GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 11 Dec 2019, 23:28

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59675
If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin  [#permalink]

Show Tags

New post 06 Sep 2019, 04:36
1
1
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

83% (01:25) correct 17% (01:59) wrong based on 58 sessions

HideShow timer Statistics

Manager
Manager
avatar
B
Joined: 18 Sep 2018
Posts: 64
Re: If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin  [#permalink]

Show Tags

New post 06 Sep 2019, 05:09
IMO D

A point on the line satisfies the equation. So substitute the point (R, R^2+1) into the equation.

We get R^2-2R-3 = 0.

Solving the above equation, we get R = 3 or -1.

From the options we have only 3.

Posted from my mobile device
GMAT Club Legend
GMAT Club Legend
User avatar
V
Joined: 18 Aug 2017
Posts: 5474
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
GMAT ToolKit User Premium Member
Re: If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin  [#permalink]

Show Tags

New post 06 Sep 2019, 11:14
Bunuel wrote:
If (R, R^2 + 1) is the (x, y) coordinate of a point located on the line: y = 2x + 4, what can be the value of R?

A. –3.
B. 2.
C. 4.
D. 3.
E. 1.


equate for given x,y values
we get
r=3 & -1
IMO D; 3
Intern
Intern
avatar
B
Joined: 21 Nov 2018
Posts: 14
Location: Nigeria
Concentration: Finance, Entrepreneurship
GPA: 4
WE: Analyst (Investment Banking)
Re: If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin  [#permalink]

Show Tags

New post 08 Sep 2019, 22:43
I'm not clear on the above solutions as the value i derived for R are -3 and 1 as opposed to the 3 and -1 derived by other users. here are my workings;
[quote][/quote]

y = 2x+4
R^2+1 = 2R+4
R^2 - 2R +1-4 = 0
R^2 - 2R -3 = 0

Thus R is -3 and 1 as only this provides the current sum of -2 and product of -3.

the other answers of 3 and - 1 result in a sum of 2 which does not agree with the equation.

Please clarify Bunuel ... kindly point out my error if any
Intern
Intern
avatar
B
Joined: 18 Jul 2019
Posts: 23
Re: If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin  [#permalink]

Show Tags

New post 24 Nov 2019, 01:09
OmotayoH wrote:
I'm not clear on the above solutions as the value i derived for R are -3 and 1 as opposed to the 3 and -1 derived by other users. here are my workings;
Quote:


y = 2x+4
R^2+1 = 2R+4
R^2 - 2R +1-4 = 0
R^2 - 2R -3 = 0

Thus R is -3 and 1 as only this provides the current sum of -2 and product of -3.

the other answers of 3 and - 1 result in a sum of 2 which does not agree with the equation.

Please clarify Bunuel ... kindly point out my error if any


R^2 - 2R -3 = 0
R^2-3R+R-3=0
R(R-3) + 1( R-3) = 0
(R-3)(R+1)=0
R=-3 /-1

Hope that helps :)
GMAT Club Bot
Re: If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin   [#permalink] 24 Nov 2019, 01:09
Display posts from previous: Sort by

If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne