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A milk vendor has 2 cans of milk. The first contains 25% water and the

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Carcass
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A milk vendor has 2 cans of milk. The first contains 25% water and the [#permalink] New post   20 May 2021, 09:47
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A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?

A. 4 litres, 8 litres
B. 6 litres, 6 litres
C. 5 litres, 7 litres
D. 7 litres, 5 litres
E 8 litres, 7litres
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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the [#permalink] New post   20 May 2021, 11:07
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Carcass wrote:
A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?

A. 4 litres, 8 litres
B. 6 litres, 6 litres
C. 5 litres, 7 litres
D. 7 litres, 5 litres
E 8 litres, 7litres


GIVEN: The first can contains 25% water, and second can contains 50% water
We want 12 liters of milk such that the ratio of water to milk is 3 : 5
In other words, we want the resulting mixture to be 3/8 water

Let x = the volume of 25% water needed
So, 12 - x = the volume of 50% water needed (since we want a total volume of 12 liters)

So, the total volume of water in the resulting 12 liters of mixture = 0.25x + 0.5(12 - x)
Since we want the resulting 12-liter mixture to be 3/8 water, we can write: [0.25x + 0.5(12 - x)]/12 = 3/8
Simplify numerator: (6 - 0.25x)/12 = 3/8
Cross multiply: 48 - 2x = 36
Solve: x = 6

So we want 6 liters of the first milk, and 6 liters of the second milk

Answer: B
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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the [#permalink] New post   04 Oct 2022, 15:49
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For milk

0.75x + 0.5y = 0.625(12)

For water

0.25x + 0.5y = 0.375(12)

Each will give

0.75x + 0.5y = 7.5
0.25x + 0.5y = 4.5

Subtract equation 2 from 1

0.5x = 3
x = 6.

Substitute for x = 6 in either equation to give y = 6.

Answer choice B is the answer.

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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the [#permalink] New post   04 Oct 2022, 19:12
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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] New post   12 Dec 2023, 14:43
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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] New post   22 Jan 2024, 06:55
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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]
22 Jan 2024, 06:55
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