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

It is currently 15 Oct 2019, 02:03

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

M14-10

  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: 58344
M14-10  [#permalink]

Show Tags

New post 16 Sep 2014, 00:52
1
6
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

46% (02:04) correct 54% (02:05) wrong based on 56 sessions

HideShow timer Statistics

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58344
Re M14-10  [#permalink]

Show Tags

New post 16 Sep 2014, 00:53
5
Official Solution:


Given two points \(A(-2, 2)\) and \(B(3, 2)\). Question asks to find the area of triangle \(ABC\), where \(C(x,y)\). Look at the diagram below:

Image

(1) \(|y-2|=1\). Either \(y=3\) or \(y=1\), hence vertex \(C\) could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the area of \(ABC\) will be the same: \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point \(C\) (see two possible locations of \(C\): \(C_1\) and \(C_2\), the heights of \(ABC_1\) and \(ABC_2\) are the same and equal to 1). So, we have that \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) Angle at the vertex \((x, y)\) equals 90 degrees. This statement says that \(ABC\) is a right triangle with hypotenuse \(AB\): consider \(AB\) to be diameter of a circle. In this case \(C\) could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of \(ABC\) will be different for different location of point \(C\), resulting the different areas (see two possible locations of \(C\): \(C_3\) and \(C_4\), heights of \(ABC_3\) and \(ABC_4\) are different). Not sufficient.


Answer: A
_________________
Intern
Intern
avatar
Joined: 02 Feb 2016
Posts: 2
Re M14-10  [#permalink]

Show Tags

New post 30 Jun 2016, 01:47
I think this the explanation isn't clear enough, please elaborate. Please explain on this
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58344
Re: M14-10  [#permalink]

Show Tags

New post 30 Jun 2016, 08:27
Manager
Manager
User avatar
Status: Keep it simple stupid!!
Joined: 02 Mar 2016
Posts: 72
Location: India
Concentration: General Management, Operations
GMAT 1: 680 Q47 V36
GPA: 3.37
Re: M14-10  [#permalink]

Show Tags

New post 18 Jul 2016, 22:59
I think this is an awesome question and Explanation is superb...!!wow!!
_________________
"Give without remembering, take without forgetting"
- Kudos
Intern
Intern
avatar
Joined: 29 Jun 2016
Posts: 11
Re: M14-10  [#permalink]

Show Tags

New post 17 Aug 2016, 23:17
A really good question after a long time
Intern
Intern
avatar
B
Joined: 13 Oct 2015
Posts: 16
Re M14-10  [#permalink]

Show Tags

New post 16 Sep 2016, 16:06
I think this is a high-quality question and I agree with explanation.
Intern
Intern
avatar
B
Joined: 28 Mar 2017
Posts: 9
GMAT 1: 550 Q43 V23
Re: M14-10  [#permalink]

Show Tags

New post 12 Feb 2018, 08:48
Very good question. Keep the good work Bunuel.. Thanks for such lovely questions
Manager
Manager
avatar
G
Joined: 31 Jul 2017
Posts: 88
Location: India
Schools: Anderson '21, LBS '21
GMAT ToolKit User Premium Member
Re: M14-10  [#permalink]

Show Tags

New post 27 Mar 2019, 02:37
This question needs to be discussed more often.. Just commenting for the same... An awesome question
GMAT Club Bot
Re: M14-10   [#permalink] 27 Mar 2019, 02:37
Display posts from previous: Sort by

M14-10

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





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