Odd Even : GMAT Data Sufficiency (DS)

Odd Even

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Odd Even [#permalink]

17 Jan 2011, 20:46

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Question Stats:

46% (02:43) correct

53% (01:47) wrong

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[Reveal] Spoiler: OA

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Re: Odd Even [#permalink]

18 Jan 2011, 00:50

Given xy+z is odd. We need to find whether x is even...

This can happen in the following situations

a) z is even and xy is odd (which implies both x and y are odd)

b) z is odd and xy is even ( which implies either x or y is even)

Now consider Statement 1: xy+xz is even

simplifying x (y+z) is even

Let us test the two situations quoted above

a) let z be even and let x & y be odd

odd ( odd + even) gives you odd contradicting x(y+z) is even therefore z being even and xy being odd is ruled out

b) let z be odd and let xy be even

Let us initially consider x to be odd and y to be even in xy

odd (even + odd) gives odd, which is not true - therefore x being odd and y being even is ruled out.

consider x to be even and y to be odd in xy

even ( odd + odd) gives even satisfying the statement and hence x is even.

Statement 1 is sufficient

Statement 2: y+xz is an odd integer

a) let z be even and xy be odd

odd + odd * even gives even as against the statement hence ruled out

b) let z be odd and xy be even

Let y be even and x be odd

even+odd * odd is odd gives y may be even and x may be odd

Let x be even and y be odd

odd+even * odd is odd which suggests x is even and y may be odd.

Since we get contradicting results in situation b Statement 2 is insufficient

Hence Answer is A

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Re: Odd Even [#permalink]

19 Jan 2011, 14:46

are u sure the OA is A..
because I find two diff answers for A..

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Re: Odd Even [#permalink]

16 Mar 2011, 01:58

It took me too much time but couldn't find the satisfactory answer.
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Re: Odd Even [#permalink]

16 Mar 2011, 11:39

xy+z=odd. What is x?

1. xy+xz=even

Subtract the stem from the statement
xy+z-xy-xz = even-odd

even-odd = odd
z-xz=odd
z(1-x) = odd

Thus; z=odd and 1-x=odd; note: only odd*odd=odd

1-x = odd
x = even
Sufficient.

2. y+xz = odd
xy+z = odd

Adding both equations;
y+xz+xy+z= odd+odd
y+z+x(y+z) = even [Note: odd+odd=even]
(y+z)(1+x) = even

But
even*even=even
even*odd=even
Thus
1+x = even will make x odd
1+x = odd will make x even
Not Sufficient.

Ans: "A"
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Re: Odd Even [#permalink]

16 Mar 2011, 20:17

@ezinis For a minute this rocked my world I was minding my x and y's Do you have more questions like this? :-)

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Re: Odd Even [#permalink]

17 Mar 2011, 19:45

It took me a while too (4min30secs), cause I was doing crazy stuff like you guys at first too. But then I saw that it can simplify:

For xy+z to be odd, one of the two terms must be odd and the other one even.

For (1), all that happens is that z is multiplied. The only way multiplication can change the even/oddness of a number is if it was odd and is multiplied by an even number. Therefore since it goes from odd->even, z was odd and x must be even.

(2) we now just need to see if it is enough in itself to solve as well. Its convoluted, I would just take 1,2,3 as possible numbers and see that both statements can stand regardless of whether x is the '2' or '1'/'3'.

This is why we have to be quick in the easier ones

Re: Odd Even [#permalink] 17 Mar 2011, 19:45

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Odd Even

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