A, B and C start from the same place travel in the same direction at the speed of 50,70 and 90m/s respectively.If B starts 10 seconds before c,how many seconds before B should A start so B,C reach the destination 8 seconds before A?

Let the time taken by B be t seconds. So, time taken by c = t - 10 seconds

Distance covered by B in 't' sec = Distance covered by C in 't-10' sec
=> 70t = 90(t - 10) => t = 45 sec

So, time taken by A to reach the destination = 45*70/50 = 63 sec

Since A should lag B and C by 8 sec when they reach the destination, A should have travelled for 63 - 8 = 55 seconds by then.

So, A should start 55 - 45 = 10 seconds before B starts