Last visit was: 13 Jun 2024, 23:26 It is currently 13 Jun 2024, 23:26
Toolkit
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

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.

# If x, y, and z are positive numbers, is x > y > z ?

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 93696
Own Kudos [?]: 631593 [32]
Given Kudos: 82279
Math Expert
Joined: 02 Sep 2009
Posts: 93696
Own Kudos [?]: 631593 [15]
Given Kudos: 82279
Manager
Joined: 23 Jun 2008
Posts: 69
Own Kudos [?]: 67 [9]
Given Kudos: 24
Location: Australia
Schools: AGSM '21
GMAT Date: 04-01-2014
General Discussion
Manager
Joined: 20 Dec 2013
Posts: 183
Own Kudos [?]: 291 [1]
Given Kudos: 35
Location: India
Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
1
Kudos
From S1:if xz>yz
and x,y,z are +ve,we can cancel out z to conclude x>y.But nothing is given about z.So insufficient.

From S2:x>z.Nothing is given about y.Insufficient.

Combining the two we get,x>y and x>z.
But whether y>/<z is not known.So insufficient.

Ans.E
Intern
Joined: 20 Nov 2013
Posts: 21
Own Kudos [?]: 30 [1]
Given Kudos: 188
Schools: LBS '17
Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
1
Kudos
Ans E

We know that all are positive integers so we can divide by the variable without changing the sign.

From1:
xz>yz --> x>y Insufficient

From2:
yx>yz --> x>z Insufficient

Combining both:
x>z and x>y but we don't know anything about how y compares to z . Insufficient.
Intern
Joined: 21 Jan 2014
Posts: 4
Own Kudos [?]: 3 [3]
Given Kudos: 0
Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
3
Kudos
Although the correct answer is E, can we apply the following approach? (if not, then why can't we use it?)
At the beginning, we conclude that (1) and (2) (if used separately) are not sufficient. Then, we combine these 2 inequalities, in other words, we form a system of 2 inequalities. Since x,y,z are variables, and we don’t know their signs, we cannot multiply. But we can subtract (2) from (1). As a result, this operation yields: xz-yx>yz-yz => xz-yx>0 => xz>yx => z>y => x>z>y => C
Math Expert
Joined: 02 Sep 2009
Posts: 93696
Own Kudos [?]: 631593 [5]
Given Kudos: 82279
Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
3
Kudos
2
Bookmarks
kronint wrote:
Although the correct answer is E, can we apply the following approach? (if not, then why can't we use it?)
At the beginning, we conclude that (1) and (2) (if used separately) are not sufficient. Then, we combine these 2 inequalities, in other words, we form a system of 2 inequalities. Since x,y,z are variables, and we don’t know their signs, we cannot multiply. But we can subtract (2) from (1). As a result, this operation yields: xz-yx>yz-yz => xz-yx>0 => xz>yx => z>y => x>z>y => C

No, that's not correct. You cannot subtract yx > yz from xz > yz because their signs are in the same direction (> and >).

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.
Intern
Joined: 28 Nov 2016
Posts: 5
Own Kudos [?]: 2 [0]
Given Kudos: 35
Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
When I plug in numbers for

X > Y > Z such as 3 > 2 > 1 or 20 > 10 > 2

the equations from (1) and (2) are always right. Any idea why the answer is still E?

Math Expert
Joined: 02 Sep 2009
Posts: 93696
Own Kudos [?]: 631593 [0]
Given Kudos: 82279
Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
TomSplashy wrote:
When I plug in numbers for

X > Y > Z such as 3 > 2 > 1 or 20 > 10 > 2

the equations from (1) and (2) are always right. Any idea why the answer is still E?

You plug the numbers that answer YES to the question and then check the equations given in (1) and (2). That#s not correct. When trying to solve an YES/NO DS question with plug-in method you should try to find two sets of numbers (which satisfy given information in the stem and the statements), that give different answer to the question (one YES and another NO). For example, you can check x=10, y=5 and z=1 for an YES answer and x=10, y=1 and z=5 for a NO answer.

Hope it's clear.
Manager
Joined: 11 Jun 2015
Posts: 73
Own Kudos [?]: 30 [0]
Given Kudos: 86
Location: India
Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
if x, y, and z are positive numbers, is x > y > z ?

(1) xz > yz
(2) yx > yz

Can you tell me if this method will work ?

divide eq 2 by eq 1

yx/xz>yz/yz

yx/xz>1
yx>xz
and since all are positive
y>z ?

Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93696
Own Kudos [?]: 631593 [0]
Given Kudos: 82279
Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
renjana wrote:
if x, y, and z are positive numbers, is x > y > z ?

(1) xz > yz
(2) yx > yz

Can you tell me if this method will work ?

divide eq 2 by eq 1

yx/xz>yz/yz

yx/xz>1
yx>xz
and since all are positive
y>z ?

Bunuel

You cannot divide equations like that. For example:

2 > 1
1 > 1/10

If you divide you'll get 2 > 10, which is not correct.
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 19013
Own Kudos [?]: 22400 [1]
Given Kudos: 286
Location: United States (CA)
Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
1
Kudos
Bunuel wrote:
If x, y, and z are positive numbers, is x > y > z ?

(1) xz > yz
(2) yx > yz

Solution:

We need to determine whether x > y > z, given that x, y, and z are all positive.

Statement One Alone:

Dividing both sides of the inequality by z, we have:

x > y

However, since we still don’t know whether y > z, statement one alone is not sufficient.

Statement Two Alone:

Dividing both sides of the inequality by y, we have:

x > z

However, since we still don’t know whether y > z, statement two alone is not sufficient.

Statements One and Two Together:

Using both statements, we see that x > y and x > z. However, since we still don’t know whether y > z, both statements together are not sufficient.

Non-Human User
Joined: 09 Sep 2013
Posts: 33549
Own Kudos [?]: 837 [0]
Given Kudos: 0
Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
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.
Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
Moderator:
Math Expert
93696 posts