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08 Feb 2013, 11:45
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
66% (03:17) correct
34%
(03:05)
wrong
based on 460
sessions
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Region R is a square in the x-y plane with vertices J = (–1, –2), K = (–1, 4), L = (5, 4), and M = (5, –2). What is the probability that a randomly selected point in region R lies below the line 3x – 5y = 10?
(A) \(\frac{5}{12}\)
(B) \(\frac{5}{18}\)
(C) \(\frac{5}{24}\)
(D) \(\frac{5}{36}\)
(E) \(\frac{5}{72}\)
For thoughts on geometric probability, as well as a complete solution with diagram to this particular problem, see:
https://magoosh.com/gmat/2013/geometric- ... -the-gmat/
Experts: are there any words of wisdom you would like to share on the topic of Geometric Probability?
Mike
(A) \(\frac{5}{12}\)
(B) \(\frac{5}{18}\)
(C) \(\frac{5}{24}\)
(D) \(\frac{5}{36}\)
(E) \(\frac{5}{72}\)
For thoughts on geometric probability, as well as a complete solution with diagram to this particular problem, see:
https://magoosh.com/gmat/2013/geometric- ... -the-gmat/
Experts: are there any words of wisdom you would like to share on the topic of Geometric Probability?
Mike
09 Feb 2013, 18:25
Kudos
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The square is formed by x=5 & x=-1 ; y=4 & y=-2
-1 to 5 or -2 to 4 = 6 which is the length of the square (if it was an rectangle we would have to find the length and breadth)
So the area of the square is 36.
To find the points where the line 3x-5y=10 cuts across the square, insert the two values of x or y into the equation
x=5
=> 3*5-5y=10
=> y=1. So (5,1) is first point of intersection of the line with the square.
x=-1
=> 3*(-1)-5y=10
=> y=-(7/5). This point (-1, -7/5) is not on any of the boundary of the square so this is not relevant. Let's try inserting the values of y to find where is the second point of intersection is
y=-2
=> 3x-5*(-2)=10
=> x=0. So (0, -2) is the second point of intersection of the line with the square.
So the line cuts the square at two adjacent sides of the square to form a triangle with base 5 and height 3. Area of this triangle is 15/2
Probability of point lying in this triangle = (15/2)/36 = 5/24
-1 to 5 or -2 to 4 = 6 which is the length of the square (if it was an rectangle we would have to find the length and breadth)
So the area of the square is 36.
To find the points where the line 3x-5y=10 cuts across the square, insert the two values of x or y into the equation
x=5
=> 3*5-5y=10
=> y=1. So (5,1) is first point of intersection of the line with the square.
x=-1
=> 3*(-1)-5y=10
=> y=-(7/5). This point (-1, -7/5) is not on any of the boundary of the square so this is not relevant. Let's try inserting the values of y to find where is the second point of intersection is
y=-2
=> 3x-5*(-2)=10
=> x=0. So (0, -2) is the second point of intersection of the line with the square.
So the line cuts the square at two adjacent sides of the square to form a triangle with base 5 and height 3. Area of this triangle is 15/2
Probability of point lying in this triangle = (15/2)/36 = 5/24
General Discussion
08 Feb 2013, 12:03
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mikemcgarry
Basically the square is formed with 4 lines x=-1, x=5, y=-2,y=4
Area of square =36
Now line is given as 5y=3x-10
We can see that the points on which this line intersects 4 lines forming square are (0,-2) and (5,1)
Thus area of triangle formed under the line =15/2
Probability of a point in this area=(15/2)/36 =5/24
Ans C it is!
