Question

What is the rate of interest, if there is a difference of Rs. $121.50$ between the simple interest and the compound interest on principle of Rs. $15,000$ for $2$ years.

Answer 1

Hint:The formula for Simple interest is, $SI = \dfrac{{\operatorname{P} \times r \times t}}{{100}}$ and for compound interest it is $CI = P{\left( {1 + \dfrac{r}{{100}}} \right)^t} - P$ where, $P = $ principal amount, $r = $ rate of interest and $t = $ time in years.

Complete step-by-step solution:
The principal amount in Rs., $P = 15,000$
Time in years is, $t = 2$
The simple interest is given by,
$SI = \dfrac{{\operatorname{P} \times r \times t}}{{100}}$
The compound interest is given by,
$CI = P{\left( {1 + \dfrac{r}{{100}}} \right)^t} - P$
Where,
$P = $ Principal amount,
$r = $ Rate of interest
$t = $ Time in years.
According to question,
The difference $D$ of simple interest and compound interest is given Rs. $121.5$
$D = P{\left( {1 + \dfrac{r}{{100}}} \right)^t} - P - \dfrac{{\operatorname{P} \times r \times t}}{{100}} \cdots \left( 1 \right)$
Substituting the value of Difference, $D = 121.5$, Principal, $P = 15,000$ and Time,$t = 2$ in equation (1),
$\Rightarrow D = 15,000{\left( {1 + \dfrac{r}{{100}}} \right)^2} - 15,000 - \dfrac{{15,000 \times r \times 2}}{{100}} \cdots \left( 2 \right)$
$\Rightarrow 15,000$ is common in all the 3 terms of equation (2). So it can be taken out common,
$\Rightarrow D = 15,000\left[ {{{\left( {\dfrac{{100 + r}}{{100}}} \right)}^2} - 1 - \dfrac{{2r}}{{100}}} \right]$
$\Rightarrow D = 15,000\left[ {\dfrac{{{{\left( {100 + r} \right)}^2}}}{{10000}} - 1 - \dfrac{{2r}}{{100}}} \right] \cdots \left( 3 \right)$
The LCM of $10000$,$1$ and $100$ is $10000$
\[
\Rightarrow D = 15,000\left[ {\dfrac{{{{\left( {100 + r} \right)}^2}}}{{10000}} - 1 - \dfrac{{2r}}{{100}}} \right] \\
\Rightarrow D = 15,000\left[ {\dfrac{{{{\left( {100 + r} \right)}^2} - 10000 - 200r}}{{10000}}} \right] \cdots \left( 4 \right) \\
 \]
Using the identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ , here $a = 100$ and $b = r$ . The equation (1) transforms as
\[\Rightarrow [D = 15,000\left[ {\dfrac{{10000 + {r^2} + 200r - 10000 - 200r}}{{10000}}} \right] \cdots \left( 5 \right)\]
Cancelling the equal positive and negative terms in equation (5),
\[\Rightarrow[121.5 = 15,000\left[ {\dfrac{{{r^2}}}{{10000}}} \right] \cdots \left( 6 \right)\]
The value of ${r^2}$ is calculated from equation (6) as,
\[{r^2} = \dfrac{{121.5}}{{15,000}} \times 10,000 \cdots \left( 7 \right)\]
Taking the square root of the equation (7),
\[
\Rightarrow r = \sqrt {\dfrac{{121.5 \times 10}}{{15}}} \\
\Rightarrow r = \sqrt {81} \\
\Rightarrow r = 9 \\
 \]
Hence, the value of rate of interest is, $r = 9\% $ .

Note:This problem can also be solved using a shortcut formula,
When the difference of simple interest and compound interest is two years.
$D = \dfrac{{\operatorname{P} \times {r^2}}}{{{{100}^2}}} \cdots \left( i \right)$
Substitute the value of difference of simple and compound interest, $D = 121.5$ and principal is, $P = 15,000$ in equation (i), and solve for the value of rate of interest .
$
\Rightarrow 121.5 = \dfrac{{15,000 \times {r^2}}}{{10,000}} \\
\Rightarrow {r^2} = \dfrac{{121.5 \times 10,000}}{{15,000}} \\
\Rightarrow r = \sqrt {81} \\
\Rightarrow r = 9 \\
 $
Hence, the rate of interest by shortcut formula is also $9\% $ percent .