HELP PLEASE! (total distance Jeffrey travels)

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cole92

Junior Member
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Mar 30, 2006
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65
Here is the exact problem:

Jeffrey visits his friend Kelly and then returns home by the same route. He always bikes at 6mph when going uphill, 12 mph going downhill, and 8 mph on level ground. Find the total distance he bikes if the total biking time is 6 hours.

I tried it one way, but then realized a big error half way through. I keep trying it, but it doesnt work out right. Please help me and thanks in advance to anybody that can help.
 
Never have I seen such a problem. It is very interesting. At first glance, there is insufficient information. The speeds and total time have been chose very carefully.

Definitions
x = Uphill miles to (6 mph)
y = Downhill miles to (12 mph)
z = Level miles to (8 mph)
x+y+z = One Way Trip.

Reasonable Conclusions/Implications
x = Downhill miles from (12 mph)
y = Uphill miles from (6 mph)
z = Level miles to (8 mph)

Common Knowledge
Distance = Rate * Time

Problem Statement
The only total informatiion we have is in hours, so let's change the formula to time.
The ANSWER to what is asked is 2(x+y+z), the round trip.

Distance / Rate = Time

Add them all up
x/6 + y/12 + z/8 + x/12 + y/6 + z/8 = 6

Now what?
 
i forgot to mention that i must use 3 different variabes. although i might be able to change those to just x,y,z....i will check out both the answers

btw, tkhunny, the first the fractions are x/6 y/12 and z/8.....shouldnt they be the same for the next 3? why did they change to x/12 y/6 and z/8?
 
to solve those, would you first make a common denominator? if so, then after adding them up you get 6x+6y+6z=6 i believe. then what would you do after that?
i would think solve for each variable, which i tried, but i dont know if they came out right....i got:
x= -y - z
y= -x - z
z= -x - y


are these correct? did i even need this? lol, im still confused i guess....but what exactly IS the answer? what was the total distance? im sry for still being confused, but i guess i still somewhat dont understand
 
Okay, letting x be miles traveled on level ground, and y be miles traveled on unlevel ground you have (x/8)+(y/6)+(x/8)+(y/12)=6.
So, that gives you (x/4)+(y/4)=6.
So, just multiply both sides by 4, and that gives you the answer!
x+y=24.
 
Opps, actually i should have used x for level ground, y for uphill, and z be downhill, giving you (x/8)+(y/12)+(z/6)+(x/8)+(y/6)+(z/12)=6, but i think it will give you same results.
 
cole92 said:
btw, tkhunny, the first the fractions are x/6 y/12 and z/8.....shouldnt they be the same for the next 3? why did they change to x/12 y/6 and z/8?
Click to expand...
cole, what happens IF uphill all the way to Kelly's?
then downhill all the way back from Kelly's, right?
so:
start: >........@ 6 .............>Kelly's joint
return: <.........@ 12 ........<Kelly's joint
 
actually, i thoughyt about that already, however, since we must use 3 variables, it cant just be uphill/downhill all the way..... :D i believe jacket81 may be correct....because it does use 3 variables, ill look again, lol, and maybe i will get it this time but thanks for your concern with the question
 
Re: HELP ON THIS PROBLEM PLEASE!

Hello, cole92!

Jeffrey visits his friend Kelly and then returns home by the same route.
He always bikes at 6mph when going uphill, 12 mph going downhill, and 8 mph on level ground.
Find the total distance he bikes if the total biking time is 6 hours.
Click to expand...
To Kelly's house:

Let \(\displaystyle U\) = number of uphill miles.
\(\displaystyle \;\;\)At 6 mph, his time uphill is: \(\displaystyle \frac{U}{6}\) hours.

Let \(\displaystyle D\) = number of downhill miles.
\(\displaystyle \;\;\) At 12 mph, his downhill time is: \(\displaystyle \frac{D}{12}\) hours.

Let \(\displaystyle L\) = number of level miles.
\(\displaystyle \;\;\)At 8 mph, his level time is: \(\displaystyle \frac{L}{8}\) hours.

His total time to Kelly's house is: \(\displaystyle \L\,\frac{U}{6}\,+\,\frac{D}{12}\,+\,\frac{L}{8}\,\) hours.


Return trip:

He bikes \(\displaystyle D\) miles uphill at 6 mph; his uphill time is: \(\displaystyle \frac{D}{6}\) hours.

He bikes \(\displaystyle U\) miles downhill at 12 mph; his downhill time is: \(\displaystyle \frac{U}{12}\) hours.

He bikes \(\displaystyle L\) level miles at 8 mph; his level time is: \(\displaystyle \frac{L}{8}\) hours.

His total time for the return trip is: \(\displaystyle \L\,\frac{D}{6}\,+\,\frac{U}{12}\,+\,\frac{L}{8}\,\) hours.


The total time for the round trip is 6 hours:
\(\displaystyle \L\;\;\left(\frac{U}{6}\,+\,\frac{D}{12}\,+\,\frac{L}{8}\right)\,+\,\left(\frac{D}{6}\,+\,\frac{U}{12}\,+\,\frac{L}{8}\right)\;= \;6\;\;\Rightarrow\;\;\frac{U}{4}\,+\,\frac{D}{4}\,+\,\frac{L}{4}\;=\;6\)

Hence: \(\displaystyle \L\;U\,+\,D\.+\,L\;=\;24\) . . . He bikes 24 miles one way.


\(\displaystyle \text{Therefore, his total distance is }48\text{ miles.}\)
 
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