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Fence X is twice as long as fence Y, and fence Y is 2 feet shorter tha [#permalink]

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28 Mar 2014, 17:58

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46% (01:38) wrong based on 125 sessions

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Fence X is twice as long as fence Y, and fence Y is 2 feet shorter than fence Z. If 3 feet were added to each fence, which of the following must be true?

I) X is twice as long as Y. II) Y is 2 feet shorter than Z. III) X is longer than Z.

A) I only B) II only C) III only D) I and II E) II and III

Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

Fence X is twice as long as fence Y, and fence Y is 2 feet shorter than fence Z. If 3 feet were added to each fence, which of the following must be true?

I) X is twice as long as Y. II) Y is 2 feet shorter than Z. III) X is longer than Z.

A) I only B) II only C) III only D) I and II E) II and III

Length of Z = z Length of Y = z-2 Length of X = 2z-4

Minimum possible length of fence Z is 2 feet, because if we take the value for length of z below 2, lengths of Z and Y would be negative which is not possible.

If Z = 2 then Y = 0 and X = 0

When we add 3 feet in each fence, we will get Z= 5 Y=3 X=3

Only Statement II holds true with above values. _________________

Re: Fence X is twice as long as fence Y, and fence Y is 2 feet shorter tha [#permalink]

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29 Mar 2014, 12:04

Thank you Narenn. I posted this problem because Kaplan suggest to use the "picking numbers" strategy. But I think that backsolving is not the best approach. Why?:

If X= 6 Y= 3 Z= 5

Then, after adding 3 feet to each:

X= 9 Y=6 Z=8

And the answer would be E, which is wrong. _________________

Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

Re: Fence X is twice as long as fence Y, and fence Y is 2 feet shorter tha [#permalink]

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31 Mar 2014, 12:50

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). _________________

Thank you Narenn. I posted this problem because Kaplan suggest to use the "picking numbers" strategy. But I think that backsolving is not the best approach. Why?:

If X= 6 Y= 3 Z= 5

Then, after adding 3 feet to each:

X= 9 Y=6 Z=8

And the answer would be E, which is wrong.

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. _________________

Re: Fence X is twice as long as fence Y, and fence Y is 2 feet shorter tha [#permalink]

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31 Mar 2014, 22:27

Narenn wrote:

Maxirosario2012 wrote:

Fence X is twice as long as fence Y, and fence Y is 2 feet shorter than fence Z. If 3 feet were added to each fence, which of the following must be true?

I) X is twice as long as Y. II) Y is 2 feet shorter than Z. III) X is longer than Z.

A) I only B) II only C) III only D) I and II E) II and III

Length of Z = z Length of Y = z-2 Length of X = 2z-4

Minimum possible length of fence Z is 2 feet, because if we take the value for length of z below 2, lengths of Z and Y would be negative which is not possible.

If Z = 2 then Y = 0 and X = 0

When we add 3 feet in each fence, we will get Z= 5 Y=3 X=3

Only Statement II holds true with above values.

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

Re: Fence X is twice as long as fence Y, and fence Y is 2 feet shorter tha [#permalink]

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29 Jan 2016, 23:47

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