A boy and his sister are to travel to a place that is 100km away, they have just one bicycle between them. On foot he can travel at 12km/hr and 20km/hr riding the bicycle. The sister's foot speed is 10km/hr and her riding speed is 16km/hr.
If both of them cannot be on the bicycle at the same time, at what point would the first rider drop the bicycle for the second person to pick up in order for the both of them to arrive at their destination in the shortest possible time, and how long would that be?

More than a year and not even one try.

dorox: More than a year and not even one try.

I wonder how many views.....but I must say that the solution is mathematical.....any attempt to do guess work will take ages.

Solution
Let the time taken for the boy to walk on foot be X;
and let the time taken for the boy to ride be Y

Based on this assumptions the girl's time will be Y for walking and X for riding.

Also, speed * time = distance

Thus we form the equations:
Boy: 12X + 20Y = 100
Girl: 16X + 10Y = 100
(X,Y) = (5,2)

Boy walks for 5 hours and rides for 2 hours while Girl rides for 5 hours and walks for 2 hours.

Total time required is 7 hours.....mathematical solution. The physical solution will be a little above 7hours.

G: 16r 16r 16r 10w 10w 16r 16r
B: 12w 12w 12w 12w 20r 12w 12w 8w

*** 16r = 16km ride and 10w = 10km walk ***

With each column representing 1hour activity, this journey will take 7hours 24mins 40mins.

The possibility of getting a lower journey time depends on the breakdown of activity schedule.

saintneo:
I wonder how many views.....but I must say that the solution is mathematical.....any attempt to do guess work will take ages.
Solution
Let the time taken for the boy to walk on foot be X;
and let the time taken for the boy to ride be Y
Based on this assumptions the girl's time will be Y for walking and X for riding.
Also, speed * time = distance
Thus we form the equations:
Boy: 12X + 20Y = 100
Girl: 16X + 10Y = 100
(X,Y) = (5,2)
Boy walks for 5 hours and rides for 2 hours while Girl rides for 5 hours and walks for 2 hours.
Total time required is 7 hours.....mathematical solution. The physical solution will be a little above 7hours.
G: 16r 16r 16r 10w 10w 16r 16r
B: 12w 12w 12w 12w 20r 12w 12w 8w
*** 16r = 16km ride and 10w = 10km walk ***
With each column representing 1hour activity, this journey will take 7hours 24mins.
The possibility of getting a lower journey time depends on the breakdown of activity schedule.

Nice try, I am impressed, but your answer is a bit off the mark especially the first one, the second one is much closer.
The idea behind your mathematical solution is good and elegant, but you modeling asumption was wrong.

dorox:

Nice try, I am impressed, but your answer is a bit off the mark especially the first one, the second one is much closer.
The idea behind your mathematical solution is good and elegant, but you modeling asumption was wrong.

OK. i hope it won't take one year before you provide the ultimate solution.

saintneo:


OK. i hope it won't take one year before you provide the ultimate solution.

I can give you the answer right now, but a mathematical solution would have to wait until much later tonight when I get home, it is kind of akward using the phone to write out a mathematical solution.
Perhaps you will find this hint usefull. When the first rider drops the bicycle, the second rider who has been on foot will continue to be on foot until he/she gets to the point where the bicycle was dropped.
The mistake you made in your model was to not include that part of the journey where the boy and the girl were both walking.

dorox:
I can give you the answer right now, but a mathematical solution would have to wait until much later tonight when I get home, it is kind of akward using the phone to write out a mathematical solution.
Perhaps you will find this hint usefull. When the first rider drops the bicycle, the second rider who has been on foot will continue to be on foot until he/she gets to the point where the bicycle was dropped.
The mistake you made in your model was to not include that part of the journey where the boy and the girl were both walking.

it is there already maybe u didn't understand the code.

G: 16r>> 16r>> 16r>> 10w>> 10w>> 16r>> 16r
B: 12w>> 12w>> 12w>> 12w>> 20r>> 12w>> 12w>> 8w
*** 16r = 16km ride and 10w = 10km walk for girl and 12w = 12km walk, 8w = 8km walk and 20r = 20km ride for boy

each ride or walk takes 1hr only the 8w takes 40mins
***
G = girl's routine
B = boy's routine

saintneo:

it is there already maybe u didn't understand the code.

G: 16r>> 16r>> 16r>> 10w>> 10w>> 16r>> 16r
B: 12w>> 12w>> 12w>> 12w>> 20r>> 12w>> 12w>> 8w
*** 16r = 16km ride and 10w = 10km walk for girl and 12w = 12km walk, 8w = 8km walk and 20r = 20km ride for boy

each ride or walk takes 1hr only the 8w takes 40mins
***
G = girl's routine
B = boy's routine

I did, my coment was with respect to your first solution. The second one is much closer to the correct answer of 7hr 21.16 minutes. The last part of your second solution is not optimum because the girl arrives at destination ahead of the boy instead of the dame time.