The population of mosquitoes in a specific area around a lake is controlled by pesticide. The rate of decrease of the number of mosquitoes is proportional to the number of mosquitoes at any time t. Given that the population decreases from ${\text{500}}\,{\text{000}}$ to ${\text{400}}\,{\text{000}}$ in a five year period, find the time it takes in years for the population of mosquitoes to decrease by half.

Let the number of mosquitoes be y.

$\frac{{{\text{d}}y}}{{{\text{d}}t}} = - ky$     M1

$\int {\frac{1}{y}{\text{d}}y = \int { - k{\text{d}}t} }$     M1

$\ln y = - kt + c$     A1

$y = {{\text{e}}^{ - kt + c}}$

$y = A{{\text{e}}^{ - kt}}$

when $t = 0,{\text{ }}y = 500\,000 \Rightarrow A = 500\,000$     A1

$y = {\text{500}}\,{\text{000}}{{\text{e}}^{ - kt}}$

when $t = 5,{\text{ }}y = {\text{400}}\,{\text{000}}$

${\text{400}}\,000 = 500\,{\text{000}}{{\text{e}}^{ - 5k}}$     M1

$\frac{4}{5} = {{\text{e}}^{ - 5k}}$

$- 5k = \ln \frac{4}{5}$

$k = - \frac{1}{5}\ln \frac{4}{5}\,\,\,\,\,{\text{( =  0.0446)}}$     A1

$250\,000 = 500\,{\text{000}}{{\text{e}}^{ - kt}}$     M1

$\frac{1}{2} = {{\text{e}}^{ - kt}}$

$\ln \frac{1}{2} = - kt$

$t = \frac{5}{{\ln \frac{4}{5}}}\ln \frac{1}{2} = 15.5{\text{ years}}$     A1

[8 marks]

Some candidates assumed that the decrease in population size was exponential / geometric and were therefore unable to gain the first 4 marks. Apart from this, reasonably good attempts were made by many candidates.