The population of a village increases @5% p.a.. If present population is 8000, after how many years the population will be 9261?A2 yearsB3 yearsC7/2 YearsD4 years

To solve this problem, we can use the formula for compound interest which is also applicable to population growth. The formula is:

A = P (1 + r/n)^(nt)

where: A = the amount of money accumulated after n years, including interest. P = the principal amount (the initial amount of money) r = annual interest rate (in decimal) n = number of times that interest is compounded per year t = time the money is invested for in years

In this case, we are looking for t (time). We know that: A = 9261 (the future population) P = 8000 (the present population) r = 5/100 = 0.05 (the annual growth rate in decimal form) n = 1 (since the population increases annually)

Substituting these values into the formula, we get:

9261 = 8000 (1 + 0.05/1)^(1*t)

Solving for t, we get:

(9261/8000) = (1 + 0.05)^t

1.157625 = 1.05^t

Taking the natural logarithm (ln) of both sides to solve for t, we get:

ln(1.157625) = t * ln(1.05)

t = ln(1.157625) / ln(1.05)

t ≈ 3 years

So, the population will be 9261 after approximately 3 years. Therefore, the answer is B) 3 years.