 2 years at 10%.

Amount = P (1 + $\frac{r}{100})^n$

Or 2420 = P (1 + $\frac{10}{100})^2$

Or P = 2420 x ($\frac{10}{11})^2$ = Rs. 2000

Amount after 3 years - Amount after 2 years = 2662 - 2420 = Rs. 242

This difference is the interest earned on the amount of the previous year, i.e. on Rs. 2420.

So, P = Rs. 2420, SI = Rs. 242 and n = 1 year

Now, SI = Prn/100 (we can apply the SI formula, as n = 1 and for one year CI = SI)

Or r = (SI x 100) / Pn = (242 x 100) / (2420 x 1) = 10%

So, P amounts to Rs. 2420 in 2 years at 10%.

Amount = P (1 + $\frac{r}{100})^n$

Or 2420 = P (1 + $\frac{10}{100})^2$

Or P = 2420 x ($\frac{10}{11})^2$ = Rs. 2000