This set of Class 8 Maths Chapter 8 Multiple Choice Questions & Answers (MCQs) focuses on “Applications of Compound Interest Formula – 1”.
1. The population of a city is increasing at the rate of 10% per annum. Calculate the population of the city on this basis after 3 years if the current population is 10000.
a) 12100
b) 13100
c) 13310
d) 1210
View Answer
Explanation: Here, P = Initial population = 10000
R = Rate of growth of population = 10%
n = number of years
Population after 3 years = P (1 + \(\frac{R}{100}\))n = 10000 (1 + \(\frac{10}{100}\))3 = 10000 (\(\frac{11}{10}\))3 = 13310.
2. The population of the town is 30000. The annual birth rate is 5% and the annual death rate is 3%. Calculate the population after 2 years.
a) 61212
b) 612
c) 312
d) 31212
View Answer
Explanation: P = Initial population = 30000
R = 5 – 3 = 2%
n = 2
Population after 2 years = P (1 + \(\frac{R}{100}\))n = 30000 (1 + \(\frac{2}{100}\))2 = 30000 (\(\frac{51}{50}\))2 = 31212.
3. The population of a village was 10000 four years ago. If it had increased by 2%, 10%, 5%, 4% in last four years. Find the present population of the village.
a) 12250
b) 17250
c) 12750
d) 15250
View Answer
Explanation: Present Population = 10000 (1 + \(\frac{2}{100}\))(1 + \(\frac{10}{100}\))(1 + \(\frac{5}{100}\))(1 + \(\frac{4}{100}\))
= 10000 (\(\frac{51}{50})(\frac{11}{10})(\frac{21}{20})(\frac{26}{25}\)) = 12252.24 ≈ 12250.
4. The population of a sate decreases every year at the rate 4% per annum. The population of the state 3 years ago was 150000. Find present population.
a) 13271
b) 132710
c) 138240
d) 13824
View Answer
Explanation: Population three years ago = 150000
Rate of decrease of population = 4% per annum
Present population = 150000 (1 – \(\frac{4}{100}\))3 = 150000 (\(\frac{24}{25}\))3 = 132710.4 ≈ 132710.
5. If the population of a city has been increasing at the rate of 10%. The present population of the city is 9680000. Find its population 2 years ago.
a) 800000
b) 8000
c) 80000
d) 8000000
View Answer
Explanation: Let the population 2 years ago be P.
Present Population = P (1 + \(\frac{10}{100}\))2
9680000 = P (1 + \(\frac{10}{100}\))2
P = 8000000.
6. The production of spare parts rose to 25000 from 16000 in 2 years. Find the rate of growth per annum.
a) 25
b) 20
c) 250
d) 200
View Answer
Explanation: Previous production = 16000
Present production = 25000
25000 = 16000 (1 + \(\frac{R}{100}\))2
\(\frac{5}{4}\) = 1 + \(\frac{R}{100}\)
R = 25.
7. The bacteria in a culture grows by 5% in the first hour, decreases by 10% in the second hour and increases by 2% in the third hour. If the original count of the bacteria is 50000, Calculate the count of bacteria at the end of 3 hours.
a) 48000
b) 48195
c) 24097
d) 24095
View Answer
Explanation: P = Original count of bacteria = 50000
Count of bacteria after 3 hours = 50000 (1 + \(\frac{5}{100}\))(1 – \(\frac{10}{100}\))(1 + \(\frac{2}{100}\)) = 48195.
8. Manish opened a cafe with an initial investment of Rs. 64000. In the first year he incurred the loss of 10%. During second year he gained the profit of 4% and in the third year he gained the profit of 15%. Calculate his net profit gained in the entire three years.
a) 68889.5
b) 4889.6
c) 68889.61
d) 68889.60
View Answer
Explanation: Initial investment = Rs. 64000
Loss in first year = 10%
Profit in second year = 4%
Profit in third year = 15%
profit = 64000 (1 – \(\frac{10}{100}\))(1 + \(\frac{4}{100}\))(1 + \(\frac{15}{100}\)) = 64000 (\(\frac{9}{10})(\frac{26}{25})(\frac{23}{20})\) = 68889.60.
Net profit = 68889.60 – 64000 = 4889.6.
9. 900 workers were employed to construct a dam in four years. At the end of first year, 5% workers were retrenched. At the end of the second year 2% of the workers were retrenched. To complete the construction in time, the number of workers were increased to 10% at the end of third year. How many workers were working at the end of fourth year.
a) 945
b) 1000
c) 921
d) 989
View Answer
Explanation: Initial number of workers = 900
Reduction of workers at the end of first year = 5%
Reduction of workers at the end of second year = 2%
Increase of workers at the end third year = 10%
Number of workers working during fourth year = 900 (1 – \(\frac{5}{100}\))(1 – \(\frac{2}{100}\))(1 + \(\frac{10}{100}\))
= 900 (\(\frac{95}{100})(\frac{98}{100})(\frac{110}{100})\) = 921.69 ≅ 921 workers.
10. An engine has initially 8000 litre oil. The amount of oil has to be increased at the rate of 10% every year. Find the time-period at the end of which the amount of oil becomes 9261 litre.
a) n = \(\frac{3}{2}\)
b) n = 3
c) n = \(\frac{2}{3}\)
d) n = 2
View Answer
Explanation: P = Initial amount of oil = 8000 l
R = rate of increase = 10% per annum.
Let the total time be n years. Then,
A = P (1 + \(\frac{R}{100})^{n}\)
9261 = 8000 (1 + \(\frac{10}{100})^{n}\)
\(\frac{9261}{8000}\) = (\(\frac{21}{20})^{n}\)
(\(\frac{21}{20}\))3 = (\(\frac{21}{20})^{n}\)
n = 3
Hence, required time-period is 3 years.
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