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GMAT Boost Challenge Question #1

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GMAT Boost Challenge Question #1

by gmatboost » Thu Aug 04, 2011 12:12 pm
Here is a GMAT Boost Challenge Question. I frequently leave out the answer choices on PS questions so that the focus is on the math in the question. If you have any questions, let me know.
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Two different vertices are randomly selected from the four in a certain square. A line is drawn connecting these two points. Then the process is repeated again, with all four vertices equally likely to be selected each time. What is the probability that the two lines that result from this process are perpendicular?
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by knight247 » Thu Aug 04, 2011 12:43 pm
Here is my take on it. Picking out any two random vertices and drawing a line joining them is the same as picking out any one side of the square or any one of the two diagonals.So 6 possible outcomes all together. In Square ABCD, picking out vertices A and B and then drawing a line is the same as picking out line AB. We have to consider different scenarios for sides and diagonals.
I'm a little confused if the second line we pick up is going to be out of the set of 6 or the set of 4 or 2 depending on whether we need to pick up a side or diagonal on the second pick.
So its either

4/6*2/6+2/6*1/6=8/36+2/36=10/36=5/18
or
4/6*2/4+2/6*1/2=8/24+2/12=12/24=1/2

Been studying all day and these numbers are just dancing in front of my eyes. Plz be gentle if i'm wrong.lol
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by Frankenstein » Thu Aug 04, 2011 8:58 pm
gmatboost wrote:Here is a GMAT Boost Challenge Question. I frequently leave out the answer choices on PS questions so that the focus is on the math in the question. If you have any questions, let me know.
------

Two different vertices are randomly selected from the four in a certain square. A line is drawn connecting these two points. Then the process is repeated again, with all four vertices equally likely to be selected each time. What is the probability that the two lines that result from this process are perpendicular?
Hi,
There are 4 sides and 2 diagonals. So, in total 6.
Total number of selections will be 6*6/2 = 18 (as order is not important)
Two perpendicular lines can be selected in 2 cases:
1) 2 adjacent sides can be picked in 4 ways
2) a pair of diagonals can be selected in 1 way
So, total 5 ways

Hence, probability is 5/18.
Cheers!

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by ColumbiaVC » Thu Aug 04, 2011 9:08 pm
[spoiler]1/3[/spoiler]
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by Plesé » Mon Aug 08, 2011 11:08 am
I think is 1/3. We need to choose 2 pairs of vertices, so for the first pair we have C4,2=6 possibilities. Since we already choose the first pair, it's left to us only two verices, hence there's actually nothing to chosse. There is only one way for the two lines (linking each of the vertices) be perpendicular: if the pairs are symmetric in relation to one diagonal. Therefore the probability is 3/6=1/3.
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by Plesé » Mon Aug 08, 2011 11:09 am
*error 2/6=1/3
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by force5 » Mon Aug 08, 2011 11:33 am
1st line can be drawn in 6 ways. (4 vertices)
2nd line can be drawn in 6 ways ( vertices can be selected again)

hence total number of ways to draw 2 lines 6x6 = 36

perpendicular lines can be chosen by 2 scenarios

choose edges:
hence 1st line can be drawn in 4 ways (4 edges)
and 2nd can be drawn in 2 ways ( 2 perpendicular sides)

total ways 8

choose diagonals:
diagonal: 2

total 10

hence prob of getting 2 perpendicular lines 10/36 = 5/18
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by rppala90 » Mon Aug 08, 2011 3:17 pm
[spoiler]2/15[/spoiler]

Let M be --> Number of ways two parallel lines can be drawn touching two vertices of square.
N be --> Number of ways two lines can be drawn such that each line touches two vertices.

M is 2 as there are only 2 ways parallel lines can be drawn touching two vertices of square. ( vertical and horizontal).

N is 6c2 as there are total 6 lines but I have to draw/choose 2 lines.
So Number of ways two lines can be choosen/drawn out of 6 possible lines is [spoiler]6c2 = 15[/spoiler].
Hence answer is [spoiler]2/15[/spoiler].
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by rppala90 » Mon Aug 08, 2011 3:27 pm
oops! the question is asking about perpendicular lines not parallel.
Hence the answer is [spoiler]4/15 not 2/15[/spoiler].
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by Plesé » Mon Aug 08, 2011 3:40 pm
I'm wrong since I consider that the pairs cannot have a vertice in common.
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by rppala90 » Mon Aug 08, 2011 8:48 pm
on further thought even the diagonals are perpendicular,
hence it should be [spoiler]5/15 i.e 1/3[/spoiler]
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by force5 » Tue Aug 09, 2011 4:07 am
gmatboost what is the OA
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by gmatboost » Thu Aug 11, 2011 9:13 am
The answer to GMAT Boost Challenge Question #1 is [spoiler]5/18[/spoiler].

force5, Frankenstein, and knight247 all did a great job of explaining it above.

Check out GMAT Boost Challenge Question #2, which will be posted momentarily.
Greg Michnikov, Founder of GMAT Boost

GMAT Boost offers 250+ challenging GMAT Math practice questions, each with a thorough video explanation, and 100+ GMAT Math video tips, each 90 seconds or less.
It's a total of 20+ hours of expert instruction for an introductory price of just $10.
View sample questions and tips without signing up, or sign up now for full access.

Also, check out the most useful GMAT Math blog on the internet here.
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