BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

There are two parallel line segments AB and CD in a plane. Ab contains 12 marked points...

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Moderator
Posts: 2599
Joined: Sun Oct 29, 2017 2:08 pm
Followed by:2 members

There are two parallel line segments AB and CD in a plane. Ab contains 12 marked points...

by AAPL » Thu Jun 04, 2020 2:04 am

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

GMAT Paper Tests

There are two parallel line segments AB and CD in a plane. Ab contains 12 marked points whereas CD contains 8 marked points. How many triangles can be formed by using these marked points as vertices?

A. (12!∗28)+(8!∗66)
B. 12!∗8!
C. 20C3
D. 864
E. 1024

OA D
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8085
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

Re: There are two parallel line segments AB and CD in a plane. Ab contains 12 marked points...

by Scott@TargetTestPrep » Tue Jun 09, 2020 4:09 am
AAPL wrote: ↑
Thu Jun 04, 2020 2:04 am
 GMAT Paper Tests

There are two parallel line segments AB and CD in a plane. Ab contains 12 marked points whereas CD contains 8 marked points. How many triangles can be formed by using these marked points as vertices?

A. (12!∗28)+(8!∗66)
B. 12!∗8!
C. 20C3
D. 864
E. 1024

OA D
Solution:

Since 3 non-collinear points form a triangle, a triangle can be form by either picking two points from AB and one point from CD OR two points from CD and one point from AB (notice that, for example, we can’t pick 3 points from AB to form a triangle since these 3 points would be linear).

Case 1: two points from AB and one point from CD

The number of ways this can be done is: 12C2 x 8C1 = (12 x 11)/2 x 8 = 66 x 8 = 528.

Case 2: two points from CD and one point from AB

The number of ways this can be done is: 8C2 x 12C1 = (8 x 7)/2 x 12 = 28 x 12 = 336.

Therefore, the total number ways a triangle can be formed is 528 + 336 = 864.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Legendary Member
Posts: 2499
Joined: Sun Oct 29, 2017 2:04 pm
Followed by:6 members

Re: There are two parallel line segments AB and CD in a plane. Ab contains 12 marked points...

by swerve » Wed Jun 10, 2020 4:16 am
AAPL wrote: ↑
Thu Jun 04, 2020 2:04 am
 GMAT Paper Tests

There are two parallel line segments AB and CD in a plane. Ab contains 12 marked points whereas CD contains 8 marked points. How many triangles can be formed by using these marked points as vertices?

A. (12!∗28)+(8!∗66)
B. 12!∗8!
C. 20C3
D. 864
E. 1024

OA D
We need 3 points in total to make a triangle.

You can select either 2 points from AB or 2 points from CD. Thus there are 2 possible scenarios
1. Choosing from AB first
12C2 AND 8C1 = 66*8
= 528

OR
2. Choosing from CD first
8C2 AND 12C1 = 336

Total up = 864 \(\Longrightarrow\) D
Top
Post new topic Post Reply
• Page 1 of 1