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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
If 4 points are indicated on a line and 5 points are indicated on
[#permalink]
Show Tags
16 May 2012, 06:52
2
26
00:00
A
B
C
D
E
Difficulty:
25% (medium)
Question Stats:
74% (00:57) correct 26% (01:46) wrong based on 593 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?
Re: If 4 points are indicated on a line and 5 points are indicated on
[#permalink]
Show Tags
16 May 2012, 07:58
24
18
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).
Re: If 4 points are indicated on a line and 5 points are indicated on
[#permalink]
Show Tags
16 May 2012, 07:04
8
4
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.
Re: If 4 points are indicated on a line and 5 points are indicated on
[#permalink]
Show Tags
29 Dec 2012, 01:27
3
2
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
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
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?
Re: If 4 points are indicated on a line and 5 points are indicated on
[#permalink]
Show Tags
06 Aug 2015, 15:29
1
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?
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 664
Location: United States (CA)
Age: 38
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47
GMAT 3: 750 Q50 V42
GRE 1: 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 6740 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/y94hlarr Date of Birth: 09 December 1979.
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
close
74% (00:57) correct 
I wonder how often I would see such a question on test day.
