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Question : The population of a city increases compounded by 10% annually. The present population of this city is 1,33,100. What was the population of the city 2 years ago?

Option 1: 1,01,000

Option 2: 1,10,000

Option 3: 1,21,000

Option 4: 1,00,000


Team Careers360 25th Jan, 2024
Answer (1)
Team Careers360 26th Jan, 2024

Correct Answer: 1,10,000


Solution : $P=P{_0}​×(1+r)^t$
where: $P$ is the present population, $P{_0}$​ is the past population, $r$ is the annual growth rate (as a decimal), and $t$ is the number of years.
Given, $P{_0} = 133,100$
$r = 10\%$
We have to find the population 2 years ago ($t=−2$ years)
$133100=P{_0}×(1+\frac{10}{100})^{2}$
$⇒P{_0}=133100×(\frac{11}{10})^{−2}$
$⇒P{_0}=133100×\frac{1}{(\frac{11}{10})^2}$​
$⇒P{_0}=133100×\frac{100}{121}$
$⇒P{_0}=1,10,000$
Hence, the correct answer is 1,10,000.

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