Hi shaparage,

With these type of problems, the key is to be able to set up the problem and translate it into an equation. One cannot just memorize the approach to these problems. The variations are too many. In this case, we can say that x liters of solution are moved from A to B, and vice versa. Let the price of oil be p1 $/liter in container A and p2 $/liter in container B. The total cost of oil in container A will be: (140-x)p1 + (x)p2

Similarly, the cost of oil in container B will be (60-x)p2 + (x)p1. Set the price per liter in the two containers to be equal and you will find that the terms involving p1 and p2 drop out and x turns out to be 42 liters.

Cheers,
Dabral

Say that each liter in A costs $a and each liter in B costs $b. Then the price of the barrels is $140a and $60b, respectively.

After we switch c liters from each barrel to the other, we've got $(140 - c)*a + cb in one barrel and $(60 - c)*b + c*a in the other. Since these create an equal price per liter, we have

((140 - c)a + bc)/140 = ((60 - c)b + ac)/60

60 * (140a - ac + bc) = 140 * (60b - bc + ac)

8400a - 60ac + 60bc = 8400b - 140bc + 140ac

8400a - 200ac = 8400b - 200bc

42a - ac = 42b - bc

42a - 42b - ac + bc = 0

(42 - c) * (a - b) = 0

So either c = 42 or a = b. We're told (in the stem) that the prices are different, so a ≠ b and c = 42 is our only solution.

By the way, that is not a trivial GMAT problem! It takes more than the normal amount of of setup and legwork to sort out, and feels a bit more like an algebra homework problem than anything you'd encounter on this test.