BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Is x/y < 0?

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members
Location: Las Vegas, USA
Thanked: 19 times
by Max@Math Revolution » Wed Jan 02, 2019 11:12 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

[Math Revolution GMAT math practice question]

Is x/y < 0?

1) |x+y| < |x|+|y|
2) x+y < 0
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members
Thanked: 59 times

by fskilnik@GMATH » Thu Jan 03, 2019 3:41 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is x/y < 0?

1) |x+y| < |x|+|y|
2) x+y < 0
$${x \over y}\,\,\mathop < \limits^? \,\,0\,\,\,\,\left( {y \ne 0} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,xy\,\,\mathop < \limits^? \,\,0\,\,\,\,,\,\,\,{\rm{with}}\,\,y \ne 0$$
$$\left( 1 \right)\,\,\,\,\left| {x + y} \right|\,\,\, < \,\,\,\left| x \right| + \left| y \right|\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,xy < 0\,\,\,\,\,\,\,\left( {{\rm{SUFF}}.} \right)$$
$$\left( 2 \right)\,\,\,\,x + y < 0\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0, - 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\,\,\,\,\,\,\left( {{\rm{INSUFF}}.} \right)$$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Thu Jan 03, 2019 7:50 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is x/y < 0?

1) |x+y| < |x|+|y|
2) x+y < 0
Question stem, rephrased:
Do x and y have different signs?

Statement 1: |x+y| < |x|+|y|
Since an absolute value cannot be negative, both sides here are NONNEGATIVE, enabling us to safely square the inequality:
(|x+y|)²< (|x|+|y|)²
x² + y² + 2xy < x² + y² + 2|x||y|
xy < |xy|
The resulting inequality is valid only if x and y have DIFFERENT SIGNS.
SUFFICIENT.

Statement 2: x+y < 0
If x=-1 and y=-1, then x and y have the same sign.
If x=-10 and y=1, then x and y have different signs.
INSUFFICIENT.

The correct answer is A.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members
Location: Las Vegas, USA
Thanked: 19 times

by Max@Math Revolution » Sun Jan 06, 2019 5:04 pm
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The question is equivalent to asking if xy < 0. This can be seen by multiplying both sides of the inequality by y^2.

Condition 1) is equivalent to xy < 0 as shown below:
|x+y|<|x|+|y|
=> |x+y|^2 < (|x|+|y|)^2
=> (x+y)^2 < |x|2+2|x||y|+|y|^2
=> x^2+2xy+y^2 < x^2+2|xy|+y^2
=> 2xy < 2|xy|
=> xy < |xy|
=> xy < 0
Thus, condition 1) is sufficient.

Condition 2)
If x = -2 and y = 1, then the answer is 'yes'.
If x = -1 and y = -1, then the answer is 'no'.
Since it does not give a unique answer, condition 2) is not sufficient.

Therefore, the correct answer is A.
Answer: A
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1