Problem related to compound interest arithmetic.

49 Views Asked by At

A certain amount of money gets a compound interest of 8800 rupees (Indian Currency) in it's first year and 10,648 rupess in third year. I have to find the amount of money(i.e principal) and rate of compound interest. Also the the interest is compounded annually.

I know the formulae of compound interest, but I am unable to find the required answers.

If $P$ is the prinicpal and $r$ is the rate of compound interest, then $$ P(1+\frac{r}{100})^{1} = P + 8800 \\\text{and} \; P(1+\frac{r}{100})^{3} = P+ 10,648. $$

I have to solve this two equation to get $ P \; \text{and} \; r$, which is quite lengthy. Any insight will be very helpful.

Thank you

1

There are 1 best solutions below

3
On

Hint:

Let $X=(1+\frac r{100})$

We have $$PX=P+8800 \ ... (1)$$ and $$PX^3=P+10648 \ ... (2)$$

Now from equation $(1)$ we get:

$$(1) \ \implies X=\frac {P+8800}P$$

Substituting into $(2)$:

$$P\left( \frac {P+8800}P \right)^3=P+10648$$

Can you take it from here?