Help Needed For An Algebra Word Problem

[Dennis In Va]

Hello,

Here is an Algebra word problem that is driving me nuts:
_____

A boy has planned a 3-hour bicycle ride. After biking at a rate of 14 mph for a while, the bike breaks down. The boy rides back with his father at a rate of 35 mph. How far did the boy ride his bike if he returns home 1 hour after he left?
_____

As I read the problem, the 3 hours goes into the trash bin as unnecessary information. For one thing, the boy's bike broke down, which was unplanned for a 3-hour ride. Finally, the last sentence states that he returned home 1 hour after he left. I read this as meaning that the entire round trip took 1 hour or 60 minutes, instead of the originally planned 3-hour round trip.
_____

Can any of you make sense out of this problem and translate it into an equation that will solve it?

Thanks in advance.

Dennis in Va.

[nmoira]

Hello,

Here is an Algebra word problem that is driving me nuts:
_____

A boy has planned a 3-hour bicycle ride. After biking at a rate of 14 mph for a while, the bike breaks down. The boy rides back with his father at a rate of 35 mph. How far did the boy ride his bike if he returns home 1 hour after he left?
_____



He went a distance, x, twice.

distance (miles)/time (hours) =speed (mph)
distance = speed x time

Let the distance be x

x = 14y
x = 35z
y+z = 1 (one hour)
y = 1-z
---
Substitute:
x = 35z
x = 14- 14z

Subtract second equation above from first:

0 = 35z + 14z - 14
49z=14
z = 2/7 of an hour

Substitute z back in:
x=35 x 2/7
x = 10

or
x = 14 x (1- 2/7)
x = 14 x 5/7
x = 10

[forty-two]

I agree that the 3 hour bit is unnecessary info. This is the eq I got:

Let x equal the time the boy rode his bike till it broke down in hours (and assuming there was no dead time b/w the breakdown and beginning to return with his father):

14x=35(1-x)
14x=35-35x
49x=35
x=35/49 hours = 5/7 hours
Distance biked = 14x = 14mph * 5/7 hours = 10 miles
Check: Distance biked should equal distance rode:
35(1-x) = 35 mph * (1 hour - 5/7 hours) = 35 mph * 2/7 hours = 10 miles

HTH

[Maverick]

Hi Dennis,

Interesting problem.:001_smile: I agree with you, the 3 hours can be thrown out. The key to remember on this type of problem is rate x time = distance. If you are a visual type, draw yourself a little sketch--he goes a certain distance at a rate of 14 mph until he breaks down. He rides back in the car with his Dad (the same distance) at a rate of 35 mph. Assuming it didn't take any time at all for his Dad to rescue him and load his bike the total trip took 1 hour. If you let x equal the amount of time it he rode before his bike broke down, the distance he rode can be expressed as 14x. The amount of time he took to ride back in the car is 35(1-x) or the rate times the time it him took to ride back--1 hour minus the time he rode there. (The two times have to add up to 1 hour) OK. So your equation is 14x = 35(1-x) because each of those is equal to the same distance so they are equal to each other. You can probably solve from there on your own. I got 5/7 of an hour there and 2/7 of an hour back. Multiple either of those times the appropriate rate and get 10 miles as the distance. :auto:

[nmoira]

Let x equal the time the boy rode his bike till it broke down in hoursOK, sure you could do it with one variable... :)

[Jackie in AR]

Hello,

Here is an Algebra word problem that is driving me nuts:
_____

A boy has planned a 3-hour bicycle ride. After biking at a rate of 14 mph for a while, the bike breaks down. The boy rides back with his father at a rate of 35 mph. How far did the boy ride his bike if he returns home 1 hour after he left?
_____

As I read the problem, the 3 hours goes into the trash bin as unnecessary information. For one thing, the boy's bike broke down, which was unplanned for a 3-hour ride. Finally, the last sentence states that he returned home 1 hour after he left. I read this as meaning that the entire round trip took 1 hour or 60 minutes, instead of the originally planned 3-hour round trip.
_____

Can any of you make sense out of this problem and translate it into an equation that will solve it?

Thanks in advance.

Dennis in Va.


The distance traveled is the same for the bike and the car. You have to assume he was picked up immediately by the car as soon as his bike broke down. So the equation should be set up as:

distance by bike = distance by car

Have x = minutes on bike; then 60-x = minutes in car

14 (speed) times x/60 (time on bike) = 35 (speed) times 60-x/60 (time in car)

The distance was approximately 10 miles.

[forty-two]

Of course, the more I think about this problem, the more the implicit assumption that his father picked him up instantly bugs me - that is just not realistic. If *I* were writing this problem, it would look something like this:

A boy heads out for a 3 hour bike ride. He rides for a while at a constant speed of 14mph before his bike breaks down unexpectedly. After he spends 2 minutes looking over his bike and realizing it is unfixable with the tools he has on him, the boy then pulls out his cell phone and calls his father for a ride, which takes 1 minute. Assuming his father leaves the house 5 minutes after hanging up, driving at a constant speed of 35mph along the same route the boy took with his bike both coming and going, and that it took 5 minutes to load the bike and boy in the car, how far did the boy bike if he and his father arrived home 1.5 hours after the boy left?

It's probably a good thing I don't write algebra problems :D