Brent@GMATPrepNow wrote:Solution X is 30% alcohol by volume, solution Y is 10% alcohol by volume, and solution Z is 40% alcohol by volume. If 41 gallons of solution X, 20 gallons of solution Y, and 20 gallons of solution Z are combined, which of the following best approximates the percentage of alcohol in the resulting mixture?
A) 20.13
B) 24.91
C) 25.07
D) 27.46
E) 27.53
Answer:
E
Difficulty level: 650 - 700
Source:
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We can avoid tedious calculations by creating this mixture in parts
First combine 20 gallons of solution Y (10% alcohol) with 20 gallons of solution Z (40% alcohol)
Since we have
EQUAL volumes of each solution, the percentage of alcohol in the resulting mixture will be the
AVERAGE of the two concentrations.
So, concentration of alcohol in new mixture = (10% + 40%)/2 =
25%
So, now have
40 gallons of a solution that's 25% alcohol.
Next, add 41 gallons of solution X (30% alcohol) to this 40-gallon mixture (25% alcohol)
KEY POINT:
If we were to add 40 gallons of solution X (
30% alcohol) to the 40-gallon mixture (
25% alcohol), then we'd have
EQUAL volumes of each solution.
In this case the percentage of alcohol in the resulting mixture
WOULD equal the AVERAGE of the two concentrations.
So, the concentration of alcohol in new mixture WOULD equal (
30% +
25%)/2 = 27.5%
In reality we have 41 gallons of solution X (
30%).
Since the volume of solution X (
30%) is greater than the volume of the existing solution (
25%), the concentration will be closer to
30% than it is to
25%
So, the concentration of the final solution will be
a bit more than 27.5%
Answer: E
Cheers,
Brent
Last edited by
Brent@GMATPrepNow on Tue May 29, 2018 5:53 am, edited 1 time in total.
Brent Hanneson - Creator of GMATPrepNow.com
