- 30 Apr '08 06:30 / 3 editsAl, Ben, Carl, and Dan need to get to the special place. They are currently 20 miles from the special place, and they can each walk at a constant 4 miles per hour. They figure they can get there in 5 hours, but it would behoove them to get there as soon as possible.

Just then, they meet a friendly motorcyclist named Ed who can travel at 56 miles per hour, but can only carry one passenger with him (neither of the other four have a motorcycle license, so only Ed can drive). They make the following arrangement:

Ed will take Al on his motorcycle a certain distance (call it*x*miles) while Ben, Carl, and Dan are walking. Then Ed will immediately turn around to pick up Ben. Al will walk the rest of the journey. Ed will carry Ben until they reach Al, drop off Ben (who then continues walking with Al), and immediately turn again to get Carl (who, along with Dan, was walking the whole time). After dropping off Carl to where Al and Ben would be, Ed would go back and get Dan (again, Dan was walking the entire time). In order to minimize the time taken, Ed and Dan would reach the special place at exactly the same time as Al, Ben, and Carl.

The four depart as soon as the arrangements are made, and they begin from exactly 20 miles away. Assuming all turns, pick-ups, drop-offs, and accelerations to either 4 or 56 mph are instantaneous, how long will it take for the five of them reach the special place, and what was the initial distance*x*? - 30 Apr '08 21:55 / 4 edits

The motocycle will make 7 trips total, call them M1, M2, M3, M4, M5, M6, M7*Originally posted by Jirakon***Al, Ben, Carl, and Dan need to get to the special place. They are currently 20 miles from the special place, and they can each walk at a constant 4 miles per hour. They figure they can get there in 5 hours, but it would behoove them to get there as soon as possible.**

Just then, they meet a friendly motorcyclist named Ed who can travel at 56 miles per hour, ...[text shortened]... t take for the five of them reach the special place, and what was the initial distance*x*?

Since the motorcycle and the walkers are moving at constant speeds, M1 = M3 = M5 = M7, and M2 = M4 = M6

so the motorcycle will be travels two different distances, M1 and M2

While the motorcycle is traveling M1, the walkers are walking W1.

W2 corresponds to the distance the walkers travel while the the motorcycle goes M2.

Now that we have our variables, we can lay out equations.

since 56/4 = 14,

M1 = 14*W1

If the motorcycle travels out and back 14 miles and meets the walkers a mile from where they started, the motorcycle can go

x + y = 14, and

x - y = 1

therefore, we know x = 7.5, so

M1 = 7.5 (W1 + W2)

solving for W2, we get W2 = 13/15 * W1

finally, we know that M1 + 3 * (W1 + W2) = 20

substituting yields

14*W1 + 3 * W1 + 39/15 * W1 = 20

therefore W1 ~= 1.021 miles, and

M1 = x = 14.286 miles

*edit* to be more precise, W1 = 50/49 miles, and M1 = 700/49 miles

*final edit*

the time taken is (15 min/mile) * (20 - 700/49 miles) = 1:25:42.86 - 30 Apr '08 23:51To make it a little bit more abstract...

There are 4 variables in the puzzle - the distances, which the cicle

carries each of the 4 guys. The 4th one will be direct to the

target, so it remains a minimazation problem with 3 variables.

The minimum will be reached, if nobody of the 4 has to wait,

so we are looking for a solution, where all reach the target at the

same time - only an equation with 3 variables remains.

But all will arrive at the same time, so the way they walk and the

way they drive is the same for all of them!

It remains an equation with just one variable!

the rest is calculation, not math ;-) - 01 May '08 18:09

I don't see why you this this problem has only one variable. Just because you know the value of a variable doesn't mean it doesn't exist. Take the speed of the walkers and the speed of the motorcycle, for example. If it changed, the answer to the problem would change entirely.*Originally posted by afx***To make it a little bit more abstract...**

There are 4 variables in the puzzle - the distances, which the cicle

carries each of the 4 guys. The 4th one will be direct to the

target, so it remains a minimazation problem with 3 variables.

The minimum will be reached, if nobody of the 4 has to wait,

so we are looking for a solution, where all reach the target a ...[text shortened]... f them!

It remains an equation with just one variable!

the rest is calculation, not math ;-)

Also, how about the distance they must travel? How about the number of people?

You could generalize the solution into a single equation solving for x in terms of all of these things.

Ignoring the ability to change these other 'variables' I mentioned, you need to be able to solve two equations for two variables when calculating how far out the motorcycle can travel before coming back to meet the walkers. - 12 May '08 10:46

It's a clever solution. I agree with all except the final edit. Seems like you've only accounted for the time spent walking, and forgot to add on the time for the bike part of the trip. The answer should be 1h;41m;1.22s*Originally posted by forkedknight***The motocycle will make 7 trips total, call them M1, M2, M3, M4, M5, M6, M7**

Since the motorcycle and the walkers are moving at constant speeds, M1 = M3 = M5 = M7, and M2 = M4 = M6

so the motorcycle will be travels two different distances, M1 and M2

While the motorcycle is traveling M1, the walkers are walking W1.

W2 corresponds to the distance t ...[text shortened]... 700/49 miles

*final edit*

the time taken is (15 min/mile) * (20 - 700/49 miles) = 1:25:42.86