Ex.8.3 Q1 Comparing Quantities Solution - NCERT Maths Class 8


Question

Calculate the amount and compound interest on

(i) \(\rm{Rs}\, 10,800\) for \(3\) years at \(\begin{align}12\frac{{1}}{{2}}\%\end{align}\) per annum compounded annually.

(ii) \(\rm{Rs}\, 18,000\) for \(\begin{align} 2\frac{{1}}{{2}}\end{align} \) years at \(10\%\) per annum compounded annually.

(iii) \(\rm{Rs}\, 62,500\) for \(\begin{align} 1\frac{{1}}{{2}}\end{align} \) years at \(8\%\) per annum compounded half yearly.

(iv) \(\rm{Rs}\, 8,000\) for \(1\) year at \(9\%\) per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).

(v) \(\rm{Rs}\, 10,000\) for \(1\) year at \(8\%\) per annum compounded half yearly.

 Video Solution
Comparing Quantities
Ex 8.3 | Question 1

Text Solution

What is Known?

Principal, Time Period and Rate of Interest

What is unknown?

Amount and Compound Interest (C.I)

Reasoning:

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

\(P = \rm{Rs}\, 10800\)
\(N = 3\;\rm{years}\) 
\(\begin{align}R = 12\frac{1}{2} \%= \frac{25}{2}\%\end{align}\)  compounded annually

Steps:

(i)

\[\begin{align}A &= P{\left( {1 + }\frac{r}{{100}} \right)^{n}}\\&= 10800{\left({1 + }\frac{{25}}{{2 \times 100}} \right)^{3}}\\&= 10800{\left( {\frac{{225}}{{200}}} \right)^{3}}\\ &= 10800 \times \frac{{225 \times 225 \times 225}}{{200 \times 200 \times 200}}\\& = 15377.34\end{align}\]

\[\begin{align}{\rm{C}}{\rm{.I}}.  &=  A - P\\&= {\rm{15377}}{\rm{.34}} - {\rm{10800}}\\ &= {\rm{4577}}{\rm{.34}}\end{align}\]

Amount \(= \rm{Rs}\, 15377.34\)

Compound Interest \(=\rm{ Rs}\, 4577.34\)

 (ii)

\(P= \rm{Rs}\, 18000\)

\(\begin{align} n= 2\frac{{1}}{{2}}\end{align} \)years

\(\rm{R}= 10\%\) p.a compounded annually

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

Since \(‘n’\) is \(\begin{align} 2\frac{{1}}{{2}}\end{align} \)years, amount can be calculated for \(2\) years and having amount

as principal S.I can be calculated for \(\begin{align} \frac{{1}}{{2}}\end{align} \) year, because C.I is only annually.

\[\begin{align}A &= P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\ &= 18000{\left( {{1 + }\frac{{{10}}}{{{100}}}} \right)^{2}}\\ &= 18000 \times \frac{{{11 \times 11}}}{{{10 \times 10}}}\\ &= 21780\end{align}\]

Amount after \(2\) years \(=\rm{ Rs}\, 21780\)

\[\begin{align}{\rm{S}}{\rm{.I}}{\rm{. for }}\frac{1}{2}{\rm{ years}} &= \frac{1}{2} \times 21780 \times \frac{{10}}{{100}}\\ &= 1089\end{align}\]

Amount after \(2\frac{1}{2} \) years

\[\begin{align}  &= 21780 + 1089\\ &= {\rm{Rs}}\;22869\end{align}\]

Compound Interest after \(2\frac{1}{2}\) years

\[\begin{align}  &= 22869 - 18000\\ &= {\rm{Rs}}\;4869\end{align}\]

 (iii)

\(P=\rm{ Rs}\, 62,500\)

\(\begin{align} n= 1\frac{{1}}{{2}}\end{align} \) years

\(R= 8\%\) p.a. compounded half yearly

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

There are \(3\) half years in \(\begin{align} 1\frac{{1}}{{2}}\end{align} \) years. Therefore, compounding has to be done \(3\) times and rate of interest will be \(4\%\)

\[\begin{align}A &= P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\&= 62500{\left( {{1 + }\frac{4}{{{100}}}} \right)^3}\\&= 62500 \times \frac{{{104 \times 104} \times {104}}}{{{100 \times 100} \times {100}}}\\&= 70304\end{align}\]

C.I. after \(1\frac{1}{2}\) years ( \(8\%\) p.a. interest half yearly )

\[\begin{align} &= 70304 - 62500\\&= 7804\end{align}\]

Amount after \(1\frac{1}{2}\)years ( \(8\%\) p.a. interest half yearly) \(= 70304\)

Amount after \(\begin{align} 1\frac{{1}}{{2}}\end{align}\) years \(= \rm{Rs}\, 70304\)

Compound Interest after\( \begin{align} 1\frac{{1}}{{2}}\end{align}\) years \(= \rm{Rs}\, 7804\)

 (iv)

\(P= \rm{Rs}\, 8000\)

\(n= 1\) year

\(R= 9\%\) p.a. compounded half yearly

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

S.I. for \(1^{\rm st}\)  \(6\) months

\[\begin{align} &= \frac{1}{2} \times 8000 \times \frac{9}{{100}}\\ &= 40 \times 9\\ &= 360\end{align}\]

Amount after \(1^{\text{st}} 6 \) months including Simple Interest

\[\begin{align} &= 8000 + 360\\ &= {\rm{Rs}}\;8360\end{align}\]

Principal for \(2\)nd \(6 \) months \(= \rm{Rs}\, 8360\)

S.I. for \(2^{\text{nd}} 6\) months

\[\begin{align} &= \frac{1}{2} \times 8360 \times \frac{9}{{100}}\\ &= \frac{{418 \times 9}}{{10}}\\ &= 376.20\end{align}\]

C.I. after \(1\) year ( \(9\%\)  p.a. interest half yearly )

\[\begin{align} &= 360 + 376.20\\ &= 736.20\end{align}\]

Amount after \(1\) year ( \(9\%\)  p.a. interest half yearly )

\[\begin{align}  &= 8000 + 736.20\\ &= 8736.20\end{align}\]

Amount after \(1\) year \(=\rm{ Rs}\, 8736.20\)

Compound Interest after \(1\) year \(= \rm{Rs}\, 736.20\)

 (v)

\(P= \rm{Rs}\, 10,000\)

\(n=1\) year

\(R= 8\%\) p.a. compounded half yearly

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

There are \(2\) half years in \(1\) years. Therefore, compounding has to be done \(2\) times and rate of interest will be \(4\%\)

\[\begin{align}A &= P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\& = 10000{\left( {{1 + }\frac{4}{{{100}}}} \right)^2}\\& = 10000 \times \frac{{{104 \times 104}}}{{{100 \times 100}}}\\ &= 10816\end{align}\]

C.I. after \(1\) year (\( 8\%\)  p.a. interest half yearly )

\[\begin{align} &= 10816 + 10000\\ &= 816\end{align}\]

Amount after \(1\) year ( \(8\%\)  p.a. interest half yearly) \(= 10816\)

Amount after \(1\) year \(= \rm{Rs}\, 10816\)

Compound Interest after \(1\) year \(= \rm{Rs}\, 816\)

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