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

It is currently 22 Feb 2019, 13:24

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
Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
  • Free GMAT RC Webinar

     February 23, 2019

     February 23, 2019

     07:00 AM PST

     09:00 AM PST

    Learn reading strategies that can help even non-voracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT
  • FREE Quant Workshop by e-GMAT!

     February 24, 2019

     February 24, 2019

     07:00 AM PST

     09:00 AM PST

    Get personalized insights on how to achieve your Target Quant Score.

Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos

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

Hide Tags

 
GMATH Teacher
User avatar
G
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 772
Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

Show Tags

New post 14 Sep 2018, 08:48
13
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

63% (02:25) correct 37% (01:47) wrong based on 46 sessions

HideShow timer Statistics

[GMATH practice question]

Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?

(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7

_________________

Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

GMATH Teacher
User avatar
G
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 772
Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

Show Tags

New post Updated on: 14 Sep 2018, 13:05
fskilnik wrote:
[GMATH practice question]

Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?

(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7

\(P = \left( {{x_P}\,,\,\,{y_P}} \right)\,\,\, \in \,\,\,\,\left\{ {\,\left( {x,y} \right)\,\,\,:\,\,\,{x^2} - 2x + {y^2} - 4y = 4\,} \right\}\)

\({y_P}\,\,\max \,\,\,,\,\,\,\,\,? = {x_P} + {y_P}\)


Let´s apply the "filling the squares" technique presented in our course!


\({x^2} - 2x + {y^2} - 4y = 4\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\underbrace {{x^2} - 2x + \underline 1 }_{{{\left( {x - 1} \right)}^{\,2}}} + \underbrace {{y^2} - 4y + \underline 4 }_{{{\left( {y - 2} \right)}^{\,2}}} = \underbrace {4 + \underline 1 + \underline 4 }_9\)

\(P\,\, \in \,\,\,\left\{ {\,\,\left( {x,y} \right)\,\,:\,\,\,{{\left( {x - 1} \right)}^2} + {{\left( {y - 2} \right)}^2} = {3^2}} \right\}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,P\,\, \in \,\,\,\, \odot \,\,\left\{ \begin{gathered}
\,{\text{Centre}}\, = \left( {1,2} \right) \hfill \\
{\text{Radius}} = 3 \hfill \\
\end{gathered} \right.\)

\(\left. \begin{gathered}
P = \left( {{x_P}\,,\,\,{y_P}} \right)\,\, \in \,\,\,\, \odot \,\, \hfill \\
{y_P}\,\,\max \,\, \hfill \\
\end{gathered} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{geometrically}}\,\,{\text{evident}}\,!} \,\,\,\,\,P = \left( {1,2 + 3} \right) = \left( {1,5} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 6\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________

Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net


Originally posted by fskilnik on 14 Sep 2018, 08:49.
Last edited by fskilnik on 14 Sep 2018, 13:05, edited 1 time in total.
Director
Director
avatar
P
Joined: 31 Jul 2017
Posts: 518
Location: Malaysia
Schools: INSEAD Jan '19
GMAT 1: 700 Q50 V33
GPA: 3.95
WE: Consulting (Energy and Utilities)
Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

Show Tags

New post 14 Sep 2018, 09:22
1
fskilnik wrote:
[GMATH practice question]

Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?

(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7


We can write the equation as -
\((x-1)^2 + (y-2)^2 = 9.\) As y is max, by differentiating we get -
\(2(x-1) = 0\) or \(x = 1, y =5\)

Another way -

As \((x-1)^2 + (y-2)^2 = 9.\)
y is max when y = 5, x = 1

Hence, \(x + y = 6.\)
Option C.
_________________

If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!

GMATH Teacher
User avatar
G
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 772
Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

Show Tags

New post 14 Sep 2018, 14:36
fskilnik wrote:
[GMATH practice question]

Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?

(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7


Alternate solution:

\(P = \left( {{x_P}\,,\,\,{y_P}} \right)\,\,\, \in \,\,\,\,\left\{ {\,\left( {x,y} \right)\,\,\,:\,\,\,{x^2} - 2x + {y^2} - 4y = 4\,} \right\}\)

\({y_P}\,\,\max \,\,\,,\,\,\,\,\,? = {x_P} + {y_P}\)


\({x^2} - 2x + {y^2} - 4y = 4\,\,\,\,\, \Leftrightarrow \,\,\,\,{y^2} - 4y = 4 - \left( {{x^2} - 2x + \underline 1 } \right) + \underline 1 = 5 - {\left( {x - 1} \right)^2}\)


\({y^2} - 4y = 5 - {\left( {x - 1} \right)^2} \leqslant 5\)


\(y\,\,\max \,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}
{x = {x_p} = 1} \\
{{y_p}^2 - 4{y_p} = 5}
\end{array}\begin{array}{*{20}{c}}
{} \\
{\,\,\,\mathop \Rightarrow \limits^{S = 4\,,\,P = - 5} \,\,\,\,{y_p} = \max \left\{ {5, - 1} \right\}\,\, = 5\,\,\,\,}
\end{array}} \right.\)


\(? = 1 + 5 = 6\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________

Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

Manager
Manager
avatar
B
Joined: 29 Jan 2018
Posts: 50
Concentration: Marketing, Strategy
WE: Business Development (Retail)
Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

Show Tags

New post 17 Sep 2018, 08:59
Can someone pls explain this question!
GMATH Teacher
User avatar
G
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 772
Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

Show Tags

New post 17 Sep 2018, 13:47
Ratnaa19 wrote:
Can someone pls explain this question!


Hi, Ratnaa19!

I suggest you try to understand my first solution, it is easier.

If you did not study the usual equation of a circle with given center and radius yet, I guess you should bookmark this problem for future reference.

On the other hand, if you know that equation and you want my help in any particular part of my solution, please ask in a very precise manner your doubt, so that I will be able to help you better.

Thank you for your understanding.

Regards and success in your studies,
Fabio.
_________________

Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

GMAT Club Bot
Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos   [#permalink] 17 Sep 2018, 13:47
Display posts from previous: Sort by

Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos

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


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.