Question

If the difference of the compound interest and the simple interest on a sum of money for $2$ years is $1\%$ of the sum of money then find the annual rate of interest.

Answer 1

$CI=P[1+r/100{]}^n-P$

$SI=pnr/100$

From the problem,

$N=2$

$CI–SI=P\ast 1/100$

We need to find $r.$

$CI–SI=P[1+r/100{]}^n-P-Pnr/100$

$=P\left[\right(1+r/100{)}^n-1-nr/100]$

Substituting the values of $n$ and $r,$ we get

$CI–SI=P\ast 1/100=P[1+r/100{)}^2-1-2\ast r/100]$

Since $P$ is common on both sides of the

equation, they get cancelled.

Let us consider the expression within the

square brackets.

Expanding and subtracting the remaining

values, we get

$1+2r/100+r^2/10000-1-2r/100=r^2/10000$

Therefore, $1/100=r^2/10000.$

$R^2=10000/100=100$

Therefore, $r=10$

Thus, the rate of interest is $10$