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
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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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