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# Is (x + y)/2 an integer?

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Math Expert
Joined: 02 Sep 2009
Posts: 54493
Is (x + y)/2 an integer?  [#permalink]

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25 Jun 2017, 03:34
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Difficulty:

65% (hard)

Question Stats:

64% (02:09) correct 36% (01:38) wrong based on 71 sessions

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

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

(2) $$xy = 9$$

Official Solution is HERE.

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

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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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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

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Posts: 54493
Re: Is (x + y)/2 an integer?  [#permalink]

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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.

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