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If x, y, and z are nonzero integers and x > yz, which of the [#permalink ]
16 Nov 2005, 17:03

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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

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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A.

unless we know if they are +ve or -ve we cannot simply this inequality.

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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...

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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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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 ]
08 Mar 2012, 20:07
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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 ]
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 ]
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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Re: Q: If X,Y, and Z are nonzero integers and X>YZ, which of [#permalink ]
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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Re: Q: If X,Y, and Z are nonzero integers and X>YZ, which of [#permalink ]
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 ]
17 Aug 2012, 04:40
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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

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.

Answer: A.

Hope it's clear.

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Re: If x, y, and z are nonzero integers and x > yz, which of the [#permalink ]
17 Aug 2012, 16:00

+1 A

We cannot infer anything. We need to know whether y and z are positive or negative.

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Re: If x, y, and z are nonzero integers and x > yz, which of the [#permalink ]
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