A town has a population growth rate of 10% per year. The population in

There are many ways of solving this problem of which the easiest approach is to use the concept of successive percentage change. Between 1990 and 1992, we know that there have been 2 successive increases of 10% each.

When there are two successive percentage changes of a% and b% respectively, the effective percentage because of the two is given by (a+b+\(\frac{ab}{100}\))%. Bear in mind that a and b can be positive or negative depending on whether they represent an increase or a decrease.

In our question, a=b=10. Therefore, effective percentage increase = 10 + 10 + \(\frac{10*10}{100}\) = 21%.

Therefore, population in 1992 = 121% of population in 1990 (since there was a 21% increase). Substituting the value of population in 1990, we have,
Population in 1992 = 121/100 * 2000 = 121 * 20 = 2420. The correct answer option is D.

When you have two percentage changes one after the other, consider using the successive percentage concept. This may come in handy for you when you are dealing with Compound Interest questions as well since calculation of compound interest is nothing but successively multiplying the Principal by the rate percent. Think about it!

Hope that helps!