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Let the speed of boat in still water be 'v' km/h. ATQ, 48/(v + 4) + 48/(v - 4) = 5 Or, 48 X (v - 4 + v + 4) = 5 X (v2 - 16) Or, 48 X 2v = 5v2 - 80 Or, 5v2 - 96v - 80 = 0 Or, 5v2 - 100v + 4v - 80 = 0 Or, 5v(v - 20) + 4(v - 20) = 0 Or, (5v + 4) (v - 20) = 0 So, 'v' = 20 or 'v' = - (4/5) But speed cannot be negative. So, 'v' = 20 Upstream speed of boat = 20 - 4 = 16 km/h Therefore, upstream distance cover by boat in 150 minutes = 16 X (150/60) = 40 km
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Statements:
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