Ex.8.3 Q5 Comparing Quantities Solution - NCERT Maths Class 8


Question

Vasudevan invested \(\rm{Rs}\, 60,000\) at an interest rate of \(12\%\) per annum compounded half yearly. What amount would he get

(i) after \(6\) months?

(ii) after \(1\) year?

 Video Solution
Comparing Quantities
Ex 8.3 | Question 5

Text Solution

What is known?

Principal, Time Period and Rate of Interest

What is unknown?

Amount and Compound Interest (C.I.)

Reasoning:

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

\(P =\rm{ Rs}\,60,000\)

\(N =6\) months and \(1\) year

\(R =12\%\) p.a. compounded half yearly

Steps:

For easy calculation of compound interest, we will put Interest Rate as \(6\%\) half yearly and \('n'\) as \(1\)

(i) Compound Interest to be paid for \(6\) month

\[\begin{align}A & = P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\& = 60000{\left( {{1 + }\frac{{6}}{{{100}}}} \right)^{1}}\\& = 60000{\left( {\frac{{100}}{{100}} + \frac{{6}}{{{100}}}} \right)^{1}}\\& = 60000\left( {\frac{{{106}}}{{{100}}}} \right)\\&= {60000} \times {1}{.06}\\&= 63600\end{align}\]

Compound Interest for \(6\) month

\[\begin{align} &= 63600 - 60000\\&= 3600\end{align}\]

(ii) Compound Interest to be paid for \(12\) months (\(1\) year) Compounded half yearly So \(n=2,r=6\%,\)

\[\begin{align}A &= P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\&= 60000{\left( {{1 + }\frac{{6}}{{{100}}}} \right)^{2}}\\& = 60000{\left( {\frac{{100}}{{100}} + \frac{{6}}{{{100}}}} \right)^{2}}\\& = 60000{\left( {\frac{{{106}}}{{{100}}}} \right)^2}\\& = 60000\left( {\frac{{{106} \times {106}}}{{{100} \times 100}}} \right)\\& = 60000\left( {\frac{{11236}}{{{100}00}}} \right)\\&= {60000} \times {1}{.1236}\\&= 67416\end{align}\]

Compound Interest for \(12\) months

\[\begin{align} &= 67416 - 60000\\&= 7416\end{align}\]

The amount that Vasudevan will get after \(6\) months \(= \rm{Rs}\,63600\)

The amount that Vasudevan will get after \(1\) year  \(= \rm{Rs}\, 67416\)

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