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TheNightKing
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GMAT 1: 600 Q46 V27
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Let AB, CD, EF, GH, and JK be the five diameters of a circle 09 Sep 2019, 10:14
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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85% (hard)

Question Stats:

49% (02:35) correct 51% (02:25) wrong based on 115 sessions

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Let AB, CD, EF, GH, and JK be the five diameters of a circle with center O. In how many ways can three points be chosen out of A,B,C,D,E,F,G,H,J,K and O so as to form a triangle?

A. 165
B. 160
C. 185
D. 200
E. 225­
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B
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Let AB, CD, EF, GH, and JK be the five diameters of a circle 09 Sep 2019, 10:29
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Imo 11C3-5 = 160

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Let AB, CD, EF, GH, and JK be the five diameters of a circle 31 Mar 2020, 01:47
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TheNightKing
Let AB, CD, EF, GH, and JK be the five diameters of a circle with center O. In how many ways can three points be chosen out of A,B,C,D,E,F,G,H,J,K and O so as to form a triangle?
A. 165
B. 160
C. 185
D. 200
E. 225
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Solution:



    • A straight line cannot make a triangle.
       We need to select any 3 points from given points but we will have to discard those cases where these 3 points make a straight line.
    • The number of ways in which 3 points can be selected = \(11C3 = \frac{11!}{3!*8!} =165\)
      o There are 5 combinations of the straight line
         AOB, COD, EOF, GOH, JOK
    • The number of ways in which triangle can be formed = \(165 – 5 =160\)
Hence, the correct answer is Option A
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Let AB, CD, EF, GH, and JK be the five diameters of a circle 02 Jan 2026, 05:44
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To find the number of ways to form a triangle from the given points, we first identify the total number of points available. There are 5 diameters, which provide 5×2=10 points on the circumference (A through K), plus the center point O, for a total of 11 points.

Solution 1: The Shortest Approach

A triangle is formed by any 3 points that are not collinear (do not lie on the same straight line).
Total combinations of 3 points: Choose 3 points from 11 = 11×10×9/3×2×1 = 165
Subtract collinear sets: The only way 3 points can be collinear in this configuration is if they lie on the same diameter (the two endpoints and the center O).
There are 5 diameters (AB, CD, EF, GH, JK).
Each diameter plus the center O forms a set of 3 collinear points (e.g., A-O-B).
There are exactly 5 such sets.
Final Calculation: 165 − 5 = 160.
Correct Answer: B. 160


Solution 2: Detailed Case-by-Case Explanation
In this approach, we categorize the triangles based on whether they include the center point O or only points from the circumference.
Case 1: Triangles not including the center O

We choose 3 points from the 10 points on the circumference (A, B, C, D, E, F, G, H, J, K). Since no three points on the circumference of a circle can be collinear, every combination of 3 points will form a triangle.

Calculation: 10×9×8/3×2×1 = 120 ways.

Case 2: Triangles including the center O
We choose the center point O and 2 points from the 10 points on the circumference.

Total ways to pick O and 2 others = 10×9/2×1 = 45 ways
Subtracting collinear cases: We must exclude cases where the two chosen points on the circumference form a diameter with O. If we pick the two endpoints of a diameter (like A and B), then A-O-B is a straight line, not a triangle.
There are 5 diameters, so there are 5 pairs that are collinear with O.

Valid triangles with O: 45 − 5 = 40 ways.

Total Number of Triangles
Summing the cases together:
Total = Case 1 + Case 2
Total = 120 + 40 = 160
Correct Answer: B. 160
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