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Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following

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Director  V
Joined: 12 Feb 2015
Posts: 917
If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 55% (02:13) correct 45% (02:28) wrong based on 98 sessions

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If xyz ≠ 0, xy > 0, and x^2 + xy + xz < 0, then which of the following must be true?

I. x(y + z) < 0

II. x + y + z < 0

III. If x < 0, then z > 0.

A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II, and III

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Manish "Only I can change my life. No one can do it for me"
Manhattan Prep Instructor G
Joined: 04 Dec 2015
Posts: 832
GMAT 1: 790 Q51 V49 GRE 1: Q170 V170 Re: If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following  [#permalink]

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CAMANISHPARMAR wrote:
If xyz ≠ 0, xy > 0, and x^2 + xy + xz < 0, then which of the following must be true?

I. x(y + z) < 0

II. x + y + z < 0

III. If x < 0, then z > 0.

A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II, and III

Fun problem! The key with these is to do as much work as you can using logic and simplification, before you start testing cases. That's especially true on this problem, because there are three variables. If there are more than two variables in a problem, I usually try to avoid testing cases as much as possible - it just takes too long to consider all of the possibilities.

The first thing I noticed is that x and y have the same sign.

I also noticed that the inequality, $$x^2 + xy + xz < 0$$, can be simplified like this: $$x(x+y+z) < 0$$. This means that either x is negative, or x+y+z is negative, but not both.

Now, go look at options I, II, and III.

Option II stood out to me the most, because I already know something about x+y+z from my reasoning above. It doesn't HAVE to be negative, since the alternative is that x could just be negative. Then, if y and z were big enough, x+y+z would be positive. Eliminate B, D, and E, because they include option II.

This would be a good point to guess! Or, we could keep going if we have enough time.

Notice that the difference between (A) and (C) is that (A) doesn't include III, while (C) does. They both include I! So, at this point, we know that I has to be true without testing it. We only have to worry about III. That's a good reason to use the process of elimination!

III says that if x < 0, z > 0. Well, if x < 0, from the above reasoning, we know that x + y + z > 0. We also know that x and y have the same sign. So, here's what we have:

If x is negative, y is also negative, but x + y + z is positive.

This should only happen if z is positive! You can't add three negative numbers together and get a positive result. So, z definitely has to be positive. III is true.

Eliminate (A) and choose (C).
_________________ Chelsey Cooley | Manhattan Prep | Seattle and Online

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Director  G
Joined: 09 Mar 2018
Posts: 994
Location: India
Re: If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following  [#permalink]

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1
CAMANISHPARMAR wrote:
If xyz ≠ 0, xy > 0, and x^2 + xy + xz < 0, then which of the following must be true?

I. x(y + z) < 0

II. x + y + z < 0

III. If x < 0, then z > 0.

A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II, and III

IMO C

xy > 0, and x^2 + xy + xz < 0

xy > 0, and x (x + y + z) < 0

Some cases before marking the options

Case 1, x = -ive and y = -ive & For x (x+y+z) < 0, z > 0 and z > x + y

Case 2, x = + ive and y = + ive & For x (x+y+z) < 0, z > -(x + y)

III. If x < 0, then z > 0
This will be definitely true, case 1 holds good.

I. x(y + z) < 0
True when it is Case 1, True when it is Case 2

II. x + y + z < 0
For case 1, this is not true

For case 2, this is true
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Intern  B
Joined: 12 Jan 2019
Posts: 34
If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following  [#permalink]

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Can someone please simplify why it cannot be E

Aren't I II III independent statements?
If yes can we make use of the 3 of them to imply whether x is negative or no? If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following   [#permalink] 31 Jul 2019, 06:29
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If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following

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