Exercise 8.3 Comparing Quantities- NCERT Solutions Class 8

Go back to  'Comparing Quantities'

Question 1

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.

Solution

Video 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 \(= \rm{Rs}\, 816\)

Question 2

Kamala borrowed \(\rm{Rs}\, 26400\) from a Bank to buy a scooter at a rate of \(15\%\) p.a. compounded yearly. What amount will she pay at the end of \(2\) years and \(4\) months to clear the loan?

(Hint: Find \(A\) for \(2\) years with interest is compounded yearly and then find S.I. on the \(2\)nd year amount for \(\begin{align}\frac{4}{{12}}\end{align}\) years.)

Solution

Video Solution

What is known?

Principal, Time Period and Rate of Interest

What is unknown?

Amount and Compound Interest (C.I.)

Reasoning:

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

\(P = \rm{Rs}\, 26400\)

\(N = 2\) years \(4\) months

\(R=15\%\) compounded annually

Steps:

First, we will calculate Compound Interest (C.I) for the period of \(2\) years

\[\begin{align}A &= P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\&= 26400{\left( {{1 + }\frac{{{15}}}{{{100}}}} \right)^{2}}\\& = 26400{\left( {\frac{{100}}{{100}} + \frac{{{15}}}{{{100}}}} \right)^{2}}\\&= 26400 \times \frac{{{115 \times 115}}}{{{100 \times 100}}}\\& = {26400}\left( {\frac{{{23}}}{{{20}}}{ \times }\frac{{{23}}}{{{20}}}} \right)\\&= {26400} \times {1}{.3225}\\
 &= 34914\end{align}\]

\[\begin{align}{\rm{C}}{\rm{.I}}.  &=  A - {\rm{P}}\\&{ = {\rm{34914}} - {\rm{26400}}}\\&{ = {\rm{8514}}}\end{align}\]

Second, we will find Simple Interest (S.I) for the period of \(4\) months

Principal for \(4\) months after C.I. for \(2\) years \(= \rm{Rs}\,34,914\)

S.I. for \(4\) months

\[\begin{align} &= \frac{4}{{12}} \times 34914 \times \frac{{15}}{{100}}\\&= \frac{1}{3} \times 34914 \times \frac{3}{{20}}\\&= \frac{{34914}}{{20}}\\&= 1745.70\end{align}\]

Total interest for \(2\) years \(4\) months

\[\begin{align} &= 8514 + 1745.70\\&= 10259.70\end{align}\]

Total amount for \(2\) years \(4\)  months

\[\begin{align}  &= 26400 + 10259.70\\&= 36659.70\end{align}\]

The amount Kamala will have to pay after \(2\) years \(4\) months \(=\rm{Rs}\;36659.70\)

Question 3

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?

Solution

Video 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

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

Question 4

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?

Solution

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

Question 5

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?

Solution

Video 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  \(= \rm{Rs}\, 67416\)

Question 6

Arif took a loan of \(\rm{Rs}\,80,000\) from a bank. If the rate of interest is \(10\%\) per annum, find the difference in amounts he would be paying after \(1\begin{align}\frac{1}{2}\end{align}\) years if the interest is

(i) Compounded annually

(ii) Compounded half-yearly

Solution

Video Solution

What is known?

Principal, Time Period and Rate of Interest

What is unknown?

Amount and Compound Interest (C.I.)

Reasoning:

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

\(P=\rm{Rs}\, 80,000\)

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

\(R=10\%\) p.a. compounded half yearly and \(10\%\) p.a. compounded yearly

Steps:

For calculation of Compound Interest (C.I.) compounded annually:

Since ‘\(n\)’ is \(1\begin{align}\frac{{1}}{{2}}\end{align}\)years, amount can be calculated for \(1\) year and having that amount as principal, S.I can be calculated for \(\begin{align}\frac{1}{2}\end{align}\) year, because C.I is only annually.

\[\begin{align} A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}  \\   &= 80000\left( {{1 + }\frac{{{10}}}{{{100}}}} \right)^{1}  \\   &= {80000 \times }\frac{{{11}}}{{{10}}} \\  &=  {80000 \times 1}{.1} \\ 
& = { 88000} \\  \end{align}\]

Amount after \(1\) years \(= \rm{ Rs}\, 88,000\)

Therefore, the principal for the \({1}\begin{align}\frac{{1}}{{2}}\end{align}\)th year \(= \rm{ Rs}\, 88,000\)

S.I. for \( \frac{1}{2} \) years

\[\begin{align} &= \frac{1}{2} \times 88000 \times \frac{{10}}{{100}}\\&= \frac{{8800}}{2}\\&= 4400\end{align}\]

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

\[\begin{align}  &= 88000 + 4400\\&={\rm{Rs}}\;92400\end{align}\]

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

\(\begin{align}  &= 92400 - 80000\\ &= {\rm{Rs}}\;12400 \dots \left\{\text{For the C.I. to be }\\{\text{charged yearly}} \right\} \end{align}\) 

For calculation of Compound Interest (C.I.) compounded half-yearly, we will consider as rate \(5\%\) p.a. and \(‘n'\) as \(3\)

\[\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}  \\  &= 80000\left( {{1 + }\frac{{5}}{{{100}}}} \right)^{3}  \\ &= 80000\left( {{1 + }\frac{{1}}{{{20}}}} \right)^{3}  \\&= 80000\left( {\frac{{21}}{{20}}} \right)^3  \\ &= 80000 \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \\ 
&= 80000 \times \frac{{9261}}{{8000}} \\  &=10 \times 9261 \\   &= 92610 \\ \end{align}\]

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

\(\begin{align} & = 92610 - 80000\\&= {\rm{Rs}}\;12610  \dots \left\{ {\text{For the C.I. to be }}\\{\text{charged half - yearly}} \right\}\end{align}\)

Difference in amounts he would be paying \(=\rm{Rs}\, 92,610 − \rm{Rs}\, 92,400 = \rm{Rs}\, 210\)

Question 7

Maria invested Rs \(8,000 \) in a business. She would be paid interest at \(5\%\) per annum compounded annually. Find:

(i) The amount credited against her name at the end of the second year

(ii) The interest for the \(3\)rd year.

Solution

Video Solution

What is known?

Principal, Time Period and Rate of Interest

What is unknown?

Amount and Compound Interest (C.I.)

Reasoning:

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

\(P=\rm{Rs}\, 8,000\)

\(N =2\) years and \(3\)rd year

\(R= 5\% \) p.a. compounded annually

Steps:

(i) For calculation of amount credited at the end of second year:

\[\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}  \\ &= 8000\left( {{1 + }\frac{{5}}{{{100}}}} \right)^{2}  \\ &= 8000\left( {{1 + }\frac{{1}}{{{20}}}} \right)^{2}  \\ &= 8000\left( {\frac{{21}}{{20}}} \right)^2  \\ & = 8000 \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \\ & = 8000 \times \frac{{441}}{{400}} \\ 
&= 20 \times 441 \\ & = 8820 \\ 
\end{align}\]

(ii) For calculating C.I. for the \(3\)rd year, the principal \(= 8820\)

\[\begin{align}{\rm{S}}{\rm{.I}}{\rm{.}} &= \frac{P \times R \times T}{100}\\ &= \left( {\frac{{8820 \times 5 \times 1}}{{100}}} \right)\\&= 441\end{align}\]

The amount credited at the end of the \(2\)nd year \(= \rm{Rs}\, 8,820\)

The interest for the \(3^\rm{rd} \) year \(= \rm{Rs}\, 441\) 

Question 8

Find the amount and the compound interest on \(\rm{Rs}\, 10,000 \) for \((1\begin{align}\frac{1}{2})\end{align}\) years at \(10\%\) per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Solution

Video Solution

What is known?

Principal, Time Period and Rate of Interest

What is unknown?

Amount and Compound Interest (C.I.)

Reasoning:

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

 \(P=\rm{Rs}\, 10,000\)

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

\(R= 10\%\) p.a. compounded annually and half-yearly

Steps:

For calculation of C.I. compounded half yearly, we will take Interest rate as \(5\% \)

\[\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}  \\ &= 10000\left( {{1 + }\frac{{5}}{{{100}}}} \right)^{3}  \\ &= 10000\left( {{1 + }\frac{{1}}{{{20}}}} \right)^{3}  \\ & = 10000\left( {\frac{{21}}{{20}}} \right)^3  \\ &= 10000\times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \\ &= 10000 \times \frac{{9261}}{{8000}} \\ &= {5} \times \frac{{9261}}{4} \\ &= 11576.25 \\ \end{align}\]

Interest earned at \(10\%\) p.a. compounded half-yearly \(= 11576.25 \,– \,10000=\rm{Rs}\, 1576.25\)

The amount earned at \(10\%\) p.a. compounded half-yearly \(= 11576.25\)

The C.I. earned at \(10\%\) p.a. compounded half-yearly \(= 1576.25\)

The above interest earned being compounded half-yearly would be more than the interest compounded annually since interest compounded half yearly is always more than compounded annually at the same rate of interest.

Question 9

Find the amount which Ram will get on \(\rm{Rs}\; 4,096\), if he gave it for \(18\) months at \( 12\begin{align}\frac{1}{2}\% \end{align}\) per annum, interest being compounded half yearly.

Solution

Video Solution

What is known?

Principal, Time Period and Rate of Interest

What is unknown?

Amount

Reasoning:

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

\(P=\rm{Rs}\,4,096\)

\(N =18\) months

\(R =12\begin{align}\frac{1}{2}\% \end{align}\) p.a. compounded half-yearly

Steps:

For calculation of C.I. compounded half yearly, we will take Interest rate as \(6\begin{align}\frac{1}{4}\% \; = \;\frac{{25}}{4}\%\end{align} \) and  \('n'\) as \(3\) \((18\div 6= 3)\)

\[\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}  \\ &= 4096\left( {{1 + }\frac{{{25}}}{{{100} \times {4}}}} \right)^{3}  \\ & = 4096\left( {{1 + }\frac{{{25}}}{{{400}}}} \right)^{3}  \\ 
 & = 4096\left( {\frac{{425}}{{400}}} \right)^3  \\ & = 4096\left( {\frac{{17}}{{16}}} \right)^3  \\ 
 & = 4096 \times \frac{{17}}{{16}} \times \frac{{17}}{{16}} \times \frac{{17}}{{16}} \\ &= 4096 \times \frac{{4913}}{{4096}} \\ &= 4913 \\ \end{align}\]

The total amount that Ram will get at the end of \(18\) months \( = \rm{Rs}\, 4,913\)

Question 10

The population of a place increased to \(\rm{}\,54,000\) in \(\rm{}\,2003\) at a rate of \(5\%\) per annum

(i) find the population in \(2001.\)

(ii) what would be its population in \(2005\)?

Solution

Video Solution

What is known?

Population in \(2003\), Time Period and Rate of population growth

What is unknown?

Population in \(2005\) and \(2001\)

Reasoning:

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

\(P=\rm{}\, 54000 \) in the year \(2003\)

\(N =2\) years

\(R =5\%\) p.a. compounded annually

Steps:

(i) Population in the year \(2001\)

\[\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}  \\ 54000 &= P\left( {{1 + }\frac{{5}}{{{100}}}} \right)^2  \\ 54000 &= P\left( {{1 + }\frac{{1}}{{{20}}}} \right)^{2}  \\ 54000 &= P\left( {\frac{{21}}{{20}}} \right)^2  \\ 54000 &= P \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \\ 
  P &= 54000 \times \frac{{400}}{{441}} \\ P &= 48979.6 \\ \end{align}\] 

The population in \(2001\) \(= \rm{}\,48980\).

(ii) Population in the year \(2005\)

\[\begin{align}A& = P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}  \\ & = 54000\left( {{1 + }\frac{{5}}{{{100}}}} \right)^{2}  \\ & = 54000\left( {{1 + }\frac{{1}}{{{20}}}} \right)^{2}  \\ & = 54000\left( {\frac{{21}}{{20}}} \right)^2  \\ &= 54000 \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \\ & = 54000 \times \frac{{441}}{{400}} \\ &= 135 \times 441 \\ &= 59535 \\ \end{align}\]

The population in \(2005 \) \(=\rm{} 59535.\)

The population in \(2001\) \(=\rm{} 48980\).

The population in \(2005\) \(= \rm{}\,59535\).

Question 11

In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of \(2.5\%\) per hour. Find the bacteria at the end of \(2\) hours if the count was initially \(\rm{}\,5, 06,000\).

Solution

Video Solution

What is known?

Original Count, Time Period and Rate of Increase

What is unknown?

The total count after \(2\) hours

Reasoning:

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

\(P= \rm{}\,5,06,000\)

\(N = 2 \) hours

\(R= 2.5\% \) hour \(=\begin{align}\frac{{25}}{{10}} \end{align}\)hours

Steps:

\[\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}  \\ &= 506000\left( {{1 + }\frac{{{25}}}{{{1000}}}} \right)^{2}  \\ &= 506000\left( {{1 + }\frac{{1}}{{{40}}}} \right)^{2}  \\ &= 506000\left( {\frac{{41}}{{40}}} \right)^2  \\ &= 506000 \times \frac{{41}}{{40}} \times \frac{{41}}{{40}} \\ &= 506000 \times \frac{{1681}}{{1600}} \\ &= {506000} \times 1.050625 \\ &= 531616 \\ \end{align}\]

The total count of bacteria after \(2\) hours \( =\rm{}\, 531616\)

Question 12

A scooter was bought at \( \rm{Rs}\, 42,000\). Its value depreciated at the rate of \(8\%\) per annum. Find its value after one year

Solution

Video Solution

What is known?

Original Value, Rate of Depreciation

What is unknown?

The value of scooter after \(1\) year

Reasoning:

Original value of the scooter \(= \rm{Rs}\, 42,000\)

Rate of depreciation \(= 8\%\)

Steps:

The value of the scooter after \(1\) year

\[\begin{align} &= 42000 - \left( {42000 \times \frac{8}{{100}}} \right) \\ &= 42000 - \left( {42000 \times \frac{2}{{25}}} \right) \\ &= 42000 - \left( {1680 \times 2} \right) \\ &= 42000 - 3360 \\ &= 38640 \\ \end{align}\]

The value of the scooter after \(1\) year (\(8\%\) depreciation rate) \(= \rm{Rs}\, 38640.\)

  
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school

0