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PS OG11 #36

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Director
Joined: 03 Sep 2006
Posts: 859

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09 Jan 2009, 21:46
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In the attached figure, the point on segment PQ that is twice as far from P as from Q is

A) (3,1)
B) (2,1)
C) (2,-1)
D) (1.5,0.5)
E) (1,0)

Reaching the answer mathematically is fine. What's the fastest approach, shall I assume that the figure drawn will always be drawn to the scale?
Attachment:

OG11-36.PNG [ 3.76 KiB | Viewed 1249 times ]
Manager
Joined: 28 Jul 2004
Posts: 135
Location: Melbourne
Schools: Yale SOM, Tuck, Ross, IESE, HEC, Johnson, Booth

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09 Jan 2009, 22:09
I think it will be mentioned that figures are not drawn to scale explicitly (check that coz I have read some instruction related to figures in GMAT but not sure what it was).

Coming to the solution of the problem:

If you draw perpendiculars from each of the point and create three small triangles such that the three segments become hypotenuses of right triangles, you will find that the height and base of all the three triangles will be same i.e. 1 unit.

This implies that the hypotenuse of each of the triangle is sqrt(2) which further implies that the three yellow circles divide PQ in three equal parts.

So, yellow circle closest to Q (2,1) is the answer. Hence B.
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kris

Director
Joined: 03 Sep 2006
Posts: 859

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10 Jan 2009, 00:12
krishan wrote:
I think it will be mentioned that figures are not drawn to scale explicitly (check that coz I have read some instruction related to figures in GMAT but not sure what it was).

Coming to the solution of the problem:

If you draw perpendiculars from each of the point and create three small triangles such that the three segments become hypotenuses of right triangles, you will find that the height and base of all the three triangles will be same i.e. 1 unit.

This implies that the hypotenuse of each of the triangle is sqrt(2) which further implies that the three yellow circles divide PQ in three equal parts.

So, yellow circle closest to Q (2,1) is the answer. Hence B.

Thanks a lot! makes more sense now.
Manager
Joined: 04 Jan 2009
Posts: 237

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10 Jan 2009, 10:54
alternate way to solve (use if it helps you).
the distance from p to q in terms of x-coordinate is 3-0=3
the distance from p to q in terms ofy-coordinate is 2-(-1) = 3
Each of the two interior points is 1 unit from the closest end point and they are 1 unit from each other.
Thus, the answer b can be derived.
LM wrote:
krishan wrote:
I think it will be mentioned that figures are not drawn to scale explicitly (check that coz I have read some instruction related to figures in GMAT but not sure what it was).

Coming to the solution of the problem:

If you draw perpendiculars from each of the point and create three small triangles such that the three segments become hypotenuses of right triangles, you will find that the height and base of all the three triangles will be same i.e. 1 unit.

This implies that the hypotenuse of each of the triangle is sqrt(2) which further implies that the three yellow circles divide PQ in three equal parts.

So, yellow circle closest to Q (2,1) is the answer. Hence B.

Thanks a lot! makes more sense now.

_________________

-----------------------
tusharvk

Re: PS OG11 #36   [#permalink] 10 Jan 2009, 10:54
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