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if x, y and z are integers, is x+y+2z even?

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by papgust » Sun Apr 04, 2010 5:56 pm
is x+y+2z even?

A. x+y is even

2 combinations:-
1. e+e = e
2. o+o = e

2z is always going to be even. x+y is even. So, x+y+2z is always even. Sufficient.

B. y+z is even

Either both y and z are odd OR both y and z are even.

Assume y and z are odd, 2z is always even.
If x is odd and y is odd, x+y is even. So, x+y+2z is even.
If x is even and y is odd, x+y is odd. So, x+y+2z is odd.
Insufficient.

A

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by thephoenix » Sun Apr 04, 2010 6:02 pm
papgust wrote:is x+y+2z even?

A. x+y is even(its x+z)

2 combinations:-
1. e+e = e
2. o+o = e

2z is always going to be even. x+y is even. So, x+y+2z is always even. Sufficient.

B. y+z is even

Either both y and z are odd OR both y and z are even.

Assume y and z are odd, 2z is always even.
If x is odd and y is odd, x+y is even. So, x+y+2z is even.
If x is even and y is odd, x+y is odd. So, x+y+2z is odd.
Insufficient.

A

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by thephoenix » Sun Apr 04, 2010 6:05 pm
bhumika.k.shah wrote:if x, y and z are integers, is x+y+2z even?

A. x + z is even

B.y + z is even


Various approaches,please!
IMO C

a) if x is odd then z is odd ; and if x is even then z is even; but in bth case it depends on y
insuff
b)same as A here it depends on x

a+b=x+y+2z is even(add bth)
suff
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by papgust » Sun Apr 04, 2010 6:07 pm
thephoenix wrote:
papgust wrote:is x+y+2z even?

A. x+y is even(its x+z)

2 combinations:-
1. e+e = e
2. o+o = e

2z is always going to be even. x+y is even. So, x+y+2z is always even. Sufficient.

B. y+z is even

Either both y and z are odd OR both y and z are even.

Assume y and z are odd, 2z is always even.
If x is odd and y is odd, x+y is even. So, x+y+2z is even.
If x is even and y is odd, x+y is odd. So, x+y+2z is odd.
Insufficient.

A

NOTE: Please attach the OA in a spoiler.
My mistake!! Didn't read the statement properly.

Yes, it should be C
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