A swimmer sims from a point A against a current for 5 minutes and then swims backwards in favour of the current for next 5 minutes and comes to the point B. If AB is 100 meters, the speed of the current (in km per hour) is :
Solution
Correct Answer: Option D
The swimmer swims from A to C in upstream, C to A in downstream and A to B in downstream
So, The swimmer swims x distance in upstream for 5 min. And, From C to B the swimmer swims ( x + 100 ) distance in downstream for next 5 min .
upstream speed, u + v = {x/(5/60)} = 12x ......(i)
downstream speed, u - v = {(x+100)/(5/60)} = 12(x+100)
= u - L = 12x + 1200 .....(ii)
(i) - (ii)
u + v - u + v = 12x - 12x - 1200
=> 2L = -1200 => v = 600 m/hr = 600/1000 km/hr = 0.6 km/hr
So, The swimmer swims x distance in upstream for 5 min. And, From C to B the swimmer swims ( x + 100 ) distance in downstream for next 5 min .
upstream speed, u + v = {x/(5/60)} = 12x ......(i)
downstream speed, u - v = {(x+100)/(5/60)} = 12(x+100)
= u - L = 12x + 1200 .....(ii)
(i) - (ii)
u + v - u + v = 12x - 12x - 1200
=> 2L = -1200 => v = 600 m/hr = 600/1000 km/hr = 0.6 km/hr
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