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# If x, y and z are non-zero integers

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Intern
Joined: 15 Sep 2017
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If x, y and z are non-zero integers [#permalink]  09 Sep 2018, 01:41
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Question Stats:

28% (01:01) correct 71% (00:43) wrong based on 7 sessions
If x, y and z are non-zero integers, and if $$x > yz$$, then which of the following statements must be true?

Indicate all such statements.

A) $$\frac{x}{y} > z$$

B) $$\frac{x}{z} > y$$

C) $$\frac{x}{yz} > 1$$

D) $$yz < x$$
[Reveal] Spoiler: OA
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Joined: 07 Jun 2014
Posts: 4856
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Kudos [?]: 1731 [1] , given: 397

Re: If x, y and z are non-zero integers [#permalink]  09 Sep 2018, 16:05
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Expert's post
WRONG SOLUTION MY MISTAKE! IGNORE THIS AND SCROLL DOWN

AchyuthReddy wrote:
If x, y and z are non-zero integers, and if $$x > yz$$, then which of the following statements must be true?

Indicate all such statements.

A) $$\frac{x}{y} > z$$

B) $$\frac{x}{z} > y$$

C) $$\frac{x}{yz} > 1$$

d) $$yz < x$$

Since it is mentioned that x, y and z are positive you can divide them without flipping the inequality.

Given: $$x > yz$$

Dividing both sides with y

$$\frac{x}{y} > \frac{yz}{y}$$ or $$\frac{x}{y} > z$$... So A is true

Dividing both sides with z

$$\frac{x}{z} > \frac{yz}{z}$$ or $$\frac{x}{z} > y$$... So B is true

Dividing both sides with yy

$$\frac{x}{yz} > \frac{yz}{yz}$$ or $$\frac{x}{yz} > 1$$... So C is true

Option D cant be true as it is a direct contradiction of the stement given in problem.
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Sandy
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Intern
Joined: 15 Sep 2017
Posts: 34
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Kudos [?]: 15 [0], given: 2

Re: If x, y and z are non-zero integers [#permalink]  09 Sep 2018, 21:13
sandy wrote:
AchyuthReddy wrote:
If x, y and z are non-zero integers, and if $$x > yz$$, then which of the following statements must be true?

Indicate all such statements.

A) $$\frac{x}{y} > z$$

B) $$\frac{x}{z} > y$$

C) $$\frac{x}{yz} > 1$$

d) $$yz < x$$

Since it is mentioned that x, y and z are positive you can divide them without flipping the inequality.

Given: $$x > yz$$

Dividing both sides with y

$$\frac{x}{y} > \frac{yz}{y}$$ or $$\frac{x}{y} > z$$... So A is true

Dividing both sides with z

$$\frac{x}{z} > \frac{yz}{z}$$ or $$\frac{x}{z} > y$$... So B is true

Dividing both sides with yy

$$\frac{x}{yz} > \frac{yz}{yz}$$ or $$\frac{x}{yz} > 1$$... So C is true

Option D cant be true as it is a direct contradiction of the stement given in problem.

Non-Zero integer means we must exclude only zero so option D is correct

Set of Non-Zero numbers{ ......-3,-2,-1,1,2,3,4....}
This reply is basing on my knowledge if I am wrong please correct me.
Intern
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Kudos [?]: 16 [0], given: 7

Re: If x, y and z are non-zero integers [#permalink]  10 Sep 2018, 08:46
We only know that the integers are non-negative. Hence, we don't know how the sign may change. Only D option is the correct one.

A: If y is negative, then it is false
B: If z is negative, then it is false
C: If x is positive and yz is negative, then it cannot be larger than 1.
Re: If x, y and z are non-zero integers   [#permalink] 10 Sep 2018, 08:46
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# If x, y and z are non-zero integers

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