How Far Does C Walk?

Date: 08/06/2001 at 07:20:33
From: pollution
Subject: Speed

Three persons A, B, and C, travel from point X to point Y. They leave 
point X at the same time and follow the same route. 

A rides a motorcycle at a speed of 56 kilometres/h, B walks at a speed 
of 8 kilometres/h, and C walks at a speed of 7 kilometres/h. At the 
beginning, A gives a lift to B at X to travel a part of the journey 
and then comes back to pick up C. They all arrive at point Y at the 
same time. 

If B walks 2.8 kilometres, how far does C walk?

Date: 08/07/2001 at 12:38:55
From: Doctor Greenie
Subject: Re: Speed
Hello --

Thanks for sending your question to us here at Dr. Math.

Let's divide the problem into three time periods:

   t1 = time A is carrying B from X toward Y (C is walking)
   t2 = time A is heading back from Y towards X to pick up C 
        (B and C are both walking)
   t3 = time A is again heading from X towards Y, carrying C 
        (B is walking)

The rates of A, B, and C in kph are 56, 8, and 7, respectively.

Then we can write some expressions and equations using the given 
information:

(1) B travels time t1 at 56kph and time (t2+t3) at 8kph; the distance 
    he travels is

    56t1 + 8(t2+t3)  (1)

(2) C travels time (t1+t2) at 7kph and time t3 at 56kph; the distance 
    he travels is

    7(t1+t2) + 56t3  (2)

(3) B and C both travel the same distance, so (1) and (2) above are 
    equal:

    56t1 + 8(t2+t3) = 7(t1+t2) + 56t3  (3)

(4) B walks 2.8 km:

    8(t2+t3) = 2.8  (4)

At this point we should stop and think about where we are going with 
this problem. The variables we have set up to solve the problem are 
t1, t2, and t3. However, there is no requirement that we find the 
values of those three variables. The question we are trying to answer 
is the distance C walks; this distance is 7(t1+t2). So, to answer the 
problem, we only need to find a value for (t1+t2) - we don't 
necessarily need to find the values for t1 and t2 individually.

Now let's continue.

From (4) above, we have

    8(t2+t3) = 2.8

    t2+t3 = 2.8/8 = 0.35  (5)

Substituting (4) in (3) and simplifying, we have

    56t1 + 8(t2+t3) = 7(t1+t2) + 56t3

         56t1 + 2.8 = 7t1 + 7t2 + 56t3

         49t1 + 2.8 = 7t2 + 56t3  (6)

Now using (5) to rewrite the right-hand side of (6), we have

         49t1 + 2.8 = 7t2 + 56t3

         49t1 + 2.8 = 7t2 + 7t3 + 49t3

         49t1 + 2.8 = 7(t2+t3) + 49t3

         49t1 + 2.8 = 7(0.35) + 49t3

         49t1 + 2.8 = 2.45 + 49t3

         2.8 - 2.45 = 49(t3-t1)

               0.35 = 49(t3-t1)

            0.35/49 = t3-t1  (7)

Now we can combine (5) and (7):

              t2+t3 = 0.35

              t1-t3 = -0.35/49

              t1+t2 = 0.35 -0.35/49

              t1+t2 = 0.35(1-1/49)

              t1+t2 = 0.35(48/49)  (8)

We now have a numerical value for (t1+t2), so we can find the distance 
C walks:

    7(t1+t2) = (7)(0.35)(48/49) = (7)(7/20)(48/49) = 12/5 = 2.4 km

While we have answered the question which was asked, we can go further 
with the problem to learn more.

If we let d be the distance between X and Y, then we can find that 
distance by equating the times that A and B traveled.

B traveled (d-2.8) km at 56 kph and 2.8 km at 8 kph; the total time he 
traveled was

    (d-2.8)/56 + 2.8/8

A traveled (d-2.8) km with B in part 1 of his trip; he traveled back 
(d-5.2) km alone in part 2; and he traveled (d-2.4) km with C in part 
3. Altogether, he traveled (d-2.8)+(d-5.2)+(d-2.4) = (3d-10.4) km, all 
at 56 kph; the total time he traveled was

    (3d-10.4)/56

Because A, B, and C all reached Y at the same time, the total travel 
time is the same for A and B:

    (d-2.8)/56 + 2.8/8 = (3d-10.4)/56

Solving this equation gives us the distance from X to Y as 13.6 km; 
once we have this information, we can determine the duration of each 
of the three parts of the trip and the distances each person traveled 
in each of those time periods.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/