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(1) z/2 is an even integer. (2) 3z is an even integer.

Kudos for a correct solution.

1) statement 1 clearly states z to be a multiple of 4.. so suff 2) statement two does not prove z to be an even integer.. insuff ans A
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For z/2 to be even, z must be a multiple of 4. Meaning z takes the following values {4,8,12,16,20 ...} so z is even integer.

Sufficient (2) 3z is an even integer.

{ Odd * Even = Even} For 3z to be even integer. z must be an even integer, Since 3 is odd. So z is even integer.

Sufficient Ans:D

-Manoj Reddy Please press +1 Kudos if this post is helpful

Hi ManojReddy and shriramvelamuri, Its very important that we check answer for every possiblity.. look at statement 2.. (2) 3z is an even integer. z will be an even integer if z is an integer. But is that mentioned anywhere.. NO.. what if z is a fraction say 2/3... 3z=2,an even integer.. but z is not an even integer, not even an integer.. B is not suff.. hope it is clear..
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Agree with you Chetan. For a second my mind flunked. Agree got tempted to the Z an integer.....

chetan2u wrote:

ManojReddy wrote:

Is z an even integer?

(1) z/2 is an even integer.

For z/2 to be even, z must be a multiple of 4. Meaning z takes the following values {4,8,12,16,20 ...} so z is even integer.

Sufficient (2) 3z is an even integer.

{ Odd * Even = Even} For 3z to be even integer. z must be an even integer, Since 3 is odd. So z is even integer.

Sufficient Ans:D

-Manoj Reddy Please press +1 Kudos if this post is helpful

Hi ManojReddy and shriramvelamuri, Its very important that we check answer for every possiblity.. look at statement 2.. (2) 3z is an even integer. z will be an even integer if z is an integer. But is that mentioned anywhere.. NO.. what if z is a fraction say 2/3... 3z=2,an even integer.. but z is not an even integer, not even an integer.. B is not suff.. hope it is clear..

The wording of this question has a tendency to bias people towards integers. After all, the “opposite” of even is odd, and odd numbers are integers, too. However, the question does not state that z must be an integer in the first place, so do not assume that it is.

(1) SUFFICIENT: The fact that z/2 is an even integer implies that z = 2 × (an even integer), which much be an even integer. (In fact, according to statement (1), z must be divisible by 4).

(2) INSUFFICENT: The fact that 3z is an even integer implies that z = (an even integer)/3, which might not be an integer at all. For example, z could equal 2/3.

One way to avoid assuming is to invoke Principle #3: Work from Facts to Question. Statement (2) tells us that 3z = even integer = –2, 0, 2, 4, 6, 8, 10, etc. No even integers have been skipped over, nor have we allowed the question to suggest z values. That is how assumptions sneak in.

Next, we divide 3z by 3 to get z, so we divide the numbers on our list by 3: z = –2/3, 0, 2/3, 4/3, 2, 8/3, 10/3, etc. Only then do we check this list against our question and see that the answer is Maybe.

The correct answer is A.

!

If we had assumed that z must be an integer, we might have evaluated statement (2) with two cases: 3 × even = even, so z could be even. 3 × odd = odd, so z is definitely not odd. We would have incorrectly concluded that Statement (2) was sufficient and therefore incorrectly selected answer (D).

Another common assumption is that a variable must be positive. Do not assume that any unknown is positive unless it is stated as such in the information given (or if the unknown counts physical things or measures some other positive-only quantity).
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(1) z/2 is an even integer. (2) 3z is an even integer.

Kudos for a correct solution.

The reason this is probably a 700 question is because the trap is trying to make you assume that Z is an even integer already- the test is trying to make you assume that when you test statement 2- but technically 3(2/3) is an even integer...remember that's what you're trying to prove

Statement 1

Has to be an even integer because z is a multiple of 2 and no fraction would work