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Bunuel
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Triangles ABC and ADE have areas 2007 and 7002, respectively, with B = 14 Mar 2019, 03:15
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95% (hard)

Question Stats:

21% (02:49) correct 79% (02:32) wrong based on 42 sessions

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TOUGH QUESTION:



Triangles ABC and ADE have areas 2007 and 7002, respectively, with B = (0, 0), C = (223, 0), D = (680, 380), and E = (689, 389). What is the sum of all possible x-coordinates of
A?


(A) 282
(B) 300
(C) 600
(D) 900
(E) 1200
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E
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Triangles ABC and ADE have areas 2007 and 7002, respectively, with B = 14 Mar 2019, 05:27
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Bunuel
Triangles ABC and ADE have areas 2007 and 7002, respectively, with B = (0, 0), C = (223, 0), D = (680, 380), and E = (689, 389). What is the sum of all possible x-coordinates of
A?


(A) 282
(B) 300
(C) 600
(D) 900
(E) 1200
Show more

chetan2u :
hello please help with this question?
are such questions really tested on GMAT?
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Triangles ABC and ADE have areas 2007 and 7002, respectively, with B = 15 Mar 2019, 07:31
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Bunuel
Triangles ABC and ADE have areas 2007 and 7002, respectively, with B = (0, 0), C = (223, 0), D = (680, 380), and E = (689, 389). What is the sum of all possible x-coordinates of
A?


(A) 282
(B) 300
(C) 600
(D) 900
(E) 1200

chetan2u :
hello please help with this question?
are such questions really tested on GMAT?
Show more


Seems Out of scope... hard to reach the solution in 2 minutes...
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Triangles ABC and ADE have areas 2007 and 7002, respectively, with B = 26 Jul 2021, 09:03
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Experts please help with the solution.
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Triangles ABC and ADE have areas 2007 and 7002, respectively, with B = 26 Jul 2021, 23:14
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Bunuel

TOUGH QUESTION:



Triangles ABC and ADE have areas 2007 and 7002, respectively, with B = (0, 0), C = (223, 0), D = (680, 380), and E = (689, 389). What is the sum of all possible x-coordinates of
A?


(A) 282
(B) 300
(C) 600
(D) 900
(E) 1200
Show more

The worse the numbers given, the more the probability that they will not get used at all.
Draw a diagram. BC lies on X axis. To make a triangle with a given area (2007) i.e. a fixed altitude (whatever that may be, it is 18 in this case), A will lie on any point on the two red lines. The two red lines are 18 units above and below the x axis. Think about this before moving ahead.

Now draw DE. To make a triangle with a given area (7002) i.e. a fixed altitude (whatever that may be), A will lie on any point on the two given dotted purple lines.

So what we need is the x co-ordinates of the 4 possible A values (A', A'', A''' and A'''')

Attachment:
Screenshot 2021-07-27 at 11.41.54.png
Screenshot 2021-07-27 at 11.41.54.png [ 27.23 KiB | Viewed 2688 times ]

Now notice the co-ordinates of D and E are D = (680, 380), and E = (689, 389). So the diff between x co-ordinates is 9 and between y co-ordinates is also 9. This means DE is a line with slope 1.
Hence, it will cut the x axis at 300 (when we move y co-ordinate 380 units down, we will move the x co-ordinate 380 units to the left so the x co-ordinate will become 680 - 380 = 300). This is the point M.

Now note that slope of purple dotted lines is also 1 since they are parallel to DE.
By symmetry, the 4 possible x co-ordinates of A
A' = 300 + a + b
A'' = 300 + a - b
A''' = 300 - (a + b)
A'''' = 300 - (a - b)

When we add them, all a and b will get cancelled to give a total of 1200.

Answer (E)

The question is based on fundamentals of co-ordinate geometry that are tested on GMAT (how co-ordinates change when lines turn by say 90 degrees etc) but it certainly seems a bit un-manageable in exam conditions.

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Triangles ABC and ADE have areas 2007 and 7002, respectively, with B = 26 Jul 2021, 23:20
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We have some Co-ordinate geometry posts here that will help you understand the concepts used in this question:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/catego ... s=geometry
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Triangles ABC and ADE have areas 2007 and 7002, respectively, with B = 30 Jul 2021, 04:48
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VeritasKarishma


The worse the numbers given, the more the probability that they will not get used at all.
Draw a diagram. BC lies on X axis. To make a triangle with a given area (2007) i.e. a fixed altitude (whatever that may be, it is 18 in this case), A will lie on any point on the two red lines. The two red lines are 18 units above and below the x axis. Think about this before moving ahead.

Now draw DE. To make a triangle with a given area (7002) i.e. a fixed altitude (whatever that may be), A will lie on any point on the two given dotted purple lines.

So what we need is the x co-ordinates of the 4 possible A values (A', A'', A''' and A'''')

Attachment:
Screenshot 2021-07-27 at 11.41.54.png

Now notice the co-ordinates of D and E are D = (680, 380), and E = (689, 389). So the diff between x co-ordinates is 9 and between y co-ordinates is also 9. This means DE is a line with slope 1.
Hence, it will cut the x axis at 300 (when we move y co-ordinate 380 units down, we will move the x co-ordinate 380 units to the left so the x co-ordinate will become 680 - 380 = 300). This is the point M.

Now note that slope of purple dotted lines is also 1 since they are parallel to DE.
By symmetry, the 4 possible x co-ordinates of A
A' = 300 + a + b
A'' = 300 + a - b
A''' = 300 - (a + b)
A'''' = 300 - (a - b)

When we add them, all a and b will get cancelled to give a total of 1200.

Answer (E)

The question is based on fundamentals of co-ordinate geometry that are tested on GMAT (how co-ordinates change when lines turn by say 90 degrees etc) but it certainly seems a bit un-manageable in exam conditions.

ShaikhMoice
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hi VeritasKarishma maam

could not get "Now note that slope of purple dotted lines is also 1 since they are parallel to DE." (above highlighted)
why purple lines need to be parallel??

thx
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