a) State or imply $\frac{{dx}}{{dt}} = k\left( {10 - x} \right)\left( {20 - x} \right)$ and show $k = 0.01$

b) Separate variables correctly and attempt integration of at least one side

Carry out an attempt to find $A$ and $B$ such that $\frac{1}{{\left( {10 - x} \right)\left( {20 - x} \right)}} \equiv \frac{A}{{10 - x}} + \frac{B}{{20 - x}}$, or equivalent

Obtain $A = \frac{1}{{10}}$ and $B =  - \frac{1}{{10}}$, or equivalent

Integrate and obtain $ - \frac{1}{{10}}ln\left( {10 - x} \right) + \frac{1}{{10}}ln\left( {20 - x} \right)$, or equivalent

Integrate and obtain term $0.01t$, or equivalent

Evaluate a constant, or use limits $t = 0$, $x = 0$, in a solution containing terms of the form $\alpha ln\left( {10 - x} \right){\text{ }}$, $bln\left( {20 - x} \right)$ and $ct$

Obtain answer in any form, e.g. $ - \frac{1}{{10}}ln\left( {10 - x} \right) + \frac{1}{{10}}ln\left( {20 - x} \right) = 0.01t + \frac{1}{{10}}ln\,2$

Use laws of logarithms to correctly remove logarithms

Rearrange and obtain $x = 20\left( {\exp \left( {0.1t} \right) - 1} \right)/\left( {2\exp \left( {0.1t} \right) - 1} \right)$, or equivalent

c) State that $x$ approaches $10$