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

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

The document RD Sharma Solutions - Chapter 14 - Compound Interest (Part - 4), Class 8, Maths Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8

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?

ANSWER:

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.

ANSWER:

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.

ANSWER:

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.

ANSWER:

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?

ANSWER:

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.

ANSWER:

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?

ANSWER:

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.

ANSWER:

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?

ANSWER:

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?

ANSWER:

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?

ANSWER:

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.

ANSWER:

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.

ANSWER:

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?

ANSWER:

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.

ANSWER:

(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.

ANSWER:

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

The document RD Sharma Solutions - Chapter 14 - Compound Interest (Part - 4), Class 8, Maths Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8
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## RD Sharma Solutions for Class 8 Mathematics

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