A boat starts from point A and goes to point C which is 200 m away. B is 80 m away from A on the way to C. All these three points are on the same bank of the river which flows at 2 m/s. The boat man moves in the direction of the river on the way from A to C and on the way back paddles against the river along the same route. The speed of the boat in still water on the way from A and B (or from B to A) is half of the same from B to C (or from C to B). The boat takes 2 min 21 s to start from A and return to its original position. If the boatman started back immediately on arrival at C, the speed of the boat in still water from A to B could be

• Distance between A to C = 200m
  distance between A to B = 80m
  hence distance between B to C = 120m(A, B and C are on same bank)
  Let the speed of boat in still water between A and B is 'x'm/s
  =>speed of boat in still water between B and C is '2x'm/s
  Actual speed of boat(upstream) between A to B: 'x+2'm/s & between B and C: '2x+2'm/s
  downstream velocities( from C to A): C to B: '2x-2'm/s & B to A: 'x-2'm/s
  total time taken for round trip=2 min 21s=141s
  using time= distance/velocity
  =>80/(x+2)+120/(2x+2)+120/(2x-2)+80/(x-2)=141 (i.e., A->B->C->B->A = 141s)
  Its becomes a very complex equation(polynomial of degree 4 with large coefficients) to solve manually. It will be easy if options are given. The option which satisfies the above equation for 'x' will be the answer!
  
• 10 years agoHelpfull: Yes(6) No(1)
• option
  a-2
  b-2.5
  c-3
  d-4
  so 3 satisfy the above equation
• 10 years agoHelpfull: Yes(3) No(0)
• Can any 1 help me in this problem
• 10 years agoHelpfull: Yes(1) No(0)
• 80/(x+2)+120/(2x+2)+120/(2x-2)+80/(x-2)=141 (i.e., A->B->C->B->A = 141s)
  substitute value of x as 3 i.e x=3.
  then the time will be
  141 seconds. ie same as 2min 21 sec.. so the ans is 3m/sec
• 8 years agoHelpfull: Yes(0) No(0)