Rate-Distance-Time word problem

Question

The problem: A man having ten hours at his disposal made an excursion, riding out at the rate of 10 miles an hour and returning on foot, at the rate of 3 miles per hour. Find the distance he rode.

The solution as per the workbook is a fraction, 300/13.

I tried working the problem out for a while, and I don't understand how that's the answer. Any clarification?

Answer 1

Let t₁: time spent riding (in hrs) ; t₂: time spent walking (in hrs)

Also recall that rate · time = distance

t₁ + t₂ = 10

10t₁ = 3t₂ (because the distance riding and walking must be the same)

t₂ = 10/3 t₁

t₁ + 10/3 t₁ = 10

13/3 t₁ = 10

t₁ = 30 / 13 hrs so the distance he rode (and the distance he walked) is 300/13 miles

Answer 2

Hi John! My first suggestion is to draw a picture showing him riding out and then riding in. We can then try to assign a distance, rate, and time to each picture. It is also helpful that we know that the entire trip took 10 hours. So if it took him t hours going out, then he has 10 - t hours to come in.

Going out:

----------------------------->

distance = d

rate = 10 mph

time = t

Riding in:

<------------------------

distance = d

rate = 3 mph

time = 10 - t

We can then write an equation for each (distance = rate * time):

Going out: d = 10t

Riding in: d = 3(10 - t)

Since we know the distance is the same, we can use substitution and say that 10t = 3(10 - t).

Simplify the equation: 10t = 30 - 3t

13t = 30

t = 30/13 or about 2.3 hours

We let t be the time going out, so it took him 30/13 (or about 2.3 hours) to go out.

The question asked you to find the distance that he rode. To do this, just use your equation d = r*t. We know that he rode at 10 mph and that it took him 30/13 hours.

d = 10 * t

d = 10 * (30/13)

d = 300/13