Question

Shrikant and Sachin start running simulataneosly from the diametrically opposite ends of a circular track towards each other at 15km/h and 25km/h respectively. After every 10 minutes their speed reduce to half of their current speeds. If the length of the circular track is 1500m, how many times will Shrikant and Sachin meet on the track?

Answer 1

If the lenght is 1500 and they're on diametrically opposite ends, the distance between them is 750 m or 0.75 km.

H = 15 km/h
A = 25 km/h

The distances they run add to 0.75. Da + Dh = 0.75. Speed = Distance /time, Distance =Speed * time (which is the same for both).

15 * t + 25 * t = 0.75
40 t = 0.75
t = 0.01875 hours or 67.5 seconds.

Answer 2

Then they'll be on the same point and will continue to start running in opposite directions, since they have the same start point on the circle, they have to run over the entire lenght this time (1.5 km) until they meet again.

Da + Dh = 1.5
15t +25t = 1.5
40t = 1.5
t = 135 seconds.
They'll continue to meet each 135 seconds until their speedsreduce after 10 minutes.

Answer 3

10 minutes * 60 seconds/min = 600 seconds
600 - 67.5 seconds = 532.5
3* 135 = 405
532.5 - 405 =127.5

So until their speeds are reduced they have meet 4 times.

Answer 4

k, thanks, but how is that they meet after 750mts, since both of them are traveling in oposite direction at 15 and 25 km/hr

Answer 5

Because it's a circle.

Answer 6

Yes, but arent they running towards each other ?

Answer 7

Since their speeds are now half, they take double the time to meet, 270 seconds.
They still have run more 127.5 seconds in their not yet reduced speeds, so:

15 * 0.035 + 25 * 0.035 = 40 * 0.035 = 1.4 km, leaving just 0.1 km at their new speeds.

7.5t + 12.5*t = 0.1
20t=0.1
t=0.005 hours or 18 seconds more and they meet again. For all other times they meet, they take 270 seconds.

Answer 8

Yes, but arent they running towards each other ?

Answer 9

Yes. But see, they run towards each other, so H's vector ->, A's vector <-, once they meet, they have the same vector at the same point. <- ->, so now they're running opposite.

Answer 10

600 - 18 = 582
582 - 2 * 270 = 582 - 540 = 42 seconds, until their speeds gets once more reduced, they met more 3 times.

4 times with the original speeds, 3 now = 7 times. You just repeat the process until you figure their speeds are so small they practically have stopped altogether.