Aim800score wrote:

Two containers contain milk and water solutions of volume x liters and y liters, respectively. What would be the minimum concentration of milk in either container so that when the entire contents of both containers are mixed, 30 liters of 80 percent milk solution is obtained?

(1) x = 2y

(2) x = y +10

This is a

Data Sufficiency word problem. It's usually best to

put these problems into mathematical terms before we try to either solve, or test cases. Otherwise it might be easy to miss a constraint or a subtle detail in the question.

The question itself gives us some constraints. We need to mix x liters and y liters, and end up with 30 liters of 80% milk. The 30 liters part corresponds to this equation:

x + y = 30

The 80% part corresponds to this equation:

x(percent of milk in x) + y(percent of milk in y) = 80%(30) = 24

So, we have two equations and four variables. We're trying to find the

minimum percent of milk in x or y. On to the statements!

Statement 1: x = 2y. In this case, we can solve for x and y, using this equation and the first equation from the question stem.

x + y = 30

2y + y = 30

3y = 30

y = 10

x = 20

Can we now find the minimum percent of milk required? 20(percent of milk in x) + 10(percent of milk in y) = 24. To minimize one value in an equation, you generally want to maximize everything else. We'd want to make the percent of milk in x as large as possible (100%), then find the resulting percent of milk in y, or vice versa. We shouldn't actually do that work, though - it's data sufficiency, so it's enough to know that we

could figure out the minimum. Sufficient.

Statement 2: This will follow the exact same solution path as the above statement, starting with finding the values of x and y. So, it's also sufficient.

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