A container is filled with liquid. The liquid is draining from the container. The rate at which the volume of liquid in the container is decreasing at time t hours is given by dV/dt = -2/3 V^(2/3) e^(-0.02t) Given that initially the volume of liquid is 1000 cm^3.

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

1. Part (a): B1 for separating variables to get dV / V^(2/3) = -2/3 e^(-0.02t) dt (o.e.)
2. Part (a): B1 for introducing integral signs on both sides: ∫ V^(-2/3) dV = ∫ -2/3 e^(-0.02t) dt (Shown)
3. Part (b): M1 for integrating V^(-2/3) to k V^(1/3).
4. Part (b): A1 for 3V^(1/3).
5. Part (b): M1 for integrating -2/3 e^(-0.02t) to k e^(-0.02t).
6. Part (b): A1 for (100/3)e^(-0.02t).
7. Part (b): M1 for equating their integrated expressions and adding a constant of integration (e.g., 3V^(1/3) = (100/3)e^(-0.02t) + C).
8. Part (b): M1 for using initial conditions (t=0, V=1000) to find the value of C.
9. Part (b): A1 for C = -10/3 (or equivalent).
10. Part (b): A1 for V = ((100/9)e^(-0.02t) - 10/9)^3 (o.e. correctly rearranged to make V the subject).
11. Part (c): M1 for substituting V=125 into their equation from part (b) and attempting to solve for t.
12. Part (c): A1 for t = ln(0.55) / -0.02 = 29.89... hours (or 29.9 hours 1 d.p.).
13. Part (c): A1 for converting their time to hours and minutes, rounded to the nearest minute. (e.g. 29 hours and 53 minutes)