gmattesttaker2 wrote:
Michael mixed a liters of a 10-percent alcohol solution with b liters of a 20-percent
alcohol solution to get c liters of a 14-percent alcohol solution, what is the value of
b?

(1) a = 12
(2) c = 20

Target question: What is the value of b?

Another approach is to use the weighted averages formula:
Weighted average = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

For more information on weighted averages, you can watch this free GMAT Prep Now video: https://www.gmatprepnow.com/module/gmat- ... ics?id=805


So, for this question, we get the equation: 14 = (a/c)(10) + (b/c)(20)
Alternatively, we can recognize that, since c = combined volume, we know that a+b = c, so an equivalent equation is 14 = [a/(a+b)](10) + [b/(a+b)](20)

Statement 1: a = 12
Take the equation 14 = [a/(a+b)](10) + [b/(a+b)](20), and replace a with 12 to get: 14 = [12/(12+b)](10) + [b/(12+b)](20)

Do we have enough information to solve this equation for b? YES.
To confirm this, we can first eliminate the fractions by multiplying both sides by (12+b) to get: 14(12 + b) = 120 + 20b
From here, we can see that we COULD solve this for b (but we won't)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: c = 20
Let's take the equation 14 = (a/c)(10) + (b/c)(20) and replace c with 20 to get: 14 = (a/20)(10) + (b/20)(20)
Do we have enough information to solve this equation for b? At first glance, it appears that we don't. Let's find out.
First, eliminate the fractions by multiplying both sides by 20 to get: 280 = 10a + 20b
Hmmm, it appears that we have an equation with two variables, which means we can't solve it for b.
HOWEVER, keep in mind that we also know that a+b = c. So, if c=20, we know that a + b = 20
So, we actually have 2 equations (280 = 10a + 20b and a + b = 20)
Since we COULD solve this system of equations for b, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent