SSC CHSL Model Paper 11

Question : 59 of 100

Marks: +1, -0

swimmer swims 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 = 100 metres, the speed of the current (in km per hour) is :

Validate

Solution:

The distance covered upstream = AC = d
AB = 100
BC = 100 + d
Rate upstream = (a - y) m/minute
Rate downstream = (a + y) m/minute

∴d/{x-y}=5

⇒ d = 5 (x - y)

....(i)
Again,

{100+d}/{x+y}=5

⇒{100+5(x-y)}/{x+y}=5

⇒ 100 + 5x-5y = 5x+5y

⇒10y = 100

⇒ y = 10

m/minute

=10/1000×60\kmph

= 0.6kmph

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