Is (x + y)/2 an integer?

Question

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

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

(2)  xy = 9

Official Solution is    HERE .

harikrish · Answer

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.

elegantm · Answer

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

Bunuel · Answer

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} 
eq 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.

SaquibHGMATWhiz · Answer

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

rohitdidige · Answer

Hi Bunuel,

What if x = y = sqrt(9) ?

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

Bunuel · Answer

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

Is (x + y)/2 an integer?

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