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

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Intern
Joined: 18 Mar 2012
Posts: 47

Kudos [?]: 273 [2], given: 117

GPA: 3.7
If 4 points are indicated on a line and 5 points are [#permalink]

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16 May 2012, 06:52
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Difficulty:

25% (medium)

Question Stats:

74% (00:58) correct 26% (01:55) wrong based on 448 sessions

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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!
[Reveal] Spoiler: OA

Kudos [?]: 273 [2], given: 117

Math Expert
Joined: 02 Sep 2009
Posts: 42270

Kudos [?]: 132821 [20], given: 12378

If 4 points are indicated on a line and 5 points are [#permalink]

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16 May 2012, 07:58
20
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Expert's post
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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: $$30+40=70$$.

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).

Hope it's clear.
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Kudos [?]: 132821 [20], given: 12378

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Re: If 4 points are indicated on a line and 5 points are indicat [#permalink]

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16 May 2012, 07:04
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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.

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Senior Manager
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Re: If 4 points are indicated on a line and 5 points are [#permalink]

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29 Dec 2012, 01:27
2
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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
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Kudos [?]: 558 [2], given: 11

Intern
Joined: 10 Apr 2012
Posts: 23

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Concentration: Finance, Economics
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Re: If 4 points are indicated on a line and 5 points are [#permalink]

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19 Jul 2014, 21:45
1
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See Image below for what my scratch paper looked like on this one
Attachments

Gmat_56_resized.jpg [ 101.22 KiB | Viewed 10111 times ]

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Current Student
Joined: 20 Mar 2014
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Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
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Re: If 4 points are indicated on a line and 5 points are [#permalink]

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06 Aug 2015, 15:29
1
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: $$30+40=70$$.

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).

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

Kudos [?]: 1773 [1], given: 794

Manager
Joined: 22 Feb 2009
Posts: 210

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Re: If 4 points are indicated on a line and 5 points are [#permalink]

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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: $$30+40=70$$.

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).

Hope it's clear.

I really like your 2nd approach
_________________

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+1 Kudos please, if you like my post

Kudos [?]: 181 [0], given: 148

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

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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: $$30+40=70$$.

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).

Hope it's clear.

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

Kudos [?]: 5 [0], given: 11

Senior Manager
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
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Location: United States (CA)
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GRE 1: 337 Q168 V169
WE: Education (Education)
If 4 points are indicated on a line and 5 points are [#permalink]

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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.
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Screen Shot 2017-02-06 at 8.46.05 AM.png [ 117.98 KiB | Viewed 2957 times ]

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If 4 points are indicated on a line and 5 points are   [#permalink] 06 Feb 2017, 09:47
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