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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

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


Answer: C.


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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Expert Reply
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.


Answer: C.


Hi Bunuel,

What if x = y = sqrt(9) ?

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


\(\sqrt{9}=3\). Hence in this case too \(\frac{x + y}{2}=3=integer\).
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