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

Marty's solution is definitely the fastest.
Here's an algebraic solution for dave13 ...

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

Another quick way to solve could be-

3:5 ratio of 12l water:milk ==>4.5l water and 7.5l milk
Now see option A-it has 1l+2l=3l water this is not the solution
Option B-1.5l+3l water=4.5l water this is the answer
(you can also check other options to confirm that they don't give the same answer,but they won't since the question has a unique solution)

Hi MartyTargetTestPrep,
Although the answer is B , are we getting 6 and 6 because I am getting a different equal proportion of the two mixtures .
Can you please confirm if the two milk cans are indeed mixed in 6:6 proportion.
Appreciate your help, Thank you.

Yes. Since the total resulting quantity is to be 12 liters, the equal parts have to be 6 liters and 6 liters.

This explanation is excellent - I started doing this problem with algebra but this method is much quicker. However, the way I read the question there are no correct answers; the question states that the target mixture has 12L of milk in it. Is the problem written incorrectly? The author must be intending to have a final solution of 12L total - otherwise it would be impossible to obtain a solution with 12L of milk using only 12L from each container...

Hi 5e3,
You are Correct, it's 12L of milk and not 12L of total solution. From the question all we can tell is that equal proportions of both the cans have been used.

Let us take x and 12-x lt. of milk, so

(5/8) = {.75x + .5(12-x)}/12
x=6

hi guys VeritasKarishma BrentGMATPrepNow can you pls help with algebraic approach ? I always find it confusing when it comes to mixture problems

dave13,

For mixtures, use weighted average. It will all become very clear and easy.

Concentration of milk in Can1 = 3/4

Concentration of milk in Can2 = 1/2

Concentration of milk needed = 5/8

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (1/2 - 5/8) / (5/8 - 3/4) = 1/1

So he needs to take equal quantities from the 2 cans.

Since he needs 12 litres, he should take 6 litres from each can.

Answer (B)

Check:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... -averages/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... -mixtures/

Can A = 0.25W + 0.75M = \(\frac{1}{4}\)W + \(\frac{3}{4}\)M

Can B = 0.5W+0.5M = \(\frac{1}{2}\)W + \(\frac{1}{2}\)M

Can C = Can A + Can B with Ratio W:M = 3:5 which means Water = \(\frac{3}{8}\)

\(\frac{1}{4}\)A + \(\frac{1}{2}\)B = \(\frac{3}{8}\)(A+B)

2A + 4B = 3A + 3B
A = B

A+B = 12
2A = 12
A = B = 6 (B)

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