Shortcut approach --Ratio/Proportion

guerrero wrote:Three containers have their volumes in the ratio of 3:4:5. They are full of mixtures of Milk and Water in the ratio of 4:1, 3:1 and 5:2 respectively. The contents of all these three containers are poured into a fourth container. The ratio of milk and water in the fourth container will be

a)4:1
b)151:48
c)157:53
d)5:2

[spoiler]c
[/spoiler]

Let the 3 containers be A, B, and C.
Plug in values that make the math easy.

A:
Since M:W = 4:1, and 4+1 = 5, the volume in A should be a multiple of 5.

C:
Since M:W = 5:2, and 5+2 = 7, the volume in C should be a multiple of 7.

Thus, we should multiply A:B:C = 3:4:5 by both 5 and 7 -- in other words, by 35 -- so that the volume in A is a multiple of 5 and the volume in C is a multiple of 7:
A:B:C = 35(3:4:5) = 105:140:175.

A = 105 liters:
Since M:W = 4:1, and 4+1 = 5, 1 of every 5 liters is water
W = (1/5)105 = 21.
M = 105-21 = 84.

B = 140 liters:
Since M:W = 3:1, and 3+1 = 4, 1 of every 4 liters is water:
W = (1/4)140 = 35.
M = 140-35 = 105.

C = 175 liters:
Since M:W = 5:2, and 5+2 = 7, 2 of every 7 liters are water:
W = (2/7)175 = 50.
M = 175-50 = 125.

(total M) : (total W) = (84+105+125) : (21+35+50) = 314 : 106 = 157:53.

The correct answer is C.

If the combined volume is 12 cubic units, then milk in the resulting solution = (4/5)*3 + (3/4)*4 + (5/7)*5 = 314/35

So, the required ratio = (314/35)/(12-(314/35)) = 157/53

Guys, would I be able to POE answers A and E since they are included in the equation and they are obvious wrong answers.