Rate of growth \( \rm R \) = 5% p.a.

Population of the town after 2 years

Initial population \( \rm P_{o} \) = 64000

Time \( \rm T \) = 2 years

Let P be the population of the town after 2 years. Using formula,

\( \rm P = P_{o} \left ( 1 + \frac{R}{100} \right ) ^{T} \)

\( \rm = 64000 \left ( 1 + \frac{5}{100} \right )^{2} \)

\( \rm = 70 560 \)

Hence, the population of the town is estimated to be 70,560 at a growth rate of 5% per annum.

Increased population in 2 years

The increased population in 2 years is the difference in the population after 2 years and the initial population. We have,

\( \rm \triangle P = P - P_{o} \)

\( \rm = 70560 - 64000 \)

\( \rm = 6560 \)

Population before 2 years at the same rate

In this condition, our initial population is \( \rm P_{i} \) and the final population is \( \rm P_{o} \). Using formula,

\( \rm P_{o} = P_{i} \left (1 + \frac{R}{100} \right )^{T} \)

\( \rm or, P_{i} = \frac{ P_{o} }{ \left ( 1 + \frac{R}{100} \right )^{T} } \)

\( \rm = \frac{64000}{ \left ( 1 + \frac{5}{100} \right )^{2}} \)

\( \rm = 58050 \)

Hence, the population before 2 years was 58049.

The increased population in 2 years was

\( \rm \triangle P_{2} = 64000 - 58050 = 5950  \)

Percentage difference in population after and before 2 years

\( \rm \% = \frac{70560 - 58050}{580501} \cdot 100 \)

\( \rm = 21.55 \% \)

Hence, the population before and after 2 years differ by 21.55%.