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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## In a plane, there are two parallel lines. One line has 5 tagged by: BTGmoderatorLU ##### This topic has 4 expert replies and 0 member replies ### Top Member ## In a plane, there are two parallel lines. One line has 5 ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Source: e-GMAT In a plane, there are two parallel lines. One line has 5 points and another line has 4 different points. How many different triangles can we form from these 9 points? A. 62 B. 70 C. 73 D. 86 E. 122 The OA is B. ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12982 messages Followed by: 1249 members Upvotes: 5254 GMAT Score: 770 BTGmoderatorLU wrote: Source: e-GMAT In a plane, there are two parallel lines. One line has 5 points and another line has 4 different points. How many different triangles can we form from these 9 points? A. 62 B. 70 C. 73 D. 86 E. 122 The OA is B. There are two ways in which we can create a triangle. #1) Select 2 points from the 5-point line and select 1 point from the 4-point line. #2) Select 2 points from the 4-point line and select 1 point from the 5-point line. #1) Select 2 points from the 5-point line and select 1 point from the 4-point line. Take this task and break it into stages. Stage 1: Select 2 points from the 5-point line Since the order of the 2 selected points does not matter, we can use combinations. We can select 2 points from 5 points in 5C2 = 10 ways. If anyone is interested, we have a video on calculating combinations (like 5C2) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789 Stage 2: Select 1 point from the 4-point line. We can complete this stage in 4 ways By the Fundamental Counting Principle (FCP) we can complete the 2 stages in (10)(4) ways (= 40 ways) #2) Select 2 points from the 4-point line and select 1 point from the 5-point line. Take this task and break it into stages. Stage 1: Select 2 points from the 4-point line We can select 2 points from 4 points in 4C2 = 6 ways. Stage 2: Select 1 point from the 5-point line. We can complete this stage in 5 ways By the Fundamental Counting Principle (FCP) we can complete the 2 stages in (6)(5) ways (= 30 ways) ------------------------------------------------------------- So, the total number of triangles = 40 + 30 = 70 Answer: B Cheers, Brent _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 BTGmoderatorLU wrote: Source: e-GMAT In a plane, there are two parallel lines. One line has 5 points and another line has 4 different points. How many different triangles can we form from these 9 points? A. 62 B. 70 C. 73 D. 86 E. 122 $? = C\left( {5 + 4,3} \right) - \left[ {C\left( {5,3} \right) + C\left( {4,3} \right)} \right]$ $\left. \begin{gathered} C\left( {9,3} \right) = \frac{{9 \cdot 8 \cdot 7}}{{3 \cdot 2}} = 84 \hfill \\ C\left( {5,3} \right) = \frac{{5 \cdot 4}}{2} = 10 \hfill \\ C\left( {4,3} \right) = \left( {4,1} \right) = 4 \hfill \\ \end{gathered} \right\}\,\,\,\, \Rightarrow \,\,\,\,\,? = 84 - \left( {10 + 4} \right) = 70$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15348 messages Followed by: 1864 members Upvotes: 13060 GMAT Score: 790 BTGmoderatorLU wrote: Source: e-GMAT In a plane, there are two parallel lines. One line has 5 points and another line has 4 different points. How many different triangles can we form from these 9 points? A. 62 B. 70 C. 73 D. 86 E. 122 One approach: Good Cases = Total Cases - Bad Cases. Total Cases: From 9 points, the number of ways to choose 3 = 9C3 = (9*8*7)/(3*2*1) = 84. Bad Case 1: Choosing 3 collinear points from the 5-point line, with the result that a triangle cannot be formed From the 5 points on this line, the number of ways to choose 3 = 5C3 = (5*4*3)/(3*2*1) = 10. Bad Case 2: Choosing 3 collinear points from the 4-point line, with the result that a triangle cannot be formed From the 4 points on this line, the number of ways to choose 3 = 4C3 = (4*3*2)/(3*2*1) = 4. Good Cases = 84 - 10 - 4 = 70. The correct answer is B. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2791 messages Followed by: 18 members Upvotes: 43 BTGmoderatorLU wrote: Source: e-GMAT In a plane, there are two parallel lines. One line has 5 points and another line has 4 different points. How many different triangles can we form from these 9 points? A. 62 B. 70 C. 73 D. 86 E. 122 The OA is B. We can let line A be the line with 5 points and line B the line with 4 points. So we can have a triangle that is formed from choosing 1 point on line A and and 2 points on line B, or, 2 points on line A and 1 point on line B. In the former case, we have 5C1 x 4C2 = 5 x 6 = 30 ways of forming such a triangle, and in the latter case, we have 5C2 x 4C1 = 10 x 4 = 40 ways of forming such a triangle. Therefore, there are a total of 30 + 40 = 70 ways to form a triangle. Alternate solution: The number of ways to choose 3 points from 9 is 9C3 = 9!/(3!6!) = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84. However, we canâ€™t have all 3 points on the same line (since the 3 points canâ€™t be collinear). The number of ways of all 3 points from the line with 5 points is 5C3 = 5!/(3!2!) = (5 x 4 x 3)/(3 x 2) = 5 x 2 = 10 and similarly, the number of ways of all 3 points from the line with 4 points is 4C3 = 4. Therefore, there are a total of 84 - 10 - 4 = 70 ways forming a triangle. Answer: B _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. 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