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So here's my problem I'm having:

Say someone runs or walks 40 kilometers in a day in 8 hours. We don't know how many kilometers in specific the person was walking or running, though that's what we'd like to know.

We do know the persons average running / walking speed though:

  • Running speed: 9 km/h
  • Walking speed: 4 km/h

Is there a way to figure out how many kilometers the person was running and how many walking, assuming that there are only the two speeds mentioned above?

Stephan Wagner
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1 Answers1

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The person walks during $t_1$ hours and makes $4t_1$ km

The person runs during $t_2$ hours and makes $9t_2$ km

The resulting distance is $4t_1+9t_2 = 40$ km

The total time is $t_1+t_2=8$ hours

Can you solve the system ?

Here is an alternating way of thinking about the problem.

First let's have a look at this one:

We need 56 biscuits to feed 10 animals (cats and dogs). Dogs each 6 biscuits and cats eat 5 biscuits. How many cats and dogs are there?

We can solve with the system $\begin{cases} 5c+6d = 56\\c+d=10\end{cases}$

See the analogy with your problem ? Replace $(c,d)$ by $(t_1,t_2)$ and this is very similar.

We can solve it this way:

  • let each animal eat $5$ biscuits
  • then $50$ biscuits are eaten and $6$ are left over
  • cats' belly are full but dogs miss $6-5=1$ biscuit
  • since there are $6$ biscuits left, there should be $\frac 61$ dogs

We conclude for $6$ dogs and $4$ cats.

Now coming back to our problem:

There are $40$ km to be travelled by me (walking me and running me). Walking me travel at $4$ km/h while running me travel at $9$ km/h. How much of walking and running is there?

  • let walk all the way at $4$ km/h
  • then $32$ km are travelled and $8$ km are left over
  • walking me is satisfied but running me wants more
  • since there are $8$ km missing, I should have increased my allure by $9-4=5$ km/h to cover this distance in $\frac 85$ hours.

So I ran for 1h 36 min and walked the rest.

zwim
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