Exercise 7.3 (Revised) - Chapter 8 - Comparing Quantities - Ncert Solutions class 8 - Maths

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NCERT Solutions for Class 8 Maths Chapter 7 - Comparing Quantities

Ex 7.3 Question 1.

The population of a place increased to 54,000 in 2003 at a rate of $5 \%$ per annum.
(i) Find the population in 2001.
(ii) What would be its population in 2005 ?

Answer.

(i) Here, $\mathrm{A}_{2003}=$ Rs. $54,000, \mathrm{R}=5 \%, n=2$ years

Population would be less in 2001 than 2003 in two years.
Here population is increasing.
$
\begin{aligned}
& \therefore A_{2003}=P_{201}\left(1+\frac{R}{100}\right)^n \\
& \Rightarrow 54000=P_{2001}\left(1+\frac{5}{100}\right)^2 \\
& \Rightarrow 54000=P_{2001}\left(1+\frac{1}{20}\right)^2 \\
& \Rightarrow 54000=\mathrm{P}_{2001}\left(\frac{21}{20}\right)^2 \\
& \Rightarrow 54000=P_{2001} \times \frac{21}{20} \times \frac{21}{20} \\
& \Rightarrow \mathrm{P}_{2001}=\frac{54000 \times 20 \times 20}{21 \times 21} \\
& =48,979.5 \\
& \Rightarrow P_{2001}=48,980 \text { (approx.) } \\
&
\end{aligned}
$

(ii) According to question, population is increasing. Therefore population in 2005,
$
\begin{aligned}
& \mathrm{A}_{2005}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n \\
& =54000\left(1+\frac{5}{100}\right)^2 \\
& =54000\left(1+\frac{1}{20}\right)^2
\end{aligned}
$

$
\begin{aligned}
& =54000\left(\frac{21}{20}\right)^2 \\
& =54000 \times \frac{21}{20} \times \frac{21}{20} \\
& =59,535
\end{aligned}
$

Hence population in 2005 would be 59,535.

Ex 7.3 Question 2.

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 $5,06,000$.

Answer.

Here, Principal $(\mathrm{P})=5,06,000$, Rate of Interest $(\mathrm{R})=2.5 \%$, Time $(\mathrm{n})=2$ hours
After 2 hours, number of bacteria,
$
\begin{aligned}
& \text { Amount (A) } \mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n \\
& =506000\left(1+\frac{2.5}{100}\right)^2 \\
& =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}
\end{aligned}
$

$
=5,31,616.25
$

Hence, number of bacteria after two hours are 531616 (approx.).
Ex 7.3 Question 3.

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

Answer.

Here, Principal $(\mathrm{P})=$ Rs. 42,000, Rate of Interest $(\mathrm{R})=8 \%$, Time $(\mathrm{n})=1$ years
$
\begin{aligned}
& \text { Amount }(\mathrm{A})=\mathrm{P}\left(1-\frac{\mathrm{R}}{100}\right)^{12} \\
& =42000\left(1-\frac{8}{100}\right)^1 \\
& =42000\left(1+\frac{2}{25}\right)^1 \\
& =42000\left(\frac{27}{25}\right)^1 \\
& =42000 \times \frac{27}{25} \\
& =\text { Rs. } 38,640
\end{aligned}
$

Hence, the value of scooter after one year is Rs. 38,640.