Compound interest for $2$ years

Question

If total compound interest for $2$ years is Rs. $315$ and that for $3$ years is Rs. $507$. Find rate of interest.

I tried solving it with the formula $A=P(1+\frac{R}{100})^T$ where $A$ is Amount, $P$ is Principal, $R$ is Rate of interest and $T$ is time period. But I got a cubic equation in $R$. I wish there was a simpler way to solve it.

EDIT: Thanks for your wonderful answers. But I guess I should have added that it's a school level question. It is supposed to be done under 1 minute.

If we are making quadratic or cubic equation, we are taking too long.

Sometimes, hit and try also works. e.g. My first instinct was to take P as 1500, R as 10%. Because then for 1st year, interest would have been 150. For 2nd year, it would have been 150+15. So, for 2 years, it would have become 315, as desired. But this assumption didn't fit for the 3rd year.

If you can answer on these lines, that would be awesome, thanks.

Answer 1

Your wish is the alleviation of my boredom.

$P\left (\left(1+\dfrac{R}{100} \right)^2 -1\right)=315$

$$P\left(2+\dfrac{R}{100} \right)\left(\dfrac{R}{100} \right)=315 \tag 1$$

$P\left(\left(1+\dfrac{R}{100} \right)^3-1 \right)=507$

$P\left(\dfrac{R}{100} \right)\left(\left(1+\dfrac{R}{100}\right)^2+ \left(1+\dfrac{R}{100}\right)+1\right)=507$

$$P\left(\dfrac{R}{100} \right)\left(\left(1+\dfrac{R}{100}\right)\left(2+\dfrac{R}{100}\right)+1 \right)=507 \tag 2$$

Dividing $(2)$ by $(1)$ yields:

$\dfrac{(1+x)(2+x)+1}{2+x}=\dfrac{507}{315}$

Where $x=\dfrac{R}{100}$ because MathJax is ever so easy to type.

$\boxed{\dfrac{x^2+3x+3}{x+2}=\dfrac{169}{105}}$

I think the quadratic here is quite visible.

And you probably won't click this but just in case:

$(7x-1)(15x+23)=0$

And $x$ is definitely not negative!

Answer 2

Exponential functions are written in the form $$f(t)=Ae^{kt}.$$

In this case, we have two equations: \begin{align} f(2)=315+A&=Ae^{2k}\tag1 \\ f(3)=507+A&=Ae^{3k}\tag2 \end{align}

Rewriting these equations, we have \begin{align} 315&=A(e^{2k}-1)\tag3 \\ 507&=A(e^{3k}-1)\tag4 \\ \end{align}

By division $(4)\div(3)$, we have \begin{align} \frac{507}{315}&=\frac{e^{3k}-1}{e^{2k}-1} \end{align}

Sub $t=e^k$. Then we have \begin{align} \frac{507}{315}&=\frac{t^3-1}{t^2-1}=\frac{t^2+t+1}{t+1} \end{align}

Cross multiply and solve for $t$, then $t-1$ will be your rate of interest.

Answer 3

Compound interest but compounded how often? With nothing being said, I would assume annually. And with that few compounding intervals, I would not use a formula but calculate it directly.

If the original amount is "P" and the interest is "r" then the interest for the first year is rP so the amount at the end of the first year is P+ rP= P(1+ r). The interest for the second year is P(1+ r)r so the amount at the end of the second year is P(1+ r)+ P(1+r)r= P(1+r)^2= P(1+ 2r+ r^2)= P+ P(2r+ r^2). The interest earned in those two years is P(2r+ r^2)= 315.

For the third year the interest is P(1+r)^2r= P(r+ 2r^2+ r^3)= 507. We can eliminate P by dividing one equation by the other: $\frac{P(2r+ r^2)}{P(r+ 2r^2+ r^3)}= \frac{315}{507}$. The "P"s cancel and we can make this a polynomial equation by "cross multiplying": $507(r^2+ 2r)= 315(r^3+ 2r^2+ r)$ $507r^2+ 1014r= 315r^3+ 630r^2+ 315r$ $315r^3+ 123r^2- 699r= 0$

Yes, that is a cubic equation (HORRORS!) but there is no "constant" term so we can immediately factor an r: $r(315r^2+ 123r- 699)= 0$.

r= 0 is a solution but probably not the one we want. The other two solutions, the roots of $315r^2+ 123r- 699= 0$ can be calculated by "completing the square" or using then "quadratic formula".