# Mathematics Questions and Answers – Applications of Compound Interest Formula – 1

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This set of Mathematics 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

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

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%, 2.5%, 3%, 2% in last four years. Find the present population of the village.
a) 267.903
b) 260
c) 266.903
d) 266

Explanation: Present Population = 10000 (1 + $$\frac{2}{100}$$)(1 + $$\frac{2.5}{100}$$)(1 + $$\frac{3}{100}$$)(1 + $$\frac{2}{100}$$)
= 10000 ($$\frac{51}{50})(\frac{51}{50})(\frac{103}{100})(\frac{41}{40}$$) = 267.903.

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) 576000
b) 138240
c) 57600
d) 13824

Explanation: Population three years ago = 150000
Rate of decrease of population = 4% per annum
Present population = 150000 (1 – $$\frac{4}{100}$$)3 = 150000 ($$\frac{24}{25}$$)2 = 138240.

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

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

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

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

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

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 24000 ml oil. The amount of oil has to be increased at the rate of 5% every six month. Find the time-period at the end of which the amount of oil becomes 27783ml.
a) n = $$\frac{3}{2}$$
b) n = $$\frac{3}{4}$$
c) n = $$\frac{5}{2}$$
d) n = $$\frac{1}{2}$$

Explanation: P = Initial amount of oil = 24000 ml
R = rate of increase = 5% every 6 months = 10% per annum.
Let the total time be n years. Then,
A = P (1 + $$\frac{R}{200})^{2n}$$
27783 = 24000 (1 + $$\frac{10}{200})^{2n}$$
$$\frac{27783}{24000}$$ = ($$\frac{21}{20})^{2n}$$
($$\frac{21}{20}$$)3 = ($$\frac{21}{20})^{2n}$$
2n = 3
n = $$\frac{3}{2}$$
Hence, required time-period is $$\frac{3}{2}$$ years.

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