Find Rate when compound interest for successive years are given.

Question

A certain sum is lent at Compound Intrest

The interest earned in 2 years is 272.

The interest earned in 3 years is 434.

Find rate of interest? This is the proper question

Please tell me shortcut(if possible) of these type of questions for competitive exams.

Answer 1

Since the results were not very comfortable it could be that the interest after 3 years are 434. Maybe you have made typo.

The equations are

$C_0\cdot (1+r)^2-C_0=272$

$C_0\cdot (1+r)^3-C_0=434$

$C_0$ is the inital capital and $r$ is the interest rate. Dividing the second equation by the first equation. $C_0$ is cancelling out.

$\frac{(1+r)^3-1}{(1+r)^2-1}=\frac{434}{272}$

Up to here there is no simple way to calculate r. You maybe can substitute:

$1+r=q, \frac{434}{272}=a$

$\frac{q^3-1}{q^2-1}=a$

$q^3-1=a\cdot q^2-a$

$q^3-a\cdot q^2+a-1=0$

Since it is a cubic equation it is still a challenge to solve it. One way is to use the Cardano's method.

At the end of the calculation you should get $r=0.125$. And therefore $C_0=\frac{272}{1.125^2-1}=1024$

Answer 2

If the principle is $p$ and the interest rate is $r$, after two years you have $p(1+r)^2$, so have earned $p((1+r)^2-1)$ in interest. After $3$ years it is the same with a $3$. Dividing the two gets rid of $p$.

Answer 3

Let R= 1+r;

Ratio of CI from formula ( assuming unit principal, a step omitted as an exercise )

$$ 272./439. = \frac{(R + 1)}{ (R^2 + R + 1) }\rightarrow R = 1.14854, $$

numerically. The interest is 14.854 percent.