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 Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 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
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget 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
Start free 5-day trial
Read 559 reviewsTarget Test Prep

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

There is an equation: x^3-ax^2+bx-5=0, a and b are positive

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 » Tue Mar 01, 2016 11:36 pm
There is an equation: x^3-ax^2+bx-5=0, a and b are positive integers, is x=1 a root of this equation?

1) b-a=4
2)x=5 is a root of this equation


* A solution will be posted in two days.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |

GMAT/MBA Expert

User avatar
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Thu Mar 03, 2016 1:34 am
Thanked: 1 times

by Dario@VinciaPrep » Thu Mar 03, 2016 2:03 am
1 alone:
The statement can be re-written as b=a+4, and then plugged into the equation.
The equation becomes: x^3-ax^2+ax+4x-5=0
If x is replaced with 1, the equation becomes: 1-a+a+4-5=0
This will work for any value of a.
Thus STATEMENT 1 ALONE IS SUFFICIENT.

2 alone:
If x=5 is a root of this equation, I can plug it back into the equation.
The equations becomes: 125-25a+5b-5=0
Which can be re-written as: 5a-b=24, or b=5a-24
Plugging this relationship into the initial equation, we get: x^3-ax^2+5ax-24x-5=0
Then we need to check if 1 is a root of this equation.
We get: 1-a+5a-24-5=0, which is true only if 4a=28, or a=7.
As 1 is not a root for any value of a, STATEMENT 2 ALONE IS NOT SUFFICIENT.
In order to show that, you can take two examples:
- a=7, b=11. In this case, 5 and 1 are both roots of the equation;
- a=5, b=1. In this case, 5 is a root, but 1 is not a root.

Thus the answer is (A).
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 » Thu Mar 03, 2016 5:15 pm
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

There is an equation: x3-ax2+bx-5=0, a and b are positive integers, is x=1 a root of this equation?

1) b-a=4
2) x=5 is a root of this equation


Modify the original condition and the question. Substitute x=1 and 1-a+b-5=0? is derived. That is, it becomes b-a=4?, which is yes and sufficient.
Thus, A is the answer.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1