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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the [#permalink]
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I used ratios to solve this difficult question.


Let x equal the number of gallons of milk that is 25% water & Let y equal the number of gallons of milk that is 50% water

percent water / percent milk = 3:5

(25x + 50y) / (75x + 50y) = 3/5

Cross multiply and divide by 25:

5(x + 2y) = 5(3x + 2y)

Simplify:

5x + 10y =9x + 6y -----> 4y = 4x ====> y = x

Since x + y =12 and y = x, then the answer must be B) 6 liters, 6 liters, as it is the only answer that satisfies both equations.
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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the [#permalink]
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Let x be eventual volume of water so volume of milk = 12-x


x/12-x = 3/5

5x = 36 - 3x

8x = 36

x= 36/8

= 9/2

Let volume of milk from can A be a and from B b


For water

a/4 + b/2 = x

For milk

3a/4 + b/2 = 12-x


Since x = 9/2


a/4 + b/2 = 9/2

3a/4 + b/2 = 15/2



3a + 2b = 30

a + 2b = 18

2a = 12

a=6

substitute for a in any equation above gives b =6

Answer B

Adewale Fasipe, GRE quant instructor from Nigeria.
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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the [#permalink]
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we can find the proportion of milk by using a weighted avg quantity ratio formula N1/N2 = M2- WA/WA - M1

Solution 1 milk is 75% and 2nd solution milk is 50% and in final solution milk is 5/8 of total thus we say 62.5%

N1/N2 = M2- WA/WA - M1 = (75 - 62.5)/(62.5 - 50) = 1/1 thus it means milk is taken in equal proportion from both the solution thus B
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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the [#permalink]
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