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M10-19

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M10-19  [#permalink]

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New post 16 Sep 2014, 00:42
1
13
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

44% (01:21) correct 56% (01:05) wrong based on 82 sessions

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Re M10-19  [#permalink]

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New post 16 Sep 2014, 00:42
2
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Re: M10-19  [#permalink]

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New post 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.

Answer: E


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?
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Re: M10-19  [#permalink]

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New post 30 Jun 2015, 23:10
Really, it is a good and helpful question, I assumed that all numbers are integer.
Great! I got it.
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Re: M10-19  [#permalink]

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New post 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
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M10-19  [#permalink]

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New post 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 :).
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Re: M10-19  [#permalink]

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New post 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.

Answer: E


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?
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Re: M10-19  [#permalink]

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New post 17 May 2016, 13:36
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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
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Re M10-19  [#permalink]

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New post 17 Aug 2016, 04:28
I think this is a high-quality question and I agree with explanation.
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New post 13 Sep 2016, 17:06
I think this is a high-quality question and I agree with explanation.
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New post 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
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Re: M10-19  [#permalink]

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New post 21 Nov 2016, 07:58
1
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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Re: M10-19  [#permalink]

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New post 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]

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New post 13 Oct 2019, 06:53
I think this is a high-quality question and I agree with explanation. Yet another easy approach to understand this..
Let say X = 1/3 & Y = 6, X*Y = 2 (Even Int)
Similarly Let say Z = 1/3 & Y = 6, Y*Z = 2 (Even Int)
But X*Y*Z = 1/3*6*1/3 = 2/3 and this is not an even int
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M10-19  [#permalink]

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New post 13 Oct 2019, 11:12
Bunuel wrote:
Is \(X*Y*Z\) an even integer?


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

(2) \(Y*Z\) is an even integer.


It's not sufficient to find whether or not one out of X,Y, or Z is an even integer to confirm that X*Y*Z is an even integer since we don't know whether or not all three numbers are integers. We either need to know that all three numbers are integers and that one of them is even or not, or we need exact values of all three numbers.

(1) Let's assume that X*Y=4 and Z=1.2, so X*Y*Z=4.8, which is not an even integer.
However, if X*Y=4, and Z is any integer, X*Y*Z is also an even integer.
So, we don't get a unique nature of X*Y*Z from this statement.
Thus, insufficient.

(2) Similar working as for (1) applies here.
Thus, insufficient.

From (1) and (2) together, we do not get the required information.
Thus, insufficient.

Therefore, the answer is option E.
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M10-19   [#permalink] 13 Oct 2019, 11:12
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