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x<y and z>0 which of the following must be true?

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Manager
Joined: 25 Feb 2008
Posts: 53
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Kudos [?]: 14 [0], given: 0

x<y and z>0 which of the following must be true? [#permalink]

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20 Mar 2008, 07:11
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

x<y and z>0 which of the following must be true?

z/x<z/y
y/z<x/z
x>y/z
z-x<z-y
xz<yz
Please explain your answer. thanks
Manager
Joined: 20 Aug 2007
Posts: 65
Followers: 1

Kudos [?]: 17 [0], given: 0

Re: x<y [#permalink]

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20 Mar 2008, 07:36
Capthan wrote:
x<y and z>0 which of the following must be true?

z/x<z/y
y/z<x/z
x>y/z
z-x<z-y
xz<yz
Please explain your answer. thanks

answer xz<yz

just plug in number and u'll get the answer
Director
Joined: 10 Sep 2007
Posts: 943
Followers: 8

Kudos [?]: 303 [0], given: 0

Re: x<y [#permalink]

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20 Mar 2008, 07:41
Please remember that z is +ve as given in the question.

A: z/x<z/y Dividing by z both side we have => 1/x < 1/y.
We are not told about x and y being +ve or -ve only thing is that x<y. So this one might or might not hold.

B: y/z<x/z, Multiplyig both sides by z, we have y<x. Not true as given in question x<y.

C: x>y/z, Multiplyig both sides by z, we have y<zx. Since we do not have fixed values for x,y, and z so there is infinite possibilities to satisfy or not satisfy this inequality.

D: z-x<z-y. Substracting z from both side, we have -x<-y => y<x. Not true as given in question x<y.

E: xz<yz Dividing by z both side we have => x<y, which is given in the question.

So answer E.
Re: x<y   [#permalink] 20 Mar 2008, 07:41
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x<y and z>0 which of the following must be true?

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