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# If x, y, and z are nonzero integers and x > yz, which of the

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Intern
Joined: 05 Nov 2005
Posts: 34
If x, y, and z are nonzero integers and x > yz, which of the  [#permalink]

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Updated on: 17 Aug 2012, 04:34
1
5
00:00

Difficulty:

55% (hard)

Question Stats:

60% (01:11) correct 40% (00:56) wrong based on 267 sessions

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If x, y, and z are nonzero integers and x > yz, which of the following must be true

I. x/y > z
II. x/z> y
III. x/(yz) > 1

A. None of the above
B. I only
C. III only
D. I and II only
E. I, II, III

Originally posted by faisalt on 16 Nov 2005, 17:03.
Last edited by Bunuel on 17 Aug 2012, 04:34, edited 1 time in total.
Edited the question and added the OA.
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Joined: 22 Aug 2005
Posts: 1074
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16 Nov 2005, 17:09
A.

unless we know if they are +ve or -ve we cannot simply this inequality.
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16 Nov 2005, 17:13
I am inclined to pick A as well...we dont know the signs

we know that X>yz

so lets say x=-1 YZ=-3

-1/-3=1/3 <1

if X is 3 YZ=2

3/2 > 1....

likewise others dont add up either...
Manager
Joined: 19 Sep 2005
Posts: 110

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16 Nov 2005, 18:09
I think A as well.
I picked a couple of numbers

1. x = 4, y = -2, z= 5
2. x= 4, y=2, z=-3

none have to work
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17 Nov 2005, 05:57
I pick C. Seems to me that it is always correct to say x/yz > 1
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Re: Q: If X,Y, and Z are nonzero integers and X>YZ, which of  [#permalink]

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08 Mar 2012, 20:07
1
None of the options satisfies all the cases
Good example for plugging in question type ..

+ kudos if U like this post
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Re: Q: If X,Y, and Z are nonzero integers and X>YZ, which of  [#permalink]

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08 Mar 2012, 20:41
Well waiting for Bunuel's reply to this question.
I am sure there will be something special in store.
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Re: Q: If X,Y, and Z are nonzero integers and X>YZ, which of  [#permalink]

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08 Mar 2012, 21:17
if u consider y & z are negative numbers and plugin cases 1 and 2 fail.

However 3 seems to be correct coz x>yz is the given condition so x/yz>1 always.

Hope this helps.
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Joined: 07 Nov 2009
Posts: 243
Re: Q: If X,Y, and Z are nonzero integers and X>YZ, which of  [#permalink]

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09 Mar 2012, 02:33
faisalt wrote:
Q: If X,Y, and Z are nonzero integers and X>YZ, which of the following must be true
1. X/Y>Z
2. X/Z>Y
3. X/YZ>1

A. None of the above
B. 1 only
C. 3 only
D. 1 and 2 only
E. 1, 2, 3

I used numbers
Case a
X = 8 , Y = -2 , Z = -3
Option 1 and 2 fails

Case b
X = -3 , Y = -3 , Z = 2
Option 2 fails

None of them satisfies. Ans should be A
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Joined: 27 May 2012
Posts: 633
Re: Q: If X,Y, and Z are nonzero integers and X>YZ, which of  [#permalink]

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14 Aug 2012, 04:46
pratikbais wrote:
Well waiting for Bunuel's reply to this question.
I am sure there will be something special in store.

True , expert solution will reenforce our solutions, I am also getting A

simple pick easy numbers

x= 16 y = 2 and Z=4 as X>YZ satisfies all

but if we pick

x= -4 y= -2 and Z=4 as X>YZ so option 1 x/y> Z need not be true and X/YZ > 1( option 3 ) also need not be true

also
x= -4 Y=2 and Z= -4 as X>YZ so option 2 , X/Z > Y also need not be true

so none of the answer choices have to be true for X>YZ condition to hold
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Re: If x, y, and z are nonzero integers and x > yz, which of the  [#permalink]

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17 Aug 2012, 04:40
2
1
If x, y, and z are nonzero integers and x > yz, which of the following must be true

I. x/y > z
II. x/z> y
III. x/(yz) > 1

A. None of the above
B. I only
C. III only
D. I and II only
E. I, II, III

Remember, when we divide both sides of an inequality by a negative number we should flip the sign.

Given: $$x > yz$$.

If $$y<0$$, then when we divide both parts by $$y$$ we'll get $$\frac{x}{y}<z$$. Thus option I is not necessarily true;

If $$z<0$$, then when we divide both parts by $$z$$ we'll get $$\frac{x}{z}<y$$. Thus option II is not necessarily true;

If $$yz<0$$, then when we divide both parts by $$yz$$ we'll get $$\frac{x}{yz}<1$$. Thus option III is not necessarily true.

So, none of the options must be true.

Hope it's clear.
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Re: If x, y, and z are nonzero integers and x > yz, which of the  [#permalink]

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21 Nov 2018, 20:11
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Re: If x, y, and z are nonzero integers and x > yz, which of the &nbs [#permalink] 21 Nov 2018, 20:11
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