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Senior Manager  G
Joined: 10 Jan 2013
Posts: 269
Location: India
Concentration: General Management, Strategy
GPA: 3.95
If x^3y<0, and y/z>0, then which of the following must be less than 1?  [#permalink]

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5 00:00

Difficulty:   45% (medium)

Question Stats: 61% (02:32) correct 39% (02:17) wrong based on 44 sessions

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If $$x^3y<0$$, and $$y/z>0$$, then which of the following must be less than 1?

A. $$\sqrt{x}$$

B. $$y/(x)^2$$

C. $$x^3z^4$$

D. $$x^2yz$$

E. $$xy^2z^3$$
Intern  B
Joined: 26 Apr 2019
Posts: 18
Re: If x^3y<0, and y/z>0, then which of the following must be less than 1?  [#permalink]

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2
as z and y will have same sign , x and y with opposite sign so only e option will have negative soln in all condition and is less than 1 in all cases.
Senior Manager  G
Joined: 10 Jan 2013
Posts: 269
Location: India
Concentration: General Management, Strategy
GPA: 3.95
Re: If x^3y<0, and y/z>0, then which of the following must be less than 1?  [#permalink]

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saurabh9gupta wrote:
If $$x^3y<0$$, and $$y/z>0$$, then which of the following must be less than 1?

A. $$\sqrt{x}$$

B. $$y/(x)^2$$

C. $$x^3z^4$$

D. $$x^2yz$$

E. $$xy^2z^3$$

Bunuel chetan2u - your approach please
Rice (Jones) School Moderator G
Joined: 18 Jun 2018
Posts: 261
Location: United States (AZ)
Concentration: Finance, Healthcare
GMAT 1: 600 Q44 V28 GPA: 3.36
Re: If x^3y<0, and y/z>0, then which of the following must be less than 1?  [#permalink]

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Alternate approach:

Given: $$x^{3}$$y < 0 and $$\frac{y}{z}$$>0

Let x = -1, y = 1, and z = 2 ==> test these numbers in the answer choices and you can quickly eliminate B (1) and D (2). A (-1), C (-16) & E (-8) are < 1 so keep them and pick new set of numbers.

Let x =1, y = -1, and z = -2 ==> test these numbers in the remaining answer choices and you will get A(1), C (16), and E (-8 which is < 1), so E is the answer.
Senior Manager  P
Joined: 16 Jan 2019
Posts: 426
Location: India
Concentration: General Management
WE: Sales (Other)
Re: If x^3y<0, and y/z>0, then which of the following must be less than 1?  [#permalink]

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$$x^3y<0$$

$$x^3$$ and $$x$$ have the same sign so $$xy<0$$, which means $$x$$ and $$y$$ have opposite signs

$$\frac{y}{z}>0$$, which means $$y$$ and $$z$$ have the same sign

Therefore $$x$$ and $$z$$ have opposite signs and so $$xz<0$$ and $$\frac{x}{z}<0$$

Note that at this point we do not know the exact sign of $$x$$, $$y$$ or $$z$$ but we do know how the sign of one affects the other

A. $$\sqrt{x}$$

$$\sqrt{x}$$ and $$x$$ have the same sign but we don't know what that sign is

B. $$\frac{y}{x^2}$$

It is possible that $$x$$ is negative and $$y$$ is positive in which case $$\frac{y}{x^2}$$ is positive

C. $$x^3z^4$$

It is possible that $$x$$ is positive and $$z$$ is negative in which case $$x^3z^4$$ is positive

D. $$x^2yz$$

Since $$y$$ and $$z$$ have same sign $$yz>0$$ and $$x^2>0$$, so $$x^2yz>0$$

E. $$xy^2z^3$$

$$xy^2z^3$$ = $$(xz)(yz)^2$$

$$(yz)^2$$ is always positive and as we established above $$xz$$ is negative

This means $$(xz)(yz)^2$$ = $$xy^2z^3$$ is negative and therefore less than 1

Answer is (E)

Hit Kudos if this helped!
Manager  B
Joined: 01 Nov 2018
Posts: 81
GMAT 1: 690 Q48 V35 GPA: 3.88
Re: If x^3y<0, and y/z>0, then which of the following must be less than 1?  [#permalink]

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Picking numbers here is a bit tedious. Just list out the 2 possible cases.
Case 1: X=pos Y=neg Z=neg
Case 2: X=neg Y=pos Z=pos

Just putting the signs in choice E shows that if X is positive, Z will be negative and hence the answer will be negative (<1)
Likewise, if X is negative, the entire expression will be negative and hence <1. Re: If x^3y<0, and y/z>0, then which of the following must be less than 1?   [#permalink] 15 Aug 2019, 17:25
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# If x^3y<0, and y/z>0, then which of the following must be less than 1?

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