If y is an integer and x + y + z is odd, is y even?

Hi vsaketram,

Your mistake is highlighted by red part above

You have assumed that 4x is even whereas for x=1/4, 4x will be odd
Please note that y is integer but there is no such information about z and z

Answer should be c....

the stem says that x+y+ z is odd , so for the sum of 3 number should odd , two of there should be odd and other should be even OR ,all of 3 numbers should be odd ...

Data 1 ) says |x| = .5 Or : x= .5 , x= -.5 this is clearly insufficient as we don't know any info about y & z , so options A & D are ruled out...

Data 2 ) says that z= 3x , as we don't know neither the value of z nor x , this data is not sufficient alone. notice here if we knew the x is integer , this data would be sufficient . because in the

x+y + z , we would have 4x+ y = an odd number and as 4x is even , so even + y = odd , therefore y has to be odd . But we don't know whether x is integer , so this option is not sufficient .


Combining both data , we have 2 cases ;

case 1 ) : if x=.5 and z=3x , so z= 1.5 and x+z = .5 +1.5 = 2 and , 2+y = odd integer so, y must be an odd integer,

case 2) : if x= -.5 and z= 3x , so z= -1.5 and x+z = -.5 + -1.5 = -2 , and , -2 +y = odd integer so, y must be an odd integer

as we see both cases satisfy that y must be an odd integer , so, the answer to the question is Surly : NO therefore is sufficient and answer is : C....

The Answer must be B.

Statement 1-
assume x=.5
if z=2.5, x=z=3 and y is even. If z=1.5 then x=z is even and y is odd
Not Suff

Statement 2-

z=3x.

hence z+x=4x and hence is even and hence y is odd

Please let me know if this is correct

you're forgetting that x + y + z is odd. if x=.5 and y is an integer, then z must have a fractional .5 in order for x + y +z to be an integer. Y must be even.
x=.5 ; y=2 ; z = .5 --> x + y + z is odd [Y even] and x=.5 ; z=2.5 ; y = 2 --> x + y + z is odd, and y is even. sufficient.

The second premise just tells us that z must be odd, and if the sum of x,y,and z is odd, then the sum of x and y must be even, which could be achieved through the sum of either two odds or two evens. Insufficient.

A

Actually, I forgot about the absolute value part of X. The first premise is NOT sufficient

x =-.5 y = 1 z=.5
x=.5 y = 2 z = .5

Hi Kitxrow,

You have missed a case that I have mentioned in my explanation above.

In first statement when you take z=1.5 then y will be odd therefore first statement is not sufficient

OA is C

Looking back now, even together, are the statements not insufficient?

combing the two two statements:
x = .5 y = 2 z = .5 [sum is odd, |x| = .5 , and y is even
x=.5 y = 3 z = 1.5 [sum is odd, |x| = .5. and y is odd


NEVERMIND!
..?

PLEASE DON'T FORGET THAT AFTER COMBINING THERE ARE TWO FACTS TO BE TAKEN INTO ACCOUNT

i.e x=+0.5 AND z=3x

SO YOU CAN'T TAKE z=0.5 now as you have taken in your patest comment (Highlighted by Red part

x+y+z = x+(y)Integer+3x = odd
i.e. 4x + y = odd
for x= 0.5, y = odd-2 = odd
for x= -0.5, y = odd+2 = odd
Consistent answer therefore
SUFFICIENT

yeeeeesh.

okay, think I got it under control.

(II) z = 3X
--> If x is odd, z is odd ; for x + y + z to odd y MUST BE ODD.
--> If x is even, z is even; for x + y + z to be odd, y MUST BE ODD.

Y IS NOT EVEN. SUFFICIENT.

B

JUST DISREGARD EVERYTHING

Well you can appreciate the clarification by pressing KUDOS

Actually, I don't know why I'm having such a hard time with this. Why is the second premise not sufficient.

If z = 3x and y is an integer and x+y+z is odd:
x=0 ; y=1 ; z = 0 Y IS ODD
x=-1 ; y = 5 ; z = -3 Y IS ODD
x=2 ; y =1 ; z=6 Y IS ODD
x=-.5 ; y = 3 ; z=-1.5 Y IS ODD
x=2.5 ; y=1 ; z=7.5 Y IS ODD

I must be missing something. Under what circumstances could y be even?

Look at the red part below

Statement 2: z = 3x
x+y+z = x+(y)Integer+3x = odd
i.e. 4x + y = odd
if x=1/4, 1+y = odd ===> y=odd-1 = even
if x=1/2, 2+y = odd ===> y=odd-2 = odd
NOT SUFFICIENT

tricky question...i fell in the trap...
the example with x=1/4 is what you need to see why C is correct.
x=1/4 -> z=3/4
1/4+3/4 = 1 -> 1+y=odd -> y is even
x=1/2, z=3/2 -> 1/2+3/2 = 4/2 = 2. 2+y=odd, y is odd.

x+y+z = odd
if x is negative
Then
y+z = odd+odd is anyway even
so only two even numbers add upto even
so y is even

when z = 3x
for x = even z is even ----(i)
for x = odd z is odd ----- (ii)

so if x+y+z = odd
considering condition i
y = odd - even
y is odd

considering condition ii
y = odd - even
y is odd

So D is the answer

Very tricky question but Thank you for the good explanation GMATinsight .

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