| View previous topic :: View next topic |
| Author |
Message |
umaa Really wants to Beat The GMAT!
Joined: 08 Jun 2008 Posts: 106
Thanks given: 2 Thanked 0 times in 0 posts
|
Posted: Tue Jul 08, 2008 10:29 pm Post subject: no of parallelograms?? |
|
|
The number of parallelograms that can be formed from a set of FOUR parallel straight line intersecting a set of THREE parallel straight lines=?
Options are,
a. 9
b. 12
c. 6
d. 18
|
|
| Back to top |
|
 |
|
|
amitdgr Rising GMAT Star
Joined: 29 Jun 2008 Posts: 73
Thanks given: 2 Thanked 2 times in 2 posts
|
Posted: Wed Jul 09, 2008 2:36 am Post subject: Re: no of parallelograms?? |
|
|
| umaa wrote: | The number of parallelograms that can be formed from a set of FOUR parallel straight line intersecting a set of THREE parallel straight lines=?
Options are,
a. 9
b. 12
c. 6
d. 18 |
Is the OA c.6 ??
Amit
|
|
| Back to top |
|
 |
umaa Really wants to Beat The GMAT!
Joined: 08 Jun 2008 Posts: 106
Thanks given: 2 Thanked 0 times in 0 posts
|
Posted: Wed Jul 09, 2008 2:48 am Post subject: |
|
|
| No my friend. I've got the same answer. The answer is 18 (d).
|
|
| Back to top |
|
 |
amitdgr Rising GMAT Star
Joined: 29 Jun 2008 Posts: 73
Thanks given: 2 Thanked 2 times in 2 posts
|
Posted: Wed Jul 09, 2008 3:16 am Post subject: |
|
|
| umaa wrote: | | No my friend. I've got the same answer. The answer is 18 (d). |
18 ? oh ..
how is it 18 ?
Amit
|
|
| Back to top |
|
 |
Ian Stewart GMAT Instructor
Joined: 02 Jun 2008 Posts: 523
Thanks given: 1 Thanked 100 times in 94 posts
Location: London
|
Posted: Wed Jul 09, 2008 5:43 am Post subject: |
|
|
You're only counting the six smallest parallelograms in the picture (nice picture, by the way!). If, for example, you take all six of the small parallelograms and consider it to be one shape, you get one large parallelogram. And there are many others:
6 parallelograms that measure 1 across, 1 vertical (smallest possible)
3 parallelograms that are 2 across and 1 vertical
4 parallelograms that are 1 across and 2 vertical
2 parallelograms that are 2 across and 2 vertical
2 parallelograms that are 1 across and 3 vertical
1 parallelogram that is 2 across and 3 vertical (all of them together)
Hope the terminology makes sense. That's a total of 18 parallelograms.
_________________ co-founder, www.gmatix.com
currently available:
-free international directory of private GMAT tutors and MBA Admissions Consultants
-more soon |
|
| Back to top |
|
|
 |
|
|
umaa Really wants to Beat The GMAT!
Joined: 08 Jun 2008 Posts: 106
Thanks given: 2 Thanked 0 times in 0 posts
|
Posted: Wed Jul 09, 2008 6:02 am Post subject: Thanks |
|
|
Thanks Ian. I missed that out.
|
|
| Back to top |
|
 |
amitdgr Rising GMAT Star
Joined: 29 Jun 2008 Posts: 73
Thanks given: 2 Thanked 2 times in 2 posts
|
Posted: Wed Jul 09, 2008 6:05 am Post subject: |
|
|
| Ian Stewart wrote: | You're only counting the six smallest parallelograms in the picture (nice picture, by the way!). If, for example, you take all six of the small parallelograms and consider it to be one shape, you get one large parallelogram. And there are many others:
6 parallelograms that measure 1 across, 1 vertical (smallest possible)
3 parallelograms that are 2 across and 1 vertical
4 parallelograms that are 1 across and 2 vertical
2 parallelograms that are 2 across and 2 vertical
2 parallelograms that are 1 across and 3 vertical
1 parallelogram that is 2 across and 3 vertical (all of them together)
Hope the terminology makes sense. That's a total of 18 parallelograms. |
Thanks for explaining it Ian
| Ian Stewart wrote: | | (nice picture, by the way!). |
Amit
|
|
| Back to top |
|
 |
cubicle_bound_misfit Rising GMAT Star

Joined: 19 May 2008 Posts: 85
Thanks given: 1 Thanked 3 times in 3 posts
Location: Texaco Gas Station Test Date: 06/30/2008 Target GMAT Score: 730
|
Posted: Wed Jul 09, 2008 11:18 pm Post subject: |
|
|
you need to choose 2 parallel lines along with 2 vertical line.
2 parallel lines can be chosen from 4 parallel line in 4C2 ways .
2 vertical lines can be chosen from 3 vertical line in 3C2 ways
together the condition can be chosen in 4C2 * 3C2 ways ==18.
Hope that helps.
Regards,
_________________ Cubicle Bound Misfit |
|
| Back to top |
|
 |
|