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y and z are nin zero integers ,is the square of (y+z) even? [#permalink]
04 May 2011, 18:20
Question Stats:
80% (01:20) correct
20% (01:10) wrong based on 1 sessions
y and z are nin zero integers ,is the square of (y+z) even? 1.y-z is odd 2.yz is even
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Re: odds and evens! [#permalink]
04 May 2011, 18:56
The question can rephrased as - Is y even and z odd or vice versa? or Are both even or both odd? Answering any of these questions will answer our question. 1) Sufficient y may be even or z odd and vice versa. The answer is always NO PS : It is given that y and z are NON ZERO. 2) Insufficient Both may be even or just one of them may be even. In the first case y + z = even. The answer is YES In the second case y + z = odd. The answer is NO Answer A AnkitK wrote: y and z are nin zero integers ,is the square of (y+z) even? 1.y-z is odd 2.yz is even
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Re: odds and evens! [#permalink]
04 May 2011, 19:04
(1) One of y or z is even, and one of y or z is odd So y+z is odd hence (y+z)^2 = odd Sufficient (2) y can be even, z can be odd y can be odd, z can be even y or z can both be even So square of (y+z) is even. Not Sufficient Answer - A
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Re: odds and evens! [#permalink]
04 May 2011, 20:40
y-z = odd means either y is even and z is odd or vice versa. Hence A. yz = even means y,z even or y is even and z odd and viceversa. Thus A.
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Re: odds and evens! [#permalink]
05 May 2011, 15:00
AnkitK wrote: y and z are nin zero integers ,is the square of (y+z) even? 1.y-z is odd 2.yz is even As I posted in another thread a moment ago, positive integer exponents never matter in an even/odd question, so we can just ignore the 'square of' part of the question: it's just asking if y+z is even. Addition and subtraction follow the same odd/even rules, so if y-z is odd, then y+z is odd, and Statement 1 is sufficient. From Statement 2, y and z can both be even, in which case y+z is even, or one can be even and the other odd, in which case y+z is odd. So the answer is A.
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Re: odds and evens! [#permalink]
07 May 2011, 17:00
1. Sufficient
y-z is odd => either y is even,z is odd or y is odd ,z is even
in both the scenarios mentioned above y+z is odd = > (y+z)^ 2 is odd
2. Not sufficient yz is even
atleast one of the above is even
when both y and z are even , given expression is even but when y is odd and z is even , given expression is odd
Answer is A.
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Re: odds and evens!
[#permalink]
07 May 2011, 17:00
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