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Manhattan GMAT problem solving - August 21, 2006

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Manhattan GMAT problem solving - August 21, 2006

by Kevin » Mon Aug 21, 2006 6:42 am
Most Manhattan GMAT students are trying to break the 700 barrier. As a result, we've developed our own math problems written at the 700+ level; these are the types of questions you'll WANT to see, when you are working at that level. Try to solve this 700+ level problem (I'll post the solution next Monday).

Question: Points of Probability
Image

A certain computer program randomly generates equations of lines in the form Image. If point P is a point on a line generated by this program, what is the probability that the line does NOT pass through figure ABCD?

(A) 3/4
(B) 3/5
(C) 1/2
(D) 2/5
(E) 1/4
Kevin Fitzgerald
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by dblazquez » Tue Aug 22, 2006 3:35 am
(A) 3/4

The key is to know the size of the angle that makes lines pass throught the figure

1) point D=(4,2) C=(8,2) and P=(6,2)
2) draw segments PD and PC
3) the small angle of x with PD is 45, the small angle of x with PC is 45
4) the inner angle PD-PC is therefore 90
5) from the 360 possible grades, we are taking as valid all of them except 90.
6) 270/360 = 3/4
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no

by aim-wsc » Tue Aug 22, 2006 5:43 am
think again.

u r on the right track. just one more step s needed.


CAN U BELIEVE IT? THIS IS MY FIRST POST IN Q EVER.!!!

ok
d good news is i m back. i m in for a month.

so can serve u ppl.

answer & explanation tomorrow!!
daniel, try once more.
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not too sure

by achal.kumble » Tue Aug 22, 2006 9:21 pm
The answer should be (c)
the slope of the line containing P is given by tan theta(tan t)
when 45<=tan t<=135, the line containing P passes thru ABCD.
that is for a total of 90 possible angles the line containing P passes thru ABCD. Therefore reqd probability is 1-(90/180)=1-1/2= 1/2
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by dblazquez » Wed Aug 23, 2006 5:45 am
Okay, i have had to draw it a couple of times but i have realized my mistake; i have to count the "forbidden angles" twice because we are dealing with lines, not segments starting at the point P.

So the answer is c) 1/2 .... i hate GMAT traps :shock: good question Kevin
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by aim-wsc » Wed Aug 23, 2006 9:18 am
ok answer is fine.
but the explanation is objectionable.... :shock:

OK i know its just an careless error made by Achal but still i must correct it

i presume you wanted to write just theta there!

so:
45< theta (t) < 135

* again 90 possible angle is not true possible angles are infinite. :lol:
anyway what matters is the answer after all :wink: so it is acceptable.

My approach was same like Daniel except i ll put one more point that:
you dont have to consider all 360 degrees; 180 degree rotation is enough since then the same lines repeat...

so it is kindda 90/180
therefore 1/2.
MY PIC : C
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by 800guy » Wed Aug 23, 2006 3:56 pm
awesome thread!!
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by Kevin » Wed Aug 30, 2006 10:51 am
ANSWER

Image

The most difficult part of this question is conceptualizing what you're asked to find. The best way to handle tricky problems like this is to break them down into their component parts, resolve each part, then reconstruct them as a whole. You start off with a coordinate plane and four points (A, B, C, and D) that form a square when joined. Beneath the square, on the x-axis, lies another point, P. Then you are asked to determine the probability that a line randomly drawn through P will not also pass through square ABCD. In any probability question, the first thing you need to determine is the number of total possibilities. In this case, the total number of possibilities will be the total number of lines that can be drawn through P. The problem, though, is that there are literally infinitely many lines that satisfy this criterion. We cannot use infinity as the denominator of our probability fraction. So what to do?

This is where some creative thinking is necessary. First, you need to recognize that there must be a limited range of possibilities for lines that pass through both P and ABCD. If that were not the case, the probability would necessarily be 1 or 0. What is this range? Well, drawing out a diagram of the problem will help enormously here. Any line that passes through both P and ABCD has to pass through a triangle formed by ABP. This triangle is an isosceles right triangle. We know this because if we drop perpendiculars from points A and B, we end up with two isosceles right triangles, with angles of 45 degrees on either side of P. Since angles along a straight line must equal 180, we know that angle APB must measure 90 degrees. So if angle APB measures 90 degrees, we can use that to figure out the proportion of lines that pass through both P and ABCD. If we draw a semicircle with P as its center, we can see that the proportion of lines passing through P and ABCD will be the same as the proportion of the semicircle occupied by sector APB. Since a semicircle contains 180 degrees, sector APB occupies 1/2 of the semicircle. So 1/2 of all possible lines through P will also pass though ABCD, which means the 1/2 will NOT pass through ABCD and we have our answer.

The correct answer is C.
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by gmatjedi » Tue May 25, 2010 4:46 pm
key steps:

establish points of (x,y) for a, b, c, d and p

realize that pd and vertical line dropped from d to x axis and realize that pc and vertical line dropped from c to axis create 45-45-90 triangles

realize that line contains 180 degrees

45 x2/180 = 1/2 probability that line through p does not pass through abcd
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by Gmat Bond » Fri May 28, 2010 1:34 am
Thanks for the post .
The name is Bond, GMAT BOND !!
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