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06 Jun 2022, 01:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
55% (02:49) correct
45%
(02:33)
wrong
based on 53
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In a triangle ABC, there are 5, 6 and 4 points, different from the vertices A, B and C of the triangle, on the three sides AB, BC and CA respectively. How many distinct triangles can be drawn using the 18 points (including the vertices A, B and C)?
(A) 111
(B) 705
(C) 816
(D) 910
(E) 955
Are You Up For the Challenge: 700 Level Questions
(A) 111
(B) 705
(C) 816
(D) 910
(E) 955
Are You Up For the Challenge: 700 Level Questions
06 Jun 2022, 02:52
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First, let's pick 3 points out of 18.
We get (18X17X16)/(3X2X1)=816.
But, any three points exclusively from AB, or BC, or CA cannot form a triangle.
There are a total of 5+2=7 points on AB. Pick three from seven, and we get (7X6X5)/(3X2X1)=35.
There are a total of 6+2=8 points on BC. Pick three from eight, and we get (8X7X6)/(3X2X1)=56.
There are a total of 4+2=6 points on CA. Pick three from six, and we get (6X5X4)/(3X2X1)=20.
Final answer: 816-35-56-20=705.
The answer is thus (B).
We get (18X17X16)/(3X2X1)=816.
But, any three points exclusively from AB, or BC, or CA cannot form a triangle.
There are a total of 5+2=7 points on AB. Pick three from seven, and we get (7X6X5)/(3X2X1)=35.
There are a total of 6+2=8 points on BC. Pick three from eight, and we get (8X7X6)/(3X2X1)=56.
There are a total of 4+2=6 points on CA. Pick three from six, and we get (6X5X4)/(3X2X1)=20.
Final answer: 816-35-56-20=705.
The answer is thus (B).
General Discussion
06 Jun 2022, 06:09
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The maximum number of triangles that could be created if all were possible is:
18!/3!15!= 816
This eliminates D&E
Assuming you recognize the problem with selecting 3 collinear points, it also eliminates C.
Reviewing A and B, does it make intuitive sense that the collinearity issue would eliminate almost 90% of the possible triangles as required by A ? Doesn't feel that way, leaving B as the likely answer without having to do further math.
Posted from my mobile device
18!/3!15!= 816
This eliminates D&E
Assuming you recognize the problem with selecting 3 collinear points, it also eliminates C.
Reviewing A and B, does it make intuitive sense that the collinearity issue would eliminate almost 90% of the possible triangles as required by A ? Doesn't feel that way, leaving B as the likely answer without having to do further math.
Posted from my mobile device
