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

Redeem

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.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
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
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start 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

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

general numbers

Expert replies
Source: — Problem Solving |

by srcc25anu » Fri May 31, 2013 5:41 pm
I would have gone for an option something like this:
b = 0 or a = c

Because b (a-c) = 0 that implies either b = 0 or a = c
Join the discussion

by Brent@GMATPrepNow » Fri May 31, 2013 6:13 pm
J N wrote:
For integers a, b, and c, if ab = bc, then which of the following must also be true?

A) a = c
B) a^2b=bc^2
C) ac= 1
D) abc > bc
E) a + b + c = 0
Many students will incorrectly choose A.
They'll arrive at this conclusion by taking ab = bc and dividing both sides by b to get a = c.
However, when they do this, they are forgetting that it's possible that b = 0, in which case they are unwittingly dividing both sides by zero.
Notice that one possible solution to the given equation is a=2, b=0 and c=3, in which case a does not equal b. So, we can eliminate A since we're looking for an answer choice that MUST ALWAYS be true.

Here's one way to handle the rest of the question . . .

Take ab = bc
Rearrange to get: ab - bc = 0
Factor to get: b(a - c) = 0
From this, we can conclude that b = 0 OR a - c = 0.
In other words b = 0 OR a = c

Now we'l check answer choice B.
Must it be true that a^2b = bc^2?
Yes!
Rearrange to get a^2b - bc^2 = 0
Factor: b(a^2 - c^2) = 0
Factor more: b(a + c)(a - c) = 0

Now we already know that b = 0 OR a = c
If b = 0, then b(a + c)(a - c) must equal 0
If a = c, then b(a + c)(a - c) must equal 0

Since both possible cases (b = 0 or a = c) result in answer choice B being true, the correct answer is B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion