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Shortcut approach --Ratio/Proportion

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by guerrero » Tue Jan 08, 2013 2:52 am
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]
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by Anurag@Gurome » Tue Jan 08, 2013 3:16 am
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 three containers contain 3x, 4x and 5x liters of mixtures respectively.
Milk in 1st container = (4/5) * 3x = (12x)/5
Water in 1st container = 3x - (12x/5) = (3x)/5
Milk in 2nd container = (3/4) * 4x = 3x
Water in 2nd container = 4x - 3x = x
Milk in 3rd container = (5/7) * 5x = (25x)/7
Water in 3rd container = 5x - (25x/7) = (10x)/7

Therefore, total milk = (12x)/5 + 3x + (25x)/7 = (314x)/35
Total water = (3x)/5 + x + (10x)/7 = (21x + 35x + 50x)/35 = (106x)/35

So, ratio of milk and water in the fourth container = (314x)/35 : (106x)/35 = 314 : 106 = [spoiler]157 : 53[/spoiler]

The correct answer is C.
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by GMATGuruNY » Tue Jan 08, 2013 5:07 am
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.
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by The Iceman » Tue Jan 08, 2013 5:13 am
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
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by bnpetteway » Tue Jan 08, 2013 4:40 pm
Guys, would I be able to POE answers A and E since they are included in the equation and they are obvious wrong answers.
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