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

Question

(1) x + z is even

(2) y + z is even

Azedenkae · Answer

Well, clearly each statement alone is insufficient. You either do not have even/oddness of x or y, and that means regardless of what the other two values are, you can change y or x (for statements (1) and (2) respectively) and you can get a different answer.

Note that what the statement is essentially saying, is that for x and z has to be both even or both odd (for statement 1), as otherwise they cannot add together to be even. The same goes for statement 2.

So when we combine the two statements, we can start testing different situations. If x is odd, then z is odd, and thus y must also be odd. We then have  x + y + 2z be odd + odd + 2x odd, which comes out to be even . If x is even, then z is even, and thus y must also be even. We then have  x + y + 2z be even + even + 2x even, which still comes out to be even . As such, both statements together are sufficient, always giving us an even answer.  Hence C.

baru · Answer

is x + y + 2z even?
we need (x + y) is even or not

(1) x + z is even ==>not sufficient as it doesn't say anything about ( x + y ) even or not

(2) y + z is even ==> not sufficient as it doesn't say anything about ( x + y ) even or not
Combined ==> (x+z) + (y+z) = Even + Even = Even = x + y +2z ==> Sufficient

Option C

NidSha · Answer

is the answer C??

basically x+y+z+z = even + even = even

i. not suff as x + z = even = not sure about any of x and y being even or odd

ii. same as above..

wardalaknaoui1 · Answer

If x, y and z are integers, is x + y + 2z even?
(1) x + z is even
(2) y + z is even 
Hi, I got answer E!!
x+y+2z------ 2Z is automatically even, because any number odd or even multiplied by even gives us even.
statement 1: x+z is even even+even=even , odd+odd= even so we are not sure about x, because for z it is not important to know even or odd.
Statement 2: y+z is even either both are even or odd, so we are not sure about y here.
Combining the 2 statements we still do not know if x and y are odd or even.
so x + y + 2z let's take case by case
E+O+E=O
O+E+E=O
E+E+E=E
O+O+E=E 
Please if i am wrong correct me . thank you

amanvermagmat · Answer

Hello

I think if you take the statements together properly, you will see that x/z are either both even or both odd. Similarly y/z are both even or both odd (you yourself have explained this)
Now if z is even, it means both x/y are also even, then x+y+2z will be E+E+E = even
And if z is odd, it means both x/y are also odd, then x+y+2z will be O+O+E = even

So it has to be even.

Or you could simply add the two: x+z is even and y+z is also even; Add the two to get x+z + y+z = E + E 
Or x+y+2z = Even

jackspire · Answer

Hello  wardalaknaoui1 ,

When we combine both statements, we will get two even numbers. When two even numbers are added we always get another even number.
So, (x+z)+(y+z) = even + even = even = x+y+2z

Question is asking whether x+y+2z is even or not and we got our answer when we combine both the statements.

Hope it helps.

chetan2u · Answer

2z will always be even, so the question basically asks us whether x+y is even.
 Now I does not deal in y and II does not deal in x, so don't waste time looking for A and B as answer by taking each statement individually.

Combined..
x+z and y+z, both are even, so their SUM should also be even..
So, x+z+y+z or x+y+2z is even
Suff

C

UserMaple5 · Answer

Question: Is x + y + 2z = even?

For the above equation to hold true, x+y would have to be even. x+y can only be even if both x and y are even or both are odd. Notice that z itself can be either even or odd, we just know that the 2z term will always be even.

Stmt 1: x + z = even. 2 cases: either x and z are both even or both are odd. In either case, we don't know what y will be so we can't figure out if x + y is even or odd. Not sufficient.

Stmt 2: y + z = even. Using the same logic as stmt 1, we don't know anything about x so we can't figure out if x+y is even or odd. Not sufficient

Together: add both statements: x + z + y + z = even because (x+z) is even and (y+z) is even. So together they will be even. 
Adding it yields: x + y + 2z = even. This directly answers the question in the stem.

GMATWhizTeam · Answer

Step 1: Analyse Question Stem

Any number of the form 2n, where n is an integer, is always even.

The question clearly says that x, y and z are integers. Therefore,  the term 2z is definitely even . This can be used to break down the question stem and rephrase it.

Is x + y + 2z = even?

Since 2z is even, question can be rephrased as, Is x + y + even = even?

By moving the even term from the LHS to the RHS, this can further be rephrased as, Is x + y = even – even?

Since even – even = even, the question now becomes,  Is x + y = even?

Essentially, we are trying to find out if x and y are both odd OR both even.

Step 2: Analyse Statements Independently (And eliminate options) – AD / BCE

Statement 1:  x + z is even

This means that x and z are both odd OR both even. Statement 1 does not provide any information about y.

The data in statement 1 is insufficient to find if x and y are of the same nature.
Statement 1 alone is insufficient. Answer options A and D can be eliminated.

Statement 2:  y + z is even

This means that y and z are both odd OR both even. Statement 2 does not provide any information about x.

The data in statement 2 is insufficient to find if x and y are of the same nature.
Statement 2 alone is insufficient. Answer option B can be eliminated.

Step 3: Analyse Statements by combining

From statement 1:  Both x and z are of the same nature.
 From statement 2:  Both y and z are of the same nature.
From this, it is clear that both x and y are of the same nature i.e. they are both odd or both even.

The combination of statements is sufficient to answer the question, Is x + y = even, with a definite YES.
Statements 1 and 2 together are sufficient. Answer option E can be eliminated.

The correct answer option is C.

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

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

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