if x,y and z are integers and xy+z is an odd integer, is x an even integer?
1) xy + xz is even
2) y + xz is odd
good question from GMAT Prep.. .worth the try!
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IMO E
1) xy + xz is even
so xy and xz both even (a)
or both odd(b)
2) y + xz is odd
y odd, xz even(c)
xz odd, y even(d)
combine 1 and 2
combine (a) and (c)
xz =even
so y = odd
combine (a) and (d)
xz = even
and xz =0dd, contradictory, so this combination is not possible
combine (b) and (c)
xz =odd and xz even
contradictory, so this combination is also not possible
combine (b) and (d)
xz odd and y even
so two case arise:
y =odd, when xz = even
y = even when xz = odd
It is given xy+z is an odd integer
so (case I )xy =odd and z = even
(case II) xy =even and z = odd
Case I: xy =odd, Y can be even as well as odd
y= even,so xz has to be odd
But x =even
so case I is not possible
Case II:
xy = even and z = odd
if y = even, so xz = odd
so x has to be odd because z = odd and xz =odd
if y = odd, then xz =even
but since in CASE II z= odd
so x has to be even
thus we are getting 2 different values of x(even as well as odd)
so answer is E
1) xy + xz is even
so xy and xz both even (a)
or both odd(b)
2) y + xz is odd
y odd, xz even(c)
xz odd, y even(d)
combine 1 and 2
combine (a) and (c)
xz =even
so y = odd
combine (a) and (d)
xz = even
and xz =0dd, contradictory, so this combination is not possible
combine (b) and (c)
xz =odd and xz even
contradictory, so this combination is also not possible
combine (b) and (d)
xz odd and y even
so two case arise:
y =odd, when xz = even
y = even when xz = odd
It is given xy+z is an odd integer
so (case I )xy =odd and z = even
(case II) xy =even and z = odd
Case I: xy =odd, Y can be even as well as odd
y= even,so xz has to be odd
But x =even
so case I is not possible
Case II:
xy = even and z = odd
if y = even, so xz = odd
so x has to be odd because z = odd and xz =odd
if y = odd, then xz =even
but since in CASE II z= odd
so x has to be even
thus we are getting 2 different values of x(even as well as odd)
so answer is E
I did not have time to go through the long explanation, esp because your answer was wrong.
Let me post the explanation for the OA is A.
The Q:
if x,y and z are integers and xy+z is an odd integer, is x an even integer?
1) xy + xz is even
2) y + xz is odd
Given: xy + z = O => xy = O - z; where O represents an odd number
=> if z is odd then xy is even; else if z is even xy is odd.
1) xy + xz = E; where E represents an even number
=> O - z + xz = E; since xy = O - z.
=>z(x-1) = E - O = O => z is O and x-1 is O => x is even. So 1 is sufficient
2) y + xz = O => xz = O - y
If y is even; xz is odd => x is odd
If y is odd; xz = even ---- does not say anything about x.
So x is not definite to be a even or odd. So 2 is insufficient.
Thus OA is A
Let me post the explanation for the OA is A.
The Q:
if x,y and z are integers and xy+z is an odd integer, is x an even integer?
1) xy + xz is even
2) y + xz is odd
Given: xy + z = O => xy = O - z; where O represents an odd number
=> if z is odd then xy is even; else if z is even xy is odd.
1) xy + xz = E; where E represents an even number
=> O - z + xz = E; since xy = O - z.
=>z(x-1) = E - O = O => z is O and x-1 is O => x is even. So 1 is sufficient
2) y + xz = O => xz = O - y
If y is even; xz is odd => x is odd
If y is odd; xz = even ---- does not say anything about x.
So x is not definite to be a even or odd. So 2 is insufficient.
Thus OA is A