Data Sufficiency

xxx

Question: Is x+y+2z even (2z is always even)

Stmt I

x+z is even (we know x is even and y is even)

No info about y (If y is odd and if both x and z are even the sum would be odd (or) If y is even then x+y+2z will be even

INSUFF

Stmt II

y+z is even

Similar explanation as above apllies to x as it applied to y

INSUFF


Stmt I and II

x+z is even

y+z is even

Add both we get x+z+y+Z ios even (since even+even = even)

= x+y+2z which is what we need ot find Therefore its even

SUFF

C)

@cramya,

i see what you did but i thought it was E because of the following -

x + z = even

so it could be odd + odd or even + even

y + z = even

so it could be odd + odd or even + even

seeing that, x and y could both be odd or even. so for x + y + 2z to be even, x and y both have to be even, and we can't conclude that.

what are the mistakes i made?

Seeing that, x and y could both be odd or even. so for x + y + 2z to be even, x and y both have to be even, and we can't conclude that.

what are the mistakes i made?

In x+z both could be odd or both even(one odd /other even not possible)
In y+z both could be odd or both even

x+z+y+z (this is nothing but x+y+2z)
odd+odd+odd+odd

= even

or

x+z+y+z(this is nothing but x+y+2z)
even+even+even+even
= even

Hope this helps!

ah yes, thanks much!