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

 It is currently 18 Jun 2019, 18:08 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

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

Author Message
TAGS:

Hide Tags

GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 936
Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

Show Tags

13 00:00

Difficulty:   55% (hard)

Question Stats: 61% (02:25) correct 39% (01:41) wrong based on 47 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 P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 936
Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

Show Tags

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, 09:49.
Last edited by fskilnik on 14 Sep 2018, 14:05, edited 1 time in total.
Director  P
Joined: 31 Jul 2017
Posts: 516
Location: Malaysia
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

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 P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 936
Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

Show Tags

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  B
Joined: 29 Jan 2018
Posts: 54
Concentration: Marketing, Strategy
Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

Show Tags

Can someone pls explain this question!
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 936
Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

Show Tags

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.

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 Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos   [#permalink] 17 Sep 2018, 14:47
Display posts from previous: Sort by

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