Atul gives Vishu a head-start of 20 seconds in a 900 m race and beats him by 135 m. While running the same race again Atul gives a start of 189 m and beats him by 8 seconds. In how much time can Vishu complete the full race of 900 m?

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Let Atul's speed be A m/s and Vishu's speed be V m/s.

First race:
- Vishu gets a 20 seconds head start.
- Total distance = 900 m.
- Atul beats Vishu by 135 m, so when Atul finishes 900 m, Vishu has run 900 - 135 = 765 m.

Let the time taken by Atul to finish the race be t seconds.

Since Vishu started 20 seconds earlier, Vishu's time to run 765 m is t + 20 seconds.

From this:
Atul's speed = 900 / t
Vishu's speed = 765 / (t + 20)

Second race:
- Atul gives Vishu a start of 189 m.
- Atul beats Vishu by 8 seconds.

When Atul finishes 900 m in time t2, Vishu runs (900 - 189) = 711 m in time t2 + 8 seconds.

So:
Atul's speed = 900 / t2
Vishu's speed = 711 / (t2 + 8)

Since Atul's speed is constant:
900 / t = 900 / t2 => t = t2

Equate Vishu's speeds from both races:
765 / (t + 20) = 711 / (t + 8)

Cross-multiplied:
765(t + 8) = 711(t + 20)
765t + 6120 = 711t + 14220
765t - 711t = 14220 - 6120
54t = 8100
t = 8100 / 54 = 150 seconds

So, Atul's time to finish the race is 150 seconds.

Vishu's speed from first race:
V = 765 / (150 + 20) = 765 / 170 = 4.5 m/s

Time taken by Vishu to finish 900 m:
900 / 4.5 = 200 seconds = 3 minutes 20 seconds.

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