A water tank is in the shape of a cuboid with base area 40000 cm². At time t minutes the depth of water in the tank is h cm. Water is pumped into the tank at a rate of 50000 cm³ per minute. Water is leaking out of the tank through a hole in the bottom at a rate of 600h cm³ per minute.

Mark scheme

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

1. [B1] (a) State that dV/dt = 50000-600h.
2. [M1] (a) Obtain dV/dh = 40000 and use dV/dt and dV/dh in the chain rule (dh/dt = (dV/dt)/(dV/dh)) to obtain an equation for dh/dt.
3. [A1] (a) Obtain 200 dh/dt = 250-3h from full and correct working. AG.
4. [M1] (b) Separate variables correctly and integrate one side correctly. E.g. ∫(1/(250-3h))dh = ∫(1/200)dt.
5. [A1] (b) Obtain -1/3 ln|250-3h| = t/200 (+C).
6. [M1] (b) Use t = 0, h = 50 in an expression containing ln(250-3h) or ln|250-3h| to find the constant of integration. Or equivalent use of limits 50 and 80.
7. [A1] (b) Obtain C = -200/3 ln(100) or equivalent.
8. [A1] (b) Obtain t = 150. (2 significant figures as required)