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TGTET Paper 1 Exam 23 Jul 2017 Paper

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Question : 111 of 150
Marks: +1, -0
The present population of a city is 157464. If the population increased by 8% annually then the population of the city 3 years ago was ______
Solution:  
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 P0P_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 P0×(1+8100)=P0×108100=P0×2725P_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 P0×(2725)×(2725)P_0 \times (\frac{27}{25}) \times (\frac{27}{25}).
After 3 years, the population would be P0×(2725)×(2725)×(2725)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:
P0×(2725)3=157464P_0 \times \left(\frac{27}{25}\right)^3 = 157464
To find P0P_0, we rearrange the formula:
P0=157464÷(2725)3P_0 = 157464 \div \left(\frac{27}{25}\right)^3
P0=157464×(2527)3P_0 = 157464 \times \left(\frac{25}{27}\right)^3
P0=157464×25×25×2527×27×27P_0 = 157464 \times \frac{25 \times 25 \times 25}{27 \times 27 \times 27}
Let's calculate the values:
27×27×27=1968327 \times 27 \times 27 = 19683
25×25×25=1562525 \times 25 \times 25 = 15625
Now, divide 157464 by 19683:
157464÷19683=8157464 \div 19683 = 8
So, the equation becomes:
P0=8×15625P_0 = 8 \times 15625
P0=125000P_0 = 125000
Therefore, the population of the city 3 years ago was 1,25,000.
Answer:
1,25,000
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