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If x, y, and z are nonzero integers and x > yz, which of the
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10 Apr 2020, 13:15

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Question Stats:

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

I. \(\frac{x}{y }> z\)

II. \(\frac{x}{z}> y\)

III. \(\frac{x}{(yz) }> 1\)

A. None of the above

B. I only

C. III only

D. I and II only

E. I, II, III

I. \(\frac{x}{y }> z\)

II. \(\frac{x}{z}> y\)

III. \(\frac{x}{(yz) }> 1\)

A. None of the above

B. I only

C. III only

D. I and II only

E. I, II, III

Re: If x, y, and z are nonzero integers and x > yz, which of the
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10 Apr 2020, 13:18

2

Expert Reply

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.

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.

Re: If x, y, and z are nonzero integers and x > yz, which of the
[#permalink]
13 Sep 2020, 10:06

1

Carcass wrote:

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.

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.

Hi Carcass,

For the 3rd option,how can we take the case of yz<0 ? Its mentioned in the question that x > yz so whatever cases we assume must first meet this condition otherwise it would violate the first condition given in the question.

case 1:

all are +ve i.e,

+x,+y,+z

in which case as per condition x>yz, x/yz would be > 1

case 2:

x is less -ve

yz together more -ve than x

in which case -ve/-ve is +ve,therefore x>yz is met but could be less than 1 for eg,

x=-5,y=2,z=-3

-5/2*-3

=0.8333 which is <1

I am not sure if your explanation for 3rd option is correct

Re: If x, y, and z are nonzero integers and x > yz, which of the
[#permalink]
13 Sep 2020, 10:56

Expert Reply

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

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

Re: If x, y, and z are nonzero integers and x > yz, which of the
[#permalink]
13 Sep 2020, 12:11

Carcass wrote:

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

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

So what is the point of question giving X > YZ?In absolute value questions,values are considered only after keeping in mind the original condition

Re: If x, y, and z are nonzero integers and x > yz, which of the
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31 May 2021, 07:20

1

nonzero integer means "positive and negative"

some answers could work with positive however, negative numbers are not working.

some answers could work with positive however, negative numbers are not working.

Re: If x, y, and z are nonzero integers and x > yz, which of the
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16 Jun 2023, 02:33

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

16 Jun 2023, 02:33
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