Question

From the same place at 7 am ‘A’ started walking in the north at the speed of 5 km/hr. After
1 hour, B started cycling in the east at a speed of 16 km/hr. At what time they will be at a
distance of 52 km apart from each other? From the same place at 7 am 'A' started walking in
the north at the speed of 5 km/hr. After 1 hour, B started cycling in the east at a speed of
16 km/hr. At what time they will be at a distance of 52 km apart from each other?​

Answer 1

Answer:

They will be at a distance of 52km between them at 11 AM

Let t = time in hrs for the walker to be 52 km from the cyclist

(t-1) = time for the cyclist to be 52km from the walker

5t = the distance walked

16(t-1) = distance cycled

, this can also be written as (16t-16)

Since the 2 people are walking towards North and East, we can use Pythagoras theorem as the angle between them is 90 degrees.

(5t)^2 + (16t-16)^2 = 52^2

25t^2+256t^2-512t+256 = 2704

281t^2 -512t-2448=0

t=4 hours. Hence, they will be at a distance of 52 km at (7+4)=11 AM