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Let’s assume the speed of boat in still water and the speed of stream are ‘B’ and ‘C’ respectively. If the speed of stream is 30 km/h. C = 30 km/h The speed of the boat in still water is ‘y’ % more than the speed of the boat in upstream. B = (100+y)% of (B-C) Put the value of ‘C’ in the above equation. B = (100+y)% of (B-30) 100B = (100+y)x(B-30) [100B/(B-30)] = (100+y) y = [100B/(B-30)] - 100 Eq.(i) The time taken by the boat to cover (2d+50) km distance in downstream is 3.5 hours less than the time taken by the same boat to cover ‘d’ km distance in upstream. [(2d+50)/(B+C)] = [d/(B-C)] - 3.5 Put the value of ‘C’ in the above equation. [(2d+50)/(B+30)] = [d/(B-30)] - 3.5 [d/(B-30)] - [(2d+50)/(B+30)] = 3.5 Eq.(ii) The same boat can cover (d+75) km distance in still water in (y/6) hours. (d+75)/B = (y/6) Put the value of ‘y’ from Eq.(i) in the above equation. (d+75)/B = [[100B/(B-30)] - 100]/6 (d+75) = B[[100B/(B-30)] - 100]/6 d = B[[100B/(B-30)] - 100]/6 - 75 Eq.(iii) Put the value of ‘d’ from Eq.(iii) to Eq.(ii). [[B[[100B/(B-30)] - 100]/6 - 75]/(B-30)] - [(2x[B[[100B/(B-30)] - 100]/6 - 75]+50)/(B+30)] = 3.5 After solving the above equation, there will be three values of ‘B’. But two of them will be negative. So these can be eliminated. Hence B = 70 which is the only possible value. Put the value of ‘B’ in Eq.(i). y = [100x70/(70-30)] - 100 y = [7000/40] - 100 y = 175 - 100 y = 75 Put the value of ‘B’ in Eq.(ii). [d/(70-30)] - [(2d+50)/(70+30)] = 3.5 [d/40] - [(2d+50)/100] = 3.5 [d/40] - [(d+25)/50] = 3.5 [5d/200] - [4(d+25)/200] = 3.5 5d - 4d - 100 = 3.5x200 d - 100 = 700 d = 700+100 d = 800 (i) The value of ‘y’ is the multiple of 5. The above given statement is true. (ii) The speed of boat in still water should not be more than 60 km/h. The above given statement is not true. (iii) The value of ‘d’ is a four digit number. The above given statement is not true.
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