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
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
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.

geometry doubt

Expert replies
Source: — Problem Solving |

by HSPA » Mon Mar 21, 2011 9:42 am
if y is the angle between a line and X-axis then tan y is the line's slope... if y is more tan y is more
Steeper means more y so slope = tan y = more

Apart from above problem:
Do you have a question or doubt :)
An unclear question is not a doubt
Join the discussion

by gmatapril » Mon Mar 21, 2011 9:45 am
HSPA wrote:if y is the angle between a line and X-axis then tan y is the line's slope... if y is more tan y is more
Steeper means more y so slope = tan y = more

Apart from above problem:
Do you have a question or doubt :)
An unclear question is not a doubt
Lines n and p lie on the xy plane. Is the slope of line n less than slope of line p
(1) Lines n and p intersect at (5, 1)
(2) The y-intercept of line n is greater than the y intercept of p


Statement 1 tells us where the lines intersect, but tells us nothing about either of the lines' slopes. If you draw a picture of two lines intersecting at (5,1), you can see that with the information given, we could label either one n. We could make the one with greater slope n, or the one with less slope n. So, we don't have enough information from statement 1.

Now let's consider statement 2, that says that the y intercept of n is greater than that of the y intercept of p. The slopes could be unequal (think about intersecting lines), or we could have parallel lines, in which case the slopes are equal. So, statement 2 is not sufficient on its own.

Now let's consider the statements together. The lines intersect at (5,1), and n has the higher y intercept. Let's look at 3 cases:

1) Both have y intercepts above y=1
Since n intersects higher, then we know n had further to descend, so its slope is steeper (but more negative) than p's. Thus, p has a greater slope.

2) n has intercept above y=1, p has intercept below
n would have a negative slope and p a positive, so p has a greater slope

3) Both have y intercepts below y=1
Both have positive slopes, but p has further to ascend. Thus, p has a greater slope.
Join the discussion

by HSPA » Mon Mar 21, 2011 9:53 am
The answer is C

Let y= m1x+c1 and y= m2x+C2 be the two lines

a) they meet at point (5,1) so both must satisfy this point
5m1+C1 = 1
5m2+C2 = 1
b) C1 > C2

if the above two equations has to be equal to 1 with C1>C2 then m2>m1 is a must

Hope this helps
Join the discussion

by gmatapril » Mon Mar 21, 2011 10:22 am
HSPA wrote:The answer is C

Let y= m1x+c1 and y= m2x+C2 be the two lines

a) they meet at point (5,1) so both must satisfy this point
5m1+C1 = 1
5m2+C2 = 1
b) C1 > C2

if the above two equations has to be equal to 1 with C1>C2 then m2>m1 is a must

Hope this helps
how have you proved this " with C1>C2 THEN M2>M1"
PLEASE EXPLAIN
Join the discussion

by HSPA » Mon Mar 21, 2011 9:58 pm
gmatapril wrote:
HSPA wrote:The answer is C

Let y= m1x+c1 and y= m2x+C2 be the two lines

a) they meet at point (5,1) so both must satisfy this point
5m1+C1 = 1
5m2+C2 = 1
b) C1 > C2

if the above two equations has to be equal to 1 with C1>C2 then m2>m1 is a must

Hope this helps
how have you proved this " with C1>C2 THEN M2>M1"
PLEASE EXPLAIN
Let C1 > C2 be C1= -4, C2= -5
Using a) m1= (1-C1)/5 = (1+4)/5 = 1 ; m2= (1-c2)/5 = 6/5 = 1.2 => m2>m1 as 1.2 > 1; Hence if C1 is > C2 => m2 is > m1
Join the discussion