D01-31

Question

How many liters of pure alcohol need to be added to a 40-liter solution, which is 10% alcohol by volume, in order to double the alcohol proportion?

A. 4
B. 5
C. 10
D. 20
E. 40

Bunuel · Accepted Answer

Official Solution:

How many liters of pure alcohol need to be added to a 40-liter solution, which is 10% alcohol by volume, in order to double the alcohol proportion?

A. 4
B. 5
C. 10
D. 20
E. 40

The original 40-liter solution contains 10% alcohol by volume, which amounts to  0.1*40=4  liters of alcohol.
 
Let's say we need to add  x  liters of pure alcohol to the 40-liter solution to create a new solution (with a total volume of  40+x  liters) that is 20% alcohol by volume.
 
Now, we can set up an equation to represent the amount of alcohol in the new solution:  4+x=0.2*(40+x) . Solving this equation, we find that  x=5 .
 
Thus, 5 liters of pure alcohol must be added to the original 40-liter solution to double its alcohol proportion.

Answer: B

mockingjay · Answer

Allegation method:

Alch in Sol(original)......Alch in Sol(added)

10%.........................100% ------> put 100% here because we are adding pure alcohol
...............20%................. -------> we require a solution that has double the alcohol proportion from original one 
80%.........................10%

therefore, we get ratio of 8:1 ----> original solution:added solution or pure alcohol

ie, 40 ltrs : 5 ltrs

so, 5 ltrs of pure alcohol must be added

Sallyzodiac · Answer

Bunuel ; I tried solving with the following equation:

\frac{4+x}{40} = \frac{1}{5} , where  \frac{1}{5}  is the desired (alcohol/total solution) ratio. Solving for  x  we get 4. Why doesn't this work?

Bunuel · Answer

Because when you add x liters of alcohol, total becomes 40+x liters not 40 liters. It should be (4+x)/(40+x)=1/5, which is the same formula as in my solution above.

mevicks · Answer

MockingJays solution is great! Heres the pictorial representation of the same:

BrentGMATPrepNow · Answer

We start with a 40-liter solution that is 10% alcohol
So, there are  4  liters of pure alcohol in the ORIGINAL 40 liters. 
Let x = the number of liters of pure alcohol that we will add to the ORIGINAL mixture.

So,  4  + x = the number of liters in the RESULTING mixture. 
Also, since we are adding x liters to the original 40 liters, the RESULTING mixture has a total volume of  40 + x

We want the resulting mixture to be 20% alcohol. 
So we can write: ( 4  + x)/( 40 + x ) = 20%
In other words: ( 4  + x)/( 40 + x ) = 20/100
In other words: ( 4  + x)/( 40 + x ) = 1/5
Cross multiply to get: 5(4 + x) = 1(40 + x)
Expand: 20 + 5x = 40 + x
So: 5x = 20 + x
So: 4x = 20
Sol x = 5

Answer: B

Cheers, 
Brent

SJ1295 · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

MBAToronto2024 · Answer

How to solve it using weighted averages quickly?  KarishmaB

KarishmaB · Answer

10% alcohol needs to be mixed with 100% alcohol (pure alcohol) to give 20% alcohol solution (double the alcohol proportion). \frac{w1}{w2} = \frac{(A2- Avg)}{(Avg - A1) }= \frac{(100 - 20)}{(20 - 10)} = \frac{8}{1} So for every 8 parts of 10% alcohol sol, we need 1 part of pure alcohol. Since we have 40 litres of 10% alcohol solution, we need 5 litres of pure alcohol (multiplier is 5) Answer (B) Discussed in these videos here: https://www.youtube.com/watch?v=_GOAU7moZ2Q https://www.youtube.com/watch?v=VdBl9Hw0HBg

bondi123 · Answer