Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Which MUST BE true? please help!!

Expert replies
by gmatpup » Sat Dec 10, 2011 4:56 pm
If xy+z = x(y+z), which MUST BE true?

A. x=0 and z=0
B. x=1 and y=1
C. y=1 and z=0
D. x=1 or y=0
E. x=1 or z=0


Please provide an explanation, I am confused :( Thank you!!

Answer is E
Join the discussion
Source: — Problem Solving |

by GmatMathPro » Sat Dec 10, 2011 5:01 pm
gmatpup wrote:If xy+z = x(y+z), which MUST BE true?

A. x=0 and z=0
B. x=1 and y=1
C. y=1 and z=0
D. x=1 or y=0
E. x=1 or z=0


Please provide an explanation, I am confused :( Thank you!!
Answer is E
xy+z=x(y+z)

xy+z=xy+xz (distribute the x)

z=xz (subtract xy from both sides)

xz-z=0 (subtract z from both sides)

z(x-1)=0 factor out a z

z=0 or x-1=0 (set both factors equal to zero)

z=0 or x=1 solve
Pete Ackley
GMAT Math Pro
Free Online Tutoring Trial
Join the discussion

by Anurag@Gurome » Sun Dec 11, 2011 4:08 am
gmatpup wrote:If xy+z = x(y+z), which MUST BE true?
--> (xy + z) = x(y + z)
--> (xy + z) = (xy + xz)
--> z = xz
--> (xz - z) = 0
--> z(x - 1) = 0

Hence, either z = 0 or x = 1.

The correct answer is E.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Join the discussion

by LalaB » Sun Dec 11, 2011 10:27 am
@Anurag@Gurome and GmatMathPro

I have got the same answer choice. But ...looking closely to answer choices, I have suddenly thought -why B must be wrong?
we have xy+z = x(y+z). if x=1 and y=1, then 1*1+z=1(1+z)

whats wrong in my logic?
Join the discussion

by kanwar86 » Sun Dec 11, 2011 11:01 am
LalaB wrote:@Anurag@Gurome and GmatMathPro

I have got the same answer choice. But ...looking closely to answer choices, I have suddenly thought -why B must be wrong?
we have xy+z = x(y+z). if x=1 and y=1, then 1*1+z=1(1+z)

whats wrong in my logic?
There is nothing wrong with your logic. It's just that option B is a subset of E.
Option E is if (x=1 or z=0), the given equation is valid.
That is, any value of y (as long as x=1) satisfies the given equation.
Option B, x=1 and y=1, is a particular case of E.
Regards

Kanwar

"In case my post helped, do care to thank. Happy learning :)"
Join the discussion