Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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]

Show Tags

04 Feb 2014, 18:21

2

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]

Show Tags

05 Feb 2014, 05:50

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]

Show Tags

28 Apr 2014, 21:36

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]

Show Tags

29 Apr 2014, 01:19

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. \(\frac{a}{b}=integer\). _________________

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

Show Tags

18 Sep 2015, 02:06

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

Show Tags

20 Sep 2015, 00:54

Expert's post

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x and y are integers, is (x−1)>y?

(1) x is a positive multiple of y (2) x/y > 1

In the original condition there are 2 variables (x,y) and we need 2 equations to match the number of variables and equations. Since there is 1 each in 1) and 2), C has high probability of being the answer. Using both 1) & 2) together (to save time), the answer is no if x=2, y=1, while the answer is yes if x=4, y=1 yes. Therefore the conditions are not sufficient. The answer is E.

Normally in case of DS inequality problems, 1) a>b>c --> a>b, b>c, a>c 2) a>b --> a+x>b+x and a-x>b-x 3) a>b and c>d --> a+c>b+d 4) a>0 --> 2a>0, a<0 --> 2a<a 5) -1<0<1 --> a-1<a<a+1 if the problem is not solved using these 5 concepts, E is usually the answer.

Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E. _________________

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...