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

• 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
  
• 11 years agoHelpfull: Yes(31) No(0)
• 10c3-nc3=110
  n=5
• 11 years agoHelpfull: Yes(9) No(1)
• 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.
• 11 years agoHelpfull: Yes(7) No(0)
• 10 points so 10c3
  collineat points reduced to nc3
  so 10c3-nc3-110
  120-nc3=110
  n=5
• 11 years agoHelpfull: Yes(4) No(1)
• 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
• 5 years agoHelpfull: Yes(0) No(0)