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
View Answer

Answer: c
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

Answer: d
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
View Answer

Answer: a
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
View Answer

Answer: b
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
View Answer

Answer: d
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

Answer: a
Explanation: Previous production = 16000
Present production = 25000
25000 = 16000 (1 + \(\frac{R}{100}\))2
\(\frac{5}{4}\) = 1 + \(\frac{R}{100}\)
R = 25.
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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

Answer: b
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

Answer: b
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.
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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

Answer: c
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.
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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 = \(\frac{1}{2}\)
View Answer

Answer: a
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.

Sanfoundry Global Education & Learning Series – Mathematics – Class 8.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter