Questions from Math FLT

Questions from Math FLT-06 :- 5.Three pipes are made of different shapes . The cross-sections of the pipes are an equilateral triangle . A hexagon and a circle. The perimeter of each of these cross- sections is equal . Flow through the pipes is proportional to the area of cross-section . If it takes 8 minutes for the triangular pipe to fill up the tank , what will be the difference in the times taken by the hexagonal and circular pipes 28.if a+b+c=0 , where a ≠ b ≠ c then [(a2)/(2a2+bc)] +[(b2)/(2b2+ac)] + [(c2)/(2c2+ab)] is equal to 29. The maximum possible value of y = min(1/2 - 3x2/4 , 5x2/4) for the range 0 < x < 1 is
for 1st one (5.) - I got answer = 1 min 12 sec (approx) - but it is not matching with the answer = 30 sec I look forward to general approach for questions like last one (29.)
Also any general approach for question like below (form FLT-M001) - 1. The lines y=3x and x+y=40 and the x axis bound a triangular area . Find the total number of points on or inside this triangle with integral coordinates? 11.Consider a triangle drawn on the X-Y plane with the vertices at (41,0) ,(0,41) and (0,0), each vertex being represented by its (X,Y) coordinates. The number of points with integer coordinates inside the triangle(excluding all the points on the boundary )is

Q5. Let the Perimeter = 1

Q5. Let the Perimeter = 1 unit Equilateral triangle: Side = 1/3 Area = √3/4 x 1/9 = √3 /36 Hexagon : Side = 1/6 Area = 6 x √3/4 x 1/36 = √3 /24 Circle : 2 π r = 1 => r = 1 / 2 π Area = π / 4 π 2 = 1/4 π Flow is proportional to Area So Time x Area = Constant => √3 /36 x 480 = √3 /24 x H => H = (24 x 480 ) /36 = 320 seconds And √3 /36 x 8 = 1/4 π x C => C = (480 √3 ) / 9π = 480 /3 √3 π = 290 seconds Difference = 30 seconds

Q1. is more complicated than

Q1. is more complicated than Q11. but you can use the same method Three vertices are (41,0) ,(0,41) and (0,0). The triangle formed will have three sides along x axis, y axis and x + y = 41. Consider either x or y (lets take y) For y = 1, possible x values are 1, 2, 3 . . . . . . 39. Total possible co-ordinates = 39 For y = 2, possible x values are 1, 2, 3 . . . . . . 38. Total possible co-ordinates = 38 . . . . . . . . . . . Similarly for y=39, possible x value is 1. Total possible co-ordinates = 1 Sum of all the possible co-ordinates = (39 x 40) / 2 = 780