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Is (x + y)/2 an integer?

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Math Expert
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Joined: 02 Sep 2009
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Is (x + y)/2 an integer?  [#permalink]

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New post 25 Jun 2017, 03:34
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

60% (02:02) correct 40% (01:35) wrong based on 82 sessions

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Concentration: Strategy, Operations
GMAT 1: 580 Q49 V21
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Re: Is (x + y)/2 an integer?  [#permalink]

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New post 25 Jun 2017, 07:00
Bunuel wrote:
Is \(\frac{x + y}{2}\) an integer?


(1) \(|x| = \sqrt{y^2}\)

(2) \(xy = 9\)



Solution :

Statement 1:
This statement is insufficient on its own.
Case 1: x=1/2 and y^2=1/4.
Case 2 : x=3 and y^2=9.

Statement 2:
Case 1: x=3 and y=3.
Case 2: x=1/3 and y=27.

Combining St1 and St2,
we get x=3 and y=3, a unique solution and the result of ((x+y)/2)= int.
Therefore the answer is Option C.
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Re: Is (x + y)/2 an integer?  [#permalink]

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New post 25 Jun 2017, 08:34
Bunuel wrote:
Is \(\frac{x + y}{2}\) an integer?


(1) \(|x| = \sqrt{y^2}\)

(2) \(xy = 9\)


Statement 1 -
\(|x| = \sqrt{y^2}\)
\(|x| = y\)
Inserting the value in question ---- 2x/2 = x
X could be non integer as well
Hence Insufficient

Statement 2 -
\(xy = 9\)
Assuming both x & y are integers,
xy = 9 .1
xy = 3.3
Before said assumption would yield different results, Hence Insufficient.

Combining both,
We know from statement 1 that,
x = y
x.y = 3.3

no other scenario is possible for x & y

Hence Answer C
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Re: Is (x + y)/2 an integer?  [#permalink]

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New post 17 Jul 2017, 05:59
Bunuel wrote:
Is \(\frac{x + y}{2}\) an integer?


(1) \(|x| = \sqrt{y^2}\)

(2) \(xy = 9\)


OFFICIAL SOLUTION:

Is \(\frac{x + y}{2}\) an integer?

First of all, notice that we are NOT told that \(x\) and \(y\) are integers.

(1) \(|x| = \sqrt {y^2}\)

\(|x| = |y|\). So, \(x=-y\) or \(x=y\). If \(x=-y\), then \(\frac{x + y}{2}=\frac{-y + y}{2}=0=integer\) but if \(x=y\), then \(\frac{x + y}{2}=\frac{x + x}{2}=x\), which will be an integer if \(x\) is an integer but won't be an integer if \(x\) is not an integer. Not sufficient.

(2) \(xy = 9\). If \(x=y=3\), then \(\frac{x + y}{2}=\frac{3 + 3}{2}=3=integer\) but if \(x=\frac{1}{3}\) and \(y=27\), then \(\frac{x + y}{2} \neq integer\)

(1)+(2) If from (1) \(x=-y\), then from (2) we'll have \(-x^2=9\), which is the same as \(x^2=-9\) but this is not possible (the square of a number cannot be negative). Thus, it must be true that \(x=y\), so \(x^2=9\), which gives \(x=y=3\) or \(x=y=-3\). In any case \(\frac{x + y}{2}=integer\). Sufficient.


Answer: C.
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Re: Is (x + y)/2 an integer?   [#permalink] 17 Jul 2017, 05:59
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