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

It is currently 17 Oct 2019, 18:31

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

M05-13

  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: 58402
M05-13  [#permalink]

Show Tags

New post 16 Sep 2014, 00:25
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

71% (01:21) correct 29% (01:27) wrong based on 171 sessions

HideShow timer Statistics

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
Re M05-13  [#permalink]

Show Tags

New post 16 Sep 2014, 00:25
Official Solution:


Notice that since we have two distinct points of line K, then we can find its equation.

(1) Coordinates of Point C are (5, -1). We know the equation of line K, hence we can find whether it passes through some particular point. Sufficient.

(2) Point C is equidistant from point A and point B. Point C may be the midpoint of the line segment AB, so on line K. But point C can also be anywhere on the line which is perpendicular to line K and passes through that midpoint. Not sufficient.


Answer: A
_________________
Intern
Intern
avatar
S
Status: in process
Joined: 18 Jun 2017
Posts: 30
Location: Uzbekistan
Concentration: Operations, Leadership
Schools: Babson '21
GMAT 1: 690 Q47 V37
GPA: 4
WE: Education (Education)
GMAT ToolKit User
Re: M05-13  [#permalink]

Show Tags

New post 16 Dec 2017, 14:00
1
Bunuel
Can you please tell me if we always assume that "the line" is straight?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
Re: M05-13  [#permalink]

Show Tags

New post 16 Dec 2017, 14:10
1
SVP
SVP
User avatar
V
Joined: 26 Mar 2013
Posts: 2345
Reviews Badge CAT Tests
Re: M05-13  [#permalink]

Show Tags

New post 17 Dec 2017, 08:44
Bunuel wrote:
Official Solution:


Notice that since we have two distinct points of line K, then we can find its equation.

(1) Coordinates of Point C are (5, -1). We know the equation of line K, hence we can find whether it passes through some particular point. Sufficient.

(2) Point C is equidistant from point A and point B. Point C may be the midpoint of the line segment AB, so on line K. But point C can also be anywhere on the line which is perpendicular to line K and passes through that midpoint. Not sufficient.


Answer: A


Bunuel

Can you expain how the highlited part can be the case?

Thanks
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
Re: M05-13  [#permalink]

Show Tags

New post 17 Dec 2017, 09:20
1
Mo2men wrote:
Bunuel wrote:
Official Solution:


Notice that since we have two distinct points of line K, then we can find its equation.

(1) Coordinates of Point C are (5, -1). We know the equation of line K, hence we can find whether it passes through some particular point. Sufficient.

(2) Point C is equidistant from point A and point B. Point C may be the midpoint of the line segment AB, so on line K. But point C can also be anywhere on the line which is perpendicular to line K and passes through that midpoint. Not sufficient.


Answer: A



Bunuel

Can you expain how the highlited part can be the case?

Thanks


Check the image below:

Image

Point (5, -1), red point, is the midpoint of (6, -7) and (4, 5). Red line is passing through that midpoint and is perpendicular to the blue line. Any point on the red line will be equidistant from (6, -7) and (4, 5).

Attachment:
Untitled.png

>> !!!

You do not have the required permissions to view the files attached to this post.


_________________
Manager
Manager
avatar
G
Joined: 26 Dec 2017
Posts: 149
Reviews Badge
Re: M05-13  [#permalink]

Show Tags

New post 14 Jan 2018, 01:15
1
Hi Bunnel,
I might have a perpendicular line passing through C but C is on the line K right which in question we need to say whether k pass through C. In the above graph which you posted the point C is on line K. pls explain.
_________________
--If you like my post pls give kudos
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
Re: M05-13  [#permalink]

Show Tags

New post 14 Jan 2018, 01:34
tejyr wrote:
Hi Bunnel,
I might have a perpendicular line passing through C but C is on the line K right which in question we need to say whether k pass through C. In the above graph which you posted the point C is on line K. pls explain.


Any point on red line is equidistant from point A and point B, so C could be anywhere on that red line. If it's on the intersection (red dot) then it'll be on blue line to but if it's anywhere else on red line then it won't be on blue line.
_________________
Manager
Manager
avatar
G
Joined: 26 Dec 2017
Posts: 149
Reviews Badge
Re: M05-13  [#permalink]

Show Tags

New post 14 Jan 2018, 07:18
I was unable to understand how any point on the red line is equidistant from A and B.
For instance I took a point on the red perpendicular line (0,-11/6).
when i calculate the distance between (0,-11/6) and A
;(0,-11/6) and B are not equal.
_________________
--If you like my post pls give kudos
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
Re: M05-13  [#permalink]

Show Tags

New post 14 Jan 2018, 07:49
tejyr wrote:
I was unable to understand how any point on the red line is equidistant from A and B.
For instance I took a point on the red perpendicular line (0,-11/6).
when i calculate the distance between (0,-11/6) and A
;(0,-11/6) and B are not equal.


The distance MUST be the same. The distance between (0, -11/6) and (4, 5) as well as the distance between (0, -11/6) and (6, -7) is \(\frac{\sqrt{2257}}{6}\).

I'll try to explain in another way. Consider points A = (0, 1) and B = (0, -1). Perpendicular bisector of AB is x-axis. Obviously any point on x-axis is equidistant from A and B.

The same way, any point on perpendicular bisector, of segment AB in our question, will be equidistant from A and B.
_________________
Manager
Manager
avatar
G
Joined: 26 Dec 2017
Posts: 149
Reviews Badge
Re: M05-13  [#permalink]

Show Tags

New post 15 Jan 2018, 21:43
My bad any point in red line is equidistant from A,b. Tq for explanation.
_________________
--If you like my post pls give kudos
Manager
Manager
User avatar
S
Joined: 25 May 2019
Posts: 68
Re: M05-13  [#permalink]

Show Tags

New post 18 Sep 2019, 02:53
I was totally OK with the concept. I found that c is not on Line AB. Got the equation right, substituted Cs coordinates, and understood it doesn't pass.

But I failed to remember that it is Yes - No question and ans, in this case, is No.

Friends keep this in mind. We often forget about it
_________________
_________________
Please give kudos, if you like my post

When the going gets tough, the tough gets going...
GMAT Club Bot
Re: M05-13   [#permalink] 18 Sep 2019, 02:53
Display posts from previous: Sort by

M05-13

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

Moderators: chetan2u, Bunuel






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