A vessel contains a mixture of milk and water in the ratio of 14 : 3.

\(\frac{milk}{water}\) = \(\frac{14}{3}\)

milk = \(\frac{14}{17}\)

water = \(\frac{3}{17}\)

25.5 litres of mixture is taken out = 25.5 = \(\frac{255}{10}\) = \(\frac{51}{2} \ litres\)

Remaining Milk = \(14x - \frac{14}{17}*\frac{51}{2}\) = \(14x-21 \ litres\)

Remaining Water = \(3x - \frac{3}{17}*\frac{51}{2}\) = \(3x-4.5 \ litres\)

Then, 2.5 litres of water and 5 litres of milk is added. Thus, final quantity of water and milk is:

Milk = \(14x-21+5\) = \(14x-16 \ litres\)

Water = \(3x-4.5+2.5\) = \(3x-2 \ litres\)

Resultant Mixture = \(14x-16+3x-2 \ litres\) = \(17x-18 \ litres\)

Given, resultant mixture contains 20% water:

\(0.2*(17x-18) = 3x-2\)
\(3.4x - 3.6 = 3x-2\)
\(0.4x = 1.6\)
\(x = 4\)

Thus, initial quantity of mixture was \(= 17x = 17*4 = 68 \ litres (C)\)

Would it be wrong to straightforward the answer to the only multiple of 17 of the ACs?
Only 68 is divisible by 17, so the initial quantity must have been 68, otherwise that initial ratio would have been impossible.

Is my reasoning wrong (even though in this case the ans is correct)?

­Yes, I think you are correct. If there is only one option which is a multiple of 17 then that must be the answer, because the original quantity must be a multiple of 17.­

Going by options, 14:3, 14+3 = 17, only 68 would be the possible value. Does GMAT give options so straightforward like this? Am I wrong in thinking that GMAT would me more careful while giving the options? Or does GMAT do have questions which when worked through options would be solved more faster?