gmat geometry : PS Archive
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# gmat geometry

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27 Mar 2010, 23:10
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7 points are marked on a straight line and another 8 points are marked on a second straight line with no points in common. How many triangles can be constructed with vertices from among the above points?

A. 91

B. 105

C. 196

D. 21

E. 14
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28 Mar 2010, 00:12
Well, I was just thinking that is this a geometry problem or a combinatorics case??

Anyhow, couldn't solved it yet.
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28 Mar 2010, 04:15
Richaarora22 wrote:
7 points are marked on a straight line and another 8 points are marked on a second straight line with no points in common. How many triangles can be constructed with vertices from among the above points?
A. 91
B. 105
C. 196
D. 21
E. 14

to form a traingle we need 3 points .. as we have 2 lines, 2 points from one line and another point from the other line.
7c2 * 8 + 8c2 * 7 ( doesnt match any of the answers though)
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28 Mar 2010, 04:33
chix475ntu wrote:
Richaarora22 wrote:
7 points are marked on a straight line and another 8 points are marked on a second straight line with no points in common. How many triangles can be constructed with vertices from among the above points?
A. 91
B. 105
C. 196
D. 21
E. 14

to form a traingle we need 3 points .. as we have 2 lines, 2 points from one line and another point from the other line.
7c2 * 8 + 8c2 * 7 ( doesnt match any of the answers though)

another way.... 15c3 = total triangles
7c3- triangles formed by line A, 8c3 - triangles formed by line B
required Triangles = 15c3-7c3-8c3 = 455 - 35 - 56 = 364

7c2 * 8 + 8c2 * 7 is also 364....??? whats OA?
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28 Mar 2010, 06:50
We have 02 lines, the first has 8 points & the second 7. If we start with the first 8 points, we have:

8! / 3! * (8-3)! possibilities to draw triangles or 56 triangles

Second line with 7 points:

7! / 3! * (7-3) ! or 35 triangles

56 + 35 = 91 Triangles - Answer A.

Because otherwise, it should be 364 but it is not in the answer choices !!!!!!

- Line 1 (8 points) & Line 2 (7 points), For Triangle: from line 1, we need 2 points: 8!/2!(8-2)! = 28 * 7 (third point on line 2) = 196

- Line 2 (7 points) & Line 1 (8 points), For Triangle: from line 2, we need 2 points: 7!/2!(7-2)! = 21 * 8 (third point on line 1) = 168

Total Number of triangles: 196 + 168 = 364 - NOT IN ANSWER CHOICE.........

Please advice why 364 is not there and therefore whether A is correct. Thanks.
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28 Mar 2010, 19:58
Albator wrote:
We have 02 lines, the first has 8 points & the second 7. If we start with the first 8 points, we have:

8! / 3! * (8-3)! possibilities to draw triangles or 56 triangles

Second line with 7 points:

7! / 3! * (7-3) ! or 35 triangles

56 + 35 = 91 Triangles - Answer A.

Because otherwise, it should be 364 but it is not in the answer choices !!!!!!

- Line 1 (8 points) & Line 2 (7 points), For Triangle: from line 1, we need 2 points: 8!/2!(8-2)! = 28 * 7 (third point on line 2) = 196

- Line 2 (7 points) & Line 1 (8 points), For Triangle: from line 2, we need 2 points: 7!/2!(7-2)! = 21 * 8 (third point on line 1) = 168

Total Number of triangles: 196 + 168 = 364 - NOT IN ANSWER CHOICE.........

Please advice why 364 is not there and therefore whether A is correct. Thanks.

Agreed. I also got the same answer. The answer choice seems to to be incorrect.
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29 Mar 2010, 09:47
For me 364 seems to be the answer too.
Are we missing something here?
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01 Apr 2010, 04:58
Richaarora22 wrote:
7 points are marked on a straight line and another 8 points are marked on a second straight line with no points in common. How many triangles can be constructed with vertices from among the above points?

A. 91

B. 105

C. 196

D. 21

E. 14

Well the question asks "how many triangles can be constructed with vertices from among the above points"

So we have 7 points above and 8 points below on two seperate lines
We can choose 2 points out 8 as 8C2 = 28
No. of triangles can be 7*28 = 196 (one point taken from above line with 2 points on line below)

IMO C

Hope this helps!!!!
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07 Oct 2010, 17:15
hardnstrong wrote:
Richaarora22 wrote:
7 points are marked on a straight line and another 8 points are marked on a second straight line with no points in common. How many triangles can be constructed with vertices from among the above points?

A. 91

B. 105

C. 196

D. 21

E. 14

Well the question asks "how many triangles can be constructed with vertices from among the above points"

So we have 7 points above and 8 points below on two seperate lines
We can choose 2 points out 8 as 8C2 = 28
No. of triangles can be 7*28 = 196 (one point taken from above line with 2 points on line below)

IMO C

Hope this helps!!!!

If you are suggesting that the phrase ' above points ' means 7 points not the total 15 points, then you are wrong.

Above points means all the above 15 points.

How you can say the 7 points are above and 8 are below? it could be vice-versa. Moreover it said vertices from above points. if I agree to what you have said, how can you create triangle with vertices on 7 collinear points?
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07 Oct 2010, 20:30
I beleive this is where its important to get questions and their answer choices correct. I found the correct question at below link

http://www.beatthegmat.com/how-many-uni ... 51280.html
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Re: gmat geometry   [#permalink] 07 Oct 2010, 20:30
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