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# M10-19

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Math Expert
Joined: 02 Sep 2009
Posts: 44586

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16 Sep 2014, 00:42
Expert's post
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Difficulty:

75% (hard)

Question Stats:

39% (00:46) correct 61% (00:34) wrong based on 125 sessions

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Is $$X*Y*Z$$ an even integer?

(1) $$X*Y$$ is an even integer.

(2) $$Y*Z$$ is an even integer.
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 44586

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16 Sep 2014, 00:42
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Expert's post
Official Solution:

Statements (1) and (2) combined are insufficient. Consider $$X = Y = Z = 2$$ (the answer is "yes") and $$X = Y = Z = \sqrt{2}$$ (the answer is "no"). Do not assume that the numbers are all integers; the stem doesn't stipulate that.

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Joined: 06 Mar 2014
Posts: 258
Location: India
GMAT Date: 04-30-2015

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21 Oct 2014, 07:14
Bunuel wrote:
Official Solution:

Statements (1) and (2) combined are insufficient. Consider $$X = Y = Z = 2$$ (the answer is "yes") and $$X = Y = Z = \sqrt{2}$$ (the answer is "no"). Do not assume that the numbers are all integers; the stem doesn't stipulate that.

While attempting this question, i tried all possible scenarios (keeping in mind that numbers may not be integers) and ultimately it boiled down to Either C or E.

Somehow could not think of $$\sqrt{2}$$ as a possible option.
I was wondering if we take out the square root scenario, is there any other possible way one could still prove that using Both statements together are insufficient?
Intern
Joined: 09 Feb 2015
Posts: 20

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30 Jun 2015, 23:10
Really, it is a good and helpful question, I assumed that all numbers are integer.
Great! I got it.
Senior Manager
Joined: 11 Nov 2014
Posts: 348
Location: India
WE: Project Management (Telecommunications)

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12 May 2016, 06:00
Bunuel Vyshak
when x y z = under-root 2
we are given that x*y & y*z = even
which is under-root 2 * under-root 2 * under-root 2 * under-root 2 = 4
so xyz has to be even
Intern
Joined: 22 Apr 2015
Posts: 12
GMAT 1: 760 Q50 V44

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12 May 2016, 06:21
earnit wrote:
While attempting this question, i tried all possible scenarios (keeping in mind that numbers may not be integers) and ultimately it boiled down to Either C or E.

Somehow could not think of $$\sqrt{2}$$ as a possible option.
I was wondering if we take out the square root scenario, is there any other possible way one could still prove that using Both statements together are insufficient?

earnit
If both X and Z are .5 and Y is 4, you will satisfy both statements and still be able to get a "No". That's what I used for solving this problem .
Intern
Joined: 07 Dec 2014
Posts: 3
Location: Australia

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12 May 2016, 19:35
Did you not think of 1/3, 6 , 1/3 either?
earnit wrote:
Bunuel wrote:
Official Solution:

Statements (1) and (2) combined are insufficient. Consider $$X = Y = Z = 2$$ (the answer is "yes") and $$X = Y = Z = \sqrt{2}$$ (the answer is "no"). Do not assume that the numbers are all integers; the stem doesn't stipulate that.

While attempting this question, i tried all possible scenarios (keeping in mind that numbers may not be integers) and ultimately it boiled down to Either C or E.

Somehow could not think of $$\sqrt{2}$$ as a possible option.
I was wondering if we take out the square root scenario, is there any other possible way one could still prove that using Both statements together are insufficient?
Intern
Joined: 04 Feb 2016
Posts: 13

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17 May 2016, 13:36
1
KUDOS
Just use X=1/2 Y=8 and Z=1/4

I played around with fractions and not square roots. Still got the E for answer
Current Student
Joined: 29 Jan 2013
Posts: 41
Location: United States
Schools: Booth PT '20 (M)
GMAT 1: 650 Q50 V26
WE: Manufacturing and Production (Manufacturing)

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17 Aug 2016, 04:28
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 02 Sep 2016
Posts: 12

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13 Sep 2016, 17:06
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 06 Feb 2016
Posts: 48
Location: Poland
Concentration: Finance, Accounting
GMAT 1: 730 Q49 V41
GPA: 3.5

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16 Sep 2016, 03:34
My approach was to find a NO answer to this question because it is easy to find a YES answer. X * Y * Z = 1/26 * 52 * 1/4 => NO
BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2576
GRE 1: 323 Q169 V154

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21 Nov 2016, 07:58
1
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Great Question
Here we need to check if xyz is even or not
Note=> x,y,z are not mentioned to be integers.
Lets go to the statements
Statement 1
xy=even
xy=2
z=2=> xyz=even
z=1/2=> xyz=odd
hence insufficient

Statement 2
yz=even
yz=2
x=2=> xyz=even
x=1/2=> xyz=odd
hence insufficient

Now lets combine the statements
using the test cases
(x,y,z)=>2,2,2=> xyz=even
(x,y,z)=> 1/4,8,1/4=> xyz=8/16=1/2 => neither even nor odd
Hence Insufficient
Hence E
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Manager
Joined: 31 May 2017
Posts: 152

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20 Mar 2018, 19:47
stonecold - Hell yeah. Good solution.

I assumed individual X , Y and Z as integer. Careless mistake, but cost me dearly
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Re: M10-19   [#permalink] 20 Mar 2018, 19:47
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# M10-19

Moderators: chetan2u, Bunuel

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