Problem related to compound interest arithmetic.

Question

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

You should end up with a quadtratic term in P which is solvable easily. What have you done so far? — Tony Hellmuth, Jun 02 '18 at 04:04
I eneded up with $r$ having power 2 and 3.. I am stuck after that — user533186, Jun 02 '18 at 04:10

score 3 · Answer 1 · answered Jun 02 '18 at 04:13

3

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?

answered Jun 02 '18 at 04:13

Tony Hellmuth

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I understand your process of solving ! Thanks for the hint. Let me try.. – user533186 Jun 02 '18 at 04:20
1

Alternatively, (1) and (2) give $P=8800/(X-1)=10648/(X^3-1)$ so $8800(X^2+X+1)=10648$, etc. – Gerry Myerson Jun 02 '18 at 04:34
Yep was about to add that. – Tony Hellmuth Jun 02 '18 at 04:36

Problem related to compound interest arithmetic.

1 Answers1