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Doubt in a DS question [#permalink]
04 Feb 2014, 15:58

I recently came across this question and wasn't sure if the answer mentioned in the source was correct. I will appreciate if someone can explain why this answer is correct.

Re: Doubt in a DS question [#permalink]
04 Feb 2014, 17:16

Expert's post

akkiankurgupta wrote:

I recently came across this question and wasn't sure if the answer mentioned in the source was correct. I will appreciate if someone can explain why this answer is correct.

If x and y are integers, is x > y ?

1. x is a positive multiple of y 2. xy > 1

Dear akkiankurgupta, I'm happy to help. I actually disagree with the OA.

The reason I disagree is that any integer is a multiple of itself. For example, 3 is a multiple of 3; in fact, 3 is a positive multiple of 3. Therefore, we could have x = y = 3, and both statements would be true by x would not be greater than y --- a "no" answer to the prompt. Of course x = 6 and y = 3 also work, and produce a "yes" answer. I'm not sure the authors considered this technicality ---- including this, the answer would be (E).

Even disregarding this technicality, there's something fishy about statement #1 --- by the statement "x is a positive multiple of y", do we mean (a) that x is a positive integer that also happens to be a multiple of y, in which cases x = +6 and y = -3 would be possible. or (b) that x is y times some positive integer, in which cases x = -6 and y = -2 Oddly enough, neither of these interpretations, nor the inclusion or exclusion of the technicality above, produce an answer of (C). Something is funny here.

Those are my thoughts. Let me know if you have any further questions.

Re: If X and y are integers, is (x−1)>y? [#permalink]
04 Feb 2014, 17:21

1

This post received KUDOS

Expert's post

guerrero25 wrote:

If X and y are integers, is (x−1)>y?

(1)x is a positive multiple of y

(2)x / y > 1

Dear guerrero25, In your understanding, or in the source, what exactly does "positive multiple" mean? In my mind, the statement "x is a positive multiple of y" lends itself to two very different interpretations: (1) x is a positive number and also happens to be a multiple of y; in this case, we could have x = +10 and y = -5 OR (b) x equals y times some positive integer; in this case, we could have x = -10 and y = -5 Of course, either interpretation would be correct with the OA in this question, because x = 15 and y = 3 satisfy both statements and give a "yes" answer to the prompt, and x = 2 and y = 1 satisfy both statements and give a "no" answer to the prompt.

Re: If x and y are integers, is (x−1)>y? [#permalink]
05 Feb 2014, 04:50

Here is the OE .

Solution: E. This is a great opportunity to play devil's advocate, picking numbers to get "yes" and "no" answers using both statements.

For statement 1, x could be 10 and y could be 1, giving you "yes". But it could also be 1 and 1, so you could get the answer "no".

For statement 2, again x could be 10 and y could be 1, giving you "yes". But you could also have the same with negative numbers, -10 and -1, giving you "no".

The statements together may look to be sufficient (again, 10 and 1 give you "yes"). But 2 and 1 also work, which would make x-1 equal to y, not greater. So because you can still get "no" and "yes", the correct answer is E.

Re: If x and y are integers, is (x−1)>y? [#permalink]
28 Apr 2014, 20:36

guerrero25 wrote:

Here is the OE .

Solution: E. This is a great opportunity to play devil's advocate, picking numbers to get "yes" and "no" answers using both statements.

For statement 1, x could be 10 and y could be 1, giving you "yes". But it could also be 1 and 1, so you could get the answer "no".

For statement 2, again x could be 10 and y could be 1, giving you "yes". But you could also have the same with negative numbers, -10 and -1, giving you "no".

The statements together may look to be sufficient (again, 10 and 1 give you "yes"). But 2 and 1 also work, which would make x-1 equal to y, not greater. So because you can still get "no" and "yes", the correct answer is E.

Hi I dont understand how C is wrong answer here? 2 is not multiple of 1 so how can we use 2 and 1 while combining two statements? What I thought, when we combine both statements then numbers could be X=4 (multiple of 2) Y=2 , 4/2 is greater than 1 so answer comes yes. The same way all multiples give same "yes" answer. We cannt use X=2, Y=2 since it does not satisfy statement 2. Please explain ..how E answer is right not C...

Re: If x and y are integers, is (x−1)>y? [#permalink]
29 Apr 2014, 00:19

Expert's post

drkomal2000 wrote:

guerrero25 wrote:

Here is the OE .

Solution: E. This is a great opportunity to play devil's advocate, picking numbers to get "yes" and "no" answers using both statements.

For statement 1, x could be 10 and y could be 1, giving you "yes". But it could also be 1 and 1, so you could get the answer "no".

For statement 2, again x could be 10 and y could be 1, giving you "yes". But you could also have the same with negative numbers, -10 and -1, giving you "no".

The statements together may look to be sufficient (again, 10 and 1 give you "yes"). But 2 and 1 also work, which would make x-1 equal to y, not greater. So because you can still get "no" and "yes", the correct answer is E.

Hi I dont understand how C is wrong answer here? 2 is not multiple of 1 so how can we use 2 and 1 while combining two statements? What I thought, when we combine both statements then numbers could be X=4 (multiple of 2) Y=2 , 4/2 is greater than 1 so answer comes yes. The same way all multiples give same "yes" answer. We cannt use X=2, Y=2 since it does not satisfy statement 2. Please explain ..how E answer is right not C...

Thanks in advance Komal

An integer \(a\) is a multiple of an integer \(b\) means that \(\frac{a}{b}=integer\).

Thus since 2/1=2=integer, then 2 IS a multiple of 1. By the way, any integer is a multiple of 1, since 1 is a factor of every integer.

Also on the GMAT when we are told that \(a\) is divisible by \(b\) (or which is the same: "\(a\) is a multiple of \(b\)", or "\(b\) is a factor of \(a\)"), we can say that: 1. \(a\) is an integer; 2. \(b\) is an integer; 3. \(\frac{a}{b}=integer\). _________________

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