 QUESTION

# The simple interest on the certain sum of money for 3 years at $8%$ per annum is half the compound interest on $Rs.4000$ for 2 years at $10%$ per annum. Then, what is the sum placed on simple interest?(a) $Rs.1550$ (b) $Rs.1650$ (c) $Rs.1750$ (d) $Rs.2000$

Hint: In this question, we will first find the value of simple interest and compound interest with the given values and then put it into the condition given in question to find the required value of principal amount.

Let, P be the principal amount that is taken or given on loan. Let, R be the percentage rate of interest per annum on P and let T be the time duration of the loan.

Then, the formula to calculate simple interest is,

Simple interest $=\dfrac{\text{P }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ T}}{100}$.

And, formula to calculate compound interest is,

Compound interest $=\text{P}{{\left( \text{1+}\dfrac{\text{R}}{\text{100}} \right)}^{\text{T}}}-\text{P}$.

Now, let the required sum of money be $p$.

Here, in case of simple interest, P = $p$, R = 8, and T = 3.

Therefore, applying the above formula of simple interest, we get,

Simple interest $=\dfrac{p\times 8\times 3}{100}$

$=\dfrac{24p}{100}\cdots \cdots \left( i \right)$ .

Also, in case of compound interest, P = Rs. 4000, R = 10, T = 2.

Therefore, applying the above formula of compound interest, we get,

Compound interest $=Rs.\,\left[ 4000{{\left( 1+\dfrac{10}{100} \right)}^{2}}-4000 \right]$
$=Rs.\,\left[ 4000{{\left( 1+\dfrac{1}{10} \right)}^{2}}-4000 \right]$

Taking LCM, we get,

$=Rs.\,\left[ 4000{{\left( \dfrac{10+1}{10} \right)}^{2}}-4000 \right]$

$=Rs.\,\left[ 4000{{\left( \dfrac{11}{10} \right)}^{2}}-4000 \right]$

Taking 4000 common, we get,

$=Rs.\,4000\left[ {{\left( \dfrac{11}{10} \right)}^{2}}-1 \right]$

Distributing square in numerator and denominator, we get,

\begin{align} & =Rs.\,4000\left[ \dfrac{{{11}^{2}}}{{{10}^{2}}}-1 \right] \\ & =Rs.\,4000\left( \dfrac{121}{100}-1 \right) \\ \end{align}

Taking LCM again, we get,

\begin{align} & =Rs.\,4000\left( \dfrac{121}{100}-\dfrac{100}{100} \right) \\ & =Rs.\,4000\left( \dfrac{121-100}{100} \right) \\ & =Rs.\,4000\left( \dfrac{21}{100} \right) \\ & \\ \end{align}

Opening the bracket and solving, we get,

\begin{align} & =Rs\text{.}\,4000\times \dfrac{21}{100} \\ & =Rs.\,40\times 21 \\ & =Rs.\,840\cdots \cdots \left( ii \right) \\ \end{align}

According to the question, simple interest with given values is half of the compound interest

with given value.

That is, $\text{Simple interest = }\dfrac{\text{compound interest}}{\text{2}}$

Putting values from equation $\left( i \right)$ and $\left( ii \right)$, we get,

$\dfrac{24p}{100}=\dfrac{Rs.840}{2}$

Cross multiplying, we get,

\begin{align} & \Rightarrow 24p\times 2=Rs.840\times 100 \\ & \Rightarrow 48p=Rs.84000 \\ \end{align}

Dividing 48 from both sides of the equation, we have,

\begin{align} & \Rightarrow p=\dfrac{Rs.84000}{48} \\ & \Rightarrow p=Rs.1750 \\ \end{align}

Hence, the sum placed on simple interest is $Rs.1750$.

Therefore, the correct answer is option (c).

Note: Another method to solve this question is that, without using compound interest formula, you can use a simple interest formula twice for both years by changing the value of P.