TGTET Paper 1 Exam 23 Jul 2017 Paper

Concept:Compound growth applied to population over time.Explanation:We are given the current population of a city and the annual growth rate. We need to find the population 3 years ago. Let's assume the population 3 years ago was $P_0$.The population increases by 8% annually. This means each year the population becomes 108% of the previous year's population.So, after 1 year, the population would be $P_0 \times (1 + \frac{8}{100}) = P_0 \times \frac{108}{100} = P_0 \times \frac{27}{25}$.After 2 years, the population would be $P_0 \times (\frac{27}{25}) \times (\frac{27}{25})$.After 3 years, the population would be $P_0 \times (\frac{27}{25}) \times (\frac{27}{25}) \times (\frac{27}{25})$.We are given that the present population (after 3 years of growth) is 157464.So, we can set up the equation:$$P_0 \times \left(\frac{27}{25}\right)^3 = 157464$$To find $P_0$, we rearrange the formula:$$P_0 = 157464 \div \left(\frac{27}{25}\right)^3$$$$P_0 = 157464 \times \left(\frac{25}{27}\right)^3$$$$P_0 = 157464 \times \frac{25 \times 25 \times 25}{27 \times 27 \times 27}$$Let's calculate the values:$27 \times 27 \times 27 = 19683$$25 \times 25 \times 25 = 15625$Now, divide 157464 by 19683:$157464 \div 19683 = 8$So, the equation becomes:$$P_0 = 8 \times 15625$$$$P_0 = 125000$$Therefore, the population of the city 3 years ago was 1,25,000.Answer:1,25,000