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!
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if x, y and z are integers, is x+y+2z even?
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bhumika.k.shah
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papgust
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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
NOTE: Please attach the OA in a spoiler.
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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thephoenix
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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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thephoenix
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IMO Cbhumika.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!
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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papgust
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My mistake!! Didn't read the statement properly.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.
Yes, it should be C
