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22 Oct 2013, 08:07
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C
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% (02:15) correct
39%
(02:23)
wrong
based on 158
sessions
History
Date
Time
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Not Attempted Yet
If x, y, and z are all positive integers, is x + y + z even?
1) \(\frac{2x}{(y+z)}\) = odd
2) \(\frac{x^2}{4yz}\) is an integer
1) \(\frac{2x}{(y+z)}\) = odd
2) \(\frac{x^2}{4yz}\) is an integer
22 Oct 2013, 09:14
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If x, y, and z are all positive integers, is x + y + z even?
(1) \(\frac{2x}{(y+z)}\) = odd --> \(2x=even=(y+z)*odd\) --> \(y+z=even\). But x can be even or odd. Not sufficient.
(2) \(\frac{x^2}{4yz}\) is an integer --> \(x^2=4yz*integer=even\) --> x must be even. We don't know about y and z. Not sufficient.
(1)+(2) \(y+z=even\) and \(x=even\) --> \(x + y + z=even\). Sufficient.
Answer: C.
(1) \(\frac{2x}{(y+z)}\) = odd --> \(2x=even=(y+z)*odd\) --> \(y+z=even\). But x can be even or odd. Not sufficient.
(2) \(\frac{x^2}{4yz}\) is an integer --> \(x^2=4yz*integer=even\) --> x must be even. We don't know about y and z. Not sufficient.
(1)+(2) \(y+z=even\) and \(x=even\) --> \(x + y + z=even\). Sufficient.
Answer: C.
General Discussion
stonecold
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20 Jan 2017, 10:52
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Great Question.
From statement 1->
y+z must be even.
x can be even/odd.
Not sufficient.
From statement 2->
x must be even as x^2 is even.
y and z can be even or odd.
Not sufficient.
Combing them -> x is even and y+z is even => x+y+z must be even too.
Hence sufficient.
Hence C.
From statement 1->
y+z must be even.
x can be even/odd.
Not sufficient.
From statement 2->
x must be even as x^2 is even.
y and z can be even or odd.
Not sufficient.
Combing them -> x is even and y+z is even => x+y+z must be even too.
Hence sufficient.
Hence C.
