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CO ordiates !

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CO ordiates !

by vishubn » Fri Oct 24, 2008 8:33 pm

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S is a set of points in the plane. How many distinct triangles can be drawn that have three
of the points in S as vertices?
(1) The number of distinct points in S is 5.
(2) No three of the points in S are collinear.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

OA C
I think DS is wat i screw up :(
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Re: CO ordiates !

by logitech » Sat Oct 25, 2008 12:40 am

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vishubn wrote:S is a set of points in the plane. How many distinct triangles can be drawn that have three
of the points in S as vertices?
(1) The number of distinct points in S is 5.
(2) No three of the points in S are collinear.

OA C
I think DS is wat i screw up :(
Vishbun:

(2) is insuf because it has no numbers in it! SO you can eliminate B and D

now you have check whether (1) is sufficient and remember even you have to guess, you increased your chances from 20 % to 33 % just by eliminating the easy statement.


(1) gives you some numbers but you don't know how those points are distributed in XY plane ? what if there are all in same line


. . . . .

So you still can't decide. You can eliminate A too.

Now your chances are 50 % - 50 %

Is it C or is it E ?

Well if you combines both statements, you have the number information and you know how they are respect to eacther

Go with the C!
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Re: CO ordiates !

by parallel_chase » Sat Oct 25, 2008 12:46 am

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vishubn wrote:S is a set of points in the plane. How many distinct triangles can be drawn that have three
of the points in S as vertices?
(1) The number of distinct points in S is 5.
(2) No three of the points in S are collinear.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

OA C
I think DS is wat i screw up :(
The answer is Indeed C.

I guess you got confused by the word "collinear"
Collinear mean when two or more points lie on the same line


Statement I
5 distinct points, triangles formed = 5c3 = 10, but we dont know how many points are collinear. Therefore Insufficient.

Statement II
no 3 points are collinear, but we dont know how many points does set S have. Insufficient.

Combining I & II

No 3 points collinear, 5 distinct points, distinct triangles = 5C3 = 10.

Hope this helps.
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Re: CO ordiates !

by lunarpower » Sat Oct 25, 2008 8:46 pm

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well done. i have only two things to add, and both of them are somewhat nitpicky.

one (emphasis mine):
parallel_chase wrote:Statement I
5 distinct points, triangles formed = 5c3 = 10, but we dont know how many points are collinear. Therefore Insufficient.
actually, that's not the number of triangles formed. that's the number of distinct sets of three points.
ironically in this case, the whole crux of this statement's being insufficient is that you don't know whether all of these sets of points actually represent triangles. in particular, if you get 3 points that are lined up, then you'll get a straight line instead of a triangle.

also, you don't actually have to compute the combinatorial formula (and indeed you shouldn't, unless such computations are straightforward and easy for you). instead, you can just consider the following 2 extreme cases:
(a) all 5 points are in a straight line**. total = 0 triangles.
(b) none of the points are lined up. total = definitely more than 0 triangles.
therefore, insufficient.

**notice that it should be easier to realize that this particular issue matters, because the other statement alludes to it.

--

two:
No 3 points collinear, 5 distinct points, distinct triangles = 5C3 = 10.
again, note that there's no need to perform the actual combinatorial computation here; indeed, it's a waste of time, unless you find such computations straightforward and easy.

you can just realize that there are 5 points, none of which are lined up, and so there must be some fixed # of combinations of those points. it's the same as picking 3 books from a set of five: even if you don't know how to compute the number right away, it's good enough just to figure out that the number exists in the first place.
Ron has been teaching various standardized tests for 20 years.

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Re: CO ordiates !

by reacher » Sun Oct 26, 2008 6:28 am

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vishubn wrote:S is a set of points in the plane. How many distinct triangles can be drawn that have three
of the points in S as vertices?
(1) The number of distinct points in S is 5.
(2) No three of the points in S are collinear.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

This question is asking you to recollect the fundamental ideas in geometry, especially the postulates. Earlier quotes have caught the fish for you but it will be better if you learn how to catch it yourself. You know that there are "postulates: a postulate is a statement that is assumed true without proof" and "theorems: a true statement that can be proven" in geometry.

There are six postulates:
1. A line contains at least two points.
2. A plane contains at least three noncollinear points.
3. Through any two points, there is exactly one line.
4. Through any three noncollinear points, there is exactly one plane.
5. If two points lie in a plane, then the line joining them lies in that plane.
6. If two planes intersect, then their intersection is a line.

The question above is giving you the hints of the postulates in disguised form. Read these postulates with the explanation given by earlier quotes and I am sure you will see the fish biting your bait.
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by [email protected] » Sun Apr 08, 2018 3:16 pm

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Hi All,

We're told that Set S is a set of points in the plane (meaning points in a graph). We're asked for the number of distinct triangles that can be drawn that have three of the points in S as vertices. To start, it's worth noting that any 2 DISTINCT points in a plane will form a line segment; to form a triangle, we need a third point that is NOT in the same line as the other two.

1) The number of distinct points in S is 5.

IF....
All 5 points are on a straight line, then the answer to the question is 0.
At least one of the points is NOT on the same line as the other 4 points, then the answer to the question is some positive integer.
Fact 1 is INSUFFICIENT

2) No three of the points in S are collinear.

With Fact 2, we don't know the actual number of points, but we know that there are at least 3 points - and no 3 points are in the same line. This means that there's at least 1 triangle, but there may be more than that.
Fact 2 is INSUFFICIENT

Combined, we know
-The number of distinct points in S is 5.
-No three of the points in S are collinear.

With exactly 5 points in the plane, and the fact that no 3 points are in the same line, we CAN determine the exact number of triangles that can be formed (you don't actually have to do the math either, although if you did, then you would find that there are 5!/3!2! = 10 possible triangles).
Combined, SUFFICIENT

Final Answer: C

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by Scott@TargetTestPrep » Thu Apr 12, 2018 3:48 pm

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vishubn wrote:S is a set of points in the plane. How many distinct triangles can be drawn that have three
of the points in S as vertices?

(1) The number of distinct points in S is 5.
(2) No three of the points in S are collinear.
We are given that S is a set of points in the plane, and we must determine how many distinct triangles can be drawn with three of the points in S as vertices. So, essentially, we must determine how many distinct triangles can be drawn with the points provided.

Statement One Alone:

The number of distinct points in S is 5.

Using the information in statement one it may be tempting to conclude that the number of triangles that can be drawn is 5C3 = (5 x 4 x 3)/3! = 10 triangles. However, because we do not know the positioning of the points, we cannot actually say that 10 distinct triangles can be created. Let's say for instance that all the points were collinear, which means that they are all located on one line. If that were the case, we would not be able to create any triangles. Thus, statement one is not sufficient to answer the question.
Note: We were able to determine that 10 triangles could be formed with 5 points if and only if no 3 points are collinear. Only then would the number of triangles be 5C3 = 10.

Statement Two Alone:

No three of the points in S are collinear.

Since we don't know the number of points in S, we cannot answer the question using the information in statement two . We can eliminate answer choice B.

Statements One and Two Together:

Using the information from statements one and two, we know that we have 5 points in the plane and that no three points are collinear. Thus, we can determine that the number of triangles that can be created in the plane is 5C3 = 10.

Answer: C
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Re: CO ordiates !

by GMATE » Sat Oct 17, 2020 7:54 am

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I am confused by the term distinct. I understand that both 1 and 2 are insufficient individually however, I don't understand why we have 10 DISTINCT triangles.
From statment 1&2 we know that we have 5C3 = 10 triangles, but why are they distinct?
If two or more of the triangles have the same angles and same distance between the points are they then still distinct?
Does distinct in this context simply mean that the triangles don't have the same three points OR does the shape have to be distinct as well?

Thank you in advance,
GMATE
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