IPMAT Rohtak 2019 (QA) - A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 5 km in stationary water? | PYQs + Solutions | AfterBoards
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IPMAT Rohtak 2019

Arithmetic
>
Time, Speed & Distance

Easy

A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 5 km in stationary water?

Correct Option: 3
Against current (upstream): \newline 22 km in 11 hour = 22 km in 6060 minutes \newline Rate upstream = 22 km/hr
Along current (downstream): \newline 11 km in 1010 minutes = 66 km/hr \newline Rate downstream = 66 km/hr
If upstream speed = UU = speed in still water - speed of current = x−yx - y \newline And downstream speed = DD = speed in still water + speed of current = x+yx + y
Then: \newline x−y=2x - y = 2 ... (1) \newline x+y=6x + y = 6 ... (2)
Adding equations: \newline 2x=82x = 8 \newline x=4x = 4
Therefore, speed in still water = 44 km/hr
For 55 km in still water:
Time = 54\frac{5}{4} hours = 1.251.25 hours = 11 hour 1515 minutes

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