# 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?

## 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 \( = \rm{Rs}\,43.20\)