Fence X is twice as long as fence Y, and fence Y is 2 feet shorter tha

X = 2Y and Z = Y+2 are given to us to be true according to the question.

1) (X+3) = 2(Y+3) => X + 3 = 2Y + 6 => X = 2Y +3, which cannot be true as X=2Y. NEVER TRUE--> REJECT.
2) (Y+3) + 2 = Z+3 => Y+2 = Z, which is true. ALWAYS TRUE --> ACCEPT.
3) X+3 > Z+3 => X>Z => 2Y > (Y+2) => Y>2. So this is true only if Y>2, else it is false. We can check this out quickly. If X=2, Y=1, Z=4, then X<Z, but if X=6,Y=3,Z=5, then X>Z. CONDITIONALLY TRUE --> REJECT.

Therefore the answer must be (B).

You are correct. Number plugging in almost never a good approach when you have to prove that something MUST BE TRUE. To prove that something will be true in every case, how many cases are sufficient? Can I check 2 cases and say that it will be true in every case? Should I check 6, or 20? Even if something is true in 20 cases, it may not hold for the 21st case! We must use logic to prove that something must be true in every case.
But we can use number plugging to disprove something. i.e. we try to find the case in which it doesn't hold and then we can say that something needn't be true in every case.

Just taking the example of X = 6 (you took one case), you cannot say that II and III must be true. All you can say is that I certainly is not true in every case.

Using logic:
"X is twice as long as fence Y"
"fence Y is 2 feet shorter than fence Z"
"3 feet were added to each fence"

I) X is twice as long as Y.
X is originally twice as long as Y. When we add 3 to both, X will no longer be twice. Not true.

II) Y is 2 feet shorter than Z.
Originally, Y was 2 feet shorter than Z. If you add equal length to both (i.e. 3 feet) , Y will still remain 2 feet shorter than Z. Always true.

III) X is longer than Z.
X is twice of Y (hence X greater than Y) and Z is 2 more than Y (Z is greater than Y too). Both X and Z are greater than Y so we don't know what is the relation between X and Z.
Say if Y = 1, Z = 3 and X = 2. So X needn't be longer than Z.

if Z=2, Y=0 and X=0 then X/Y not equal 2, so minimal cases is Z=3, Y=1 and X=2. The critical is that if X<Z or X=Z these facts make III wrong

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