Bunuel wrote:
Tough and Tricky questions: Mixture Problems.
A bartender mixes drink A with drink B. If drink A contains 24 percent alcohol, what percentage of the mixture of drink A and B is alcohol?
(1) The ratio of drink A and drink B is 1:4
(2) The alcohol concentration of drink B is 1.5 times greater than the alcohol concentration of drink A.
Kudos for a correct solution. OFFICIAL SOLUTION:Statement (1) gives us the quantity of each of the drinks in the new cocktail. Statement (2) gives us the connection between the alcohol concentration of Drink B and the mixed cocktail. To solve this question, try using the statements and see what happens.
Let us use the statement (1) by itself. If the ratio of A to B is 1:4, then there is one part A to 4 parts B. This means if there are five parts total, one part is A and 4 parts are B, or 1/5 is A and 4/5 is B. A = .2 and B = .8. This information doesn’t tell us enough to solve the question because we don’t know the alcohol concentration of drink B. Statement (1) is not sufficient (meaning choices A and D can be ruled out).
Let us use the statement (2) by itself. It tells us the concentration in drink B. It is 1.5 times greater than drink A. The concentration is drink A is 24%, therefore the concentration in drink B = .24 × 1.5 = .36 (36%) alcohol. This information doesn’t tell us enough to solve the question because we don’t know the quantity of each drink in the new cocktail. Statement (2) is not sufficient (meaning choice B can be ruled out).
If we use the both statements together, we can set up an equation for this, using the following amounts:
Suppose the new mixture is 1. Then Alcohol in Drink A = .24 × .2, Alcohol in Drink B = .36 × .8. We can get the percentage of alcohol in the new mixture: (.048 + .288) / 1 = .336 (33.6%).
Since we required both statements to answer this question, the correct answer is (C).
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