M07-28

Official Solution:

Lemonade A has a lemon acid to water ratio of 1:4, and Lemonade B has a lemon acid to water ratio of 1:5. In a mixture combining the two lemonades, the lemon acid to water ratio is 9:41. What is the ratio of the amount of Lemonade A to the amount of Lemonade B in this combined mixture?

A. 1:3
B. 16:25
C. 2:3
D. 3:4
E. 3:2


Assuming \(x\) amounts of Lemonade A and \(y\) amounts of Lemonade B were mixed, we can calculate the amount of lemon acid in each lemonade and in the combined mixture.

The amount of lemon acid in \(x\) amounts of Lemonade A is (lemon acid)/(total) \(= \frac{1}{1 + 4} = \frac{1}{5}\) of \(x\).

The amount of lemon acid in \(y\) amounts of Lemonade A is (lemon acid)/(total) \(= \frac{1}{1 + 5} = \frac{1}{6}\) of \(y\).

The amount of lemon acid in \(x+y\) amounts of the combined mixture is (lemon acid)/(total) \(= \frac{9}{9 + 41} = \frac{9}{50}\) of \(x+y\).

We can equate the amounts of lemon acid in the lemonades and the combined mixture to obtain: \(\frac{1}{5}*x + \frac{1}{6}*y = \frac{9}{50}*(x+y)\)

Multiplying both sides by the least common multiple of 4, 6, and 50, which is 150, we get: \(30x +25y = 27(x + y)\). This simplifies to \(3x = 2y\), giving \(x:y=2:3\)


Answer: C