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
GMATBootcamp Starts Sep 21
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE BOOTCAMP

Live Online Bootcamp Class with Top GMAT Expert Chris Peckover

Sep 21 to Oct 9, 2026

Schedule
Mon to Fri · 7:00 to 10:00 PM ET
Included
Live classes + 6 months of TTP OnDemand
  • Boost your GMAT score in less than one month in a live online class
  • 6 months access to TTP OnDemand video courses included
View bootcamp & 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.

Please explain

Expert replies
by ketkoag » Tue Mar 24, 2009 5:15 am
What is the greatest common factor of the positive integers j and k?
(1) k = j + 1
(2) jk is divisible by 5

OA d??

17. If x ≠ -1, which is greater, 1/(x+1) or x/2 ?
(1) x ≥ 0
(2) x < 3

Is the answer e?? Please explain
Join the discussion
Source: — Data Sufficiency |

by DanaJ » Tue Mar 24, 2009 5:27 am
Q1. 1 is sufficient to answer the question at hand, since there is a general rule that the greatest common factor of two consecutive numbers is 1.

2 however is not enough, IMHO. Think of it this way:
a. j = 4 and k = 5 - in this case, jk = 20 and is divisible by 5, but the greatest common factor of j and k is 1.
b. j = 10 and k = 5 - now you get that jk = 50, which is again divisible by 5. However, the greatest common factor is 5.
This is why I believe that 1 is insufficient, so my answer is A. I may be missing smth here...


Q2. 1. I'd use picking numbers for this one.
a. take x = 0 and you get that 1/(x+1) = 1 and x/2 = 0, making 1/(x+1) > x/2
b. take x = 2 and you have 1/(x+1) = 1/3 = 0.33 and x/2 = 1. In this case, 1/(x+1) < x/2.
Since both cases comply with the rule that x is greater than equal to zero, 1 is insufficient.

2. The examples at 1 can be used to demonstrate that 2 is insufficient as well.

Put both stmts together and you get nothing: again, the examples at 1 are consistent with 0 <= x < 3. IMHO, it's E.
Join the discussion

by karmayogi » Tue Mar 24, 2009 5:41 am
DanaJ wrote:Q1. 1 is sufficient to answer the question at hand, since there is a general rule that the greatest common factor of two consecutive numbers is 1.

2 however is not enough, IMHO. Think of it this way:
a. j = 4 and k = 5 - in this case, jk = 20 and is divisible by 5, but the greatest common factor of j and k is 1.
b. j = 10 and k = 5 - now you get that jk = 50, which is again divisible by 5. However, the greatest common factor is 5.
This is why I believe that 1 is insufficient, so my answer is A. I may be missing smth here...
You are right. 2 statement is in-sufficient.
Each soul is potentially divine. The goal is to manifest this divine within.
--By Swami Vivekananda
Join the discussion

by sanjay_dce » Tue Mar 24, 2009 9:23 am
DanaJ wrote:Q1. 1 is sufficient to answer the question at hand, since there is a general rule that the greatest common factor of two consecutive numbers is 1.

2 however is not enough, IMHO. Think of it this way:
a. j = 4 and k = 5 - in this case, jk = 20 and is divisible by 5, but the greatest common factor of j and k is 1.
b. j = 10 and k = 5 - now you get that jk = 50, which is again divisible by 5. However, the greatest common factor is 5.
This is why I believe that 1 is insufficient, so my answer is A. I may be missing smth here...


Q2. 1. I'd use picking numbers for this one.
a. take x = 0 and you get that 1/(x+1) = 1 and x/2 = 0, making 1/(x+1) > x/2
b. take x = 2 and you have 1/(x+1) = 1/3 = 0.33 and x/2 = 1. In this case, 1/(x+1) < x/2.
Since both cases comply with the rule that x is greater than equal to zero, 1 is insufficient.

2. The examples at 1 can be used to demonstrate that 2 is insufficient as well.

Put both stmts together and you get nothing: again, the examples at 1 are consistent with 0 <= x < 3. IMHO, it's E.
I Think DanaJ has answered both the questions correctly
Join the discussion