GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Feb 2019, 23:58

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT Prep Hour

February 20, 2019

February 20, 2019

08:00 PM EST

09:00 PM EST

Strategies and techniques for approaching featured GMAT topics. Wednesday, February 20th at 8 PM EST

February 21, 2019

February 21, 2019

10:00 PM PST

11:00 PM PST

Kick off your 2019 GMAT prep with a free 7-day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.

# If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following

Author Message
TAGS:

### Hide Tags

Director
Joined: 11 Feb 2015
Posts: 717
If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following  [#permalink]

### Show Tags

25 Jan 2019, 08:37
1
1
00:00

Difficulty:

55% (hard)

Question Stats:

57% (02:11) correct 43% (02:29) wrong based on 48 sessions

### HideShow timer Statistics

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

_________________

"Please hit +1 Kudos if you like this post"

_________________
Manish

"Only I can change my life. No one can do it for me"

VP
Joined: 09 Mar 2018
Posts: 1000
Location: India
Re: If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following  [#permalink]

### Show Tags

25 Jan 2019, 09:16
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
_________________

If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Manhattan Prep Instructor
Joined: 04 Dec 2015
Posts: 689
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]

### Show Tags

26 Jan 2019, 14:18
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

My latest GMAT blog posts | Suggestions for blog articles are always welcome!

Re: If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following   [#permalink] 26 Jan 2019, 14:18
Display posts from previous: Sort by