SOLUTION: A bus with 12 soccer players broke down in a town 20 miles from its destination. The coach’s car was available but could carry only 4 players at a time. Also, it could travel only

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Question 184982: A bus with 12 soccer players broke down in a town 20 miles from its destination. The coach’s car was available but could carry only 4 players at a time. Also, it could travel only 20 miles per hour because of the traffic. The players said they could walk at 4 miles per hour when they were not riding. Suppose the coach took 4 of them part way, came back for 4 more and took them part way, and then came back for the last 4. How could they all get to their scheduled soccer game at the same time?

Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!

We make this diagram, where the broken-down bus is at A and the
destination is at F.  B,C,D, and E are points along the way
with unknown distances v,w,x,y,z miles between them. 

|  v miles  |  w miles  | x miles  |  y miles  |  z miles  |      
------------------------------------------------------------
A           B           C          D           E           F

We of course have the first equation as

             v + w + x + y + z = 20

Here are the 5 trips the car makes.

1. Car drives from A to D. (1st 4 get out and walk from D to F.)
2. Car drives from D to B. (picks up 2nd 4, who have walked from A to B.) 
3. Car drives from B to E. (2nd 4 get out and walk from E to F.) 
4. Car drives from E to C. (picks up 3rd 4, who have walked from A to C.) 
5. Car drives from C to F.  

Now we'll fill in the details:

1. Car takes 1st 4 from A to D, a distance of v+w+x miles.
   When car gets to D, it has traveled v+w+x miles and therefore
   (v+w+x)/20 hours have passed, using TIME = DISTANCE/RATE. 

   [Then the 1st 4 have y+z miles yet to walk to F, which will take them
    (y+z)/4 hours.  So total time = (v+w+x)/20 + (y+z)/4 hours, or
    simplifying, total time = (v+w+x+5y+5z)/20 hours.]  

2. Car drives from D to B, a distance of x+w miles, which takes 
   (x+w)/20 hours more. When car gets to B, (v+w+x)/20 + (x+w)/20,
   or (v+2w+2x)/20 hours have passed. During this time the 2nd and
   3rd 4 have walked from A to B and using Distance=Rate·Time, this 
   is 4(v+2w+2x)/20 or (v+2w+2x)/5 miles.  Since A to B is also v, 
   we have the equation v = (v+2w+2x)/5 which when simplified becomes
   the equation 5v = v+2w+2x or 4v = 2w+2x or 2v = w+x 

3. Car drives from B to E, a distance of w+x+y miles, which takes
   (w+x+y)/20 miles.  When car gets to E, (v+2w+2x)/20 + (w+x+y)/20,
   or (v+3w+3x+y)/20 hours have passed. 

   [Then the 2nd 4 have z miles yet to walk from E to F, which will 
    take them z/4 hours.  So total time = (v+3w+3x+y)/20 + z/4 hours,
    or simplifying, total time = (v+3w+3x+y+5z)/20 hours.] 

    Now we have two expressions for the total time, so we can
    set them equal
    (v+w+x+5y+5z)/20 = (v+3w+3x+y+5z)/20
         v+w+x+5y+5z = v+3w+3x+y+5z
                  4y = 2w+2x
                  2y = w+x
    
4. Car drives from E to C, a distance of y+x miles, which takes 
   (y+x)/20 hours more. When car gets to B, (v+3w+3x+y)/20 + (y+x)/20,
   or (v+3w+4x+2y)/20 hours have passed. During this time the 3rd 4 
   have walked from A to C and using DISTANCE=RATExTIME, this 
   is 4(v+3w+4x+2y)/20 or (v+3w+4x+2y)/5 miles.  Since A to C is also v+w, 
   we have the equation v+w = (v+3w+4x+2y)/5 which when simplified becomes
   the equation 5v+5w = v+3w+4x+2y or 4v+2w=4x+2y or 2v+w = 2x+y.

5. Car drives from C to F, the destination, a distance of x+y+z miles, 
   which takes (x+y+z)/20 hours more. So the total time that has passed
   is (v+3w+4x+2y)/20 + (x+y+z)/20 or (v+3w+5x+3y+z)/20.  So this is a
   third expression for the total time, and we set it equal to one of the
   above expressions for the total time, and get

   (v+3w+5x+3y+z)/20 = (v+w+x+5y+5z)/20
        v+3w+5x+3y+z = v+w+x+5y+5z   
               2w+4x = 2y+4z
                 w+x = y+z

So the equations we have are

v+w+x+y+z=20    
2v = w+x
2y = w+x
2v+w = 2x+y 
w+x = y+z

To solve it by matrices, we write it:

 v + w +  x +  y + z = 20
2v - w -  x          =  0
   - w -  x + 2y     =  0
2v + w - 2x -  y     =  0
     w +  x -  y - z =  0

So we get v=4, w=4, x=4, y=4, z=4

So    

|  4 miles  |  4 miles  | 4 miles  |  4 miles  |  4 miles  |      
------------------------------------------------------------
A           B           C          D           E           F

1. Car drives from A to D. That's 12 miles. 
2. Car drives from D to B. That's 8 miles.  
3. Car drives from B to E. That's 12 miles 
4. Car drives from E to C. That's 8 miles. 
5. Car drives from C to F. That's 12 miles.

So the car went a total of 12+8+12+8+12 or 52 miles.
At 20 miles per hour, that took the car 52/20 or 13/5 hours or 2 3/5 hours

The 1st 4:
Rode 12 miles from A to D, which took them 12/20 or 3/5 hours,
Then walked 8 miles from D to F, which took them 8/4 or 2 hours.
So it took the 1st 4 also 2 3/5 hours to get from A to F     
    
The 2nd 4:
Walked 4 miles from A to B, which took them 4/4 or 1 hour
Rode 12 miles from B to E, which took them 12/20 or 3/5 hour.
Walked 4 miles from E to F, which took them 4/4 or 1 hour.
So it took the 2nd 4 also 2 3/5 hours to get from A to F

The 3rd 4:
Walked 8 miles from A to C, which took them 8/4 or 2 hours
Rode 12 miles from C to F, which took them 12/20 or 3/5 hours
So it took the 3rd 4 also 2 3/5 hours to get from A to F

Edwin
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