A drink holding 6 ounces of an alcoholic drink that...

Hi AAPL,

We're told that a drink holding 6 ounces of an alcoholic drink that is 1 part rum to 2 parts coke is added to a jug holding 32 ounces of an alcoholic drink that is 1 part rum to 3 parts coke. We're asked for the ratio of rum to coke in the resulting mixture.

Your logic to approaching this question is perfect. Since this question can be viewed as a Weighted Average, you can also use the averages of the two individual mixtures to eliminate answers. When combining a mix that is 1:2 with a mix that is 1:3, the OVERALL ratio will have to be 'between' those two ratios. For example, Answer C would be the equivalent of 1:1.6666, but that's NOT between 1:2 and 1:3, so that ratio is not a possible result; you can eliminate Answers D and E for the same reason. IF you were combining EQUAL amounts of the two mixes, then the ratio would end up 'in the middle' - 1:2.5 (which is the equivalent of 2:5). That doesn't happen here though - there's LOTS more of the 1:3 mix, so 2:5 cannot be the answer either. That leaves us with just one option: the correct answer.

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

AAPL wrote:A drink holding 6 ounces of an alcoholic drink that is 1 part rum to 2 parts coke is added to a jug holding 32 ounces of an alcoholic drink that is 1 part rum to 3 parts coke. What is the ratio of rum to coke in the resulting mixture?

A. 2:5
B. 5:14
C. 3:5
D. 4:7
E. 14:5

(1/3 x 6 + 1/4 x 32)/2/3 x 6 + 3/4 x 32)

(2 + 8)/(4 + 24)

10/28 = 5/14

Alternate Solution:

We start with 6 ounces of a drink that is 1/3 rum. We add it to 32 ounces that is1/4 rum. The result is a mixture of 6 + 32 = 38 ounces of x percent of rum.

Putting this information into an equation, we have:

6(1/3) + 32(1/4) = 0.38x

2 + 8 = 0.38x

10/0.38 = x

Since the percent of rum is 10/0.38, then the percent of coke is 28/0.38.

Thus, we can express the ratio of rum to coke as:

(10/0.38)/(28/0.38) = 10/28 = 5/14

Answer: B