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15 Jan 2023, 09:02
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Schachfreizeit
Hello Schachfreizeit,
I will respond to both your queries here so please bear with me as the response may be long. 😊
Let me begin by telling you that just like you, I am also not a native speaker. I can understand your pain and will try to help you understand the question and the solution in as simple terms as possible.
UNDERSTANDING THE QUESTION (Not solving yet)
- Question Stem: We will take small chunks of the stem and understand it as we go.
- “S is a set of points in the plane.”
- This sentence tells us that there are some points in a plane (a plane is any 2D surface, say a wall). All these points form a group represented as S.
- “How many distinct triangles can be drawn that have three of the points in S as vertices?”
- This sentence asks us how many triangles we can draw using the points from set S only.
- Now, to draw a triangle, we always need 3 points. These 3 points act as the vertices of our triangle.
- So, the question basically asks, “in how many ways we can select 3 points from group S”. Because each new combination of 3 points will form a unique triangle. However, there is an *EXCEPTION to this that we will cover later.
- This gives us the total number of points in set S.
- To understand this statement, you need to understand the term “collinear”. Collinear means “lying on the same line”. So, this statement says that no three points in set S lie on a straight line.
- Now, why is this information relevant? Let’s see. Suppose we select 3 points and these 3 points are collinear. Then, when you join these 3 points, you would form a line instead of a triangle as needed. So, such a collection of 3 points will not be valid for us. *This is the exception that we had referred to earlier. That is, in general, each combination of 3 points will not give us a triangle. But each combination of 3 non-collinear points will definitely give us a triangle!
- Now that we understand the term ‘collinear’, let’s come back to Statement 2. This statement confirms that every collection of 3 points of set S will give us a triangle. Hence, we must consider every possible combination.
I hope you now understand the question clearly.
SOLVING THE QUESTION AND COMING TO 5C3
We need to find number of all possible combinations of 3 points. Each such combination would form a triangle.
To find number of ways in which we can select 3 points from set S, we must know:
- The total number of points in set S.
- And whether any 3 points are collinear. (Because if they are, they would not form a triangle and hence that combination of 3 points must be rejected)
From Statement 1: “The number of distinct points in S is 5”
This gave us (i) as we needed. But we still do not know about collinearity of these points – that is, we have no idea about (ii). So, statement 1 alone is insufficient.
From Statement 2: “No three of the points in S are collinear”
This tells us all about (ii) but nothing about (i). (Make sure you never drag anything from statement 1 into 2) So, statement 2 alone is insufficient.
Since, individually each statement is insufficient, we must combine the statements. Now, Statement 1 and 2 together give us both - (i) and (ii) - that we needed. Thus, the answer is C.
We will still do the Math and show you the final calculation.
- The total number of triangles that can be formed = Number of ways in which we can select 3 points from a total of 5 points.
- From our conceptual understanding, we know that we can select ‘r’ things from a total of ‘n’ things in nCr ways.
- Using the same, we can select 3 points from a total of 5 points in 5C3 ways. This is precisely how you get 5C3 as the number of distinct triangles that can be drawn using 3 points in the set as vertices.
Hope this helps!
Best Regards,
Ashish
Quant Expert, e-GMAT
31 Jul 2025, 23:48
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I chose E considering if you combine (1) and (2), it is still not possible to ascertain if the resulting pentagon is regular or not. If it is regular, there are only 2 distinct congruence classes of triangles, but if the pentagon is not regular, it may contain up to 5C3 distinct congruence classes of triangles.
31 Jul 2025, 23:54
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robin36520
It does not matter. We have 5 distinct points on a plane. The number of distinct triangles that can be formed is simply 5C3. Do not overcomplicate it.
