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

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

a data from gmatclub

Expert replies
by diebeatsthegmat » Fri Apr 15, 2011 11:30 pm
In the xy-plane, if line k has negative slope, is the y-intercept of line k positive?

(1) The x-intercept of line k is less than the y-intercept of line k.

(2) The slope of line k is less than -2.
Join the discussion
Source: — Data Sufficiency |

deleted

by clock60 » Sat Apr 16, 2011 12:09 am
deleted
Last edited by clock60 on Sat Apr 16, 2011 11:44 am, edited 1 time in total.
Join the discussion

by srcc25anu » Sat Apr 16, 2011 3:01 am
stat 1: not sufficient because line with -ve slope and x-intercepy < y-intercept can pass through (1,2) or (-1,-2)

stat 2: slope of k < -2 still not sufficient

together also, not sufficient
IMO it should be E
Join the discussion

by force5 » Sat Apr 16, 2011 3:37 am
Good question can be done in many ways. i find this the best.....

Stmnt1 - x intercept is < y intercept. ( not since slope is negative you can make lines passing positive y axis and negative y axis.

stmnt 2 - slope of line k is < -2 (this statement alone is insufficient)

combining.

First i will do the usual way.

y= mx+c ( m is given less than -2)

y intercept is c
x intercept is c/m

its given that from stmnt 1 that c/m < c
or c/m-c<0
c(1-m/m)<0
now m is -ve (if you want take a value say m= -3)

-c4/3<0 hence c > 0

hence C

Besides this - if we analyse

slope is Y/X
since slope is -ve either of y and x are negative.
since stmnt 1 says x intercept is smaller that means y has to be positive.
Join the discussion

by clock60 » Sat Apr 16, 2011 11:42 am
hi guys again in this post, i changed my mind, and the answer here is C
k=mx+b, if m<0 is b>0
(1) if y=0 then x intercept, mx+b=0,mx=-b, x=-b/m. and it is less then b
-b/m<b.
-b/m-b<0. b/m+b>0, b(1/m+1)>0
it is possible in two cases
b>0 and (1/m+1)>0 (both are +ve)
1/m+1>0 if m<-1., so if we prove that m<-1, then b>0
the second case
b<0 and 1/m+1<0 (both are -ve)
1/m+1<0 if -1<m<0 so if we find that m>-1 then b<0
but we are not given any restrictions on m
and it comes that 1 st insuff
(2)m<-2, insuff but provides important info
together
if m<-2 then as we proved in the 1 st (if m<-1) then b>0
so both suff
the answer is C
Join the discussion

by MAAJ » Mon Apr 18, 2011 8:36 am
IMO [spoiler](C)[/spoiler], what's the OA?

1) The x-intercept of line k is less than the y-intercept of line k.

0 = mx+b
-b = mx
-b/m = x Hence -b/m < b

-b/m - b < 0
(-b-mb)/m < 0
b(-1-m)/m < 0
Insufficient

(2) The slope of line k is less than -2
Insufficient

(3) Combined:
If b(-1-m)/m < 0 and m < -2 Then (-1-m) is always positive, so:
b(positive)/(negative) < 0
Thus, b must be positive
"There's a difference between interest and commitment. When you're interested in doing something, you do it only when circumstance permit. When you're committed to something, you accept no excuses, only results."
Join the discussion