In a 100 metre race, A can allow B to start the race either 10 metres

We are given that the distance of the race is 100 meters and that A can allow B to start the race either 10 metres ahead of the starting point or 2 seconds before A so that both reach the finish line at the same time.

Let’s denote the time it takes A to finish 100 meters as t. Then, the speed of A in terms of t is 100/t meters per second. If B can start the race 2 seconds before A and they both finish at the same time, then it takes B t + 2 seconds to finish 100 meters. Therefore, the speed of B in terms of t is 100/(t + 2) meters per second.

Furthermore, we are given that if B starts the race 10 meters ahead of the starting point, then A and B finish at the same time, thus it takes B t seconds to run 100 - 10 = 90 meters. So, the speed of B in terms of t is also equal to 90/t.

Let’s equate the two expressions for the speed of B:

100/(t+2) = 90/t

100t = 90t + 180

10t = 180

t = 18.

So it takes A 18 seconds to finish the 100-meter race. Since B takes 2 seconds longer than A to finish the race, it takes B 18 + 2 = 20 seconds to finish 100 meters, and therefore, the speed of B is:

100/20 = 5 m/s.

Answer: B