We have three parallel Lines, AB, CD, and EF. If line AB hs 6 points,

Question

We have three parallel Lines, AB, CD, and EF. If line AB hs 6 points, Line CD has 3 points and Line EF has 5 points but not three points such that one point from each line is on a straight line. Then how many triangles can be formed whose vertices lie on the points available on lines?

A) 90
B) 153
C) 243
D) 333
E) 364

Source: https://www.GMATinsight.com

OliveOyl2005 · Answer

Choosing 2 parallel lines at a time, and choosing 2 points and 1 point from either lines, will give final answer as 243.
IMO ans is C

GMATinsight · Answer

Hey OliveOyl2005 You might want to find a small error. The answer is not Option C

https://www.youtube.com/watch?v=dhQeCIsp0Gk https://gmatclub.com/forum/download/file.php?mode=view&id=55229 Screenshot 2020-06-13 at 1.28.07 PM.png

Kinshook · Answer

Given: 
1. We have three parallel Lines, AB, CD, and EF. 
2. Line AB hs 6 points, Line CD has 3 points and Line EF has 5 points but not three points such that one point from each line is on a straight line.

Asked: How many triangles can be formed whose vertices lie on the points available on lines?

Triangles formed by lines AB & CD = 6C2 * 3C1 + 3C2 * 6C1 = 15*3 + 3*6 = 45 + 18 = 63
Triangles formed by lines CD & EF = 3C2 * 5C1 + 5C2 * 3C1 = 3*5 + 10*3 = 15 + 30 = 45
Triangles formed by lines AB & EF = 6C2 * 5C1 + 6C1 * 5C2 = 15*5 + 6*10 = 75 + 60 = 135
Triangles formed by lines AB, CD & EF = 6C1 * 3C1 * 5C1 = 6*3*5 = 90

Total triangles formed by lines AB, CD & EF = 63 + 45 + 135 + 90 = 333

IMO D

Archit3110 · Answer

given that lines ab,cd,ef are ll
and AB has 6 points ; CD has 3 points and EF has 5 points
total 14 points are given and for a ∆ we need 3 points
chosing a point from each of line ; 6*5*3 ; 90
two points AB and 1 point CD , EF ; 6c2*8c1 ; 120
2 points CD and 1 point AB , EF ; 3c2*11c1; 33
2 point EF and 1 point AB , CD ; 5c2*9c1 ; 90 
total 90+120+33+90 ; 333
OPTION D

GT1944 · Answer

can GMAT ask such a lengthy question?

GMATinsight · Answer

1) GMAT has asked a question on these lines already however there were two parallel lines to choose the points from 2) The Advanced official guide GMAT has already got us acquainted with such harder and lengthy questions. So it would be advisable to be ready for questions of this level. To practice a few questions of Advanced Guide please click the LINK To practice OG-2020 Quant Review for FREE CLICK HERE All questions have Video explanations

manikmehta95 · Answer

Hi  GT1944 ,

An easier way to solve this question is :

Total triangles formed from 14(6+3+5) distinct points(right now ignore that 3 points can be on the same line) = 14C3 = 364
Total no. of combinations in which the 3 points belong to the same line: 6C3 + 3C3 + 5C3 = 31

hence, 364 - 31 = 333

Thanks,
Manik

bumpbot · Answer

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We have three parallel Lines, AB, CD, and EF. If line AB hs 6 points,

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We have three parallel Lines, AB, CD, and EF. If line AB hs 6 points,

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