Two friends are passing out flyers along an oval-shaped boulevard by starting at the same spot and walking in opposite directions. The total distance of the route is 4 mi. If one friend distributes them at 2.4 mph, how long until they meet? answer in minutes.

Let be Two friends are passing out flyers in common time t:

Step-by-step Solution-:

Let be velocity or rate of 1st and 2nd friends are v1 and v2 respectively:

The given data is:

v_{1}=4~\text{mph} , v_{2}=2.4~\text{mph}

Let be Distance covered by 1st and 2nd friends be x and y respectively:

Distance covered by 1st is:

x=v_{1}\cdot t

Substitute the v1=4 mph into the above:

x=4\cdot t

Similarly, the Distance covered by 2nd is:

y=2.4\times{t}

Let be the total distance Denoted by S:

Total distance is defined as sum of distance covered by 1st and distance covered by 2nd:

It means,

S=x+y

Substitute the x=4t, y=2.4t, S=4~miles into the above :

4=4t+2.4t

After adding we get,

4=6.4t

Both sides divided by 6.4 we get,

\frac{4}{6.4}=\frac{\cancel{6.4}t}{\cancel{6.4}}

After canceling, we get

t=0.625~ hour

We know that 1 hour=60 minute

Conver the unit’s Hour to minutes:

t=0.625\times 60

The required value of t is 37.5 minutes.

Hence, Two friends are passing out flyers in a common time 37.5 minutes: