relation between x>y>z
This topic has expert replies
- Morgoth
- Master | Next Rank: 500 Posts
- Posts: 316
- Joined: Mon Sep 22, 2008 12:04 am
- Thanked: 36 times
- Followed by:1 members
x>y>z ?ifthyder wrote:how do we know about y ?
1) xz>yz, since x y and z are positive numbers we can simplify this
x>y. we dont know anything about z. Insufficient.
2) yx>yz, since x y and z are positive numbers we can simplify this
x>z, we dont know anything about y. Insufficient.
Combining 1 & 2
We still dont know about the relationship between z and y. Insufficient.
Thus, E is the answer.
-
- Senior | Next Rank: 100 Posts
- Posts: 35
- Joined: Thu Aug 21, 2008 2:55 pm
- Thanked: 1 times
-
- Master | Next Rank: 500 Posts
- Posts: 165
- Joined: Wed May 28, 2008 11:25 pm
- Thanked: 9 times
- GMAT Score:730
Subtract 1 from 2, you shall get
xz-yx>0
x(z-y)>0
therefore, either both x and z-y>0 or x and z-y <0
but we know X>0 ( x is given as positive) hence z-y>0, z>y.
Hence C
xz-yx>0
x(z-y)>0
therefore, either both x and z-y>0 or x and z-y <0
but we know X>0 ( x is given as positive) hence z-y>0, z>y.
Hence C
-
- Master | Next Rank: 500 Posts
- Posts: 106
- Joined: Thu May 22, 2008 10:50 am
- Thanked: 3 times
- gabriel
- Legendary Member
- Posts: 986
- Joined: Wed Dec 20, 2006 11:07 am
- Location: India
- Thanked: 51 times
- Followed by:1 members
You can subtract inequality equations. But there are some operations like multiplication, division and squaring that you can do only after considering the sign of the variables in question.EricLien9122 wrote:sumithshah:
I don't think you can subtract inequality equations.
Please share your thoughts.
Cheers.
-
- Master | Next Rank: 500 Posts
- Posts: 377
- Joined: Wed Sep 03, 2008 9:30 am
- Thanked: 15 times
- Followed by:2 members
I can subract the other way around, in which case I would get y>z.sumithshah wrote:Subtract 1 from 2, you shall get
xz-yx>0
x(z-y)>0
therefore, either both x and z-y>0 or x and z-y <0
but we know X>0 ( x is given as positive) hence z-y>0, z>y.
Hence C
I think it is E.