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

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16 May 2012, 05:52

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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!

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 indicat [#permalink]

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16 May 2012, 06:04

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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.

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

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29 Dec 2012, 00:27

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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

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 [#permalink]

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19 Jul 2014, 21:47

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

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

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31 Jul 2015, 00:02

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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, 13:54

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 [#permalink]

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30 Aug 2016, 18:06

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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