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01 Feb 2020, 10:06
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B
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Difficulty:
45%
(medium)
Question Stats:
58% (01:13) correct
42%
(01:19)
wrong
based on 370
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What is the maximum possible number of intersections among 7 lines that are all on the same plane?
(A) 14
(B) 21
(C) 28
(D) 42
(E) 56
Source: GMAT Quantum Math Course
(A) 14
(B) 21
(C) 28
(D) 42
(E) 56
Source: GMAT Quantum Math Course
03 Feb 2020, 04:18
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Explanation:
Theory: If you have N lines on the plain, you can always add one more that will intersect all of them, and thus create N new intersection points. all that you need to do for this to occur is to ensure that the added line is not parallel to any of the existing lines and that it doesn’t intersect any of the lines at an existing intersection. Just setting a new line at random will almost always comply with these requirements.
OK, with that understanding, we can easily calculate the maximum number of intersection for N lines:
0+1+2+3+…+(N−1)=N(N−1)/2=(N^2−N)/2
As, N=7, then the maximum number of intersections is:
By Formula : N(N−1)/2= 7x6/2= 21
IMO-B
Theory: If you have N lines on the plain, you can always add one more that will intersect all of them, and thus create N new intersection points. all that you need to do for this to occur is to ensure that the added line is not parallel to any of the existing lines and that it doesn’t intersect any of the lines at an existing intersection. Just setting a new line at random will almost always comply with these requirements.
OK, with that understanding, we can easily calculate the maximum number of intersection for N lines:
0+1+2+3+…+(N−1)=N(N−1)/2=(N^2−N)/2
As, N=7, then the maximum number of intersections is:
By Formula : N(N−1)/2= 7x6/2= 21
IMO-B
General Discussion
20 Aug 2020, 20:05
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We need 2 points for a line to be drawn. Please refer to the diagram below.
When 2 lines are drawn, we can have only 1 point of intersection.
When 3 lines are, we have 3 points of intersection
When 4 lines are, we have 6 points of intersection
Therefore we derive that the number of intersection point for n lines (None of them being collinear) = \(^nC_2\)
So 7 lines, give us \(^7C_2 = \frac{7*6}{2}\) = 21 lines
Option B
Arun Kumar
When 2 lines are drawn, we can have only 1 point of intersection.
When 3 lines are, we have 3 points of intersection
When 4 lines are, we have 6 points of intersection
Therefore we derive that the number of intersection point for n lines (None of them being collinear) = \(^nC_2\)
So 7 lines, give us \(^7C_2 = \frac{7*6}{2}\) = 21 lines
Option B
Arun Kumar
Attachments
Intersecting Lines.jpg [ 457.67 KiB | Viewed 12935 times ]
