Rates Question (Speed/Distance/Time) This is straight out of a maths competition. One of the few questions I can't manage to get a grip on.
Luisa cycled to and from a destination using the same route. The route included some flat roads and hills.
On the flat road she averaged 40km/h, uphill she averaged 30km/h and downhill she averaged 60km/h.
The whole journey took six hours.
What was the total distance travelled by Luisa, in km?
 A: The key part of the question is that Luisa cycles to and from the destination using the same route. 
Let $x$ be the total flat distance, $y$ the total uphill distance and $z$ the total downhill distance from her start point to the destination. The objective is to find $2(x+y+z)$.
Thus the time from start to destination is
$$\frac{x}{40}+\frac{y}{30}+\frac{z}{60}$$
On the way back, the flat distance stays the same ($x$), but the uphill distance is now $z$ and the downhill distance is $y$. Thus the time from destination to start is
$$\frac{x}{40}+\frac{y}{60}+\frac{z}{30}$$
The total time taken is $6$ hours, so we have an equation 
$$\frac{x}{40}+\frac{y}{30}+\frac{z}{60}+\frac{x}{40}+\frac{y}{60}+\frac{z}{30}=6\\\Rightarrow x+y+z=120$$
Therefore the total distance travelled is $240$ km.
A: She travelled 240 km.
Let the times spent on flat, uphill and downhill be $t_f$, $t_u$ and $t_d$ (in hours).
As she cycled to and from a destination using the same route, the total downhill and uphill distances are equal: $30t_u=60t_d$.
Also, $t_f+t_u+t_d=6$.
Solving these two equations gives $t_u/2=t_d=2-t_f/3$.
Thus the total distance covered was
$$
40t_f+30t_u+60t_d
=
40t_f+30\times2(2-t_f/3)+60(2-t_f/3)
=
240.
$$
The result is independent of the value of $t_f$.
All we know about it is $0\leq t_f\leq\frac32$, since all three times are positive.
