GMAT prep question pack 1 hard question!

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GMAT prep question pack 1 hard question!

by rakeshd347 » Mon Sep 09, 2013 3:49 am
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

This is gmatprep question pack 1 question even though there is explanation given but that is beyond my understanding. Can someone please provide simple explanation.

OA is [spoiler][D][/spoiler]
Last edited by rakeshd347 on Mon Sep 09, 2013 2:49 pm, edited 2 times in total.
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by Brent@GMATPrepNow » Mon Sep 09, 2013 6:28 am
rakeshd347 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
When posting questions, please use the spoiler function to hide the correct answer. This will allow others to attempt the question without seeing the final answer

There are two ways in which we can create a triangle.
#1) Select 2 points from the 5-point line and select 1 point from the 4-point line.
#2) Select 2 points from the 4-point line and select 1 point from the 5-point line.

#1) Select 2 points from the 5-point line and select 1 point from the 4-point line.
Take this task and break it into stages.

Stage 1: Select 2 points from the 5-point line
Since the order of the 2 selected points does not matter, we can use combinations.
We can select 2 points from 5 points in 5C2 = 10 ways.

If anyone is interested, we have a free video on calculating combinations (like 5C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Select 1 point from the 4-point line.
We can complete this stage in 4 ways

By the Fundamental Counting Principle (FCP) we can complete the 2 stages in (10)(4) ways (= 40 ways)

#2) Select 2 points from the 4-point line and select 1 point from the 5-point line.
Take this task and break it into stages.

Stage 1: Select 2 points from the 4-point line
We can select 2 points from 4 points in 4C2 = 6 ways.

Stage 2: Select 1 point from the 5-point line.
We can complete this stage in 5 ways

By the Fundamental Counting Principle (FCP) we can complete the 2 stages in (6)(5) ways (= 30 ways)
-------------------------------------------------------------
So, the total number of triangles = 40 + 30 = [spoiler]70 = D[/spoiler]

Cheers,
Brent
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by GMATGuruNY » Mon Sep 09, 2013 12:27 pm
rakeshd347 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
Alternate approach:

To form a triangle, we must select a COMBINATION of 3 points from the 9 options contained in the two lines.

Good combinations = total possible combinations - bad combinations.

Total possible combinations:
Number of ways to choose a combination of 3 points from 9 options = 9C3 = (9*8*7)/(3*2*1) = 84.

Bad combinations:
A BAD combination consists of 3 points selected from the SAME LINE (since these 3 points are collinear and thus cannot be used to form a triangle).
Number of ways to choose a combination of 3 points from the 4 options on line one = 4C3 = (4*3*2)/(3*2*1) = 4.
Number of ways to choose a combination of 3 points from the 5 options on line two = 5C3 = (5*4*3)/(3*2*1) = 10.

Thus:
Good combinations = 84-4-10 = 70.

The correct answer is D.
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by rakeshd347 » Mon Sep 09, 2013 2:51 pm
Brent@GMATPrepNow wrote:
rakeshd347 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
When posting questions, please use the spoiler function to hide the correct answer. This will allow others to attempt the question without seeing the final answer

There are two ways in which we can create a triangle.
#1) Select 2 points from the 5-point line and select 1 point from the 4-point line.
#2) Select 2 points from the 4-point line and select 1 point from the 5-point line.

#1) Select 2 points from the 5-point line and select 1 point from the 4-point line.
Take this task and break it into stages.

Stage 1: Select 2 points from the 5-point line
Since the order of the 2 selected points does not matter, we can use combinations.
We can select 2 points from 5 points in 5C2 = 10 ways.

If anyone is interested, we have a free video on calculating combinations (like 5C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Select 1 point from the 4-point line.
We can complete this stage in 4 ways

By the Fundamental Counting Principle (FCP) we can complete the 2 stages in (10)(4) ways (= 40 ways)

#2) Select 2 points from the 4-point line and select 1 point from the 5-point line.
Take this task and break it into stages.

Stage 1: Select 2 points from the 4-point line
We can select 2 points from 4 points in 4C2 = 6 ways.

Stage 2: Select 1 point from the 5-point line.
We can complete this stage in 5 ways

By the Fundamental Counting Principle (FCP) we can complete the 2 stages in (6)(5) ways (= 30 ways)
-------------------------------------------------------------
So, the total number of triangles = 40 + 30 = [spoiler]70 = D[/spoiler]

Cheers,
Brent
Thank you Brent. It looks so easy now. moreover there is a takeaway with this which I can apply to make quadrilateral as well. I have full membership of gmatprepnow and I went through counting module but couldn't find this trick in there. Moreover, I think GMAC should give good explanation with these questions.

Thanks

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by rakeshd347 » Mon Sep 09, 2013 2:53 pm
GMATGuruNY wrote:
rakeshd347 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
Alternate approach:

To form a triangle, we must select a COMBINATION of 3 points from the 9 options contained in the two lines.

Good combinations = total possible combinations - bad combinations.

Total possible combinations:
Number of ways to choose a combination of 3 points from 9 options = 9C3 = (9*8*7)/(3*2*1) = 84.

Bad combinations:
A BAD combination consists of 3 points selected from the SAME LINE (since these 3 points are collinear and thus cannot be used to form a triangle).
Number of ways to choose a combination of 3 points from the 4 options on line one = 4C3 = (4*3*2)/(3*2*1) = 4.
Number of ways to choose a combination of 3 points from the 5 options on line two = 5C3 = (5*4*3)/(3*2*1) = 10.

Thus:
Good combinations = 84-4-10 = 70.

The correct answer is D.
Thank you mate. Its always good to have 2-3 different strategies in your pocket to tackle future similar problem.

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by Brent@GMATPrepNow » Mon Sep 09, 2013 8:39 pm
rakeshd347 wrote: Thank you Brent. It looks so easy now. moreover there is a takeaway with this which I can apply to make quadrilateral as well. I have full membership of gmatprepnow and I went through counting module but couldn't find this trick in there. Moreover, I think GMAC should give good explanation with these questions.
Hey rakeshd347,

Video #3 of our Counting module covers the Fundamental Counting Principle (FCP): https://www.gmatprepnow.com/module/gmat-counting?id=775

Cheers,
Brent
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by mcdesty » Sat Jul 19, 2014 8:48 pm
Here is what my scratch paper looked like on this one.
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by GMATinsight » Sat Jul 19, 2014 10:39 pm
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
Triangles are possible If Randomly select 3 Points from 9 - (Select 3 point out of 4 in straight Line + Select 3 point out of 4 in straight Line)

Total Favorable cases = 9C3 - ( 4C3 + 5C3 ) = 84 - (4+10) = 84-14 = 70

Answer: Option D
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by GMATinsight » Sat Jul 19, 2014 10:58 pm
Alternately:

We can select the points in following ways as well

1) 2 Points out of 4 Collinear points and 1 point out of 5 Collinear points

4C2 x 5C1 = 6x5 = 30

2) 1 Points out of 4 Collinear points and 2 point out of 5 Collinear points

4C1 x 5C2 = 4x10 = 40

Total Such Numbers = 30+40 = 70

Answer: Option D
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