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If a real number x is chosen at random in the interval [0,3] and a re

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Intern
Joined: 16 Aug 2015
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Location: India
Concentration: General Management, Entrepreneurship
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If a real number x is chosen at random in the interval [0,3] and a re  [#permalink]

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13 Dec 2015, 19:29
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85% (hard)

Question Stats:

34% (01:53) correct 66% (02:12) wrong based on 91 sessions

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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
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Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
If a real number x is chosen at random in the interval [0,3] and a re  [#permalink]

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Updated on: 14 Dec 2015, 07:57
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bablu1234 wrote:
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

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.

Hope this helps.
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12-14-15 9-53-57 AM.jpg [ 26.2 KiB | Viewed 7951 times ]

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.
Updated the solution
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If a real number x is chosen at random in the interval [0,3] and a re  [#permalink]

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Updated on: 21 Dec 2015, 01:12
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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.

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" 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. CEO Joined: 20 Mar 2014 Posts: 2620 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Re: If a real number x is chosen at random in the interval [0,3] and a re [#permalink] Show Tags 14 Dec 2015, 07:46 MathRevolution wrote: 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 Hi! Engr I'm afraid that you made a mistake to understand the problem because x and y are not integers but real numbers. So we cannot count the all the cases of (x, y) satisfying x<y. The 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, 4). The area of the trapezoid above the line(joining (0, 0) and (3, 4)) 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). MathRevolution, thank you. That is indeed an incorrect assumption on my end. Also, I think the text in red above is not correct. I believe you wanted to write the line joining (0,0) and (3,3) and NOT (3,4). The line joining (0,0) and (3,4) will not leave a trapezoid at the top. Intern Joined: 16 Aug 2015 Posts: 20 Location: India Concentration: General Management, Entrepreneurship GMAT 1: 510 Q22 V19 GPA: 3.41 Re: If a real number x is chosen at random in the interval [0,3] and a re [#permalink] Show Tags 14 Dec 2015, 11:54 MathRevolution wrote: 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 Hi! Engr I'm afraid that you made a mistake to understand the problem because x and y are not integers but real numbers. So we cannot count the all the cases of (x, y) satisfying x<y. The 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, 4). The area of the trapezoid above the line(joining (0, 0) and (3, 4)) 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). Thanks MathRevolution. It was tough one and I couldn't crack it. Manager Joined: 09 Jul 2013 Posts: 109 If a real number x is chosen at random in the interval [0,3] and a re [#permalink] Show Tags 17 Dec 2015, 12:53 1 Another way to look at the problem: Break up the range of y into two parts. 1. $$y>3$$ 2. $$0\leq{y}\leq{3}$$ We can consider the probability of $$y>x$$ for each part, then add them up. For part 1, if $$y>3$$, then it will always be greater than x no matter what x is. y will be in this range 1/4 of the time. Probability = 1*1/4 = 1/4. For part 2, both x and y are in the range of [0,3], so we can logically conclude that $$y>x$$ half the time (similarly $$x>y$$ half the time). y will be in this range 3/4 of the time. 1/2*3/4 = 3/8 Add up both parts: 1/4 + 3/8 = 5/8 Answer C _________________ Dave de Koos GMAT aficionado Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7758 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If a real number x is chosen at random in the interval [0,3] and a re [#permalink] Show Tags 21 Dec 2015, 01:20 That is indeed an incorrect assumption on my end. Also, I think the text in red above is not correct. I believe you wanted to write the line joining (0,0) and (3,3) and NOT (3,4). The line joining (0,0) and (3,4) will not leave a trapezoid at the top. Hi! Engr Thank you for pointing out my error. As you pointed out the line is joining (0,0) and (3,3) and NOT (3,4). That makes the diagram a trapezoid. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: If a real number x is chosen at random in the interval [0,3] and a re  [#permalink]

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27 Jul 2018, 19:55
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