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 →

integer!

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 407
Joined: Tue Jan 25, 2011 9:19 amFollowed by: 7 members
Thanked: 25 times
by Ozlemg » Mon Jul 04, 2011 1:08 pm
If xyz < 0 and yz > 0, which of the following must be positive?
(A) xy
(B) xz
(C) (x^2)yz
(D) x(y^2)z
(E) xy(z^2)
The more you suffer before the test, the less you will do so in the test! :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 401
Joined: Tue May 24, 2011 1:14 amFollowed by: 5 members
Thanked: 37 times

by MBA.Aspirant » Mon Jul 04, 2011 2:35 pm
Ozlemg wrote:If xyz < 0 and yz > 0, which of the following must be positive?
(A) xy
(B) xz
(C) (x^2)yz
(D) x(y^2)z
(E) xy(z^2)
xyz < 0 and yz > 0, means either x is -ve or y & z are both -ve

the only scenario when the product must be +ve, is (x^2)yz

if x is -ve, squaring will make it +ve, and yz is already +ve
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 143
Joined: Sat Nov 06, 2010 9:06 pmFollowed by: 1 member
Thanked: 10 times

by edvhou812 » Mon Jul 04, 2011 3:29 pm
xyz < 0

yz > 0

I picked x to be positive and both y and z to be negative, which makes the above statements true.

After going through the options, and seeing how each works with the rules of positive and negative integers, (x^2)yz is the only one that works because x^2 remains a positive, and yz becomes negative*negative=positive, so (x^2)yz becomes a positive*positive, which is always positive.

Answer: C
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1