Three friends A, B and C are walking at the speed of 3,4,5 km/hr respectively. They start at same place at 1,2,3 O’clock respectively. When B reaches A, he sends him back with a msg for C. At what time will C get the msg?

B reaches A after x hours, when

3x = 4(x-1)
x = 4 hours, when A has gone 12 miles, at 5:00
So, at 5:00, C has gone 10 miles, and A is only 2 miles ahead.
It will take 2/(3+5) = 1/4 hours till they meet, at 5:15

Did you know?

Did you know that the time it takes for a message to reach someone can be determined by their respective speeds? In a scenario where friends A, B, and C are walking at speeds of 3, 4, and 5 km/hr, starting at 1, 2, and 3 O'clock respectively, if B reaches A and sends a message for C, we can calculate when C will receive the message. Since B is faster than A, he will catch up to him at some point. If we use their speeds as a reference, B is moving at a rate of 4 - 3 = 1 km/hr faster than A. Therefore, the time it takes for B to reach A would be the distance between them, divided by 1 km/hr. This means that when B reaches A, he will have traveled the same distance that A has walked in 1 hour. Consequently, if B reaches A at 4 O'clock, C can expect to receive the message at the time it takes him to walk the distance between A and C, which is also 1 hour. Therefore, C will receive the message at 5 O'clock.