In a field there are 300 plants of a certain species, all of which can be infected by a particular disease. At time t after the first plant is infected there are x infected plants. The rate of change of x is proportional to the product of the number of plants infected and the number of plants that are not yet infected. The variables x and t are treated as continuous, and it is given that dx/dt = 0.2 and x = 1 when t = 0.

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

1. [M1] State or imply equation of the form dx/dt = kx(300-x) and use dx/dt = 0.2 and x = 1 when t = 0.
2. [A1] Obtain k = 1/1495 and rearrange to the given answer 1495 dx/dt = x(300-x).
3. [B1] Separate variables correctly
4. [B1] Correct integration of t term
5. [B1] State or imply partial fractions of the form A/x + B/(300-x)
6. [M1] Correct method to find A or B
7. [A1] Obtain terms (1495/300)lnx - (1495/300)ln(300-x)
8. [M1] Use t = 0, x = 1 to evaluate a constant or as limits in a solution containing terms of the form ln x, ln(300-x) and t.
9. [A1] Obtain correct answer in any form
10. [M1] Use law of logarithms twice to obtain an expression for t
11. [A1] Obtain final answer t = (299/60)ln(299x/(300-x)) or equivalent single logarithm