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 = R_{1} = 5.5%
Annual death rate = R_{2} = 3.5%
Net growth rate, R = (R_{1} − R_{2}) = 2%
n = Number of years = 3
∴ Population after three years = P(1 + R/100)^{n }
= 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
R_{1} = 4%
R_{2} = 5%
R_{3} = 8%
n = Number of years = 3
∴ 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
∴ 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)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: R_{1} = 10%
R_{2 } = − 8%
R_{3} = 12%
P = Original count of bacteria = 13,125,000
We know that:P(1 + R_{1}/100)(1 − R_{2}/100)(1 + R_{3}/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 + R_{1}/100)(1 − R_{2}/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 − R_{1}/100)(1 + R_{2}/100)(1 + R_{3}/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 + R_{1}/100)²(1 − R_{2}/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
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