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![If a real number x is chosen at random in the interval [0,3] and a re](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If a real number x is chosen at random in the interval [0,3] and a re"){kind=icon}
13 Dec 2015, 19:29
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C
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Difficulty:
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35% (02:00) correct
65%
(01:58)
wrong
based on 275
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If a real number x is chosen at random in the interval [0,3] and a real number y is chosen at random in the interval [0,4], what is the probability that x < y?
A) 1/2
B) 7/12
C) 5/8
D) 2/3
E) 3/4
A) 1/2
B) 7/12
C) 5/8
D) 2/3
E) 3/4
ENGRTOMBA2018
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Last visit: 01 Dec 2021
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Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
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Updated on: 14 Dec 2015, 07:57
Originally posted by ENGRTOMBA2018 on 13 Dec 2015, 20:14.
Last edited by ENGRTOMBA2018 on 14 Dec 2015, 07:57, edited 2 times in total.
Last edited by ENGRTOMBA2018 on 14 Dec 2015, 07:57, edited 2 times in total.
Updated the solution
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bablu1234
Refer to the attached picture for the figure. We are given that \(0 \leq x \leq 3\) and \(0 \leq y \leq 4\)
Thus the region x-y when plotted gives you a rectangle with base = 3 units and height 4 units (as shown in the picture), giving you a total area = 4*3=12 \(units^2\)
Now, realize that y=x is a line that passes through (0,0) and (3,3) and divides the above rectangle into a trapezoid (ABCD) and a triangle. The area y>x will belong to the trapezoidal area.
Thus, the area of the trapezoid = 0.5*(4+1)*3 = 15/2
Finally, the required probability = trapezoid area / total area = (15/2)/12 = 12/24 = 5/8.
C is the correct answer.
Hope this helps.
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12-14-15 9-53-57 AM.jpg [ 26.2 KiB | Viewed 34072 times ]
Updated on: 21 Dec 2015, 01:12
Originally posted by MathRevolution on 14 Dec 2015, 01:17.
Last edited by MathRevolution on 21 Dec 2015, 01:12, edited 2 times in total.
Last edited by MathRevolution on 21 Dec 2015, 01:12, edited 2 times in total.
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.
If a real number x is chosen at random in the interval [0,3] and a real number y is chosen at random in the interval [0,4], what is the probability that x < y?
A) 1/2
B) 7/12
C) 5/8
D) 2/3
E) 3/4
Following is my approach.... check this out...
Since x is in [0,3] and y is in [0,4] we can put them in the coordinate plane.
That means in the coordinate plane x and y satisfy 0<= x <=3 and 0<= y <=4.
A point (x, y) can be in the rectangle with having (0, 0), (3, 0), (0, 4), (3, 4) as its 4 vertices.
Moreover the points with x<y should be in the trapezoid above the line through (0, 0) and (3, 3).
The area of the trapezoid above the line(joining (0, 0) and (3, 3)) is 15/2 and the area of the rectangle is 12.
So the probability is (15/2)/12 = 15/24=5/8.
The answer is, therefore, (C).
If a real number x is chosen at random in the interval [0,3] and a real number y is chosen at random in the interval [0,4], what is the probability that x < y?
A) 1/2
B) 7/12
C) 5/8
D) 2/3
E) 3/4
Following is my approach.... check this out...
Since x is in [0,3] and y is in [0,4] we can put them in the coordinate plane.
That means in the coordinate plane x and y satisfy 0<= x <=3 and 0<= y <=4.
A point (x, y) can be in the rectangle with having (0, 0), (3, 0), (0, 4), (3, 4) as its 4 vertices.
Moreover the points with x<y should be in the trapezoid above the line through (0, 0) and (3, 3).
The area of the trapezoid above the line(joining (0, 0) and (3, 3)) is 15/2 and the area of the rectangle is 12.
So the probability is (15/2)/12 = 15/24=5/8.
The answer is, therefore, (C).
