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12 Nov 2019, 03:56
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In the rectangular coordinate plane, a rectangular region has vertices (2.5, 4.5), (2.5, −1.5), (−7.5, -1.5), and (−7.5, 4.5). If a point from within the rectangle is to be randomly selected such that both the x- and y-coordinates are integers, what is the probability that both coordinates of the selected point are negative?
A. 1/11
B. 7/60
C. 2/15
D. 16/77
E. 4/15
Are You Up For the Challenge: 700 Level Questions
A. 1/11
B. 7/60
C. 2/15
D. 16/77
E. 4/15
Are You Up For the Challenge: 700 Level Questions
TheGuyDangerous
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Joined: 10 Sep 2019
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Concentration: Social Entrepreneurship, Healthcare
Schools: Fuqua '24 (D) ISB '23 (A) Darden '24 (D) Johnson '24 (D) Kenan-Flagler '24 (I) McDonough '24 (S) Tepper '24 (S)
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Schools: Fuqua '24 (D) ISB '23 (A) Darden '24 (D) Johnson '24 (D) Kenan-Flagler '24 (I) McDonough '24 (S) Tepper '24 (S)
GMAT 2: 720 Q50 V37
GRE 1: Q167 V159
Posts: 137
12 Nov 2019, 09:52
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Given that the randomly chosen point should have X- & Y- coordinates as negative, the point should lie in the III Quadrant. => The point has to lie between X - (0 to -7.5) & Y - (0 to -1.5).
Also given that the coordinates for the randomly selected points are 'Negative integers'.
=>X can be any one of (-7,-6,-5,-4,-3,-2,-1) => 7C1
=> Y can be any one of (-1) => 1C1
=> Total combinations = 7C1*1C1 = 7
Total combinations for a point to lie anywhere in the rectangle =
X- (-7,-6,-5,-4,-3,-2,-1,0,1,2) & Y- (-1,0,1,2,3,4)
=> 10C1*6C1 = 10*6 = 60
=> Probability = 7/60
IMO B
Also given that the coordinates for the randomly selected points are 'Negative integers'.
=>X can be any one of (-7,-6,-5,-4,-3,-2,-1) => 7C1
=> Y can be any one of (-1) => 1C1
=> Total combinations = 7C1*1C1 = 7
Total combinations for a point to lie anywhere in the rectangle =
X- (-7,-6,-5,-4,-3,-2,-1,0,1,2) & Y- (-1,0,1,2,3,4)
=> 10C1*6C1 = 10*6 = 60
=> Probability = 7/60
IMO B
General Discussion
TumharaFaizal
GMAT 1: 710 Q50 V35
Posts: 35
19 May 2020, 14:10
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Integers in length of rect. ( -7 to 2) = 10
Integers in breadth of rect. ( -1 to 4) = 6
total integer cordinates = 10* 6 = 60
negative cordinate X (-1 to -7) =7
negative cordinate Y (-1 to -1) = 1
Total negative cordinates possible= 7*1 = 7
required probability = 7/60
IMO (B).
Thanks
Integers in breadth of rect. ( -1 to 4) = 6
total integer cordinates = 10* 6 = 60
negative cordinate X (-1 to -7) =7
negative cordinate Y (-1 to -1) = 1
Total negative cordinates possible= 7*1 = 7
required probability = 7/60
IMO (B).
Thanks
