Latest Updated : November 2023

If you are in search of **NCERT solutions for Class 8 Maths Chapter** **7 Comparing Quantities**, then you are at the right place. When it comes to **Mathematics**, it is a subject that holds value for students across various academic streams, be it science, biology, or commerce and having a very good understanding of basic math concepts will help you in achieving your goal and make it much more easier.

Here we provide detailed **NCERT solutions for Class 8 Maths** of all the chapters, **exercise wise** with **updated syllabus by NCERT**. These NCERT solutions for Class 8 Maths have been carefully crafted to provide detailed explanations, ensuring that students can grasp the concepts effectively. We also provide extra practice questions for the same, so feel free to bookmark **ncertforclass8.com** to boost your preparation.

## NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities Ex 7.3

**Class 8 Maths Chapter** 7 Comparing Quantities **Exercise 7.**3

**Exercise 7.3**

**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?**

Solution:

(i) It is given that the population in the year 2003 = 54,000

ATQ, Amount (A) = 54,000

Rate of Interest (R)= 5%

Time(n) = 2 years

Let population in 2001 be P

`\[Amount\left(A\right)\ =\ P\left(1-\frac{R}{100}\right)^n\]`

`\[54,000=P\left(1+\frac{5}{100}\right)^2\ \]`

`\[54,000=P\left(1+\frac{1}{20}\right)^2\ =\ 54,000=P\left(\frac{21}{20}\right)^2\ \]`

`\[∴\ Population\ in\ 2001\ \left(P\right)\ =\ 54000\ \times\frac{20}{21\ }\times\ \frac{20}{21}=48,979.59\ =\ 48,980\ \left(approx\right)\]`

(ii) Let population in 2001 be A

`\[Amount\left(A\right)\ =\ P\left(1-\frac{R}{100}\right)^n\]`

`\[Amount\left(A\right)\ =\ 54000\left(1+\frac{5}{100}\right)^2\]`

`\[Amount\left(A\right)\ =\ 54000\left(1+\frac{1}{20}\right)^2=54,000\times\ \left(\frac{21}{20}\right)^2\]`

`\[∴\ Population\ in\ 2005=\ 54000\ \times\frac{21}{20\ }\times\ \frac{21}{20}=59,535.\]`

**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.**

Solution:

ATQ, Principal (P) = 5,06,000

Rate of Interest (R)= 2.5%

Time(n) = 2 Hours

After 2 hours the number of bacteria will be –

`\[Amount\left(A\right)\ =\ P\left(1-\frac{R}{100}\right)^n\]`

`\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =506000\left(1+\frac{2.5}{100}\right)^2\]`

`\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =506000\left(1+\frac{25}{1000}\right)^2\]`

`\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =506000\left(1+\frac{1}{40}\right)^2\]`

`\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =506000\left(\frac{41}{40}\right)^2\]`

`\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =506000\times\ \frac{41}{40}\times\ \frac{41}{40}=5,31,616.25\]`

Hence, the number of bacteria after two hours are 5,31,616.25 ( 5,31,616 ) approx.

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

Solution:

ATQ, Principal (P) = ₹ 42,000

Rate of Interest (R)= 8%

Time(n) = 1 years

`\[Amount\left(A\right)\ =\ P\left(1-\frac{R}{100}\right)^n\]`

`\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =42000\left(1-\frac{8}{100}\right)^1\]`

`\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =42000\left(1+\frac{2}{25}\right)^1\]`

`\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =42000\left(\frac{27}{25}\right)^1\]`

`\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =42000\ \times\frac{27}{25\ }=\ ₹\ 38640\]`

`\[Therefore,\ the\ value\ of\ scooter\ after\ one\ year\ is\ ₹\ \ 38,640\]`

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