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07 Feb 2018, 21:21
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
61% (02:01) correct
39%
(02:26)
wrong
based on 134
sessions
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If x and y are non-negative, is (x + y) greater than xy?
(1) x = y
(2) x + y is greater than x^2 + y^2
(1) x = y
(2) x + y is greater than x^2 + y^2
08 Feb 2018, 04:49
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Bunuel
As we're not asked about general logical properties (such as positive/negative), well go for an equation-driven approach.
This is a Precise methodology.
We'll write an 'inequality' with a question mark:
\(x + y (?) xy\)
(1) Substituting into our question stem gives \(2x (?) x^2\) --> \(0 (?) x^2 - 2x\) --> \(0 (?) x(x - 2).\)
Since we don't know if \(x(x - 2)\) is positive or not, this is insufficient.
(2) gives \(x + y > x^2 + y^2\). Looking at this expression we should be reminded of the core identity: \((x + y)^2 = x^2 + 2xy + y^2\)
Using this, we can write:
\(x + y - 2xy > x^2 + y^2 - 2xy\) which gives \(x + y - 2xy > (x - y)^2\) and then \(x + y > (x - y)^2 + 2xy\)
This essentially says that (x + y) is larger than a positive number plus 2xy.
Then (x + y) is larger than 2xy meaning it is larger than xy.
Sufficient!
(B) is our answer.
General Discussion
07 Feb 2018, 22:02
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If x and y are non-negative, is (x + y) greater than xy?
(1) x = y
If x=y, the question stem becomes \(2x > x^2\)
If x=y=\(\frac{1}{2}\), x+y is greater than xy
If x=y=5, x+y is less than xy (Insufficent)
(2) x + y is greater than x^2 + y^2
If x=y=\(\frac{1}{2}\), x+y is greater than xy
If x=\(\frac{1}{2}\), y=\(\frac{3}{4}\), x+y is greater than xy (Sufficient - Option B)
(1) x = y
If x=y, the question stem becomes \(2x > x^2\)
If x=y=\(\frac{1}{2}\), x+y is greater than xy
If x=y=5, x+y is less than xy (Insufficent)
(2) x + y is greater than x^2 + y^2
If x=y=\(\frac{1}{2}\), x+y is greater than xy
If x=\(\frac{1}{2}\), y=\(\frac{3}{4}\), x+y is greater than xy (Sufficient - Option B)
