SOLUTION PLZ

Question

An office goers sees a bus trundling past at 10 km/hr and makes a run for it at 14 km/hr.If he manages to board the bus after 9 seconds ,how far was the bus's entrance when the office goer started running.

Answer 1

We need to get everything in the same units first:

1hr=3600s

This gives: 10km/hr=10km/3600s=1km/360s

                 14km/hr=14km/3600s=7km/1800s

In 9 seconds the bus traveled 1km/360s X 9s = 9km/360=1km/40 or 0.025km.

In 9 seconds the person traveled 7km/1800s X 9s = 7km/200 or 0.035km.

Subtracting the 2 shows that the person had to run 0.01km farther to catch up, so the answer is 0.01km.

Answer 2

 

 T is the delay time

 

 The bus travels for 9+T seconds at 10 km/hr

 The office goer travels for 9 seconds at 14 km/hr

 

Next the speeds are changed to km/sec to match the times.

1 hour = 60 min and 1 min = 60 sec, so 1 hour = 3600 sec

 

So the bus speed is 1/360 km/sec

and the office goer speed is 14/3600 = 7/1800 km/sec

 

  Their distances are the same:

 

  (T+9)(1/360) = 9(7/1800)

 

 (T+9)/360 = 7/200

 

200(T+9) = 360*7

 

T = 360*7 /200 - 9

   = 3.6

 

 

So the bus went  (1/360)(12.6) = 0.035 km = 35 meters

The office goer ran (7/1800)(9) = 7/200 km = 0.035 km = 35 meters

 

So in 3.6 seconds the bus was 3.6 * 1/360 km = 3.6/360 km = 36/3600 km = 1/100 km = 10 meters ahead