# Ex.8.3 Q3 Comparing Quantities Solution - NCERT Maths Class 8

## Question

Fabina borrows \(\rm{Rs}\,12,500\) at \(12\%\) per annum for \(3 \) years at simple interest and Radha borrows the same amount for the same time period at \(10\%\) per annum, compounded annually. Who pays more interest and by how much?

## 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:

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

\(P = \rm{Rs}\, 12,500\)

\(N=3\) years

\(R=12\%\) simple interest

For Compound Interest:

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

\(P = \rm{Rs}\, 12,500\)

\(N=3\) years

\(R=10\%\) compounded annually

**Steps:**

Simple Interest paid by Fabina for \(3 \) years at the rate of \(12\%\) per annum

\[\begin{align}{\rm{S}}{\rm{.I}}{\rm{. for\; }}3\;{\rm{years}} &= 3 \times 12500 \times \frac{{12}}{{100}}\\&= 3 \times 125 \times 12\\&= 4500\end{align}\]

Amount paid by Radha for \(3 \) years at the rate of \(10\%\) p.a. compounded annually

\[\begin{align}A& = P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\ &= 12500{\left( {{1 + }\frac{{{10}}}{{{100}}}} \right)^{3}}\\ &= 12500{\left( { \frac{{11}}{{{10}}}} \right)^{3}}\\& = 12500\left( {\frac{{{11 \times 11 \times 11}}}{{{10 \times 10 \times 10}}}} \right)\\ &= 12500\left( {\frac{{{1331}}}{{{1000}}}} \right)\\&= {12500} \times {1}{.331}\\&= 16637.50\end{align}\]

Compound Interest

\[\begin{align} &= 16637.50 - 12500\\&= 4137.50\end{align}\]

Since \(4500 > 4137.50\), Fabina paid more interest than Radha

Additional Interest paid by Fabina

\(= 4500 \,– 4137.50 \\ = \rm Rs \;362.50\)