Latest TCS Aptitude Question SOLUTION: The number of triangles that can be formed with 10 points as vertices, n of them being collinear, is 110. Then n is a) 3 b) 4 c) 5 d) 6

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The number of triangles that can be formed with 10 points as vertices, n of them being collinear, is 110. Then n is
a) 3 b) 4 c) 5 d) 6

Asked In TCS Subrat Baral (11 years ago)

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Kiran Pradhan  (11 years ago)

10 points -> num of triangles= 10C3
n collinear points reduce nC3 triangles.
total num of triangles=10C3 - nC3 =110.
Solving above we will get n=5
n=5

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sksingh  (11 years ago)

10c3-nc3=110
n=5

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vivek tomar  (11 years ago)

10c3-nc3=110
120-nc3=110
n.(n-1.(n-2)/3.2=10
n.(n-1).(n-2)=60
if we put n=5 then
5.(5-1).(5-2)=5.4.3=60
so c)5 is correct ans.

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Josiya ancy  (11 years ago)

10 points so 10c3
collineat points reduced to nc3
so 10c3-nc3-110
120-nc3=110
n=5

Is this solution Helpfull? Yes (4) | No (1)  

Ankita Talukdar  (5 years ago)

frac{L/6}{v} = frac{x}{35} => v = frac{35 L}{6 x}, - equation 1

2) Anoop runs 5L/6 to the farther end of the bridge.

frac{L+x}{35} = frac{5L/6}{v}, => v = frac{35*5 L}{6 (L+x)}, - equation 2

divide equation 2 by equation 1 =>
1 = frac{5 x}{L+x}, => L + x = 5 x, L=4 x, - equation3


substitute in equation 1,
v = frac{35 * 4 x}{6 * x} = frac{140}{6} = 23.66 mph

Anoop can run at 23.66 mph

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