Question

A tourist can bicycle 26 miles in the same time as he can walk 6 miles. If he can ride 10 mph faster than he can walk, how much time ( in hr) should he allow to walk a 35-mile trail?

Answer 1

We are given that a person can bicycle 26 miles in the same time it takes to walk 6 miles. If we say that "x" is the walking speed in mph then since he can ride 10 mph faster than we can walk we have the following relationship:

`(26)/(x+10)=(6)/(x)`

this relationship comes from the fact that the velocity is defined as the distance over time, like this:

`v=(d)/(t)`

Since we are given that times are equal, then if we solve for the time we get:

`t=(d)/(v)`

Therefore, the distance over the velocity gives us the time and since the times are equal, we get the relationship. Now we can solve for "x" by cross multiplying:

`26x=6(x+10)`

Now we apply the distributive property on the right side:

`26x=6x+60`

Now we subtract 6x from both sides:

`\begin{gathered} 26x-6x=6x-6x+60 \\ 20x=60 \end{gathered}`

Now we divide both sides by 20:

`\begin{gathered} (20x)/(20)=(60)/(20) \\ x=3 \end{gathered}`

Therefore, the walking speed is 3 mph. Now we need to determine the time it takes to walk 35 miles. We do that applying the formula for the time we got previously:

`t=(d)/(v)`

Plugging in the values we get:

`t=(35)/(3)=11.7`

therefore, the time is 11.7 hours.