You might think of "work" as equivalent to "distance", and "efficiency" as "speed". You could restate the problem in those terms, but with the twist that it's how fast three people cover the same total distance between them (equivalent to having two people start at opposite ends of a distance, travelling at different speeds, and asking where they'll meet, but now we're adding this third person travelling along the same line).
Now, you solve this by reducing the number of variables. A, B and C are all unknowns, but are related:
$A = .5B \equiv 2A = B$ $C = \dfrac{A+B}2 \equiv 2C = A+B$
We can thus replace any of the letters with the other half of the equation in which the term(s) appear, reducing the scope of the problem to fewer variables. But, what's the problem?
Well, consider the body of work to be done, W. It has an unknown number of "units" of work to be performed, presumably in any order by anybody. C can do all of this work W in 40 days, so in one day, C does $W/40$ the work, or equivalently, 40 days times C's working pace equals the volume of work W.
$C=W/40 \equiv W=40C$
Now, we are asked how fast A, B and C could finish W; combining all three of their working paces, how many days does it take?
$(A+B+C)X = W$
Solve for X, by substituting A, B, and C for equivalent statements in terms of W:
$(A+B+C)X = W \\ (2C+C)X = W\ \ \ \text{substitute 2C for A+B } \\ 3CX = W\ \ \ \text{combine like terms} \\ \dfrac{3WX}{40} = W\ \ \ \text{replace C with W/40 and arrange} \\ 3WX = 40W\ \ \ \text{multiply both sides by 40} \\ X = \dfrac{40W}{3W}\ \ \ \text{divide both sides by 3W} \\ X = \dfrac{40}{3}\ \ \ \text{W cancels out} \\ X = 13.\bar{3}\ \ \ \text{calculator work} \\ $