The only completely correct solution to this question came, with a very nice diagram, from Vassil Vassilev, age 14, Lawnswood High School, Leeds.

If nobody hiked, the car would need two return journeys plus one single journey, making a total distance of $5\times 60$ miles and hence a total time of $(5\times 60)/40 = 7\,{\rm hrs}\ 30\,{\rm mins}$. You have to find out how much time is saved by the students hiking whenever they are not in the car.

- The distance 10 children moved when the car was going towards the destination point. | ||

- The distance the car moved towards the destination point. | ||

- The distance 15 children moved when the car was going back. | ||

- The distance the car moved back to pick up the rest of the children. |

You could work out all the distances but that involves a lot of
equations. It is best to focus on the time taken. Let $T$ hours be
the total time in which 5 people (plus the driver) are in the car
going towards the destination point, and$t$ hours be
the total time in which the driver only is in the car going back.
We assume that the optimal solution is such that
**all** of the people reach the destination **at
the same time** , that is after$T+t$ hours.

In total, the fifteen people cover $15\times 60$ ($=900$ miles, and this is made up from the miles hitched in the car, and miles hiked on foot. The number of person-miles hitched in the car is$5\times 40T$ , and the number of person-miles hiked is

$$10\times 4T + 15 \times 4t,$$

because 10 people are hiking for $T$ hours (when 5 other
people are in the car), and all 15 people are hiking when only the
driver is in the car. Thus

$$5\times 40T + (10\times 4T + 15 \times 4t) = 900.$$

This simplifies to give

$$12T + 3t = 45.$$

The car spends $T$ hours travelling forwards, and$t$ hours travelling back towards the start. It therefore travels forwards a distance of$40T$, and backwards a distance of $40t$. Hence$40(T-t)=60$ , or

$$2T-2t=3.$$

Solving these two simultaneous equations for $T$ and $t$, we
get$T = 1\,{\rm hr}\ 48\,{\rm mins}$, $t = 3\,{\rm hr}\
18\,{\rm mins}$, and the total time is $5\,{\rm hr}\ 6\,{\rm
mins}$.

As the time would be $7\,{\rm hr}\ 30\,{\rm mins}$ if everyone
did the whole journey by car, the time saved is $2\,{\rm hrs}\
24\,{\rm mins}$.