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

Question

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

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

MartyTargetTestPrep · Accepted Answer

To answer this question, let's match up the percentages of water in the two milk mixtures with the ratio that we want to achieve.

The ratio that we want to achieve is 3:5, water to milk.

So, we want 3/8 water and 5/8 milk.

To answer the question, let's focus on the water, since, if 3/8 of the mixture is water, the other 5/8 will have be milk.

The mixes that we have are 25% water and 50% water. Let's express them in 8ths as well.

The first mix is 2/8 water. The second mix is 4/8 water.

We want 3/8 water, and since 3/8 is halfway between 2/8 and 4/8, this question is turning out to be pretty easy.

By using allegation, we can determine that we have to use equal quantities of the two mixtures.

The correct answer is (B).

gracie · Answer

let x=total liters in first can
.75x+.5(12-x)=5/8*12→
x=6
12-6=6 liters in second can
6 liters, 6 liters
B

BrentGMATPrepNow · Answer

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

anmit123 · Answer

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)

stne · Answer

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.

MartyTargetTestPrep · Answer

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

5e3 · Answer

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...

stne · Answer

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.

LeoN88 · Answer

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

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

dave13 · Answer

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

KarishmaB · Answer

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/

bv8562 · Answer

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)

Abdullah9097 · Answer

isn't the wording of the question flawed?

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards