Find all School-related info fast with the new School-Specific MBA Forum

It is currently 22 Oct 2014, 14:39

Close

GMAT Club Daily Prep

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

w, x, y, and z are integers. If w > x > y > z > 0, is y a co

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 05 Mar 2011
Posts: 155
Followers: 0

Kudos [?]: 8 [0], given: 3

w, x, y, and z are integers. If w > x > y > z > 0, is y a co [#permalink] New post 10 Dec 2011, 16:51
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

27% (02:54) correct 73% (02:02) wrong based on 22 sessions
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
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Jul 2013, 12:44, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
1 KUDOS received
GMAT Instructor
avatar
Joined: 24 Jun 2008
Posts: 978
Location: Toronto
Followers: 261

Kudos [?]: 707 [1] , given: 3

Re: MGMAT [#permalink] New post 10 Dec 2011, 17:48
1
This post received
KUDOS
ashiima 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


This is a pretty awkwardly designed question. From Statement 1 we know:

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

Now we know from the stem that w > x, or, since w and x are integers, that w > x + 1. Substituting in the expression we just found for w, we have that

(x/z) + 1 > x + 1
x/z > x

If z is an integer, this could only be true if z = 1 (if z is bigger than 1, clearly the left side of the inequality above will be smaller than the right side). So we know that z=1, and plugging that into our expression for w, we know that

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

So w is 1 greater than x, and w and x are therefore consecutive integers. The Greatest Common Divisor of any two consecutive positive integers is *always* equal to 1. Since y cannot be equal to 1 (since y > x > 0, and x and y are integers, the smallest possible value of y is 2), y cannot be a common divisor of x and w. So Statement 1 is sufficient.


From Statement 2 we can factor out a w:

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

For this product to be 0, one of the factors on the left side must be 0. We know that w is not zero, so w - y - 2 must be zero:

w - y - 2 = 0
w = y + 2

But if w > x > y, and each of these quantities are integers, then if w is exactly 2 greater than y, it must be that w, x and y are three consecutive integers. So again we know that w and x are consecutive integers, and just as we saw in Statement 1, it is impossible for y to be a common divisor of both.

So the answer is D, since each statement is sufficient to give a 'no' answer to the question.

I find it an awkward question because it tries to confuse the fundamental concept behind the question (GCD of consecutive integers is 1) by introducing distractions like negative exponents. That's not something you see in real GMAT questions. It also is a question where the statements are sufficient to give a 'no' answer, and such questions are very rare on the real GMAT - in most yes/no DS questions, if statements are sufficient, the answer is 'yes'. The basic concept in this question is difficult enough without it needing to be complicated further.
_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Private GMAT Tutor based in Toronto

Manager
Manager
avatar
Joined: 12 Oct 2011
Posts: 133
GMAT 1: 700 Q48 V37
GMAT 2: 720 Q48 V40
Followers: 3

Kudos [?]: 67 [0], given: 23

Re: MGMAT [#permalink] New post 11 Dec 2011, 03:27
IanStewart wrote:
ashiima 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


This is a pretty awkwardly designed question. From Statement 1 we know:

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

Now we know from the stem that w > x, or, since w and x are integers, that w > x + 1. Substituting in the expression we just found for w, we have that

(x/z) + 1 > x + 1
x/z > x

If z is an integer, this could only be true if z = 1 (if z is bigger than 1, clearly the left side of the inequality above will be smaller than the right side). So we know that z=1, and plugging that into our expression for w, we know that

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

So w is 1 greater than x, and w and x are therefore consecutive integers. The Greatest Common Divisor of any two consecutive positive integers is *always* equal to 1. Since y cannot be equal to 1 (since y > x > 0, and x and y are integers, the smallest possible value of y is 2), y cannot be a common divisor of x and w. So Statement 1 is sufficient.


From Statement 2 we can factor out a w:

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

For this product to be 0, one of the factors on the left side must be 0. We know that w is not zero, so w - y - 2 must be zero:

w - y - 2 = 0
w = y + 2

But if w > x > y, and each of these quantities are integers, then if w is exactly 2 greater than y, it must be that w, x and y are three consecutive integers. So again we know that w and x are consecutive integers, and just as we saw in Statement 1, it is impossible for y to be a common divisor of both.

So the answer is D, since each statement is sufficient to give a 'no' answer to the question.

I find it an awkward question because it tries to confuse the fundamental concept behind the question (GCD of consecutive integers is 1) by introducing distractions like negative exponents. That's not something you see in real GMAT questions. It also is a question where the statements are sufficient to give a 'no' answer, and such questions are very rare on the real GMAT - in most yes/no DS questions, if statements are sufficient, the answer is 'yes'. The basic concept in this question is difficult enough without it needing to be complicated further.

Why is it impossible for y to be a common divisor of both in statement 2, there's no restriction to the value of y given in statement 2, so y could be 1, which would make it a common divisor of w and x, wouldnt it?
GMAT Instructor
avatar
Joined: 24 Jun 2008
Posts: 978
Location: Toronto
Followers: 261

Kudos [?]: 707 [0], given: 3

Re: MGMAT [#permalink] New post 11 Dec 2011, 13:36
BN1989 wrote:

Why is it impossible for y to be a common divisor of both in statement 2, there's no restriction to the value of y given in statement 2, so y could be 1, which would make it a common divisor of w and x, wouldnt it?


There is a restriction on the value of y, in the question. We know that z and y are integers, and that y > z > 0. So the smallest possible value of z is 1, and the smallest possible value of y is 2. It is not possible, just from the stem alone, that y is 1 here.
_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Private GMAT Tutor based in Toronto

Senior Manager
Senior Manager
avatar
Joined: 12 Oct 2011
Posts: 276
Followers: 0

Kudos [?]: 24 [0], given: 110

Re: MGMAT [#permalink] New post 04 Jan 2012, 12:24
First of all, Ian, wonderful approach. I learned a lot from your approach. Thanks.

I followed the following crude approach to solve the problem:

We know that w > x > y > z > 0. Thus the least value of z is 1. Least value of y is 2.

I considered the values of z, y, x and w as
Case 1. z = 1, y = 2, x = 3, w = 4. In this case, y is NOT a factor of both x and w
Case 2. z = 1, y = 2, x = 4, w = 8. In this case, y is a factor of both x and w

Statement 1: Substitute the values for each case in the equation given (w/x) = (1/z) + (1/x)
Case 1: The values satisfy the equation.
Case 2: The values do not satisfy the equation.

Statement 2: w^2 - wy - 2w = 0 => w(w - y - 2) = 0 => either w = 0 or (w - y - 2) = 0.
w cannot be 0 because w > 0. Thus, w - y = 2.
Substitute the values for each case in the equation above.
Case 1: The values satisfy the equation.
Case 2: The values do not satisfy the equation.

As seen from both statements, only 1 case satisfies the given statements. Thus, the given numbers are consecutive integers. We can see that y is, thus, not a factor of both x and w.
Thus, D is the answer.
_________________

Consider KUDOS if you feel the effort's worth it

Manager
Manager
User avatar
Joined: 29 Jul 2011
Posts: 111
Location: United States
Followers: 3

Kudos [?]: 36 [0], given: 6

Re: MGMAT [#permalink] New post 04 Jan 2012, 12:37
Tough one and calculation intensive - obviously being from MGMAT.

w > x > y > z > 0
are w/y or x/y integers

I am going to use AD/BCE strategy here. S2 appears simple..

2. Rephrase gives us w = 0, w = y+2. But w>0, so w = y+2.
That means that w,x,y are a sequence. There is no way y becomes factor of w and x. Sufficient
B removed, now D.

1. Rephrase gives wz = x+z. Lets plug numbers here
4,3,2,1 satisfies the above, y is not a factor of w and x
5,4,2,1 or 5,4,3,1 satisfies, but y is not a factor of w and x
Sufficient.

D it is.
_________________

I am the master of my fate. I am the captain of my soul.
Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution.
PS - Always look at the answers first
CR - Read the question stem first, hunt for conclusion
SC - Meaning first, Grammar second
RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: MGMAT   [#permalink] 04 Jan 2012, 12:37
    Similar topics Author Replies Last post
Similar
Topics:
10 Experts publish their posts in the topic If w>x>y>z>0, is z>4? BN1989 3 01 Apr 2012, 02:47
33 Experts publish their posts in the topic w, x, y, and z are integers. If w >x>y>z>0, is y a common enigma123 12 26 Jan 2012, 21:19
30 Experts publish their posts in the topic w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 enigma123 18 01 Dec 2011, 00:52
1 If W>X>Y>Z>0, is Z>4 1) 1/W +1/X +1/Y +1/Z = dreambeliever 7 20 Mar 2011, 13:14
w, x, y, and z are integers. If w > x > y > z > vshaunak@gmail.com 11 20 May 2007, 05:35
Display posts from previous: Sort by

w, x, y, and z are integers. If w > x > y > z > 0, is y a co

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.