It is currently 20 Apr 2018, 09:03

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If 4 points are indicated on a line and 5 points are indicated on

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Intern
Joined: 18 Mar 2012
Posts: 47
GPA: 3.7
If 4 points are indicated on a line and 5 points are indicated on [#permalink]

Show Tags

16 May 2012, 06:52
2
This post received
KUDOS
24
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

74% (00:56) correct 26% (01:50) wrong based on 549 sessions

HideShow timer Statistics

If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20
B. 30
C. 40
D. 70
E. 90
[Reveal] Spoiler: OA
Senior Manager
Joined: 13 May 2011
Posts: 276
WE 1: IT 1 Yr
WE 2: Supply Chain 5 Yrs
Re: If 4 points are indicated on a line and 5 points are indicated on [#permalink]

Show Tags

16 May 2012, 07:04
6
This post received
KUDOS
4
This post was
BOOKMARKED
say line A with 4 points and line B with 5 points. and triangle is PQR.
If P is on line A and QR on line B we can have 4C1*5C2=40 triangles.
If P is on line B and QR on line A we can have 4C2*5C1=30 triangles.
Total =70 triangles.
Math Expert
Joined: 02 Sep 2009
Posts: 44586
Re: If 4 points are indicated on a line and 5 points are indicated on [#permalink]

Show Tags

16 May 2012, 07:58
20
This post received
KUDOS
Expert's post
20
This post was
BOOKMARKED
alexpavlos wrote:
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20
B. 30
C. 40
D. 70
E. 90

Tried doing a few options, but couldn't get it right and the answer provided left me more confused!

Approach #1:

There are two types of triangles possible:
With two vertices on the line with 4 points and the third vertex on the line with 5 points --> $$C^2_4*C^1_5=30$$;
With two vertices on the line with 5 points and the third vertex on the line with 4 points --> $$C^2_5*C^1_4=40$$;

Total: $$30+40=70$$.

Answer: D.

Approach #2:

All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear.
$$C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70$$ (where $$C^3_4$$ and $$C^3_4$$ are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).

Answer: D.

Hope it's clear.
_________________
Senior Manager
Joined: 13 Aug 2012
Posts: 444
Concentration: Marketing, Finance
GPA: 3.23
Re: If 4 points are indicated on a line and 5 points are indicated on [#permalink]

Show Tags

29 Dec 2012, 01:27
3
This post received
KUDOS
2
This post was
BOOKMARKED
Not that fun this question I wonder how often I would see such a question on test day.

Solution:

* * * * * (line B)

* * * * (line A)

Choose 2 points from line A and a point from line B: = 4!/2!2! * 5!/1!4! = 30
Choose 1 point from line A and 2 points from line B: = 4!/1!3! * 5!/2!3! = 4 * 10 = 40

Combine possiblities: 30 + 40 = 70
_________________

Impossible is nothing to God.

Intern
Joined: 10 Apr 2012
Posts: 22
Concentration: Finance, Economics
GMAT 1: 760 Q50 V44
Re: If 4 points are indicated on a line and 5 points are indicated on [#permalink]

Show Tags

19 Jul 2014, 21:45
2
This post received
KUDOS
1
This post was
BOOKMARKED
See Image below for what my scratch paper looked like on this one
Attachments

Gmat_56_resized.jpg [ 101.22 KiB | Viewed 12154 times ]

Manager
Joined: 22 Feb 2009
Posts: 196
Re: If 4 points are indicated on a line and 5 points are indicated on [#permalink]

Show Tags

19 Jul 2014, 22:47
Bunuel wrote:
alexpavlos wrote:
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20
B. 30
C. 40
D. 70
E. 90

Tried doing a few options, but couldn't get it right and the answer provided left me more confused!

Approach #1:
There are two types of triangles possible:
With two vertices on the line with 4 points and the third vertex on the line with 5 points --> $$C^2_4*C^1_5=30$$;
With two vertices on the line with 5 points and the third vertex on the line with 4 points --> $$C^2_5*C^1_4=40$$;

Total: $$30+40=70$$.

Answer: D.

Approach #2:

All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear.
$$C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70$$ (where $$C^3_4$$ and $$C^3_4$$ are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).

Answer: D.

Hope it's clear.

I really like your 2nd approach
_________________

.........................................................................
+1 Kudos please, if you like my post

Intern
Joined: 29 Sep 2013
Posts: 8
Re: If 4 points are indicated on a line and 5 points are indicated on [#permalink]

Show Tags

06 Aug 2015, 14:54
Bunuel wrote:
alexpavlos wrote:
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20
B. 30
C. 40
D. 70
E. 90

Tried doing a few options, but couldn't get it right and the answer provided left me more confused!

Approach #1:
There are two types of triangles possible:
With two vertices on the line with 4 points and the third vertex on the line with 5 points --> $$C^2_4*C^1_5=30$$;
With two vertices on the line with 5 points and the third vertex on the line with 4 points --> $$C^2_5*C^1_4=40$$;

Total: $$30+40=70$$.

Answer: D.

Approach #2:

All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear.
$$C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70$$ (where $$C^3_4$$ and $$C^3_4$$ are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).

Answer: D.

Hope it's clear.

Thanks Bunuel. That was great. Do you by any chance have other problems similar to this that you know of?
Current Student
Joined: 20 Mar 2014
Posts: 2652
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If 4 points are indicated on a line and 5 points are indicated on [#permalink]

Show Tags

06 Aug 2015, 15:29
1
This post received
KUDOS
dukenukem wrote:
Bunuel wrote:
alexpavlos wrote:
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20
B. 30
C. 40
D. 70
E. 90

Tried doing a few options, but couldn't get it right and the answer provided left me more confused!

Approach #1:
There are two types of triangles possible:
With two vertices on the line with 4 points and the third vertex on the line with 5 points --> $$C^2_4*C^1_5=30$$;
With two vertices on the line with 5 points and the third vertex on the line with 4 points --> $$C^2_5*C^1_4=40$$;

Total: $$30+40=70$$.

Answer: D.

Approach #2:

All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear.
$$C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70$$ (where $$C^3_4$$ and $$C^3_4$$ are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).

Answer: D.

Hope it's clear.

Thanks Bunuel. That was great. Do you by any chance have other problems similar to this that you know of?

Combination problems are located at search.php?search_id=tag&tag_id=52
Director
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 595
Location: United States (CA)
Age: 38
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47
GMAT 3: 750 Q50 V42
GRE 1: 337 Q168 V169
WE: Education (Education)
Re: If 4 points are indicated on a line and 5 points are indicated on [#permalink]

Show Tags

06 Feb 2017, 09:47
Top Contributor
Attached is a visual that should help. Bunuel's method is the most elegant, but here is an alternative strategy if the other solution is not clear.
Attachments

Screen Shot 2017-02-06 at 8.46.05 AM.png [ 117.98 KiB | Viewed 5001 times ]

_________________

Harvard grad and 99% GMAT scorer, offering expert, private GMAT tutoring and coaching, both in-person (San Diego, CA, USA) and online worldwide, since 2002.

One of the only known humans to have taken the GMAT 5 times and scored in the 700s every time (700, 710, 730, 750, 770), including verified section scores of Q50 / V47, as well as personal bests of 8/8 IR (2 times), 6/6 AWA (4 times), 50/51Q and 48/51V (1 question wrong).

You can download my official test-taker score report (all scores within the last 5 years) directly from the Pearson Vue website: https://tinyurl.com/y8zh6qby Date of Birth: 09 December 1979.

GMAT Action Plan - McElroy Tutoring

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 2447
Location: United States (CA)
Re: If 4 points are indicated on a line and 5 points are indicated on [#permalink]

Show Tags

14 Mar 2018, 16:09
alex1233 wrote:
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20
B. 30
C. 40
D. 70
E. 90

We have two scenarios:

The first scenario is that the 1 vertex is on the first line and the 2 other vertices are on the second line.
So the number of ways to create such a triangle is:

4C1 x 5C2 = 4 x (5 x 4)/2 = 40

The second scenario is that 2 vertices are on the first line and the 1 vertex is on the second line. So the number of ways to create such a triangle is:

4C2 x 5C1 = (4 x 3)/2 x 5 = 6 x 5 = 30

Thus, the total number of ways to create the triangle is 40 + 30 = 70.

Answer: D
_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Re: If 4 points are indicated on a line and 5 points are indicated on   [#permalink] 14 Mar 2018, 16:09
Display posts from previous: Sort by

If 4 points are indicated on a line and 5 points are indicated on

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.