A boy, a girl and a dog go for a 10 mile hike.
The boy and girl can walk at 2 mph and the dog can trot at 3 mph.
They also have a bicycle which only one of them (including the dog!) can use at a time.
When riding, the boy and girl can travel at 12 mph while the dog can pedal at 16 mph.
What is the shortest time in which all three can complete the trip?
Solution:
Here is my original answer, but it neglects the physical location of the bike:The shortest time in which the boy, girl, and dog can make the ten mile hike is 3.1 hours or 3 hours 6 minutes.
Since all the three contestants must finish the trip, the three contestants must act in a manner to ensure that they finish at the same time.
Since the respective walking speed and the cycling speed of the boy and the girl are equal, it follows that the respective distances traversed by the boy and the girl in walking and cycling must be equal.
Let the distance traveled by the boy and the girl in cycling be x miles so that the respective distances traveled by walking is (10-x) miles.
Since the boy and girl each ride the bike for x miles, then the dog rides the bike for 10 - 2x miles
since x + x + (10 - 2x) = 10 miles.
Accordingly, the time taken by each of the boy and the girl to reach the finish line is (x/12 + (10-x)/2) hours, while the total time taken by the dog to reach the finish line is (2x/3 + (10 - 2x)/16) hours.
Set these two times equal to each other to find the time that it would take for the three of them to get to the finish line at the same time.
(x/12 + (10-x)/2) = (2x/3 + (10 - 2x)/16)
Multiply through by the LCM 48 to obtain:
4x + 240 - 24x = 32x + 30 - 6x
240 - 30 = 20x + 26x
210 = 46x
x = 4.57 miles.
So the boy and girl each ride the bike for 4.57 miles. Now substitute this value in each of the expressions for time above and you obtain 3.1 hours.
Consequently, the shortest time in which all three can complete the trip is 3.1 hours or, 3 hours 6 minutes.
However, several problem solvers did take the location of the bike into consideration. Here is Seth Cohen's solution:
The basic idea:
1. Boy bikes to some distance x. Girl walks and dog runs.
2. Boy leaves bike, starts walking. Dog runs to bike. Girl keeps walking.
3. Dog bikes back to girl.
4. Girl bikes to boy. Dog runs.
5. Dog runs to bike, boy and girl walk.
6. Boy and girl walk, dog bikes, so they all arrive at the end at the same time.
So we have to figure things out in terms of x, and then solve for x so that everyone arrives at the same time at the end.
1. Boy bikes x miles in x/12 hours. In this time, girl walks 2*(x/12)=x/6 miles, dog runs 3*(x/12)=x/4 miles.
2. Time for dog to get to bike: (3x/4)/3 = x/4 hours. In this time, boy walks 2*(x/4)=x/2 miles to x+(x/2) = 3x/2. Girl walks x/2 miles to (x/6)+(x/2) = 2x/3.
3. Time for dog to bike back to girl: (x/3)/(2+16) = x/54 hours. In this time, boy walks 2*(x/54)=x/27 miles to (3x/2)+(x/27) = 83x/54. Girl walks x/27 miles to (2x/3)+(x/27) = 19x/27.
4. Time for girl to bike to boy: ((83x/54)-(19x/27))/(12-2) = x/12 hours. In this time, boy walks 2*(x/12) = x/6 miles to (83x/54)+(x/6) = 46x/27. Dog runs 3*(x/12) = x/4 miles to (19x/27)+(x/4) = 103x/108.
5. Time for dog to run to bike: ((46x/27)-(103x/108))/3 = x/4 hours. In this time, boy and girl walk 2*(x/4) = x/2 miles to (46x/27)+(x/2) = 119x/54.
6. Time for boy and girl to walk to end: (10-(119x/54))/2 = (540-119x)/108 hours. Time for dog to bike to end: (10-(46x/27))/16 = (135-23x)/216 hours. Set these equal to each other to get x = 189/43. Plug this into this last time stretch to get (540-119x)/108 = 27/172 hours.
Total time = 545/172 = 3.1686 hours.
Correctly solved by:
| 1. Seth Cohen | Concord, New Hampshire, USA |
| 2. Ivy Joseph | Pune, Maharashtra, India |
| 3. K. M. Thomas | Mumbai, Maharashtra, India |
| 4. Kamal Lohia |
Holy Angel School, Hisar, Haryana, India |
| 5. Kelly Stubblefield | Mobile, Alabama, USA |