Question

20. Jogger A can run laps at the rate of 2 minutes per lap. Jogger B can run laps on the same track at the rate of 150 seconds per lap. If they start at the same place and time and run in the same direction, how long (in time) will it be before they are at the starting place again at the same time?

A) 5 minutes

B) 300 seconds

C) 10 minutes

Answer 1

Solution:

We are given that:

Jogger A can run laps at the rate of 2 minutes per lap. That is: in 2*60 = 120 second per lap.

Jogger B can run laps on the same track at the rate of 150 seconds per lap.

Lets suppose Lap distance is 100 meter.

Then:

for Jogger A it took 120 seconds for 100 meter, then for rate meter per second = 100 / 120 = 0.833333 m/s.

and for Jogger B it took 150 seconds for 100 meter, then for rate meter per second = 100 / 150 = 0.666667 m/s.

Jogger A took 120 seconds to complete a lap of 100 meter and for Jogger B it took 150 seconds.

So when Jogger B completes Lap, Jogger A covers extra distance of 150-120=30 seconds.

That means Jogger A is 30 second ahead of Jogger B.

So in that 30 seconds Jogger A covers distance = Rate * Time = 0.833333 * 30 = 25 meter.

So in one lap , when Jogger B completes Lap in 150 seconds, Jogger A covers extra 25 meter distance in the same time.

In second Lap , when Jogger B finishes his lap in 150 seconds, in that time Jogger A covers extra 25 meters , so total extra distance = 25 + 25 = 50 meters

In third lap, when Jogger B finishes his lap in 150 seconds, in that time Jogger A covers extra 25 meters , so total extra distance = 25 + 25 +25 = 75 meters

In fourths lap. when Jogger B finishes his lap in 150 seconds, in that time Jogger A covers extra 25 meters , so total extra distance = 25 + 25 +25 + 25 = 100 meters

So it took 4 laps of 150 seconds to meet Jogger A and B at starting point.

Thus total time = 4 * 150 = 600 second = 600 / 60 = 10 minutes.

Thus it took 10 minutes to meet Jogger A and B at starting point.