Fuel efficiency (European version)

The answer depends on how fast was the car is moving in each five minute segments, and that is unknown. Therefore we cannot answer the question.

To be more specific, let us assume that the car moves with velocity $v_1$ , $v_2$ and $v_3$ in the first, second and third five minutes, respectively. The total distance traveled is

$d= (v_1+v_2+v_3)t$

where $t=5min$ . The fuel burned during the first five minutes is $M_1=\gamma_1 v_1 t$ , where $\gamma_1=5 lpk$ . There are similar expressions for the other time intervals. The total amount of fuel used during the trip is

$M=M_1+M_2+M_3= t (v_1\gamma_1+v_2\gamma_2+v_3\gamma_3)$

The overall efficiency is

$\gamma= \frac{M}{d}=\frac{v_1\gamma_1+v_2\gamma_2+v_3\gamma_3}{v_1+v_2+v_3}$

If, for example, the car moves with constant velocity, we get

$\gamma= \frac{M}{d}=\frac{\gamma_1+\gamma_2+\gamma_3}{3}= 11.66lpk$

However, all kinds of other answers are possible with different choices of velocities.