Bicycling and Walking
Two friends decide to take a bicycle trip. During their journey, one of the bikes malfunctions and has to be left behind. The friends decide to continue their trip and share the remaining bicycle. They start again, one riding and one walking. At some point, the rider gets off the bike, leaves it and continues on foot. When the boy, whose bicycle failed, reaches the abandoned bicycle, he gets on and rides until he catches up with his buddy, who then takes the bike again. This plan continues until they reach their destination.
The distance from the point, where the bicycle malfunctioned, to the destination is 60 miles. When walking, each walks at 5 miles per hour, and when riding, each rides at 15 miles per hour.
How far from the destination should the last boy to ride, abandon the bike so that he and his buddy reach the destination simultaneously?
(amendment: neither boy is wearing a watch)
The distance from the point, where the bicycle malfunctioned, to the destination is 60 miles. When walking, each walks at 5 miles per hour, and when riding, each rides at 15 miles per hour.
How far from the destination should the last boy to ride, abandon the bike so that he and his buddy reach the destination simultaneously?
(amendment: neither boy is wearing a watch)
posted by Zaux at
10:36 PM
14 Comments:
5 miles from the end. I suggest that they ride for 20 minuts and walk for 1 hour - each mode of transport moving them forward by 5 miles, and therefore every 10 miles catching up with each other.
Karl ... they are not together after the first bike malfunctions .. and they are not wearing watches
added an amendment to the problem statement indicating that neither traveler is wearing a watch
The friends decide to continue their trip and share the remaining bicycle. They start again, one riding and one walking....
This indicates they start out at the same time, doesn't it?
Still, then I would suggest they first person to end up within sight of the finish line stops and awaits the other to enable them to sychronise their efforts...?
A solution is, first boy cycles half way and walks half. Second boy walks half way and cycles half way. Half way = 30 miles. Simple symmetry.
If they swap n times, then last swap must be at 30/(n+1) miles from end.
If above is wrong, I'm not really Chris.
A small flaw. They have no means of measuring the distance. I'll have another think.
Aaaarrgh. I meant 60/(n+1).
I'm not sure that this question can be answered uniquely. Gotta go out now.
who is impersonating Chris?
yes there is a unique answer
It is easy to devise a plan of action to make them arrive at the same time if you deviate from the problem statement ... but as stated, the last boy to ride, leaves the bike and begins walking... then his friend arrives at the bike ... the point at which the first boy leaves the bike is the point in question ... that point is such that they arrrive at the destination at the same time.
I sort of mis-interpreted the published solution. As it turns out, it does not matter how many times they change between riding and walking. They will arrive at the same time.
Example:
consider the situation in hours:
After 1 hour:
1st boy - Rider - 15 miles
2nd boy - Walker - 5 miles
After hour 2:
1st boy - Walker - 20 miles
2nd boy - Rider - 20 miles
After hour 3:
1st boy - Rider - 35 miles
2nd boy - Walker - 25 miles
After hour 4:
1st boy - Walker - 40 miles
2nd boy - Rider - 40 miles
After hour 5:
1st boy - Rider - 55 miles
2nd boy - Walker 45 miles
So, after 5 hours:
1st boy has 5 miles to walk @ 5mph = 1 hour
2nd boy has 15 miles to ride @ 15mph = 1 hour
So after 6 hours, they arrive at the same time
Chris, you are you ... not an imposter ... there is no unique answer.
Only got a few mins spare. The 30 miles (half-way) works. But how do measure distance; have they got a ruler?
My formula was wonky, they must swap an odd number of times.
Another answer is: when the second (to ride) boy catches up with his friend, he could abandon the bike, and they both then walk the rest of the way together a your last paragraph implies. I hadn't read it properly before, I had been aiming for the cyclist and the walker should exactly arrive together - I now see the both walk the last bit.
They could of course swap a few times, but walk the last bit from allmost anywhere.
I think that there are an infinite number of solutions.
Hi Zaux. I am Chris after all. Posts crossing over. The first stopping point must be, 1/2, 1/4, 1/6, 1/8, etc of 60 miles. My 60/(n+1), n odd is the first stopping position and the distance before the end.
Got to go for dinner. I'll be gone for at least 4 hours.
The (next to) last rider should abandon the bike half way between where he left his friend and the destination.
Suppose they had 6 miles to go.
one walks 1 mile while the other rides 3. Then they both walk 2 at which point the first walker reaches the bike. In the last segment the new rider rides the last 3 miles while the other walks the last one mile.
Watches are not required. An odometer or mile markers, and knowledge of where the end is, would be needed to do this accurately.
Mitch
I'd got that one covered. Any even divider of 60 will do the job.
But the question actually says that they walk the last bit together (possibly by mistake). So when they are "close" to the end, the last rider ditches the bike when he catches up with his friend, and they both walk the last bit together.
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