[unreality]

(1) Avory and Ben are at Point A (their neighborhood) along a linear bike trail and want to get to Point B (the town, in which resides the ice cream shop where they work). However, they only have 1 bike! It's decided that Avory will start with the bike, ride a certain distance (while Ben walks), then drop off the bike and walk. When Ben reaches the bike, Ben will ride the bike until he catches up with Avory, who takes the bike, and so on. The distance between Point A and Point B is 60 miles; they both walk at 5 mph and ride at 15 mph.

How far from Point B should the bicicyle be left behind for the last time to ensure that Avory and Ben reach Point B at the exact same time?

~~~

(2) Being genuises, Avory and Ben show up to work simultaneously and on time. Today is a long 9-hour shift for both of them (ice cream is popular in the sweltering summer heat) so their kind boss Charlie has given them plenty of breaks. However, none of their breaks coincide and they cannot NOT scoop ice cream when it's not breaktime. During his first break, Avory decides to use his ice cream tip money to buy his friend Ben a bike. He does so within time and brings the bike back to the ice cream shop, where there is a pole he can lock the bike to (the bike MUST be locked at all times due to the recent Grand Theft Bicicyle gang that's roaming around). However, Avory is painfully aware that he and Ben each have one bike lock, but of course neither can open the other's bike lock... and what's worse, their workbreaks never coincide! And at that same time, Charlie calls out to Avory that Ben is leaving early today, so Avory can't just unlock it for Ben at the end of the workday. Despaired, Avory decides to wheel the bike back to the bike shop and get his money back... unless he thinks of a solution first. Is there a solution to this dilemma?

~~~

(3) After work, Avory and Ben meet up with their bikes at Point Z along the trail and want to get to Point Y, a friend's house. Avory sees a wind tunnel that's being built parallel to the stretch of trail between Z and Y, and wants to ride in there, but Ben decides to ride on the normal trail. The first half of the wind tunnel has an air current pushing against Avory with speed c, the second half has an air current pushing with Avory at speed c. Remember that Ben and Avory both ride normally at the same rate. Who reaches Point Y first?

[Guest]

Option 1: Put Bens bike lock on the last time

Option 2: Avery puts his bike lock on during his break. Next break Ben puts his on. Last break Avery takes his off. Bens is the only one left when he leaves.

Ben. Speed is a function of time so the two don't cancel. You spend longer going slower so it affects you more then the increased speed.

[Guest]

it should be left halfway (30 miles) since they both ride and walk at the same pace. it will take 6 hours to walk halfway and 2 hours to ride halfway.

They need a closer ice cream shop.

[Guest]

A: 15 miles, having the switch being made every 15 miles

B: lock the new bike to the old bike already locked to the pole, when it is the other guys break, he unlocks the bikes from the pole, and locks the new bike to the pole, then on the next break for avery he unlocks the bikes from each other and locks his bike to the pole

C: don't know this one for sure yet, but I want to say they arrive at the same time

Note: my C answer is wrong, wakesnow got it

Edited June 25, 2009 by perk8504

[unreality]

nobody has yet fully gotten the first one, and wakesnow's Option 1 for #2 cannot work because the bike is left unlocked for most of the day. perk's solution for #2 is also wrong because they can't unlock each other's bike locks. However, wake's option 2 for #2 works if the other bike is taken into account:

* Avory locks his bike and the new bike to the pole

* Ben comes out and locks just his bike to the pole

* Avory comes out and unlocks his bike lock and then relocks it over his own bike

* now Ben's bike is locked the pole with Ben's lock and Avory's bike is locked to the poll with Avory's lock, so we've succeeded... Ben leaves first, unlocking his bike and riding away, then Avory, unlocking his own bike and riding away

edit: as for #3, wake is correct.

Normally they ride at speed r.

Say the distance they have to go is 2x.

d=rt, so t=d/r, so Ben takes up 2x/r hours

for Avory, he goes the first half at speed r-c, the second half at speed r+c

x/(r-c) + x/(r+c) is his total time

if you divide out x from Ben's and Avory's time, Ben takes a time of 2/r

and Avory takes a time of 1/(r-c) + 1/(r+c) which equals (r+c)/(r^2 - c^2) + (r-c)/(r^2-c^2)

add those together and you get 2r/(r^2 - c^2)

ie, Ben's time is 2*1/r, aka, 2*r/r^2

while Avory's time is 2*r/(r^2 - c^2)

The denominator on Avory's time is smaller due to the subtraction of c^2, thus the quantity itself is larger. Avory takes more time.

Thus, Ben gets there sooner.

Edited June 25, 2009 by unreality

[unreality]

so good job to wakes and perk, see my above post for the answers to (2) and (3)... only #1 remains unsolved

edit - well it's been solved but there's a higher realization that hasn't come to light yet

Edited June 25, 2009 by unreality

[Guest]

15 mi. is what I get:

Avory bikes 15 mi.

While Ben walks 5 mi.

Ben walks the 10 mi. to the bike.

While Avory walks 10 mi.

Now Avory is at mile 25, and Ben at mile 15.

Ben bikes 30 mi. to mile 45.

While Avory walks 10 mi. to mile 35.

Avory walks 10 mi. to bike.

While Ben walks 10 mi. to mile 55.

At this point Avory is at the bike at mile 45, and Ben is at mile 55 on foot.

Avory bikes the rest of the 15 mi. to point B. In the time it takes him to bike that 15 mi., Ben will walk the 5 mi., both arriving at the same time. And the bike was last left by Ben 15 miles from B.

[Guest]

... it wont be left behind at all since it will be at the finish line.

[Guest]

nobody has yet fully gotten the first one, and wakesnow's Option 1 for #2 cannot work because the bike is left unlocked for most of the day. perk's solution for #2 is also wrong because they can't unlock each other's bike locks. However, wake's option 2 for #2 works if the other bike is taken into account:

* Avory locks his bike and the new bike to the pole

* Ben comes out and locks just his bike to the pole

* Avory comes out and unlocks his bike lock and then relocks it over his own bike

* now Ben's bike is locked the pole with Ben's lock and Avory's bike is locked to the poll with Avory's lock, so we've succeeded... Ben leaves first, unlocking his bike and riding away, then Avory, unlocking his own bike and riding away

edit: as for #3, wake is correct.

Normally they ride at speed r.

Say the distance they have to go is 2x.

d=rt, so t=d/r, so Ben takes up 2x/r hours

for Avory, he goes the first half at speed r-c, the second half at speed r+c

x/(r-c) + x/(r+c) is his total time

if you divide out x from Ben's and Avory's time, Ben takes a time of 2/r

and Avory takes a time of 1/(r-c) + 1/(r+c) which equals (r+c)/(r^2 - c^2) + (r-c)/(r^2-c^2)

add those together and you get 2r/(r^2 - c^2)

ie, Ben's time is 2*1/r, aka, 2*r/r^2

while Avory's time is 2*r/(r^2 - c^2)

The denominator on Avory's time is smaller due to the subtraction of c^2, thus the quantity itself is larger. Avory takes more time.

Thus, Ben gets there sooner.

I guess I was figuring it was irrelevant who rode which bike, seeing as how they shared the old one to work, and would inevitably see each other again if they felt it necessary to switch bikes.

[Guest]

... it wont be left behind at all since it will be at the finish line.

Dang it, I believe you would be right, those ones hurt so unbearably bad when you don't recognize them...

[unreality]

I guess I was figuring it was irrelevant who rode which bike, seeing as how they shared the old one to work, and would inevitably see each other again if they felt it necessary to switch bikes.

right it wouldn't matter, however either way yours doesn't work since you said "lock the new bike to the old bike already locked to the pole, when it is the other guys break, he unlocks the bikes from the pole"

the other guy can't do that since he can only unlock his own bike lock.

But you were very close to the correct solution And the correct solution allows both to get their own bike back too

~~~

edit: solutions to 2 and 3 in my post here, however nobody has truly gotten #1 yet... yes answers have been proposed, but there's something more...

Edited June 25, 2009 by unreality

[Guest]

right it wouldn't matter, however either way yours doesn't work since you said "lock the new bike to the old bike already locked to the pole, when it is the other guys break, he unlocks the bikes from the pole"

the other guy can't do that since he can only unlock his own bike lock.

But you were very close to the correct solution And the correct solution allows both to get their own bike back too

~~~

edit: solutions to 2 and 3 in my post here, however nobody has truly gotten #1 yet... yes answers have been proposed, but there's something more...

by unlocking the "bikes" from the pole I mean unlocking the first bike that was locked to the pole, which effectively unlocks both of them from the pole since the new bike is locked to the first bike, he doesnt need to unlock the lock between teh bikes

[unreality]

by unlocking the "bikes" from the pole I mean unlocking the first bike that was locked to the pole, which effectively unlocks both of them from the pole since the new bike is locked to the first bike, he doesnt need to unlock the lock between teh bikes

edit: Ohhhh I see what the deal is. You assumed that Avory's bike was INITIALLY locked with Ben's bike lock. From that standpoint, your solution makes sense:

B: lock the new bike to the old bike already locked to the pole, when it is the other guys break, he unlocks the bikes from the pole, and locks the new bike to the pole, then on the next break for avery he unlocks the bikes from each other and locks his bike to the pole

ah, I see now Very clever. #2 needs to be split into two parts. Yours works if Ben's lock is initially locking Avory's bike, my answer was for Avory's lock locking Avory's bike

Edited June 25, 2009 by unreality

[Guest]

edit: Ohhhh I see what the deal is. You assumed that Avory's bike was INITIALLY locked with Ben's bike lock. From that standpoint, your solution makes sense:

B: lock the new bike to the old bike already locked to the pole, when it is the other guys break, he unlocks the bikes from the pole, and locks the new bike to the pole, then on the next break for avery he unlocks the bikes from each other and locks his bike to the pole

ah, I see now Very clever. #2 needs to be split into two parts. Yours works if Ben's lock is initially locking Avory's bike, my answer was for Avory's lock locking Avory's bike

ya, that's what I was going for, besides, who would buy a new bike for the other guy, I would buy myself a new bike and let him have the old one...lol

[unreality]

haha well since he bought the bike with ice cream scooping tips it's probably of inferior quality

[Prof. Templeton]

Any distance between halfway and point B could work. Lets say A gets on the bike, rides 5 feet and gets off and starts walking. B walks 5 feet then gets on the bike rides 30 miles and 5 feet at which point he gets off and starts walking. They both arrive at B at the same time.

[Guest]

Any distance between halfway and point B could work. Lets say A gets on the bike, rides 5 feet and gets off and starts walking. B walks 5 feet then gets on the bike rides 30 miles and 5 feet at which point he gets off and starts walking. They both arrive at B at the same time.

The point at which the total distance A has ridden in the second half is the same distance he walked in the first half

[Prof. Templeton]

The point at which the total distance A has ridden in the second half is the same distance he walked in the first half

both A and B walk and ride the same distance (30 miles), it all depends on how you want to divide it up.

[unreality]

PT got it If you graph distance vs time, the two different line slopes create parallelograms

All three have been solved ;D