Q. The simple interest in 3 yrs and the compound interest in 2 yrs on the same principal and at the same rate are 1200 and 832 respectively. Calculate the rate.
What I've done:
SI in 3 years = 1200
therefore, SI in 1 year = 400
=>CI in 1 year = 400
CI for 2 years - CI for 1 year = 832-400 =432
Here's where the problem comes..
=>432 is the interest on 400 for 1 year at the same rate.
$432=400*{R\over100}*1$
=>R=108 % whereas the real answer's 8%. Where am I going wrong?
On
Following AN's suggestion to "use algebra":
Let $A$ be the original amount, and $i$ be the annual interest rate.
For the simple interest scenario:$$3Ai=1200$$ $$Ai=400$$
For the compound interest scenario:$$A(1+i)^2=A+832$$If you multiply out the squared expression, eliminate the term $A$ on both sides, and substitute for $Ai$ whereever it appears, the solution for $i$ is apparent.
An intuitive way of solving the problem has been dealt with in comments. We make a more formal calculation. Let $P$ be the principal, and $r$ the rate of interest. Note that $r=0.06$, for example, means that the rate is $6\%$.
For the simple interest calculation, we have $$P(3r)=1200. \tag{1}$$
With compound interest at nominal rate $r$, a debt of $P$ grows to $P(1+r)^2$ in $2$ years. The principal is $P$, so the interest is $P(1+r)^2-P=P(2r+r^2)$. It follows that $$P(2r+r^2)=832.\tag{2}$$
We use Equations (1) and (2) to solve for $r$. Divide. There is nice cancellation, and we get $$\frac{2+r}{3}=\frac{832}{1200}.$$ Solve the linear equation for $r$. We get $r=\frac{832}{400}-2=\frac{32}{400}=0.08$.
Remark: The informal calculation is quite a bit simpler, and it is important to grasp it so that one understands how compound interest ("interest on interest") works. For more complicated problems of the same general nature, informal reasoning might not be sufficient, so it is also important to know how the algebraic machinery works.