It is currently 21 Oct 2017, 05:36

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Odds and Evens..

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
VP
Joined: 20 Sep 2005
Posts: 1016

Kudos [?]: 38 [0], given: 0

Odds and Evens.. [#permalink]

### Show Tags

03 Apr 2006, 22:07
1
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x,y and z are integers and xy + z is odd, is x even ?

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

Kudos [?]: 38 [0], given: 0

Manager
Joined: 10 Jan 2006
Posts: 114

Kudos [?]: 7 [0], given: 0

### Show Tags

03 Apr 2006, 22:33
E ??.. let me know if this is correct. Will post explanation.

Kudos [?]: 7 [0], given: 0

Manager
Joined: 13 Dec 2005
Posts: 224

Kudos [?]: 8 [0], given: 0

Location: Milwaukee,WI

### Show Tags

04 Apr 2006, 03:35
I think A .

Keeping in mind xy +z is odd -- eqn 1

From first xy +xz is even if x is even irrespective of combination of y and z that satisfies the eqn 1

from second y+xz is odd irrespective of x being odd or even .

I hope my logic is correct but i wont be surprise if i messed up somewhere. Very good question .

Kudos [?]: 8 [0], given: 0

Manager
Joined: 21 Dec 2005
Posts: 102

Kudos [?]: 4 [0], given: 0

### Show Tags

04 Apr 2006, 03:46
I picked numbers and both A and B are not correct so,

E

Kudos [?]: 4 [0], given: 0

Director
Joined: 13 Nov 2003
Posts: 788

Kudos [?]: 61 [0], given: 0

Location: BULGARIA

### Show Tags

04 Apr 2006, 06:49
If we substract second equation from first we get x*(y-1)=odd, since even-odd=odd then both X and (y-1) should be odd
so think C) is correct

Kudos [?]: 61 [0], given: 0

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5034

Kudos [?]: 438 [0], given: 0

Location: Singapore

### Show Tags

04 Apr 2006, 07:59
1) x(y+z) = even

x can be even or odd. Insufficient.

2) y+xz = odd
x can be even or odd. Insufficient.

Using 1) and 2)

We have y+xz = odd.

If y = odd, we know xz = even. If x = even, z = odd or even since x(y+z) will always be even if x is even.

If y = even, xz = odd. z cannot be even because y+z would be be odd, and if x is odd, then st1 cannot be satisfied. so x must be even.

Ans C

Kudos [?]: 438 [0], given: 0

VP
Joined: 21 Sep 2003
Posts: 1057

Kudos [?]: 82 [0], given: 0

Location: USA
Re: Odds and Evens.. [#permalink]

### Show Tags

04 Apr 2006, 14:22
lhotseface wrote:
If x,y and z are integers and xy + z is odd, is x even ?

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

I think it is A.

Given
xy+z = Odd -(I)
xy+xz = Even -(II)

Subtract (I) from (II)

z(x-1) = Even-Odd = Odd
=> z(x-1) must be Odd
=> both z and x-1 are odd.

If x-1 is Odd => x is Even

A it is!
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

Kudos [?]: 82 [0], given: 0

Manager
Joined: 25 Mar 2006
Posts: 128

Kudos [?]: 1 [0], given: 0

Location: Stonybrook Univ

### Show Tags

04 Apr 2006, 15:56
i think its C ... ywilfred is right from the stem z could be anything even or odd i f thats the case and x(y+z) is even ... x could be even or odd ... you need II) also ... good job

Kudos [?]: 1 [0], given: 0

VP
Joined: 20 Sep 2005
Posts: 1016

Kudos [?]: 38 [0], given: 0

Re: Odds and Evens.. [#permalink]

### Show Tags

04 Apr 2006, 17:39
OA is A. Giddi also gave the OE.

giddi77 wrote:
lhotseface wrote:
If x,y and z are integers and xy + z is odd, is x even ?

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

I think it is A.

Given
xy+z = Odd -(I)
xy+xz = Even -(II)

Subtract (I) from (II)

z(x-1) = Even-Odd = Odd
=> z(x-1) must be Odd
=> both z and x-1 are odd.

If x-1 is Odd => x is Even

A it is!

Kudos [?]: 38 [0], given: 0

Senior Manager
Joined: 24 Jan 2006
Posts: 251

Kudos [?]: 5 [0], given: 0

### Show Tags

04 Apr 2006, 20:29
I posted answer C. I have read giddi's reply but I do not understand OAs answer. I don't understand if we use both equations to reach a solution, how come answer is A. Any help is appreciated.

Kudos [?]: 5 [0], given: 0

VP
Joined: 21 Sep 2003
Posts: 1057

Kudos [?]: 82 [0], given: 0

Location: USA

### Show Tags

04 Apr 2006, 20:48
gmatacer wrote:
I posted answer C. I have read giddi's reply but I do not understand OAs answer. I don't understand if we use both equations to reach a solution, how come answer is A. Any help is appreciated.

May be you could put some values and check the equations.
Remember:

O+/-E = O
O+/-O = E
E+/-E = E

and

O*O = O
O*E = E
E*E = E

if X is even and we are given that X-Y is odd, then Y has be odd.
Eg: If 10 - Y = ODD (say 3), then Y has to be ODD (= 7 in this eg).. HTH
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

Kudos [?]: 82 [0], given: 0

04 Apr 2006, 20:48
Display posts from previous: Sort by

# Odds and Evens..

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.