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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