# Ex.8.3 Q4 Comparing Quantities Solution - NCERT Maths Class 8

## Question

I borrowed $$\rm{Rs} \,12000$$ from Jamshed at $$6\%$$ per annum simple interest for $$2$$ years. Had I borrowed this sum at $$6\%$$ per annum compound interest, what extra amount would I have to pay?

Video Solution
Comparing Quantities
Ex 8.3 | Question 4

## Text Solution

What is known?

Principal, Time Period and Rate of Interest

What is unknown?

Simple Interest and Compound Interest (C.I)

Reasoning:

For Simple Interest:

\begin{align}{A = }\frac{{{\rm{P \times R \times T}}}}{{100}}\end{align}

$$P = \rm{Rs}\,12,000$$

$$N = 2$$ years

$$R = 6\%$$ simple interest

For Compound Interest:

\begin{align}{A = P}{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{\rm{n}}}\end{align}

$$P = \rm{Rs}\,12,000$$

$$N =2$$ years

$$R = 6\%$$ compounded annually

Steps:

Simple Interest to be paid for $$2$$ years at the rate of $$6\%$$ per annum

\begin{align}{\rm{S}}{\rm{.I}}{\rm{. for \;}}2\;{\rm{years}} &= 2 \times 12000 \times \frac{6}{{100}}\\&= 2 \times 120 \times 6\\&= 1440\end{align}

Compound Interest to be paid for $$2$$ years at the rate of $$6\%$$ per annum

\begin{align}A& = P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\& = 12000{\left( {{1 + }\frac{{6}}{{{100}}}} \right)^{2}}\\& = 12000{\left( {\frac{{100}}{{100}} + \frac{{6}}{{{100}}}} \right)^{2}}\\& = 12000\left( {\frac{{{106 \times 106}}}{{{100 \times 100}}}} \right)\\& = 12000\left( {\frac{{{11236}}}{{{10000}}}} \right)\\&= {12000} \times {1}{.1236}\\&= 13483.20\end{align}

Compound Interest

\begin{align} &= 13483.20 - 12000\\&= 1483.20\end{align}

Compound Interest - Simple Interest

\begin{align} &= 1483.20 - 1440\\&= 43.20\end{align}

The extra amount that would have been paid $$= \rm{Rs}\,43.20$$

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