Word Problem: Travel and Distance?

Question

A cyclist traveled at a rate of 24 mi/h to visit a nearby town . The cyclist averaged 18 mi/h on the return trip. If the round trip took 4.9 hours, find the one way distance to the nearby town. Thanks

Answer 1

Doing word problems is simply the technique of turning the words into equations.  Define variables for the quantities you need.

"find the one way distance to the nearby town"

Let D be the distance to the nearby town.

"A cyclist traveled at a rate of 24 mi/h to visit a nearby town"

distance equal velocity times time.

Let To be the time of the trip out.

D = 24 * To

"The cyclist averaged 18 mi/h on the return trip"

Let Tr be the time to return.

The distance is still D

D = 18 * Tr

"the round trip took 4.9 hours"

To + Tr = 4.9

You have defined three variables D, To, Tr. and

have made three independent equations.

Since the number of variables equal he number of equations, the problem can be solved.

1) D = 24*To

2) D = 18*Tr

3) 4.9 = To + Tr

Solve by substitution and equating 1) and2)

Tr = 4.9 - To

24To = 18Tr

24To = 18(4.9 - To)

24To = 88.2 - 18To

42To = 88.2

To = 2.1

Substituting back into 1)

1) D = 24*To

D = 24*2.1

D = 50.4

It usually is worth checking your work.

Substituting back into 3)

3) 4.9 = To + Tr

4.9 = 2.1 + Tr

2.8 = Tr

Then back to 2)

D = 18* 2.8

D = 50.4

The distance agrees, problem solved.

Sorry for the long answer, for a reasonable question, although, it seems that teachers never say it, and maybe they do not realize it, mathematics is very much determined with vocabulary.  If you learn the vocabulary of math, math becomes much more straight forward.

Answer 2

Using d = st:

Outward: s = 24, let time = t, so d = 24t.

Return: s = 18, time = (4.9 - t), so d = 18(4.9.- t), so, since the outward and return distances are the same, then:

24t = 18(4.9 - t), so

24t = 88.2 - 18t, so

42t = 88.2, so

t = 88.2/42, so

t = 2.1 hours.

d = st, so

d = 24 x 2.1, so

d = 50.4 miles.