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 350,000 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:

Re: If x/(x+y) = n, and x/(x-y) = m, then x/y = ? (|x| does not [#permalink]
19 Feb 2014, 21:36

2

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

MrWallSt wrote:

If x / (x+y) = n, and x / (x-y) = m, then x/y=? (|x| does not equal |y|, xy does not equal 0)

A. 3mn/2 B. (3m)/(2n) C. (n * (m+2))/2 D. 2nm / (m-n) E. (n^2 - m^2) / nm

You can do it either algebraically or by assuming numbers:

Algebra: Note that we are happier with (x+y)/x rather than x/(x+y) since in the former case we can make manipulations easily. So let's take the inverse of both n and m

Re: If x/(x+y) = n, and x/(x-y) = m, then x/y = ? (|x| does not [#permalink]
19 Feb 2014, 21:59

@Karishma thanks for the post. Your blogs are extremely helpful btw. I am not sure if you ever followed up about your CFA, but I hope that went well for you. _________________

Any and all kudos are greatly appreciated. Thank you.

with algebric solution, i am getting many answers in terms of m & n. so the problem needed to be solved by picking numbers. when we pick x = 3 & y = 1. I am getting B & D both as answer and but with x = 2 & y= 3. I am getting D as an answer. shouldnt it be unique for all numbers, since there is no restriction (like x, y = consequetive or x>y or viceversa) on picking numbers.

with algebric solution, i am getting many answers in terms of m & n. so the problem needed to be solved by picking numbers. when we pick x = 3 & y = 1. I am getting B & D both as answer and but with x = 2 & y= 3. I am getting D as an answer. shouldnt it be unique for all numbers, since there is no restriction (like x, y = consequetive or x>y or viceversa) on picking numbers.

When you take one of the numbers as 1 or 0, or when you take numbers to be equal, you could get multiple options satisfying your conditions.

For example, if n = 1 3n/(m+n) will give the same result as 3/(m+n).

Sometimes, you will get the same answer from multiple options if the options are written intelligently. So you could need to try out 2 or even 3 sets of values. _________________

Re: If x/(x+y) = n, and x/(x-y) = m, then x/y = ? (|x| does not [#permalink]
25 May 2015, 03:57

VeritasPrepKarishma wrote:

When you take one of the numbers as 1 or 0, or when you take numbers to be equal, you could get multiple options satisfying your conditions.

For example, if n = 1 3n/(m+n) will give the same result as 3/(m+n).

Sometimes, you will get the same answer from multiple options if the options are written intelligently. So you could need to try out 2 or even 3 sets of values.

well the question states xy != 0, so x or y cant be 0. also |x| != |y| so they cant be equal either.

I was hoping for a more bullet proof solution for this question. as 2 sets will take more time. we will get answer but at a huge expense of time.

Re: If x/(x+y) = n, and x/(x-y) = m, then x/y = ? (|x| does not [#permalink]
25 May 2015, 19:25

Expert's post

Alaukik wrote:

VeritasPrepKarishma wrote:

When you take one of the numbers as 1 or 0, or when you take numbers to be equal, you could get multiple options satisfying your conditions.

For example, if n = 1 3n/(m+n) will give the same result as 3/(m+n).

Sometimes, you will get the same answer from multiple options if the options are written intelligently. So you could need to try out 2 or even 3 sets of values.

well the question states xy != 0, so x or y cant be 0. also |x| != |y| so they cant be equal either.

I was hoping for a more bullet proof solution for this question. as 2 sets will take more time. we will get answer but at a huge expense of time.

Note that it is a generic comment about number plugging - not just specific to this question.

Also, I have given the algebra solution above - as "bulletproof" as you can get. _________________

Re: If x/(x+y) = n, and x/(x-y) = m, then x/y = ? (|x| does not [#permalink]
02 Aug 2015, 05:55

1

This post received KUDOS

VeritasPrepKarishma your solution is excellence. I am wondering, how do we know that one form of fraction (you mentioned it as case) is easier to manipulate than another? I tried to solve this problem using the original form of fraction but I ended up nowhere. Do you have the solution using original form of the fraction?

Re: If x/(x+y) = n, and x/(x-y) = m, then x/y = ? (|x| does not [#permalink]
02 Aug 2015, 22:30

2

This post received KUDOS

Expert's post

evdo wrote:

VeritasPrepKarishma your solution is excellence. I am wondering, how do we know that one form of fraction (you mentioned it as case) is easier to manipulate than another? I tried to solve this problem using the original form of fraction but I ended up nowhere. Do you have the solution using original form of the fraction?

When you need to separate the variables, multiple terms in the numerator are easy to handle since you can split them:

(x+y)/ x = x/x + y/x = 1 + y/x

But if they are in the denominator, you cannot separate them.

You can start with the equations as they are but you will eventually cross multiply to simplify the denominator. If you do different things with the equations, you will get different but equivalent expressions. So a case can be made to do the question by number plugging instead. _________________

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

Last week, hundreds of first-year and second-year students traversed the globe as part of KWEST: Kellogg Worldwide Experience and Service Trip. Kyle Burr, one of the student-run KWEST executive...