I cannot find the rate of interest on the given sum Given Sum:
At what rate % per annum will ₹12000 yield ₹13891.50 as a compound interest in 3 years?
According to the formula of compound interest, to find the interest: $$\text{C.I} = \text{P}([1+\frac{\text{R}}{100}]^n - 1)$$
After substitution, I am getting something like:
$$13891.50 = \text{12000}([1+\frac{\text{R}}{100}]^3 - 1)$$
$$\frac{13891.50}{12000} = [1+\frac{\text{R}}{100}]^3 - 1$$
$$\frac{13891.50+12000}{12000} = [1+\frac{\text{R}}{100}]^3$$
That's all I could do.
But to find the rate, I need to get rid of the exponent first. But neither of the numbers are a perfect cube. I was told that the rate is a whole number.
I found this post where the question is almost the same, except that it is $1891.50$ instead of $13891.50$.
I am starting to doubt if the question is wrong.
Do you have any tips? Thanks in advance!
 A: The results of my first answer seems correct for the given formula but the results seem like nonsense. The formula I have always seen is
$$A=P(1+R)^{\Large{n}}\quad\text{ where }
 A=13891.50,\space P=12000,\space n=3$$
$$A=P(1+R)^{\Large{n}}\\
\implies \frac{A}{P}=(1+R)^{\Large{n}}\\
\implies R=\sqrt[\Large{n}]{\frac{A}{P}}-1
=\sqrt[\Large{3}]{\frac{13891.50}{12000}}-1\\
=0.05\\ =\space 5\%$$$
The problem with this formula is that compounding takes place only once a year. A more general formula is
$
A =\space P\big(1+\dfrac{r}{n}\big)^{nt}\space
\text{  where}$
$
\text{A = final amount after interest}\\
\text{P = original principal}\\
\text{n is the number of payments per year}\\
\text{r = annual interest rate}\\
\text{t = number of years}
$
and the results are the same if
$\space t=year\space $ and $\space n=1$
A: One way to solve this would be to find the ratio between your starting and ending values. This gives your percent increase for the total time period. $\dfrac{13891.50}{12000}=1.157625$ or a $15.7625\%$ increase over three years. To then find the yearly interest, simply take the cube root of this number. $^3\sqrt{1.157625}=1.05$ This means that for each year there was a 5% interest rate.
A: You just didn't go far enough and it's often easier to keep things in algebraic form until the last minute.
\begin{align*}
C &= P\bigg(\bigg[1+\frac{R}{100}\bigg]^{\Large{n}} - 1\bigg)\\
\implies \frac{C}{P}+1&=\bigg(\frac{R}{100}+1\bigg)^{\Large{n}}\\
\implies \frac{R}{100}+1&=\sqrt[\huge{n}]{\frac{C}{P}+1}\\
\implies R&=100\bigg(\sqrt[\huge{n}]{\frac{C}{P}+1}-1\bigg)\\
&=100\bigg(\sqrt[\huge{3}]{\frac{₹13891.50}{₹12000}+1}-1\bigg)\\
\end{align*}
A: Firstly, the compound interest formula you have here strikes me as odd. I'm used to seeing something like this:
$\begin{align*}
(1) A = P(1 + \frac{r}{n})^{nt}
\end{align*}$
The formula you gave is a variant on that: subtracting the principal from both sides, C.I. represents A-P, or "the money you made on your principal".
$\begin{align*}
(2)    A-P=C.I. = P[(1 + \frac{r}{n})^{nt} -1]
\end{align*}$
Which is just the formula I gave (1) with P taken from both sides.
A great free online resource for getting step-by-step guides to solving algebraic problems like this is symbolab.com (The link has the step-by-step solution for your exact problem; note that R = $100\sqrt[3]{1.157625} \approx 5\%$).
Another great resource (but step-by-step is behind a paywall) is: wolframalpha.com
That being said, here is how to answer your question:
At what rate % per annum will ₹12000 yield ₹13891.50 as a compound interest in 3 years?
According to the formula of compound interest, to find the interest:
$\begin{align*}
C.I=P([1+\frac{R}{100}]^n−1)
\end{align*}$
Actually, the problem doesn't contain complete information, because it doesn't say the period of compounding. I'll compute the answer for a compounding interval of every year (it's probably this), every quarter, every month, and (not realistic, but sometimes used in high school algebra:) "continuously" (i.e. the result if compounding is done at infinitesimal intervals; similar to if the interest was compounded every hour).
Every year:
$\begin{align*}
C.I=P([1+\frac{R}{100}]^n−1) \\
\implies ₹13891.50 - ₹12,000= (₹12,000)*([1 + \frac{R}{100}]^3 -1) \\
\implies \frac{₹1,891.50}{₹12,000}= 0.157625 = ([1 + \frac{R}{100}]^3 -1) \\
\implies 1.157625 = [1 + \frac{R}{100}]^3 \\
\implies \sqrt[\huge{3}]{1.157625} \approx{} 1.05 \approx {1 + \frac{R}{100}} \\
\implies 0.05 \approx \frac{R}{100}
\implies R \approx 5\%
\end{align*}$
I'd also like to shoutout PiGuy for getting this right in much fewer words. However, I hope that the step-by-step approach of my answer helps you, Sujay.
A: First derive a general formula just as compound interest formula is readily available to you.
$$\text{CI} = \text{P}([1+\frac{\text{R}}{100}]^n - 1)$$
$$ (\frac{CI}{P}+1)=[1+\frac{\text{R}}{100}]^n ;\text{  for short let } M= (1+r)^n $$
$$ n=\frac{\log M}{\log (1+r)}$$
It calculates to $1.05$ or 5 percent
$$ \log (1+r)=\frac{\log M}{n}$$
$$ (1+r)=e^{\dfrac{\log M}{n}}$$
Or
$$ r=e^{\dfrac{\log M}{n}} -1\quad \to r= M^{1/n}-1\quad $$
$$\boxed{ r= \left(\frac{CI}{P}+1\right)^{\frac{1}{n}}-1}$$
and now plug in known quantities to calculate $r$. It calculates to $ 5 $ percent.
