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# RD Sharma Solutions - Chapter 14 - Compound Interest (Part - 4), Class 8, Maths Class 8 Notes | EduRev

## RD Sharma Solutions for Class 8 Mathematics

Created by: Abhishek Kapoor

## Class 8 : RD Sharma Solutions - Chapter 14 - Compound Interest (Part - 4), Class 8, Maths Class 8 Notes | EduRev

The document RD Sharma Solutions - Chapter 14 - Compound Interest (Part - 4), Class 8, Maths Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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PAGE NO 14.27:

Question 1:

The present population of a town is 28000. If it increases at the rate of 5% per annum, what will be its population after 2 years?

Here,P = Initial population = 28,000

R = Rate of growth of population = 5% per annum

n = Number of years = 2

∴ Population after two years = P(1 + R/100)n

= 28,000(1 + 5/100)²

= 28,000(1.05)²

= 30,870

Hence, the population after two years will be 30,870.

PAGE NO 14.27:

Question 2:

The population of a city is 125000. If the annual birth rate and death rate are 5.5% and 3.5% respectively, calculate the population of city after 3 years.

Here, P = Initial population = 125,000

Annual birth rate = R1 = 5.5%

Annual death rate = R2 = 3.5%

Net growth rate, R = (R1 − R2) = 2%

n = Number of years = 3

∴ Population after three years = P(1 + R/100)

= 125,000(1 + 2/100)³

= 125,000(1.02)³

= 132,651

Hence, the population after three years will be 132,651.

PAGE NO 14.27:

Question 3:

The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.

Here,P = Initial population = 25,000

R1 = 4%

R2 = 5%

R3 = 8%

n = Number of years = 3

∴ Population after three years = P(1 + R1/100)(1 + R2/100)(1 + R3/100)

= 25,000(1 + 4/100)(1 + 5/100)(1 + 8/100)

= 25,000(1.04)(1.05)(1.08)

= 29,484

Hence, the population after three years will be 29,484.

PAGE NO 14.27:

Question 4:

Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 4%, 5% and 3% respectively, find the present population.

Here,P = Initial population = 50,000

R1 = 4%

R2 = 5%

R3 = 3%

n = Number of years = 3

∴ Population after three years = P(1 + R1/100)(1 + R2/100)(1 + R3/100)

= 50,000(1 + 4/100)(1 + 5/100)(1 + 3/100)

= 50,000(1.04)(1.05)(1.03)

= 56,238

Hence, the population after three years is 56,238.

PAGE NO 14.27:

Question 5:

There is a continuous growth in population of a village at the rate of 5% per annum. If its present population is 9261, what it was 3 years ago?

Population after three years = P(1 + R/100)n

9,261 = P(1 + 5/100)³

9,261 = P(1.05)³

P = 9,261/1.157625

= 8,000

Thus, the population three years ago was 8,000.

PAGE NO 14.27:

Question 6:

In a factory the production of scooters rose to 46305 from 40000 in 3 years. Find the annual rate of growth of the production of scooters.

Let the annual rate of growth be R.

∴ Production of scooters after three years =  P(1 + R/100)n

46,305 = 4,000(1 + R/100)³

(1 + 0.01R)³ = 46,30540,000

(1 + 0.01R)³ = 1.157625

(1 + 0.01R)³ = (1.05)³

1 + 0.01R = 1.05

0.01R = 0.05

R = 5

Thus, the annual rate of growth is 5%.

PAGE NO 14.27:

Question 7:

The annual rate of growth in population of a certain city is 8%. If its present population is 196830, what it was 3 years ago?

Population after three years = P(1 + R/100)n

196,830 = P(1 + 8/100)³

196,830 = P(1.08)³

P = 196,830/1.259712 = 156,250

Thus, the population three years ago was 156,250.

PAGE NO 14.27:

Question 8:

The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050. Find its present population.

Population after two years = P(1 + R/100)n

22,050 = P(1 + 50/1000)²

22,050 = P(1.05)²

P = 22,050/1.1025 = 20,000

Thus, the population two years ago was 20,000.

PAGE NO 14.27:

Question 9:

The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours?

Given: R1 = 10%

R = − 8%

R3 = 12%

P = Original count of bacteria = 13,125,000

We know that:P(1 + R1/100)(1 − R2/100)(1 + R3/100)

∴ Bacteria count after three hours = 13,125,000(1 + 10/100)(1 − 8/100)(1 + 12/100)

= 13,125,000(1.10)(0.92)(1.12)

= 14,876,400

Thus, the bacteria count after three hours will be 14,876,400.

PAGE NO 14.27:

Question 10:

The population of a certain city was 72000 on the last day of the year 1998. During next year it increased by 7% but due to an epidemic it decreased by 10% in the following year. What was its population at the end of the year 2000?

Population at the end of the year 2000 = P(1 + R1/100)(1 − R2/100)

= 72,000(1 + 7100)(1 − 10/100)

= 72,000(1.07)(0.9) = 69,336

Thus, the population at the end of the year 2000 was 69,336.

PAGE NO 14.27:

Question 11:

6400 workers were employed to construct a river bridge in four years. At the end of the first year, 25% workers were retrenched. At the end of the second year, 25% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25% at the end of the third year. How many workers were working during the fourth year?

Number of workers = 6,400

At the end of the first year, 25% of the workers were retrenched.

∴ 25% of 6,400 = 1,600

Number of workers at the end of the first year =  6,400 − 1,600 = 4,800

At the end of the second year, 25% of those working were retrenched.

∴ 25% of 4,800 = 1,200

Number of workers at the end of the second year = 4,800 − 1,200 = 3,600

At the end of the third year, 25% of those working increased.

∴ 25% of 3,600 = 900

Number of workers at the end of the third year = 3,600 + 900 = 4,500

Thus, the number of workers during the fourth year was 4,500.

PAGE NO 14.27:

Question 12:

Aman started a factory with an initial investment of Rs 100000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.

Aman's profit for three years = P(1 − R1/100)(1 + R2/100)(1 + R3/100)

= 100,000(1 − 5/100)(1 + 10/100)(1 + 12/100)

= 100,000(0.95)(1.10)(1.12) = 117,040

∴ Net profit = Rs 117,040 − Rs 100,000

= Rs 17,040

PAGE NO 14.28:

Question 13:

The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago.

Population after three years = P(1 + R/100)³

175,760 = P(1 + 401000)³

175,760 = P(1.04)³

P = 175,7601/124864 = 156,250

Thus, the population three years ago was 156,250.

PAGE NO 14.28:

Question 14:

The production of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will be its production after 3 years?

Production after three years = P(1 + R1/100)²(1 − R2/100)

= 8,000(1 + 151,000)²(1 − 5/100)

= 8,000(1.15)²(0.95)

= 10,051

Thus, the production after three years will be 10,051.

PAGE NO 14.28:

Question 15:

The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (i) 2001 (ii) 1997.

(i)Population of the city in 2001 = P(1 + R/100)²

= 6760000(1 + 4/100)²

= 6760000(1.04)²

= 7311616

Thus, Population of the city in 2001 is 7311616.

(ii)Population of the city in 1997 = P(1 + R/100) ²

= 6760000(1 + 4/100) ²

= 6760000(1.04) ² = 6250000

Thus,Population of the city in 1997 is 6250000.

PAGE NO 14.28:

Question 16:

Jitendra set up a factory by investing Rs 2500000. During the first two successive years his profits were 5% and 10% respectively. If each year the profit was on previous year's capital, compute his total profit.

Profit at the end of the first year = P(1 + R/100)

= 2,500,000(1 + 5/100)

= 2,500,000(1.05)

= 2,625,000

Profit at the end of the second year = P(1 + R/100)

= 2,625,000(1 + 10/100)

= 2,625,000(1.10)

= 2,887,500

Total profit  =  Rs 2,887,500 − Rs 2,500,000

= Rs 387,500

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