Last visit was: 24 Apr 2026, 20:27 It is currently 24 Apr 2026, 20:27
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,733
Own Kudos:
36,459
 [23]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,733
Kudos: 36,459
 [23]
w, x, y, and z are different integers. If w/x = y/z, which Updated on: 14 Oct 2019, 18:28
Originally posted by BrentGMATPrepNow on 13 Oct 2019, 08:36.
Last edited by chetan2u on 14 Oct 2019, 18:28, edited 1 time in total.
Corrected the OA
1
Kudos
Add Kudos
22
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

22% (02:05) correct 78% (02:09) wrong based on 215 sessions

History

Date
Time
Result
Not Attempted Yet
\(w\), \(x\), \(y\), and \(z\) are different integers. If \(\frac{w}{x}= \frac{y}{z}\), which of the following COULD be true?

i) \(wy ≠0\)
ii) \(w + x = y + z\)
iii) \(w+y = x+z\)


A) i only
B) i & ii only
C) i & iii only
D) ii & iii only
E) I, ii & iii
ShowHide Answer
Official Answer
E
Every tricky question should trigger a follow up quiz. Build that habit with GMAT Club Forum Quiz →.
Most Helpful Reply
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 24 Apr 2026
Posts: 11,229
Own Kudos:
45,009
 [5]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
My Reviews
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 45,009
 [5]
w, x, y, and z are different integers. If w/x = y/z, which 13 Oct 2019, 08:59
2
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
\(w\), \(x\), \(y\), and \(z\) are different integers. If \(\frac{w}{x}= \frac{y}{z}\), which of the following COULD be true?

Since \(\frac{w}{x}= \frac{y}{z}\), we can take w=a*y and then x=a*z, where a could be any number - integer or fraction, positive or negative, but \(a\neq{1}\), as then w and y will become same integer.

i) \(wy ≠0\)
It is not only COULD be true, but also MUST be true too as w and y are different integers and if any of w or y is 0, the other will have to be 0..YES
Example = \(\frac{w}{x}= \frac{y}{z}\)....\(\frac{1}{2}= \frac{2}{4}\)..\(wy=1*2=2\neq{0}\)

ii) \(w + x = y + z\)
when \(w=-x......w+x=0\) and \(y=-z.....\) \(y+z=0\), so YES
example = \(\frac{w}{x}= \frac{y}{z}\)....\(\frac{-2}{2}= \frac{-3}{3}\).[/m]

iii) \(w+y = x+z\)
when \(a=-1, w=-1*y=-y\) and \(x=-1*z=-z.....w+y=-y+y=0\) and \(x+z=-z+z=0\), so YES
example = \(\frac{w}{x}= \frac{y}{z}\)....\(\frac{1}{2}= \frac{-1}{-2}\).[/m]

So i, ii and iii

E
General Discussion
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 24 Apr 2026
Posts: 5,986
Own Kudos:
5,859
 [2]
Given Kudos: 163
Location: India
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
My Reviews
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
Posts: 5,986
Kudos: 5,859
 [2]
w, x, y, and z are different integers. If w/x = y/z, which 13 Oct 2019, 09:22
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATPrepNow
\(w\), \(x\), \(y\), and \(z\) are different integers. If \(\frac{w}{x}= \frac{y}{z}\), which of the following COULD be true?

i) \(wy ≠0\)
ii) \(w + x = y + z\)
iii) \(w+y = x+z\)


A) i only
B) i & ii only
C) i & iii only
D) ii & iii only
E) I, ii & iii
Show more

i) \(wy ≠0\)
If w=0; y=0; which is not possible since w & y are different integers.
MUST BE TRUE

ii) \(w + x = y + z\)
Let w/x = y/z = k
w + x = (k+1)x
y + z = (k+1)z
For w + x = y + z
(k+1)x = (k+1)z
x = z ; or k = -1; x= z is not possible since x & z are different integers
k = -1
w = -x & y = -z
w + x = y + z = 0
COULD BE TRUE


iii) \(w+y = x+z\)
Let w/x = y/z = k
For w+y = x + z
kx + kz = x + z
k (x+z) = x + z
x = -z or k = 1; k = 1 is not possible since x, y, z & w are different integers
x = -z => x + z = 0 = w + y; w =-y
COULD BE TRUE

IMO E
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,733
Own Kudos:
36,459
 [1]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,733
Kudos: 36,459
 [1]
w, x, y, and z are different integers. If w/x = y/z, which Updated on: 14 Oct 2019, 15:52
Originally posted by BrentGMATPrepNow on 14 Oct 2019, 14:56.
Last edited by BrentGMATPrepNow on 14 Oct 2019, 15:52, edited 2 times in total.
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
GMATPrepNow
\(w\), \(x\), \(y\), and \(z\) are different integers. If \(\frac{w}{x}= \frac{y}{z}\), which of the following COULD be true?

i) \(wy ≠0\)
ii) \(w + x = y + z\)
iii) \(w+y = x+z\)


A) i only
B) i & ii only
C) i & iii only
D) ii & iii only
E) I, ii & iii
Show more

i) \(wy ≠0\)
This, in fact, MUST be true.
We know this, because if one variable (w or y) = 0, then the other variable must also be 0.
Since we're told all 4 integers are DIFFERENT, statement i MUST be true.
-----------------

ii) \(w + x = y + z\)
Key concept: If \(\frac{w}{x}= \frac{y}{z}\), then \(\frac{w}{x}\) and \(\frac{y}{z}\) are equivalent fractions.

This means that \(y = wk\) and \(z = xk\) for some constant value of k (this is how we create equivalent fractions)
IMPORTANT: Since w, x, y and z are DIFFERENT, we can be certain that k ≠ 1

So, we can take: \(w + x = y + z\)
And replace y and z to get: \(w + x = wk + xk\)
Factor to get: \(w + x = k(w + x)\)
This tells us that EITHER k = 1, OR (w+x) = 0
We already know that k ≠ 1, but it COULD be the case that (w+x) = 0
One possible case is that w = 1, x = -1, y = 2 and z = -2
In this case, \(w + x = y +z\)

So, statement ii CAN be true
-----------------

iii) \(w+y = x+z\)
We'll use (again) the fact that \(y = wk\) and \(z = xk\) for some constant value of k (this is how we create equivalent fractions)

So, we can take: \(w+wk = x+xk\)
Factor each side to get: \(w(1+k) = x(1+k)\)
Notice that this equation is true if EITHER \(w=x\) OR \((1+k)=0\)
We know that \(w≠k\), but what happens if \((1+k)=0\)?

If \((1+k)=0\), then \(k = -1\)
If \(k = -1\), then we get: \(y = w(-1)\) and \(z = x(-1)\)

So, one possible case is that w = 1, x = -2, y = -1 and z = 2

We get: \(\frac{w}{x}= \frac{y}{z}=\frac{1}{-2}= \frac{-1}{2}\). PERFECT!

In this case, \(w+y = x+z\)

So, statement iii CAN be true
----------------------------

Answer: E

Cheers,
Brent
User avatar
Mo2men
User avatar
Joined: 26 Mar 2013
Last visit: 09 May 2023
Posts: 2,426
Own Kudos:
1,508
 [1]
Given Kudos: 641
Concentration: Operations, Strategy
Schools: Erasmus (II)
Products:
My Reviews
Schools: Erasmus (II)
Posts: 2,426
Kudos: 1,508
 [1]
w, x, y, and z are different integers. If w/x = y/z, which 14 Oct 2019, 15:04
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATPrepNow

iii) \(w+y = x+z\)
We'll use (again) the fact that \(y = wk\) and \(z = xk\) for some constant value of k (this is how we create equivalent fractions)

So, we can take: \(w+wk = x+xk\)
Factor each side to get: \(w(1+k) = x(1+k)\)
Notice that this equation is true if EITHER\(w=k\)OR \((1+k)=0\)
We know that \(w≠k\), but what happens if \((1+k)=0\)?

If \((1+k)=0\), then \(k = -1\)
If \(k = -1\), then we get: \(y = w(-1)\) and \(z = x(-1)\)

So, one possible case is that w = 1, x = -2, y = -1 and z = 2

We get: \(\frac{w}{x}= \frac{y}{z}=\frac{1}{-2}= \frac{-1}{2}\). PERFECT!

In this case, \(w+y = x+z\)

So, statement iii CAN be true
----------------------------

Answer: C

Cheers,
Brent
Show more

Hi Brent,

Do you mean in the highlighted either w= x ? If no, can you elaborate how w = k?

Thanks
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 24 Apr 2026
Posts: 11,229
Own Kudos:
45,009
 [1]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
My Reviews
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 45,009
 [1]
w, x, y, and z are different integers. If w/x = y/z, which 14 Oct 2019, 15:17
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATPrepNow
GMATPrepNow
\(w\), \(x\), \(y\), and \(z\) are different integers. If \(\frac{w}{x}= \frac{y}{z}\), which of the following COULD be true?

i) \(wy ≠0\)
ii) \(w + x = y + z\)
iii) \(w+y = x+z\)


A) i only
B) i & ii only
C) i & iii only
D) ii & iii only
E) I, ii & iii

i) \(wy ≠0\)
This, in fact, MUST be true.
We know this, because if one variable (w or y) = 0, then the other variable must also be 0.
Since we're told all 4 integers are DIFFERENT, statement i MUST be true.
-----------------

ii) \(w + x = y + z\)
Key concept: If \(\frac{w}{x}= \frac{y}{z}\), then \(\frac{w}{x}\) and \(\frac{y}{z}\) are equivalent fractions.

This means that \(y = wk\) and \(z = xk\) for some constant value of k (this is how we create equivalent fractions)
IMPORTANT: Since w, x, y and z are DIFFERENT, we can be certain that k ≠ 1

So, we can take: \(w + x = y + z\)
And replace y and z to get: \(w + x = wk + xk\)
Factor to get: \(w + x = k(w + x)\)
This tells us that k = 1, but we already know that k ≠ 1
So, statement ii cannot be true
-----------------

iii) \(w+y = x+z\)
We'll use (again) the fact that \(y = wk\) and \(z = xk\) for some constant value of k (this is how we create equivalent fractions)

So, we can take: \(w+wk = x+xk\)
Factor each side to get: \(w(1+k) = x(1+k)\)
Notice that this equation is true if EITHER \(w=k\) OR \((1+k)=0\)
We know that \(w≠k\), but what happens if \((1+k)=0\)?

If \((1+k)=0\), then \(k = -1\)
If \(k = -1\), then we get: \(y = w(-1)\) and \(z = x(-1)\)

So, one possible case is that w = 1, x = -2, y = -1 and z = 2

We get: \(\frac{w}{x}= \frac{y}{z}=\frac{1}{-2}= \frac{-1}{2}\). PERFECT!

In this case, \(w+y = x+z\)

So, statement iii CAN be true
----------------------------

Answer: C

Cheers,
Brent
Show more

Even ii could be true.
Let w=2, x=-2, y=3 and z=-3....w+x=2-2=0=3-3=y+z.
w/x=y/z=-1..
So w+x=y+z
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,733
Own Kudos:
36,459
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,733
Kudos: 36,459
w, x, y, and z are different integers. If w/x = y/z, which 14 Oct 2019, 15:53
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Mo2men

Hi Brent,

Do you mean in the highlighted either w= x ? If no, can you elaborate how w = k?

Thanks
Show more

You're right. It should have read w = x.

Cheers,
Brent
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,733
Own Kudos:
36,459
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,733
Kudos: 36,459
w, x, y, and z are different integers. If w/x = y/z, which 14 Oct 2019, 15:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chetan2u


Even ii could be true.
Let w=2, x=-2, y=3 and z=-3....w+x=2-2=0=3-3=y+z.
w/x=y/z=-1..
So w+x=y+z
Show more

You're absolutely right! I initially didn't account for the fact that (w+x) could equal 0

I've edited my response.

Kudos for everyone!

Cheers,
Brent
User avatar
GMATGuruNY
User avatar
Joined: 04 Aug 2010
Last visit: 02 Apr 2026
Posts: 1,347
Own Kudos:
3,905
 [2]
Given Kudos: 9
Schools:Dartmouth College
Expert
Expert reply
Posts: 1,347
Kudos: 3,905
 [2]
w, x, y, and z are different integers. If w/x = y/z, which 30 Mar 2021, 03:42
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
BrentGMATPrepNow
\(w\), \(x\), \(y\), and \(z\) are different integers. If \(\frac{w}{x}= \frac{y}{z}\), which of the following COULD be true?

i) \(wy ≠0\)
ii) \(w + x = y + z\)
iii) \(w+y = x+z\)


A) i only
B) i & ii only
C) i & iii only
D) ii & iii only
E) I, ii & iii
Show more

i) \(wy ≠0\)
Given \(\frac{w}{x}= \frac{y}{z}\), \(wy = 0\) only if \(w=y=0\).
Since w and y must be different integers, it is not possible that \(w=y=0\).
Thus, it must be true that \(wy ≠0\).
Eliminate D.

ii) \(w + x = y + z\)
\(\frac{w}{x}= \frac{y}{z}\)
\(\frac{w}{x} + \frac{x}{x}= \frac{y}{z} + \frac{z}{z}\)
\(\frac{w+x}{x}= \frac{y+z}{z}\)
The resulting equation is valid if \(w+x=0\) and \(y+z=0\).
Thus, it could be true that \(w+x = y+z\).
Eliminate A and C.

iii) \(w+y = x+z\)
\(\frac{w}{x}= \frac{y}{z}\)
\(\frac{w}{y}= \frac{x}{z}\)
\(\frac{w}{y} + \frac{y}{y}= \frac{x}{z} + \frac{z}{z}\)
\(\frac{w+y}{y}= \frac{x+z}{z}\)
The resulting equation is valid if \(w+y=0\) and \(x+z=0\).
Thus, it could be true that \(w+y = x+z\).
Eliminate B.

Show Spoiler
E
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,976
Own Kudos:
1,117
Posts: 38,976
Kudos: 1,117
w, x, y, and z are different integers. If w/x = y/z, which 24 Apr 2025, 23:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109818 posts
Krunaal
Tuck School Moderator
853 posts