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Solution X is 30% alcohol by volume, solution Y is 10%

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Solution X is 30% alcohol by volume, solution Y is 10%

by Brent@GMATPrepNow » Wed May 23, 2018 5:55 am

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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: www.gmatprepnow.com

 
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Solution X is 30% alcohol by volume, solution Y is 10%

by Brent@GMATPrepNow » Fri May 25, 2018 11:56 am
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: www.gmatprepnow.com  
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.
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by deloitte247 » Fri May 25, 2018 12:25 pm
41 gallons of solution x contains
41 * 0.3 = 12.3 gallons of alcohol
20 gallons of solution Y contains 20 * 0.1
= 2 gallons of alcohol
20 gallons of solution 2 contains 20 * 0.4
= 8 gallons of alcohol
Percentage of alcohol in the resulting mixture.
$$\left(\frac{12.3\ +\ 2\ +\ 8}{81}\right)\ \cdot\ 100\%$$
$$\left(\frac{12.3\ +\ 10}{81}\right)\ \cdot\ 100\%$$
$$\left(\frac{22.3}{81}\right)\ \cdot\ 100\%$$
0.2753 * 100
27.53

Option E is the correct answer
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