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

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Re: Is (x + y)/2 an integer? [#permalink]
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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Re: Is (x + y)/2 an integer? [#permalink]
Bunuel wrote:
Is $$\frac{x + y}{2}$$ an integer?

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

(2) $$xy = 9$$

Solution:
Pre Analysis:
• We are asked if x + y is an integer or not

Statement 1: $$|x| = \sqrt{y^2}$$
• According to this statement, |x| = y
• 2 possible solution to this:
• x = y or x - y = 0. However, no idea of x + y
• -x = y or x + y = 0 is an integer
• Thus, statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2: $$xy = 9$$
• We do not know that nature if x and y
• xy = 9 is possible when x = y = 3 and also when x = 0.9, y = 10
• Therefore, x + y can be an integer and also cannot be
• Thus, statement 2 alone is also sufficient

Combining:
• From statement 1, we get either x = y or -x = y
• From statement 2, we get xy = 9
• xy = 9 means x and y have to be of the same sign. Either both positive or both negative
• Therefore, -x = y can be discarded and x = y is the only possible scenario
• This means $$x^2=9$$ or x = y = 3 or x = y = -3
• In both the cases, x + y is integer

Hence the right answer is Option C
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Re: Is (x + y)/2 an integer? [#permalink]
Bunuel wrote:
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.

Hi Bunuel,

What if x = y = sqrt(9) ?

then (x+y)/2 wont be an integer?
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Re: Is (x + y)/2 an integer? [#permalink]
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rohitdidige wrote:
Bunuel wrote:
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.

Hi Bunuel,

What if x = y = sqrt(9) ?

then (x+y)/2 wont be an integer?

$$\sqrt{9}=3$$. Hence in this case too $$\frac{x + y}{2}=3=integer$$.
Re: Is (x + y)/2 an integer? [#permalink]
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