# DAV Class 8 Maths Chapter 6 Worksheet 4 Solutions

The DAV Class 8 Maths Book Solutions Pdf and DAV Class 8 Maths Chapter 6 Worksheet 4 Solutions of Compound Interest offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 6 WS 4 Solutions

Question 1.
The population of a town is increasing at the rate of 8% p.a. What will be the population of the town after 2 years, if the present population is 12,500?
Solution:
Population after 2 years = P $$\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}$$
= 12500 $$\left[1+\frac{8}{100}\right]^2$$
= 12500 $$\left[\frac{27}{25}\right]^2$$
= 12500 $$\frac{729}{625}$$
= 20 × 729
= 14580
Hence, the required population of the town = 14580.

Question 2.
Three years ago, the population of a town was 50,000. If the annual increase during three successive years was at the rate 4%, 5% and 4% p.a. respectively, find the present population.
Solution:
The present population = Population 3 years ago × $$\left[1+\frac{P_1}{100}\right]\left[1+\frac{P_2}{100}\right]\left[1+\frac{P_3}{100}\right]$$
= 50000 $$\left[1+\frac{4}{100}\right]\left[1+\frac{5}{100}\right]\left[1+\frac{4}{100}\right]$$
= 50000 $$\left[\frac{26}{25}\right]\left[\frac{21}{20}\right]\left[\frac{26}{25}\right]$$
= 50000 × $$\frac{14196}{12500}$$
Hence, the present population = 56784. Question 3.
Madhu bought a house for ₹ 1,31,25,000. If its value depreciates at the rate of 10% p.a., what will be its sale price after 3 years?
Solution:
Sale price = Present price $$\left[1-\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}$$
= 13125000 $$\left[1-\frac{10}{100}\right]^3$$
= 13125000 × $$\left[\frac{9}{10}\right]^3$$
= 13125000 × $$\frac{729}{1000}$$
= 13125 × 729 = ₹ 9568125
Hence, the depreciated value = ₹ 95,68,125.

Question 4.
The profits of a firm were ₹ 72,000 in the year 2014. During the next year, it increased by 7% and it decreased by 5% in the following year. What are the profits of the firm after 2 years?
Solution:
Profit after 2 years = Present profit × $$\left[1+\frac{\mathrm{R}_1}{100}\right]\left[1-\frac{\mathrm{R}_2}{100}\right]$$
= 72000 $$\begin{equation} \left[1+\frac{7}{100}\right]\left[1-\frac{5}{100}\right] \end{equation}$$
= 72000 $$\begin{equation} \frac{107}{100} \times \frac{19}{20} \end{equation}$$
= 36 × 107 × 19 = ₹ 73188
Hence, the required profits = ₹ 73188. Question 5.
The population of a town is 64000. If the annual birth rate is 10.7% and the annual death rate is 3.2%, calculate the population after 3 years.
[Hint: Net growth rate = (Birth rate – Death rate)%]
Solution:
The net growth = Birth rate – Death rate
= 10.7% – 3.2% = 7.5%
∴ Population after 3 years = Present population $$\left[1+\frac{\mathrm{R}}{100}\right]^3$$
= 64000 $$\left[1+\frac{7.5}{100}\right]^3$$
= 64000 $$\left[\frac{43}{40}\right]^3$$
= 64000 × $$\frac{79507}{64000}$$
= 79507
Hence, the population after 3 years = 79507.

Question 6.
In a factory, the production of motor bikes was 40,000 in a particular year, which rose to 48,400 in 2 years. Find the rate of growth per annum, if it was uniform during 2 years.
Solution:
Population after 2 years = Present population $$\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}$$
⇒ 48,000 = 40,000 $$\left[1+\frac{R}{100}\right]^2$$
⇒ $$\frac{48400}{40000}=\left[1+\frac{R}{100}\right]^2$$
⇒ $$\frac{121}{100}=\left[1+\frac{\mathrm{R}}{100}\right]^2$$
⇒ $$\left(\frac{11}{10}\right)^2=\left[1+\frac{\mathrm{R}}{100}\right]^2$$
⇒ $$\frac{11}{10}=1+\frac{R}{100}$$
⇒ $$\frac{R}{100}=\frac{11}{10}-1=\frac{1}{10}$$
⇒ R = 10%
Hence, the required rate of growth = 10% p.a. ### DAV Class 8 Maths Chapter 6 Value Based Questions

Question .
Donating blood is a humanitarian act. As per a survey, 16,000 blood donors are registered with ‘Red Cross’ in Delhi and this number of donors increases at the rate of 5% every six months.

(i) How many donors would be there at the end of 1$$\frac{1}{2}$$ years?
Solution:
Initial donors = 16,000
Number of half years = 1$$\frac{1}{2}$$ × 2 = 3.
Rate of growth = 5%.
Number of donors at the end of 3 years = 16000 $$\left(1+\frac{5}{100}\right)^3$$
= 16000 × $$\left(\frac{105}{100}\right)^3$$
= 16000 × $$\frac{105}{100} \times \frac{105}{100} \times \frac{105}{100}$$
= 18522

(ii) Why is it important to donate blood?
Solution:
Donating blood is very important as the blood saves someone’s life.

(iii) What steps would you take to encourage people to donate blood?
Solution:
Steps to encourage people to donate blood:
(a) To convince people that it is a generous act as blood may save someone’s life.
(b) By supplying donors monetory support in case of need.
(c) By providing donors identity card and a certificate with a promise free medical check up to them and their kins. Question 2.
A town has a population of 2,50,000. The growth rate of the population of the town is 4% per annum.

(i) What would be the population after three years?
Solution:
Initial population = 2,50,000
Rate of growth = 4% p.a.
Time = 3 years
A = P $$\left(1+\frac{r}{100}\right)^n$$
= 250000 $$\left(1+\frac{4}{100}\right)^3$$
= 250000 $$\frac{104}{100} \times \frac{104}{100} \times \frac{104}{100}$$
= 2,81,216

(ii) What are the harmful effects of such a rapid increase in the population?
Solution:
Harmful effects of rapid increase in population are poverty, unemployment, etc.