odd /even ds

Question

Is x + y +z even?

1) xyz is even

2) xz is odd

BG · Answer

A) is insufficient by itself.X,Y,Z can all be even or two of them even and the other odd.
B) is also insufficient .It tells that X and Z are both odd, but reveals nothing about Y
From both A) and B) we know that X and Z are odd and Y is even
seems like C)

pmenon · Answer

i am a C on this one, following the same logic as BG

in statement 1, xyz can be even but that can be from a combination of odd and even numbers, and we dont know which variable is which.

in statement 2, we know that x and z must be odd, but we dont know about y, which will determine whether the sum is odd or even.

putting the two together, we know that x and z must be odd, and from 1, y must be even, since the only way to get xyz even is if y is even.

coolfish1103 · Answer

x + y + z = even if one the following condition is true

E E E, E O O, O E O, O O E

(1) xyz is even

Insufficient cause examples such as E E O works, 2x2x1 = 4, and 2+2+1 = 5

(2) xz is odd

Insufficient cause we do not know the value of y.

(1) & (2)

For xyz to be even while xz is odd, x and z must be odd, which leaves y to be even in order to fulfill the argument of xyz being even.

(C)

ps_dahiya · Answer

C

St1: xyz is even say that atleast one of them is even but other two can be anything. INSUFF

St2: xz is odd means both x and z are odd. So there is no information about y. If y is odd then x+ y+ z is odd, if y is even then x+y+z is even. So INSUFF

Combined:

x and z are both odd from st2. If this is true then from st1 y must be even. So sum of two odds and an even is always even. So SUFF.

andy_gr8 · Answer

I would go with C
For x+y+z to be even either a)All should be even or b) 2 of them should be odd

a) xyz is even=> i)All are even or ii)2 are even or iii) 2 are odd
clearly in when 2 are even then the sum wont be even.so INSUFF

b) xz is odd means bith are odd.We dont know abt y so INSUFF
combina and we get
x is odd, z is odd and y is even means the sum is even
so C

krisrini · Answer

From stmt 1 xyz = 0 mod 2
From stmt 2 xz = 1 mod 2, either x or z is odd, and not both.

Dividing stmt1 / stmt 2 we get y = 0 mod 2, impliying y is even. Also from stmt we get one of x,z is odd. So we have 2 possibilities.
odd+even+even (x+y+z) or
even+even+odd

Both of them are odd. Hence x+y+z is not odd. Hence C.

Mr. Krinkle · Answer

this board is making me smarter

odd x odd = odd

odd x even = even

odd + odd + even == even

Bhai · Answer

I missed this one and got A, but agree with C.

odd /even ds

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