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The current population of a town is 16,000. It increases by 40% and 25% in two successive years but decreases by 62% in the third year. What is the po
Question

The current population of a town is 16,000. It increases by 40% and 25% in two successive years but decreases by 62% in the third year. What is the population of the town at the end of the third year?

A.

10,640

B.

10,638

C.

10,639

D.

10,642

Correct option is A

Given:
Initial population = 16,000
Increase in first year = 40%
Increase in second year = 25%
Decrease in third year = 62%

Formula Used:
Final Value = Initial Value × (1+R1100)×(1+R2100)×(1−R3100) (1 + \frac{R_1}{100}) × (1 + \frac{R_2}{100}) × (1 - \frac{R_3}{100})​​

Solution:
Population after first year = 16000 ×140100 \frac{140}{100}​ = 22400

Population after second year = 22400 ×125100 \frac{125}{100}​ = 28000

Population after third year = 28000 × 100−62100 \frac{100 - 62}{100}​​

= 28000 × 38100 \frac{38}{100}​​

= 280 × 38
= 10640

Final Answer
So the correct answer is (a) 

Exam Hall Method: 

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