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If \(x^2+2xwy+3wy^2=144,\) and if \(x, y\) and \(w\) are all positive, then what is the value \(x + 3y?\)

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by M7MBA » Thu Oct 01, 2020 2:40 am

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If \(x^2+2xwy+3wy^2=144,\) and if \(x, y\) and \(w\) are all positive, then what is the value \(x + 3y?\)

(1) \(x+2y=11\)
(2) \(w=3\)

Answer: B

Source: Magoosh
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Source: — Data Sufficiency |
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Re: If \(x^2+2xwy+3wy^2=144,\) and if \(x, y\) and \(w\) are all positive, then what is the value \(x + 3y?\)

by alishagmat » Thu Oct 01, 2020 6:17 am
Using statement 1, we have x+2y = 11, still the variable w is unknown to us and even after squaring we have x^2 +4xy + 4y^2 =121, which tells nothing about x+3y. Hence, 1 is insufficient.

Using statement 2, we simplify the equation as x^2+ 6xy + 9y^2, which equals (x+3y)^2=144
=>|x+3y| =12. But we have x and y both positive, therefore the expression x+3y is positive hence 12.
since we have the unique value of x+3y, statement 2 is sufficient alone hence answer is B
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