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

 It is currently 21 Oct 2019, 11:54 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.  For any triangle T in the xy–coordinate plan, the center of T is defin

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

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58396
For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

Show Tags

1
7 00:00

Difficulty:   25% (medium)

Question Stats: 73% (01:39) correct 27% (01:58) wrong based on 471 sessions

HideShow timer Statistics

Tough and Tricky questions: Coordinate Geometry.

For any triangle T in the xy–coordinate plan, the center of T is defined to be the point whose x–coordinate is the average (arithmetic mean) of the x–coordinates of the vertices of T and whose y–coordinate is the average of the y–coordinates of the vertices of T. If a certain triangle has vertices at the points (0,0) and (6,0) and center at the point (3,2), what are the coordinates of the remaining vertex?

A. (3,4)
B. (3,6)
C. (4,9)
D. (6,4)
E. (9,6)

_________________
Manager  Joined: 21 Jul 2014
Posts: 119
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

Show Tags

3
2
Hi Bunuel,

Great question. The best mode of attack is to Dive In.

Let's start by labeling the three vertices of triangle T as a, b, and c. We know pt a is at (0,0), b is at (6,0), and c is unknown, (cx, cy).

The center, as defined in the problem, is the arithmetic mean of the x and y coordinates individually. Let's write that out as a formula, where the center of triangle T is labeled as pt m at (mx, my).

mx = (ax + bx + cx)/3
my = (ay + by + cy)/3

We know from the problem that pt m is at (3,2), so let's plug in first for the x-coordinate, cx:
3 = (0 + 6 + cx)/3
With some arithmetic, we can solve this to see that cx = 3.

Now, let's do the same for the y-coordinate, cy:
2 = (0 + 0 + cy)/3
So... cy = 6.

Putting these together, we now have that the coordinate of the missing vertex is at (cx, cy), or (3,6)... Answer Choice B.

Hope this helps!
General Discussion
Intern  Joined: 04 Jul 2014
Posts: 45
Schools: Smeal" 20
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

Show Tags

1
Bunuel wrote:

Tough and Tricky questions: Coordinate Geometry.

For any triangle T in the xy–coordinate plan, the center of T is defined to be the point whose x–coordinate is the average (arithmetic mean) of the x–coordinates of the vertices of T and whose y–coordinate is the average of the y–coordinates of the vertices of T. If a certain triangle has vertices at the points (0,0) and (6,0) and center at the point (3,2), what are the coordinates of the remaining vertex?

A. (3,4)
B. (3,6)
C. (4,9)
D. (6,4)
E. (9,6)

Notice that (3,2) lies on one the median which divides the side connecting [0,0] & [6,0]....and it is give [3,2] is the centre...we know that centriod will divide the median in the ratio 2:1 .....the smaller part of the ratio is 2 here, so the bigger must be 4....and hence the point is [3,6]. Now add the x of all vertices and y of all vertices , find the averages and cross check.

Thus answer B [3,6].

------------
Please give kudos if this helps.
Manager  B
Joined: 11 Nov 2011
Posts: 59
Location: United States
Concentration: Finance, Human Resources
GPA: 3.33
WE: Consulting (Non-Profit and Government)
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

Show Tags

Can I solve this way?

let the vertices be A, B, and C
A=(0,0), B=(6,0), and since the central is (3,2) then C=(3,n)

sum of x=0+6+3=9; avg=9/3=3
sum of y=0+0+n=; avg=n/3=2 (since the center is (3,2))
thus n=3*2=6

So the answer is (3,6) B
Director  B
Joined: 04 Jun 2016
Posts: 556
GMAT 1: 750 Q49 V43 Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

Show Tags

Bunuel wrote:

Tough and Tricky questions: Coordinate Geometry.

For any triangle T in the xy–coordinate plan, the center of T is defined to be the point whose x–coordinate is the average (arithmetic mean) of the x–coordinates of the vertices of T and whose y–coordinate is the average of the y–coordinates of the vertices of T. If a certain triangle has vertices at the points (0,0) and (6,0) and center at the point (3,2), what are the coordinates of the remaining vertex?

A. (3,4)
B. (3,6)
C. (4,9)
D. (6,4)
E. (9,6)

Seems like a difficult question , but actually it can be solved in less than 30 seconds.
First plot all the given (X,Y) pair.
You will get the base of the triangle with length of 6 and mid point of triangle at 3,3
Now the vortex will be double at the Y coordinate of the midpoint
Double of 3= 6 (y=3)
Average of X is already 3
Therefore the X,Y pair will become 3,6 _________________
Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.
FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.
Current Student B
Joined: 26 Aug 2015
Posts: 32
Concentration: Strategy, Economics
GMAT 1: 570 Q40 V28 GMAT 2: 740 Q49 V41 Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

Show Tags

1
LogicGuru1 wrote:
Bunuel wrote:

Tough and Tricky questions: Coordinate Geometry.

For any triangle T in the xy–coordinate plan, the center of T is defined to be the point whose x–coordinate is the average (arithmetic mean) of the x–coordinates of the vertices of T and whose y–coordinate is the average of the y–coordinates of the vertices of T. If a certain triangle has vertices at the points (0,0) and (6,0) and center at the point (3,2), what are the coordinates of the remaining vertex?

A. (3,4)
B. (3,6)
C. (4,9)
D. (6,4)
E. (9,6)

Seems like a difficult question , but actually it can be solved in less than 30 seconds.
First plot all the given (X,Y) pair.
You will get the base of the triangle with length of 6 and mid point of triangle at 3,3
Now the vortex will be double at the Y coordinate of the midpoint
Double of 3= 6 (y=3)
Average of X is already 3
Therefore the X,Y pair will become 3,6 The center is 3,2.

Actually the average is of 3 data points, that is why it is 6 because (0+0+y)/3=2 y=6.

X remains at 3 because
(0 + 6 + x) /3 = 3
6+x=9
x=3

At least it made sense to me this way.
_________________
Send some kudos this way if I was helpful! !
Director  B
Joined: 04 Jun 2016
Posts: 556
GMAT 1: 750 Q49 V43 Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

Show Tags

Ilomelin wrote:
LogicGuru1 wrote:
Bunuel wrote:

Tough and Tricky questions: Coordinate Geometry.

For any triangle T in the xy–coordinate plan, the center of T is defined to be the point whose x–coordinate is the average (arithmetic mean) of the x–coordinates of the vertices of T and whose y–coordinate is the average of the y–coordinates of the vertices of T. If a certain triangle has vertices at the points (0,0) and (6,0) and center at the point (3,2), what are the coordinates of the remaining vertex?

A. (3,4)
B. (3,6)
C. (4,9)
D. (6,4)
E. (9,6)

Seems like a difficult question , but actually it can be solved in less than 30 seconds.
First plot all the given (X,Y) pair.
You will get the base of the triangle with length of 6 and mid point of triangle at 3,3
Now the vortex will be double at the Y coordinate of the midpoint
Double of 3= 6 (y=3)
Average of X is already 3
Therefore the X,Y pair will become 3,6 The center is 3,2.

Actually the average is of 3 data points, that is why it is 6 because (0+0+y)/3=2 y=6.

X remains at 3 because
(0 + 6 + x) /3 = 3
6+x=9
x=3

At least it made sense to me this way.

Yup , you are right

Average should should be 0+3+6=9/3
= 3
Double of 3=6
so vortex will be (x,y)=(3,6)
_________________
Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.
FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.
Manager  S
Joined: 24 Dec 2016
Posts: 95
Location: India
Concentration: Finance, General Management
WE: Information Technology (Computer Software)
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

Show Tags

I think its a very simple averages question. Median point (3,2) should be equal to {(x1+x2+x3)/3, (y1+y2+y3)/3}.
We have (x1,y1) and (x2,y2) as (0,0) and (6,0).
Thus, 3= (0+6+x3)/3 => x3 = 3
& 2= (0+0+y3)/3 => y3 = 6.
The point, therefore, is (3,6). Option B.
Manager  B
Joined: 20 Jun 2017
Posts: 91
GMAT 1: 570 Q49 V19 Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

Show Tags

let the co-ordinates of the 3rd vertex be (x,y)

3 = $$\frac{(0+6+x)}{3}$$

x = 3

2 = $$\frac{(0+0+y)}{3}$$

y = 6

Hence the 3rd vertex is (3,6)
VP  D
Joined: 09 Mar 2016
Posts: 1230
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

Show Tags

LighthousePrep wrote:
Hi Bunuel,

Great question. The best mode of attack is to Dive In.

Let's start by labeling the three vertices of triangle T as a, b, and c. We know pt a is at (0,0), b is at (6,0), and c is unknown, (cx, cy).

The center, as defined in the problem, is the arithmetic mean of the x and y coordinates individually. Let's write that out as a formula, where the center of triangle T is labeled as pt m at (mx, my).

mx = (ax + bx + cx)/3
my = (ay + by + cy)/3

We know from the problem that pt m is at (3,2), so let's plug in first for the x-coordinate, cx:
3 = (0 + 6 + cx)/3
With some arithmetic, we can solve this to see that cx = 3.

Now, let's do the same for the y-coordinate, cy:
2 = (0 + 0 + cy)/3
So... cy = 6.

Putting these together, we now have that the coordinate of the missing vertex is at (cx, cy), or (3,6)... Answer Choice B.

Hope this helps!

niks18 can you pls explain based on what rule do we plug in coordinates of centroid into equation to find third vertex ? i cant undestand logic. We need the coordinates of the last vertex, but we are plugging in corrdinates of centroid thank you Retired Moderator D
Joined: 25 Feb 2013
Posts: 1178
Location: India
GPA: 3.82
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

Show Tags

1
dave13 wrote:
LighthousePrep wrote:
Hi Bunuel,

Great question. The best mode of attack is to Dive In.

Let's start by labeling the three vertices of triangle T as a, b, and c. We know pt a is at (0,0), b is at (6,0), and c is unknown, (cx, cy).

The center, as defined in the problem, is the arithmetic mean of the x and y coordinates individually. Let's write that out as a formula, where the center of triangle T is labeled as pt m at (mx, my).

mx = (ax + bx + cx)/3
my = (ay + by + cy)/3

We know from the problem that pt m is at (3,2), so let's plug in first for the x-coordinate, cx:
3 = (0 + 6 + cx)/3
With some arithmetic, we can solve this to see that cx = 3.

Now, let's do the same for the y-coordinate, cy:
2 = (0 + 0 + cy)/3
So... cy = 6.

Putting these together, we now have that the coordinate of the missing vertex is at (cx, cy), or (3,6)... Answer Choice B.

Hope this helps!

niks18 can you pls explain based on what rule do we plug in coordinates of centroid into equation to find third vertex ? i cant undestand logic. We need the coordinates of the last vertex, but we are plugging in corrdinates of centroid thank you Hi dave13
the question says that the x & y coordinates of the center of triangle is equal to the averages of x & y coordinates of three vertices.
Now you know the average of x coordinate and you know two x coordinates of the vertices, you need to find out the 3rd x coordinate
Similarly for y coordinate.
Take this question as an average question and try to solve
Director  D
Joined: 24 Oct 2016
Posts: 535
GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38 Re: For any triangle T in the xy-coordinate plane, the center of  [#permalink]

Show Tags

galiya wrote:
For any triangle T in the xy-coordinate plane, the center of T is defined to be the point whose x-coord.is the avr. of the x coord-s of the vertices T and whose y-coord. is the avr. of the y-coord-s of the vert.T. If a certain triangle has vertices at the points (0;0) and (6;0) and the center at the point (3;2), what are the coord-s of the remain.vertex"

(3;4)
(3;6)
(4;9)
(6;4)
(9;6)

3 = (0 + 6 + x)/3
x = 9 - 6 = 3

2 = (0 + 0 + y)/3
y = 6

Therefore, 3rd Vertex = (3, 6) => B
_________________

If you found my post useful, KUDOS are much appreciated. Giving Kudos is a great way to thank and motivate contributors, without costing you anything.
Non-Human User Joined: 09 Sep 2013
Posts: 13383
Re: For any triangle in the xy-coordinate plane, the center of T  [#permalink]

Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: For any triangle in the xy-coordinate plane, the center of T   [#permalink] 10 Sep 2019, 20:42
Display posts from previous: Sort by

For any triangle T in the xy–coordinate plan, the center of T is defin

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

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