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# edited: sorry, no actual gmat questions. Its against the Non

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Intern
Joined: 28 May 2005
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edited: sorry, no actual gmat questions. Its against the Non [#permalink]

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28 Jun 2006, 10:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

edited: sorry, no actual gmat questions. Its against the Non Disclosure Agreement.

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Director
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28 Jun 2006, 10:58

All the points (x,y) with x>=y are below and on the y=5 line. The points above this line are x<y.

Now area above the y=5 line is 1/4th the area of the entire triangle (Simple geometry). Hence probability of finding a point over there is 1/4.
_________________

Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?

Last edited by paddyboy on 28 Jun 2006, 11:18, edited 1 time in total.

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Director
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28 Jun 2006, 11:15
The answer is 1/4. You can break up the triangle into a square and 2 triangle. Each of the triangles is now a 1/2 the area of the square. If you pick a point where y > x it has to be a point in this upper triangle. The probability of picking a point in a given area is proportional to the area.

P = (area of upper triangle) /(area ouf upper triangle + area of square + area of lower triangle)
=1/(1+2+1)
=1/4

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Manager
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28 Jun 2006, 11:39

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Intern
Joined: 28 May 2005
Posts: 14

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28 Jun 2006, 12:07
I don't have the answer, it was on the GMAT, i remember it cuz I spent some time on it and couldn't figure it out. How did you get 2/3?

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Manager
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28 Jun 2006, 12:25
From the given description, I drew a right traingle in XY-grid with the following 3 points: (0,0), (0,10), and (5,0). I suggest you draw this one too so follow my logic. The area of this trainagle, let's call it S1, is

S1 = 0.5*10*5 = 25

Now, all points with y>x will be above y=x line. This line divides original triangle into 2 trinagles, where the "upper" one (let's call it S2) will have all points where y>x. So, if we find the area of S2, then Prob=S2/S1.

To find S2, I first found the intersection point on the original (S1's) hypotenuse by solving for intersection point given two quations for the straight lines: y=-2x+10 and y=x.

From here, -2x+10=x

or, x=10/3.

So, the height of the S2 is 10/3 and the base is 10 => S2=0.5*10*(10/3)=50/3

Finally, Prob(y>x)=S2/S1 = (50/3)/25 = 2/3

Unfortunately, it took me way over 2 minutes to get to this point, because I originally assumed that one of the angles should be 30 degrees, which sent me on a wrong path for a while

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CEO
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28 Jun 2006, 13:11
Please do not post actual GMAT questions here.

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28 Jun 2006, 13:11
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# edited: sorry, no actual gmat questions. Its against the Non

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