Is z even? (1) 5z is even. (2) 15z is even.

Question

Is z an even integer?

(1) 5z is an even integer.
(2) 15z is an even integer.

chetan2u · Answer

Do we know z is integer?.....NO
Now, let us tackle each statement.

1) 5z is even 
If z is integer yes. 
If z=2/5, then no.

2) 15z is even. 
Same examples as above.

Combined 
Nothing new. 
Insufficient

E

mdsaddamforgmat · Answer

IMO: E
Problem Statement: is “Z” even?
Pre-thinking: “Z” may be odd, Even or Fraction

St. 1: 5z is even.
From the above two scenarios are possible i.e. Z is either even or maybe a fraction that can divide 5 and leave an even integer. Hence, Insufficient.
St 2: 15z is even.
Again insufficient (same as above reasoning)

Combining also we can’t arrive.
Hence, E

srm91 · Answer

Both 5 and 15 are odd numbers.
We are well aware that 
O*O= O {eg. 3*7= 21 (odd)}
O*E = E {eg. 3*4= 12 (even)}
This implies that for 5z to be even, z has to be even and for 15z to be even, z has to be even.
Thus from each statement, we can independently conclude that z is even. 
Hence correct option is (D)

PgRaul · Answer

1) 5z is even
z= 2/5 or 2 etc
z ={2,4, 6, 8, so on} or z=(2/5,4/5 and so on}
NS
2)15z is even
z=2/15 or 2 etc
z ={2,4, 6, 8, so on} or z=(2/5, 4/5,2/15,4/15 and so on}
NS
On combining, we get possible values of z
z can be {2,4 , 6 and so on } or 2/5 4/5 etc 
so NS
 ANS: E

carouselambra · Answer

Is z even?

(1) 5z is even.
"z" needs to be even if we want 5z to be even.
However, if z= 6/5 , the value of 'z' is not even here.

Hence A is not sufficient.

(2) 15z is even.
Using the same logic as above, B is not sufficient.

Combining, 5z and 15z are even.
We can form the same cases :
Case 1: "z" is not a fraction and an even number, hence both the numbers (5z and 15z) are even.
Case 2: If "z" is fraction, the case will differ.
 
IMO E.

IanStewart · Answer

The only answer I'd consider reasonable here is D, since that's the answer you get when you assume z represents the kind of quantity (an integer) about which it makes sense to ask "Is z even?" The question "Is z even?" is meaningless if z is, say, 2/5, because the concept of 'evenness' is not defined for fractions. We do *not* get a "no" answer to the question when z = 2/5; the answer we get is "this question is nonsensical". When a statement is insufficient for a yes/no DS question on the actual GMAT, you can always get a "yes" answer and a "no" answer, but in this question, you can't get a "no" answer using Statement 1 or Statement 2 alone. You can only get a "yes" answer and a "I don't know what the question even means" answer. So you can't see questions like this on the test.

And taking things a step further, if one's willing to let z represent things that make the original question "Is z even?" nonsensical, then you might as well say "maybe z represents a potato, or a line in the coordinate plane", so of course we can't say if it's "even" because it doesn't make sense to ask if a potato or a line is even, and then there's no reason to even read the statements. It only makes sense to ask the question in the first place if z is the type of thing that can be even or not even (i.e. odd).

So this kind of question setup doesn't make logical sense. I see this "trap" ("maybe z isn't an integer at all!") in even/odd and divisibility questions all the time, but only ever in prep company materials, never on the real GMAT. If you ever saw a question like this on the GMAT, it would need to ask "Is z an even  integer ?", and not just "Is z even?", because then you're answering two questions: is z an integer, and is that integer even. But prep company questions avoid that wording, because it gives away the "trap", the possibility that z is not an integer at all.

Anyway, while I'm sure the "OA" is "E" here, D is the only answer I think is justifiable, but it's not really important either way, since it's not logical as a DS question.

Pankaj0901 · Answer

akadiyan · Answer

Is z an even integer?

(1) 5z is an even integer.
5z will be an even integer only when z is an even integer. But there is no information where Z is an even integer. Not sufficient.

(2) 15z is an even integer.
15z will be an even integer only when z is an even integer. But there is no information where Z is an even integer. Not sufficient.

(1) and (2) together
Combining both the options does not say whether Z is an integer or not.

Ans   E

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Is z even? (1) 5z is even. (2) 15z is even.

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Is z even? (1) 5z is even. (2) 15z is even.

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