The rate at which the population of a city increases varies as the population. In a period of 20 years, the population increased from 4 lakhs to 6 lakhs. In another 20 years the population will be

Step-by-step solution

The population growth follows the differential equation dP dt = kP where k is the constant of proportionality. Solving this yields P(t) = Ae^ kt , where A is the initial population. Given P(0) = 4 lakhs, we have P(t) = 4e^ kt . Using P(20) = 6 lakhs, we find 6 = 4e^ 20k , which simplifies to e^ 20k = 3 2 . The population after 40 years is P(40) = 4e^ 40k = 4(e^ 20k )^2 = 4 ( 3 2 )^2 = 4 9 4 = 9 lakhs. Final answer: 9 lakhs