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.

The first and second numbers in a sequence of numbers are plotted as the \(x\) and \(y\) coordinates, respectively, of a

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 2276
Joined: Sat Oct 14, 2017 6:10 amFollowed by: 3 members
by VJesus12 » Thu Jul 30, 2020 8:48 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

The first and second numbers in a sequence of numbers are plotted as the \(x\) and \(y\) coordinates, respectively, of a point on the coordinate plane, as are the third and fourth numbers, and all subsequent pairs in the sequence and a line is formed by connecting these points, what would be the slope of the line?

(1) Except for the first number, the sequence of numbers is formed by doubling the previous number and then subtracting 1.
(2) The first number in the sequence is 3.

[spoiler]OA=A[/spoiler]

Source: Veritas Prep
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
Legendary Member
Posts: 2214
Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

Re: The first and second numbers in a sequence of numbers are plotted as the \(x\) and \(y\) coordinates, respectively,

by deloitte247 » Sat Aug 01, 2020 12:04 pm
Let the first number = x, then we have (x, x+1) coordinate, (x+2, x+3) coordinates.
Target question=> A line is formed by connecting the coordinate, what would be the slope of the line?
$$Slope=\frac{\triangle y}{\triangle x}=\frac{y2-y1}{x2-x1}$$
Statement 1: Except for the first number, the sequence of numbers is formed by doubling the previous number and then subtracting 1.
Therefore, considering the first 4 numbers for (x1,y1) and (x2,y2) coordinates.
[x, (2x-1)] and [(4x-3), (8x-7)]
x1=x, x2=4x-3, y1=2x-1, y2=8x-7
$$Therefore,\ slope\ \left(m\right)=\frac{\left(8x-7\right)-\left(2x-1\right)}{\left(4x-3\right)-x}$$ $$=\frac{8x-7-2x+1}{4x-3-x}=\frac{6x-6}{3x-3}$$
$$=\frac{2\left(3x-3\right)}{3x-3}=2$$
Statement 1 is SUFFICIENT

Statement 2: The first number in the sequence is 3.
This does not tell us how the sequence is formed and we do not know if it is consecutive or not. Statement 2 is NOT SUFFICIENT.
Since only statement 1 is SUFFICIENT, then the correct answer is option A.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1