Two containers contain milk and water solutions of volume x liters and

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.

By the Rule of Alligations

\(\frac{x}{y}\) = \(\frac{(80-b)}{(a-80)}\) ..... Eqn 1

Where a and b are the concentrations of milk in the first and second container respectively

To minimize b , We maximize a and make it 100%

Each Statement alone gives x/y = 2 ( Since, x + y = 30)

Now, there is only one unknown b in the Eqn 1 which can be calculated

Each Statement alone is Sufficient

Choice D

x+y=30
x/y=0.8-a/b-0.8… maximize b=1 to find minimum a
x/y=0.8-a/1-0.8…x/y=0.8-a/0.2…0.2x=0.8y-ay…ay=0.8y-0.2x

(1) x=2y: x+y=30…2y+y=30…y=10; a=0.8y-0.2x…ay=0.8y-0.2(2y)…a(10)=0.4(10)…a=0.4, sufic.
(1) x=y+10: x+y=30…(y+10)+y=30…y=10; a=0.8y-0.2x…ay=0.8y-0.2(2y)…a(10)=0.4(10)…a=0.4, sufic.

Answer (D)

I have one fundamental question with all the solutions given, how can the Milk Concentration in solution A be maximized to 100%, as given in the question "Two containers contain milk and water solutions of volume x liters and y liters"

Am I taking this sentence too literally ?

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