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:

w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]

Show Tags

27 Nov 2015, 01:11

Given to us : w > x > y > z > 0

In such type of questions, we have to start by reducing the given equations to a simple form and then test the simplified equations with a random number, so to quickly come to a required answer.

Now from the given equation \(w>x>y>z>0\), we definitely know that z minimum will be equal to 1. Substituting z=1 in the simplified equation, we get \(w=x+1\). From this equation, we can thus deduce that w and x are consecutive integers and it also satisfies w>x criteria. So if two numbers are consecutive to each other there can only be one number common to them (because these numbers become co-prime to each other, example 4 & 5) i.e. 1. but we definitely know y cannot be 1. It will definitely be greater than 1 because z=1 and y>z. So, y is not a common divisor of x and w. Now, in case if z>1, then from the simplified equation, we will have x>w, which will contradict the given condition of (w>x>y>z>0). Hence, statement 1 is sufficient is sufficient to answer the question that y is not a common divisor of x and w.

Using Statement 2

\(w^2-wy-2w=0\)

Simplifying the equation

\(w(w-y-2)=0\) So, either \(w=0\) or \(w-y-2=0\)

w=0 is not possible because \(w>x>y>z>0\) is given to us.

Considering, \(w-y-2=0\), we get \(w=y+2\).

Using this equation, we can conclude that there will be an integer between w and y, and that is definitely going to be x, because of the equation w>x>y>z>0. For example, if w=6, then y becomes 4, then x has to be 5 to satisfy the equation.

So, from this statement we definitely know, that w,x and y are three consecutive integers. Now three consecutive integers can only be co-prime to each other and can never have any common divisor in between them. There can only be one number common in between these numbers i.e. 1

So from this statement we can conclude that y is not the common divisor in between x and w. Hence, Statment 2 is also sufficient to answer the Question.

w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]

Show Tags

11 Mar 2016, 09:51

gmacforjyoab wrote:

enigma123 wrote:

w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

1 ) w/x = 1/z + 1/x since W is greater than x , w/x should be greater than 1 hence 1/z +1/x should also be greater than 1. since x and z are integers grater than 0 , and z<x , the only way , 1/z +1/x can be >1 is if z=1 hence equation becomes - w/x= 1/1 + 1/x ======> w-1=x .. hence x and w are consecutive integers and since y is not 1 and is >1 ( since z is 1) , the answer is NO Sufficient

2) w^2-wy-2w=0 w(w-y-2) =0 since w != 0 w=y+2 .. w, x and y are consecutive integers . Same as statement 1 , the answer is NO. Suff

-Jyothi

Thanks for the nice explanation !! Only this enabled me to understand the logic... ++ kudos

Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]

Show Tags

13 Mar 2016, 23:05

Bunuel wrote:

ronr34 wrote:

PiyushK wrote:

Another concept, two consecutive numbers are co-prime thus they have only 1 as a common factor.

1> After reducing option 1 we have Z = X/(W-1) for Z to be an integer (1) and X<W, W and Z must be consecutive numbers. And two consecutive numbers have only one common factor that is 1, bcz Y>Z>0 and Z is also an integer then Y can not be 1. Therefore, Y can not be a common divisor for both.

2> After reducing option 2. W=Y+2 it means W,X,Y all are consecutive integers and co prime to each other. thus, again Y can not be common divisor for W and X and neither it is equal to one as restricted by given data.

Therefore, Ans D.

PiyushK - I think you meant "W and X must be consecutive numbers"... not "W and Z" Bunuel - In your solution, for st. 1, you wrote

Bunuel wrote:

since w>x>z then z=1 (if z>1 then x>w which contradicts given condition)

. How did you reach this conclusion?

w > x > y > z > 0, is y a common divisor of w and x?

We have that z(w-1)=x and w>x>z>0. If z>1, say if z=2, then 2(w-1)=x. Now tell me can in this case x be less than w?

I did not understand one concept here What if W=-10 w-1=-11 z=100 => z*(w-1) = -200 so x=-200 also x<w is satisfied..

Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]

Show Tags

15 Mar 2017, 01:44

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.
_________________

What is tested is whether one is able to identify a specific set of consecutive integers.

Statement 1: z has to be 1 because w/x is greater than 1. z is the smallest integer and greater than 0. If z=1, we can see that w=x+1. A number cannot be a common divisor of two consecutive integers unless it is 1. y is not 1. Sufficient.

Statement 2: We can see W=y+2 and so w=x+1. Now the same reasoning as that for statement 1 applies. Sufficient.
_________________

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...