Question

Two dogs run around a circular track 300 m long. One dog runs at a steady rate of 15m per second, the other at a steady rate of 12 m per second. Suppose they'd tart at the same point and time. What is the least number of seconds that will elapse before they are again together at the starting point?

Answer 1

Answer:

300 seconds

Step-by-step explanation:

The first dog run at v₁ = 15 m/sec     the second one run at  v₂ = 12 m/sec

we know that  d = v*t  then  t  =  d/v

Then the first dog will take  300/ 12  =  25 seconds to make a turn

The second will take   300 / 15   =   20 seconds to make a turn

Then the first dog in 12 turns 12*25 will be at the start point, and so will the second one at the turn 15.

To check  first dog    12 *  25 = 300

And the second dog 15 * 20 = 300

That means that time required for the two dogs to be at the start point together is

300 seconds, in that time the first dog finished the 12 turns, and the second had ended the 15.

Another procedure to solve this problem is as follows:

between  12  m/sec and  15 m/sec   the minimum common multiple  is 300 ( 300 is the smaller number that accept 12 and 15 as factors         12*15 = 300) Then when time arrives at 300 seconds the two dogs will be again in the starting point