The population of town B is 300%more than that of town A.For the next two years. the population of A increases by x% per year and that of B decreases by the same percentage per year. After 2 years, if the population of A and B become equal, then the value of x is _____.

$33\frac{1}{3}$

Let the population of town A be 100

The population of B = 100 + 100 × \(\frac{300}{100}\) = 400

According to the question

100 × \(\frac{100+x}{100}\) × \(\frac{100+x}{100}\) = 400 ×\(\frac{100+x}{100}\) × \(\frac{100+x}{100}\)

(100 + x)² = 4 (100 – x)²

 \(\frac{(100+x)^2)}{(100-x)^2}\) = 4

 \(\frac{100+x}{100-x}\) =  \(\frac{2}{1}\)

Using dividendo and componendo,

 \(\frac{100}{x}\) =\(\frac{2+1}{2-1}\)

\(\frac{100}{x}\) = \(\frac{3}{1}\)

x = \(\frac{100}{3}\)

 = 33\(\frac{1}{3}\)%