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?**

**ANSWER:**

Here,P = Initial population = 28,000

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

n = Number of years = 2

âˆ´ Population after two years = P(1 + R/100)^{n}

= 28,000(1 + 5/100)Â²

= 28,000(1.05)Â²

= 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 = R_{1} = 5.5%

Annual death rate = R_{2} = 3.5%

Net growth rate, R = (R_{1} âˆ’ R_{2}) = 2%

n = Number of years = 3

âˆ´ Population after three years = P(1 + R/100)^{n }

= 125,000(1 + 2/100)Â³

= 125,000(1.02)Â³

= 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

R_{1} = 4%

R_{2} = 5%

R_{3} = 8%

n = Number of years = 3

âˆ´ Population after three years = P(1 + R_{1}/100)(1 + R_{2}/100)(1 + R_{3}/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

R_{1} = 4%

R_{2} = 5%

R_{3} = 3%

n = Number of years = 3

âˆ´ Population after three years = P(1 + R_{1}/100)(1 + R_{2}/100)(1 + R_{3}/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)Â³

9,261 = P(1.05)Â³

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.

âˆ´ Production of scooters after three years = P(1 + R/100)^{n}

46,305 = 4,000(1 + R/100)Â³

(1 + 0.01R)Â³ = 46,30540,000

(1 + 0.01R)Â³ = 1.157625

(1 + 0.01R)Â³ = (1.05)Â³

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)Â³

196,830 = P(1.08)Â³

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)Â²

22,050 = P(1.05)Â²

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: R_{1} = 10%

R_{2 } = âˆ’ 8%

R_{3} = 12%

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

We know that:P(1 + R_{1}/100)(1 âˆ’ R_{2}/100)(1 + R_{3}/100)

âˆ´ Bacteria count after three hours = 13,125,000(1 + 10/100)(1 âˆ’ 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 + R_{1}/100)(1 âˆ’ R_{2}/100)

= 72,000(1 + 7100)(1 âˆ’ 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.

âˆ´ 25% of 6,400 = 1,600

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

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

âˆ´ 25% of 4,800 = 1,200

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

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

âˆ´ 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 âˆ’ R_{1}/100)(1 + R_{2}/100)(1 + R_{3}/100)

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

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

âˆ´ Net profit = Rs 117,040 âˆ’ 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)Â³

175,760 = P(1 + 401000)Â³

175,760 = P(1.04)Â³

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 + R_{1}/100)Â²(1 âˆ’ R_{2}/100)

= 8,000(1 + 151,000)Â²(1 âˆ’ 5/100)

= 8,000(1.15)Â²(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)Â²

= 6760000(1 + 4/100)Â²

= 6760000(1.04)Â²

= 7311616

Thus, Population of the city in 2001 is 7311616.

(ii)Population of the city in 1997 = P(1 + R/100) ^{âˆ’}Â²

= 6760000(1 + 4/100) ^{âˆ’}Â²

= 6760000(1.04) ^{âˆ’}Â² = 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 âˆ’ Rs 2,500,000

= Rs 387,500

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