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# If z=(x+y)/2xy and 0<y<z<1 which of the following

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Intern
Joined: 05 Dec 2017
Posts: 20
GPA: 3.35
If z=(x+y)/2xy and 0<y<z<1 which of the following  [#permalink]

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Updated on: 04 Apr 2018, 21:51
1
00:00

Difficulty:

45% (medium)

Question Stats:

63% (01:56) correct 38% (01:55) wrong based on 70 sessions

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If $$z = \frac{(x+y)}{2xy}$$ and $$0<y<x<1,$$ which of the following must be true about range of z?

A. $$z<0.5$$
B. $$z>0.5$$
C. $$z\geq{1}$$
D. $$z<1$$
E. $$z>1$$

Originally posted by Jamil1992Mehedi on 04 Apr 2018, 20:50.
Last edited by chetan2u on 04 Apr 2018, 21:51, edited 1 time in total.
updated the question
Manager
Joined: 24 Mar 2018
Posts: 246
Re: If z=(x+y)/2xy and 0<y<z<1 which of the following  [#permalink]

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04 Apr 2018, 21:42
It is already given z < 1 how can option E be correct ?
Manager
Joined: 05 Feb 2016
Posts: 168
Location: India
Concentration: General Management, Marketing
WE: Information Technology (Computer Software)
Re: If z=(x+y)/2xy and 0<y<z<1 which of the following  [#permalink]

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04 Apr 2018, 21:49
teaserbae wrote:
It is already given z < 1 how can option E be correct ?

Question seems to be wrong..since nothing is given for x..

what if x=0.Then it will be undefined..

Might be x<1 instead of Z
Math Expert
Joined: 02 Aug 2009
Posts: 8015
If z=(x+y)/2xy and 0<y<z<1 which of the following  [#permalink]

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04 Apr 2018, 21:50
Jamil1992Mehedi wrote:
If $$z = \frac{(x+y)}{2xy}$$ and $$0<y<z<1,$$ which of the following must be true?

A. $$z<0.5$$
B. $$z>0.5$$
C. $$z\geq{1}$$
D. $$z<1$$
E. $$z>1$$

z<0.5 is a part of z<1, so cannot be the answer...
z>1 or $$z\geq{1}$$ is a part of z>0.5, so cannot be the answer...
Actually answer should be B or D...

But here it seems the question wants to know for what range of z does it hold true
also it will be $$0<y<x<1$$ and not $$0<y<z<1,$$

$$z = \frac{(x+y)}{2xy}$$...
now since x and y are both <1, their ADDITION will always be MORE than their PRODUCT*2, so $$x+y>2xy..........\frac{x+y}{2xy}>1$$..... z>1
E
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If z=(x+y)/2xy and 0<y<z<1 which of the following   [#permalink] 04 Apr 2018, 21:50
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