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Stumped

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by muzali » Tue Dec 09, 2008 9:25 pm
If x, y, and z are integers and xy+z is an odd integer, is x an even integer?

1. xy+xz is an even integer
2. y+xz is an odd integer

Correct OA=A

Would appreciate a detailed solution.
Last edited by muzali on Tue Dec 09, 2008 10:14 pm, edited 2 times in total.
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Source: — Data Sufficiency |

by cramya » Tue Dec 09, 2008 9:52 pm
Stumped would be right! Can u please confirm OA and source?

My approach

I am getting [spoiler]A)[/spoiler] (may be I am missing something and doing something incorrect in Stmt I ) but the OA is different

Given:- xy+z is an odd integer

Stmt I

xy+xz is an even integer

even integer - odd integer = odd integer
(xy+xz) - (xy+z) = odd integer
xy+xz-xy-z =odd
xz-z = odd
z(x-1) = odd

Both z and x-1 have to be odd

x-1 is odd then x has to be even

Definite YES

SUFF


Given: xy+z is an odd integer 1)
Stmt II

y+xz is an odd integer 2)
odd integer + odd intger = even integer

xy+z +y+xz = even
x(y+z) + y + z = even 3)
x->odd y->odd z->even

x-> even y->odd z->odd

2 possibilites still 1), 2) and 3) hold good

INSUFF
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Re: Stumped

by muzali » Tue Dec 09, 2008 10:12 pm
muzali wrote:If x, y, and z are integers and xy+z is an odd integer, is x an even integer?

1. xy+xz is an even integer
2. y+xz is an odd integer

OA=A

Would appreciate a detailed solution.
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by muzali » Tue Dec 09, 2008 10:13 pm
Cramya, you are right. I had unwittingly put the incorrect OA. Source is GMATPrep.
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by cramya » Tue Dec 09, 2008 10:21 pm
Cramya, you are right. I had unwittingly put the incorrect OA. Source is GMATPrep.
No problem, Muzali!

IMO it could have got a little trickier and time consumung if u looked at it as x,y,z seperately without manipulations.

I am sure there are other better/equally good approaches and mine being just one of them.

Good question!
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by ronniecoleman » Wed Dec 10, 2008 9:28 am
IMO A

Cramya..
buddy nice explanation!!!
Admission champion, Hauz khaz
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by muzali » Wed Dec 10, 2008 10:55 am
Totally agree, very methodical approach.
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by ravikirancheni » Fri Dec 12, 2008 8:21 pm
Cramya,
Just amazed by your comprehending skills....
How on the earth could you get the idea of solving this monster ...kudos !!!
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by pbanavara » Fri Dec 12, 2008 11:04 pm
cramya wrote:Stumped would be right! Can u please confirm OA and source?

My approach

I am getting [spoiler]A)[/spoiler] (may be I am missing something and doing something incorrect in Stmt I ) but the OA is different

Given:- xy+z is an odd integer

Stmt I

xy+xz is an even integer

even integer - odd integer = odd integer
(xy+xz) - (xy+z) = odd integer
xy+xz-xy-z =odd
xz-z = odd
z(x-1) = odd

Both z and x-1 have to be odd

x-1 is odd then x has to be even

Definite YES

SUFF


Given: xy+z is an odd integer 1)
Stmt II

y+xz is an odd integer 2)
odd integer + odd intger = even integer

xy+z +y+xz = even
x(y+z) + y + z = even 3)
x->odd y->odd z->even

x-> even y->odd z->odd

2 possibilites still 1), 2) and 3) hold good

INSUFF
Bingo same approach same answer.
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