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Let the upstream and downstream speed of boat be 'U' km/h and 'D' km/h respectively. ATQ: (140/D) + (80/U) = 4.5 --------- (I) And, (224/D) + (50/U) = 5.25 -------- (II) On solving, 5 X equation (I) - 8 X equation (II) , we get, 5 X [(140/D) + (80/U) ] - 8 X [(224/D) + (50/U) ] = 5 X 4.5 - 8 X 5.25 Or, (700/D) + (400/U) - (1,792/D) - (400/U) = 22.5 - 42 Or, (1,092/D) = 19.5 Or, 'D' = (1,092/19.5) = 56 On putting value of 'D' in equation (I) , We get, (140/56) + (80/U) = 4.5 Or, 2.5 + (80/U) = 4.5 Or, (80/U) = 2 So, 'U' = 40 Speed of boat in still water = (1/2) X (downstream speed + upstream speed) = (1/2) x (56 + 40) = (96/2) = 48 km/h Therefore, required distance = 48 X 2.4 = 115.2 km
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