Viral swims for 13 hrs hours while going 24 km down stream and 36 km upstream. But he takes 12 hrs hours to swim 36 km down steam and 24 km up stream. At what rate is the river flowing ?

1 km/hr

Let the speed of boat = u km/hr, and 

the speed of current = v km/hr

Speed of boat in downstream = u + v

Speed of boat in upstream =  u - v

Time = \(\frac{Distance}{speed}\)

Now, ATQ

⇒ \(\frac{24}{u\;+\;v}\) + \(\frac{36}{u\;-\;v}\) = 13 hrs    ....................(i)

& \(\frac{36}{u\;+\;v}\) + \(\frac{24}{u\;-\;v}\) = 12 hrs    ....................(ii)

⇒ (i) × 2 & (ii) × 3 

⇒ \(\frac{48}{u\;+\;v}\) + \(\frac{72}{u\;-\;v}\) = 26 hrs    ....................(iii)

& \(\frac{108}{u\;+\;v}\) + \(\frac{72}{u\;-\;v}\) = 36 hrs    ....................(iv)

⇒ (iv) - (iii)

⇒ \(\frac{108}{u\;+\;v}\) - \(\frac{48}{u\;+\;v}\) = 36 - 26

⇒ \(\frac{108\;-\;48}{u\;+\;v}\) = 10

⇒ \(\frac{60}{u\;+\;v}\) = 10

⇒ u + v = 6            ...................(v)

Put this value in equation (i)

⇒ \(\frac{24}{6}\) + \(\frac{36}{u\;-\;v}\) = 13 hrs

⇒ 4 + \(\frac{36}{u\;-\;v}\) = 13 hrs

⇒ \(\frac{36}{u\;-\;v}\) = 9 hrs

⇒ u - v = 4            ...................(vi)

Now, (v) - (iv)

(u + v) - (u - v) = 6 - 4

2v = 2

v = 1 km/hr

Hence, river flowing at the rate of 1 km/hr.