RD Sharma solution class 8 chapter 14 Compound Interest Exercise 14.4

Exercise 14.4

Page-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 1:

$$\begin{gathered} \mathrm{Here}, \\ \mathrm{P}=\mathrm{Initial} \mathrm{population}=28,000 \\ \mathrm{R}=\mathrm{Rate} \mathrm{of} \mathrm{growth} \mathrm{of} \mathrm{population}=5\% \mathrm{per} \mathrm{annum} \\ \mathrm{n}=\mathrm{Number} \mathrm{of} \mathrm{years}=2 \\ \therefore \mathrm{Population} \mathrm{after} \mathrm{two} \mathrm{years}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ =28,000{\left(1+\frac{5}{100}\right)}^2 \\ =28,000{\left(1.05\right)}^2 \\ =30,870 \\ \mathrm{Hence}, \mathrm{the} \mathrm{population} \mathrm{after} \mathrm{two} \mathrm{years} \mathrm{will} \mathrm{be} 30,870. \end{gathered}$$

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 2:

$$\begin{gathered} \mathrm{Here}, \\ \mathrm{P}=\mathrm{Initial} \mathrm{population}=125,000 \\ \mathrm{Annual} \mathrm{birth} \mathrm{rate}={\mathrm{R}}_1=5.5\% \\ \mathrm{Annual} \mathrm{death} \mathrm{rate}={\mathrm{R}}_2=3.5\% \\ \mathrm{Net} \mathrm{growth} \mathrm{rate}, \mathrm{R}=\left({\mathrm{R}}_1-{\mathrm{R}}_2\right)=2\% \\ \mathrm{n}=\mathrm{Number} \mathrm{of} \mathrm{years}=3 \\ \therefore \mathrm{Population} \mathrm{after} \mathrm{three} \mathrm{years}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ =125,000{\left(1+\frac{2}{100}\right)}^3 \\ =125,000{\left(1.02\right)}^3 \\ =132,651 \\ \mathrm{Hence}, \mathrm{the} \mathrm{population} \mathrm{after} \mathrm{three} \mathrm{years} \mathrm{will} \mathrm{be} 132,651. \end{gathered}$$

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 3:

$$\begin{gathered} \mathrm{Here}, \\ \mathrm{P}=\mathrm{Initial} \mathrm{population}=25,000 \\ {\mathrm{R}}_1=4\% \\ {\mathrm{R}}_2=5\% \\ {\mathrm{R}}_3=8\% \\ \mathrm{n}=\mathrm{Number} \mathrm{of} \mathrm{years}=3 \\ \therefore \mathrm{Population} \mathrm{after} \mathrm{three} \mathrm{years}=\mathrm{P}\left(1+\frac{{\mathrm{R}}_1}{100}\right)\left(1+\frac{{\mathrm{R}}_2}{100}\right)\left(1+\frac{{\mathrm{R}}_3}{100}\right) \\ =25,000\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\left(1+\frac{8}{100}\right) \\ =25,000\left(1.04\right)\left(1.05\right)\left(1.08\right) \\ =29,484 \\ \mathrm{Hence}, \mathrm{the} \mathrm{population} \mathrm{after} \mathrm{three} \mathrm{years} \mathrm{will} \mathrm{be} 29,484. \end{gathered}$$

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 4:

$$\begin{gathered} \mathrm{Here}, \\ \mathrm{P}=\mathrm{Initial} \mathrm{population}=50,000 \\ {\mathrm{R}}_1=4\% \\ {\mathrm{R}}_2=5\% \\ {\mathrm{R}}_3=3\% \\ \mathrm{n}=\mathrm{Number} \mathrm{of} \mathrm{years}=3 \\ \therefore \mathrm{Population} \mathrm{after} \mathrm{three} \mathrm{years}=\mathrm{P}\left(1+\frac{{\mathrm{R}}_1}{100}\right)\left(1+\frac{{\mathrm{R}}_2}{100}\right)\left(1+\frac{{\mathrm{R}}_3}{100}\right) \\ =50,000\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\left(1+\frac{3}{100}\right) \\ =50,000\left(1.04\right)\left(1.05\right)\left(1.03\right) \\ =56,238 \\ \mathrm{Hence}, \mathrm{the} \mathrm{population} \mathrm{after} \mathrm{three} \mathrm{years} \mathrm{is} 56,238. \end{gathered}$$

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 5:

$$\begin{gathered} \mathrm{Population} \mathrm{after} \mathrm{three} \mathrm{years}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ 9,261=\mathrm{P}{\left(1+\frac{5}{100}\right)}^3 \\ 9,261=\mathrm{P}{\left(1.05\right)}^3 \\ \mathrm{P}=\frac{9,261}{1.157625} \\ =8,000 \\ \mathrm{Thus}, \mathrm{the} \mathrm{population} \mathrm{three} \mathrm{years} \mathrm{ago} \mathrm{was} 8,000. \end{gathered}$$

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 6:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{annual} \mathrm{rate} \mathrm{of} \mathrm{growth} \mathrm{be} \mathrm{R}. \\ \therefore \mathrm{Production} \mathrm{of} \mathrm{scooters} \mathrm{after} \mathrm{three} \mathrm{years}= \mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ 46,305=4,000{\left(1+\frac{\mathrm{R}}{100}\right)}^3 \\ {\left(1+0.01\mathrm{R}\right)}^3=\frac{46,305}{40,000} \\ {\left(1+0.01\mathrm{R}\right)}^3=1.157625 \\ {\left(1+0.01\mathrm{R}\right)}^3={\left(1.05\right)}^3 \\ 1+0.01\mathrm{R}=1.05 \\ 0.01\mathrm{R}=0.05 \\ \mathrm{R}=5 \\ \mathrm{Thus}, \mathrm{the} \mathrm{annual} \mathrm{rate} \mathrm{of} \mathrm{growth} \mathrm{is} 5\%. \end{gathered}$$

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 7:

$$\begin{gathered} \mathrm{Population} \mathrm{after} \mathrm{three} \mathrm{years}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ 196,830=\mathrm{P}{\left(1+\frac{8}{100}\right)}^3 \\ 196,830=\mathrm{P}{\left(1.08\right)}^3 \\ \mathrm{P}=\frac{196,830}{1.259712} \\ =156,250 \\ \mathrm{Thus}, \mathrm{the} \mathrm{population} \mathrm{three} \mathrm{years} \mathrm{ago} \mathrm{was} 156,250. \end{gathered}$$

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 8:

$$\begin{gathered} \mathrm{Population} \mathrm{after} \mathrm{two} \mathrm{years}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ 22,050=\mathrm{P}{\left(1+\frac{50}{1000}\right)}^2 \\ 22,050=\mathrm{P}{\left(1.05\right)}^2 \\ \mathrm{P}=\frac{22,050}{1.1025} \\ =20,000 \\ \mathrm{Thus}, \mathrm{the} \mathrm{population} \mathrm{two} \mathrm{years} \mathrm{ago} \mathrm{was} 20,000. \end{gathered}$$

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 9:

$$\begin{gathered} \mathrm{Given}: \\ {\mathrm{R}}_1=10\% \\ {\mathrm{R}}_2=-8\% \\ {\mathrm{R}}_3=12\% \\ \mathrm{P}=\mathrm{Original} \mathrm{count} \mathrm{of} \mathrm{bacteria}=13,125,000 \\ \mathrm{We} \mathrm{know} \mathrm{that}: \\ \mathrm{P}\left(1+\frac{{\mathrm{R}}_1}{100}\right)\left(1-\frac{{\mathrm{R}}_2}{100}\right)\left(1+\frac{{\mathrm{R}}_3}{100}\right) \\ \therefore \mathrm{Bacteria} \mathrm{count} \mathrm{after} \mathrm{three} \mathrm{hours}=13,125,000\left(1+\frac{10}{100}\right)\left(1-\frac{8}{100}\right)\left(1+\frac{12}{100}\right) \\ =13,125,000\left(1.10\right)\left(0.92\right)\left(1.12\right) \\ =14,876,400 \\ \mathrm{Thus}, \mathrm{the} \mathrm{bacteria} \mathrm{count} \mathrm{after} \mathrm{three} \mathrm{hours} \mathrm{will} \mathrm{be} 14,876,400. \end{gathered}$$

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 10:

$$\begin{gathered} \mathrm{Population} \mathrm{at} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{year} 2000=\mathrm{P}\left(1+\frac{{\mathrm{R}}_1}{100}\right)\left(1-\frac{{\mathrm{R}}_2}{100}\right) \\ =72,000\left(1+\frac{7}{100}\right)\left(1-\frac{10}{100}\right) \\ =72,000\left(1.07\right)\left(0.9\right) \\ =69,336 \\ \mathrm{Thus}, \mathrm{the} \mathrm{population} \mathrm{at} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{year} 2000 \mathrm{was} 69,336. \end{gathered}$$

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 11:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{workers}=6,400 \\ \mathrm{At} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{year}, 25\% \mathrm{of} \mathrm{the} \mathrm{workers} \mathrm{were} \mathrm{retrenched}. \\ \therefore 25\% \mathrm{of} 6,400=1,600 \\ \mathrm{Number} \mathrm{of} \mathrm{workers} \mathrm{at} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{year}= 6,400-1,600=4,800 \\ \mathrm{At} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{second} \mathrm{year}, 25\% \mathrm{of} \mathrm{those} \mathrm{working} \mathrm{were} \mathrm{retrenched}. \\ \therefore 25\% \mathrm{of} 4,800=1,200 \\ \mathrm{Number} \mathrm{of} \mathrm{workers} \mathrm{at} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{second} \mathrm{year}=4,800-1,200=3,600 \\ \mathrm{At} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{third} \mathrm{year}, 25\% \mathrm{of} \mathrm{those} \mathrm{working} \mathrm{increased}. \\ \therefore 25\% \mathrm{of} 3,600=900 \\ \mathrm{Number} \mathrm{of} \mathrm{workers} \mathrm{at} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{third} \mathrm{year}=3,600+900=4,500 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{workers} \mathrm{during} \mathrm{the} \mathrm{fourth} \mathrm{year} \mathrm{was} 4,500. \end{gathered}$$

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 12:

$$\begin{gathered} \mathrm{Aman}\text{'}\mathrm{s} \mathrm{profit} \mathrm{for} \mathrm{three} \mathrm{years}=\mathrm{P}\left(1-\frac{{\mathrm{R}}_1}{100}\right)\left(1+\frac{{\mathrm{R}}_2}{100}\right)\left(1+\frac{{\mathrm{R}}_3}{100}\right) \\ =100,000\left(1-\frac{5}{100}\right)\left(1+\frac{10}{100}\right)\left(1+\frac{12}{100}\right) \\ =100,000\left(0.95\right)\left(1.10\right)\left(1.12\right) \\ =117,040 \\ \therefore \mathrm{Net} \mathrm{profit}=\mathrm{Rs} 117,040-\mathrm{Rs} 100,000 \\ =\mathrm{Rs} 17,040 \end{gathered}$$

Page-14.28

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 1:

$$\begin{gathered} \mathrm{Here}, \\ \mathrm{P}=\mathrm{Initial} \mathrm{population}=28,000 \\ \mathrm{R}=\mathrm{Rate} \mathrm{of} \mathrm{growth} \mathrm{of} \mathrm{population}=5\% \mathrm{per} \mathrm{annum} \\ \mathrm{n}=\mathrm{Number} \mathrm{of} \mathrm{years}=2 \\ \therefore \mathrm{Population} \mathrm{after} \mathrm{two} \mathrm{years}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ =28,000{\left(1+\frac{5}{100}\right)}^2 \\ =28,000{\left(1.05\right)}^2 \\ =30,870 \\ \mathrm{Hence}, \mathrm{the} \mathrm{population} \mathrm{after} \mathrm{two} \mathrm{years} \mathrm{will} \mathrm{be} 30,870. \end{gathered}$$

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 2:

$$\begin{gathered} \mathrm{Here}, \\ \mathrm{P}=\mathrm{Initial} \mathrm{population}=125,000 \\ \mathrm{Annual} \mathrm{birth} \mathrm{rate}={\mathrm{R}}_1=5.5\% \\ \mathrm{Annual} \mathrm{death} \mathrm{rate}={\mathrm{R}}_2=3.5\% \\ \mathrm{Net} \mathrm{growth} \mathrm{rate}, \mathrm{R}=\left({\mathrm{R}}_1-{\mathrm{R}}_2\right)=2\% \\ \mathrm{n}=\mathrm{Number} \mathrm{of} \mathrm{years}=3 \\ \therefore \mathrm{Population} \mathrm{after} \mathrm{three} \mathrm{years}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ =125,000{\left(1+\frac{2}{100}\right)}^3 \\ =125,000{\left(1.02\right)}^3 \\ =132,651 \\ \mathrm{Hence}, \mathrm{the} \mathrm{population} \mathrm{after} \mathrm{three} \mathrm{years} \mathrm{will} \mathrm{be} 132,651. \end{gathered}$$

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 3:

$$\begin{gathered} \mathrm{Here}, \\ \mathrm{P}=\mathrm{Initial} \mathrm{population}=25,000 \\ {\mathrm{R}}_1=4\% \\ {\mathrm{R}}_2=5\% \\ {\mathrm{R}}_3=8\% \\ \mathrm{n}=\mathrm{Number} \mathrm{of} \mathrm{years}=3 \\ \therefore \mathrm{Population} \mathrm{after} \mathrm{three} \mathrm{years}=\mathrm{P}\left(1+\frac{{\mathrm{R}}_1}{100}\right)\left(1+\frac{{\mathrm{R}}_2}{100}\right)\left(1+\frac{{\mathrm{R}}_3}{100}\right) \\ =25,000\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\left(1+\frac{8}{100}\right) \\ =25,000\left(1.04\right)\left(1.05\right)\left(1.08\right) \\ =29,484 \\ \mathrm{Hence}, \mathrm{the} \mathrm{population} \mathrm{after} \mathrm{three} \mathrm{years} \mathrm{will} \mathrm{be} 29,484. \end{gathered}$$

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 4:

$$\begin{gathered} \mathrm{Here}, \\ \mathrm{P}=\mathrm{Initial} \mathrm{population}=50,000 \\ {\mathrm{R}}_1=4\% \\ {\mathrm{R}}_2=5\% \\ {\mathrm{R}}_3=3\% \\ \mathrm{n}=\mathrm{Number} \mathrm{of} \mathrm{years}=3 \\ \therefore \mathrm{Population} \mathrm{after} \mathrm{three} \mathrm{years}=\mathrm{P}\left(1+\frac{{\mathrm{R}}_1}{100}\right)\left(1+\frac{{\mathrm{R}}_2}{100}\right)\left(1+\frac{{\mathrm{R}}_3}{100}\right) \\ =50,000\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\left(1+\frac{3}{100}\right) \\ =50,000\left(1.04\right)\left(1.05\right)\left(1.03\right) \\ =56,238 \\ \mathrm{Hence}, \mathrm{the} \mathrm{population} \mathrm{after} \mathrm{three} \mathrm{years} \mathrm{is} 56,238. \end{gathered}$$

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 5:

$$\begin{gathered} \mathrm{Population} \mathrm{after} \mathrm{three} \mathrm{years}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ 9,261=\mathrm{P}{\left(1+\frac{5}{100}\right)}^3 \\ 9,261=\mathrm{P}{\left(1.05\right)}^3 \\ \mathrm{P}=\frac{9,261}{1.157625} \\ =8,000 \\ \mathrm{Thus}, \mathrm{the} \mathrm{population} \mathrm{three} \mathrm{years} \mathrm{ago} \mathrm{was} 8,000. \end{gathered}$$

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 6:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{annual} \mathrm{rate} \mathrm{of} \mathrm{growth} \mathrm{be} \mathrm{R}. \\ \therefore \mathrm{Production} \mathrm{of} \mathrm{scooters} \mathrm{after} \mathrm{three} \mathrm{years}= \mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ 46,305=4,000{\left(1+\frac{\mathrm{R}}{100}\right)}^3 \\ {\left(1+0.01\mathrm{R}\right)}^3=\frac{46,305}{40,000} \\ {\left(1+0.01\mathrm{R}\right)}^3=1.157625 \\ {\left(1+0.01\mathrm{R}\right)}^3={\left(1.05\right)}^3 \\ 1+0.01\mathrm{R}=1.05 \\ 0.01\mathrm{R}=0.05 \\ \mathrm{R}=5 \\ \mathrm{Thus}, \mathrm{the} \mathrm{annual} \mathrm{rate} \mathrm{of} \mathrm{growth} \mathrm{is} 5\%. \end{gathered}$$

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 7:

$$\begin{gathered} \mathrm{Population} \mathrm{after} \mathrm{three} \mathrm{years}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ 196,830=\mathrm{P}{\left(1+\frac{8}{100}\right)}^3 \\ 196,830=\mathrm{P}{\left(1.08\right)}^3 \\ \mathrm{P}=\frac{196,830}{1.259712} \\ =156,250 \\ \mathrm{Thus}, \mathrm{the} \mathrm{population} \mathrm{three} \mathrm{years} \mathrm{ago} \mathrm{was} 156,250. \end{gathered}$$

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 8:

$$\begin{gathered} \mathrm{Population} \mathrm{after} \mathrm{two} \mathrm{years}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}} \\ 22,050=\mathrm{P}{\left(1+\frac{50}{1000}\right)}^2 \\ 22,050=\mathrm{P}{\left(1.05\right)}^2 \\ \mathrm{P}=\frac{22,050}{1.1025} \\ =20,000 \\ \mathrm{Thus}, \mathrm{the} \mathrm{population} \mathrm{two} \mathrm{years} \mathrm{ago} \mathrm{was} 20,000. \end{gathered}$$

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 9:

$$\begin{gathered} \mathrm{Given}: \\ {\mathrm{R}}_1=10\% \\ {\mathrm{R}}_2=-8\% \\ {\mathrm{R}}_3=12\% \\ \mathrm{P}=\mathrm{Original} \mathrm{count} \mathrm{of} \mathrm{bacteria}=13,125,000 \\ \mathrm{We} \mathrm{know} \mathrm{that}: \\ \mathrm{P}\left(1+\frac{{\mathrm{R}}_1}{100}\right)\left(1-\frac{{\mathrm{R}}_2}{100}\right)\left(1+\frac{{\mathrm{R}}_3}{100}\right) \\ \therefore \mathrm{Bacteria} \mathrm{count} \mathrm{after} \mathrm{three} \mathrm{hours}=13,125,000\left(1+\frac{10}{100}\right)\left(1-\frac{8}{100}\right)\left(1+\frac{12}{100}\right) \\ =13,125,000\left(1.10\right)\left(0.92\right)\left(1.12\right) \\ =14,876,400 \\ \mathrm{Thus}, \mathrm{the} \mathrm{bacteria} \mathrm{count} \mathrm{after} \mathrm{three} \mathrm{hours} \mathrm{will} \mathrm{be} 14,876,400. \end{gathered}$$

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 10:

$$\begin{gathered} \mathrm{Population} \mathrm{at} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{year} 2000=\mathrm{P}\left(1+\frac{{\mathrm{R}}_1}{100}\right)\left(1-\frac{{\mathrm{R}}_2}{100}\right) \\ =72,000\left(1+\frac{7}{100}\right)\left(1-\frac{10}{100}\right) \\ =72,000\left(1.07\right)\left(0.9\right) \\ =69,336 \\ \mathrm{Thus}, \mathrm{the} \mathrm{population} \mathrm{at} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{year} 2000 \mathrm{was} 69,336. \end{gathered}$$

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 11:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{workers}=6,400 \\ \mathrm{At} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{year}, 25\% \mathrm{of} \mathrm{the} \mathrm{workers} \mathrm{were} \mathrm{retrenched}. \\ \therefore 25\% \mathrm{of} 6,400=1,600 \\ \mathrm{Number} \mathrm{of} \mathrm{workers} \mathrm{at} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{year}= 6,400-1,600=4,800 \\ \mathrm{At} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{second} \mathrm{year}, 25\% \mathrm{of} \mathrm{those} \mathrm{working} \mathrm{were} \mathrm{retrenched}. \\ \therefore 25\% \mathrm{of} 4,800=1,200 \\ \mathrm{Number} \mathrm{of} \mathrm{workers} \mathrm{at} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{second} \mathrm{year}=4,800-1,200=3,600 \\ \mathrm{At} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{third} \mathrm{year}, 25\% \mathrm{of} \mathrm{those} \mathrm{working} \mathrm{increased}. \\ \therefore 25\% \mathrm{of} 3,600=900 \\ \mathrm{Number} \mathrm{of} \mathrm{workers} \mathrm{at} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{the} \mathrm{third} \mathrm{year}=3,600+900=4,500 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{workers} \mathrm{during} \mathrm{the} \mathrm{fourth} \mathrm{year} \mathrm{was} 4,500. \end{gathered}$$

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 12:

$$\begin{gathered} \mathrm{Aman}\text{'}\mathrm{s} \mathrm{profit} \mathrm{for} \mathrm{three} \mathrm{years}=\mathrm{P}\left(1-\frac{{\mathrm{R}}_1}{100}\right)\left(1+\frac{{\mathrm{R}}_2}{100}\right)\left(1+\frac{{\mathrm{R}}_3}{100}\right) \\ =100,000\left(1-\frac{5}{100}\right)\left(1+\frac{10}{100}\right)\left(1+\frac{12}{100}\right) \\ =100,000\left(0.95\right)\left(1.10\right)\left(1.12\right) \\ =117,040 \\ \therefore \mathrm{Net} \mathrm{profit}=\mathrm{Rs} 117,040-\mathrm{Rs} 100,000 \\ =\mathrm{Rs} 17,040 \end{gathered}$$