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:

If X and Y are integers such that x<y<0m what is x-y?

1) (x+y)(x-y)=5 2) xy= 6

How would you do this question?

also is it wrong to make statement 1) (x+y)(x-y)=5 into x^2-y^2=5

and then square root both sides to make x-y=√5 ??

please advise... I know another method of getting it right, but I want to confirm if the above can be mathematically done

This portion in the above highlighted section is absolutely wrong because (x-y)^2 = x^2 - 2xy + y^2 and you cannot take square root of x^2 - y^2 as x-y.

I'm not solving it, but it is obvious that the answer is C or E. What is the method you're having in your mind ?

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

A.) First distribute (x+y)(x-y)=5 to get (x^2)-xy+xy-(y^2)=5. The -xy and +xy cancel each other out so that you now have (x^2)-(y^2)=5. The question tells us that both x and y are negative (x<y<0) and are integers. From that point you can easliy determine two squares whose difference is 5. And since the question tells us the absolute value of x>y, then x=-3 and y=-2. Sufficient.

B.) The question tells us that the two variables represent negative integers(x<y<0). B tells us that xy=6, which gives us two differnt pairs of integers for x and y: (-1x-6) and (-2x-3). Insufficient.

If x and y are integers such that x<y<0 what is x-y?

(1) (x+y)(x-y)=5. x and y are integers means that both x+y and x-y are integers. So, we have that the product of two integer factors equal to 5. There are only two combination of such factors possible: (1, 5) and (-1, -5). Since given that x and y are both negative then the first case is out, so x-y is either -1 or -5, but it can not be -5, because in this case x+y must be -1 and no sum of two negative integers yields -1. Hence x-y=-1. Sufficient.

(2) xy= 6. If x=-3 and y=-2 then x-y=-1 but if x=-6 and y=-1 then x-y=-5. Not sufficient.

Re: If X and Y are integers such that x<y<0 what is x-y? [#permalink]

Show Tags

16 Apr 2012, 19:49

great answer guys. It is important to look at it a different way other than just trying to go brute algebra. It's hard to go out of that mindset once you are so used to just solving all other problems that way. Thanks
_________________

Re: If X and Y are integers such that x<y<0 what is x-y? [#permalink]

Show Tags

09 Mar 2013, 07:29

Bunuel wrote:

If x and y are integers such that x<y<0 what is x-y?

(1) (x+y)(x-y)=5. x and y are integers means that both x+y and x-y are integers. So, we have that the product of two integer factors equal to 5. There are only two combination of such factors possible: (1, 5) and (-1, -5). Since given that x and y are both negative then the first case is out, so x-y is either -1 or -5, but it can not be -5, because in this case x+y must be -1 and no sum of two negative integers yields -1. Hence x-y=-1. Sufficient.

(2) xy= 6. If x=-3 and y=-2 then x-y=-1 but if x=-6 and y=-1 then x-y=-5. Not sufficient.

Answer: A.

Hope it's clear.

In these type of questions such how do we know that in statement A we must have only 2 possible combinations? My GMAT "instinct" lead me to choose A but it I can not think of a logical way to prove that there must be for sure only 2 combinations.

If x and y are integers such that x<y<0 what is x-y?

(1) (x+y)(x-y)=5. x and y are integers means that both x+y and x-y are integers. So, we have that the product of two integer factors equal to 5. There are only two combination of such factors possible: (1, 5) and (-1, -5). Since given that x and y are both negative then the first case is out, so x-y is either -1 or -5, but it can not be -5, because in this case x+y must be -1 and no sum of two negative integers yields -1. Hence x-y=-1. Sufficient.

(2) xy= 6. If x=-3 and y=-2 then x-y=-1 but if x=-6 and y=-1 then x-y=-5. Not sufficient.

Answer: A.

Hope it's clear.

In these type of questions such how do we know that in statement A we must have only 2 possible combinations? My GMAT "instinct" lead me to choose A but it I can not think of a logical way to prove that there must be for sure only 2 combinations.

Given that (x+y)(x-y)=5. Since x and y are integers, then we have that the product of 2 multiples is equal to 5.

Now, 5 can be broken into a product of 2 multiples only in 2 ways: 5=1*5 or 5=(-1)*(-5). After that you can refer to the solution above to see how it comes that x-y=-1.

Re: If x and y are integers such that x < y < 0 what is x - y? [#permalink]

Show Tags

21 Sep 2015, 16:46

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.
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...