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 :

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:

[tex]\frac{26}{x+10}=\frac{6}{x}[/tex]

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

[tex]v=\frac{d}{t}[/tex]

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

[tex]t=\frac{d}{v}[/tex]

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:

[tex]26x=6(x+10)[/tex]

Now we apply the distributive property on the right side:

[tex]26x=6x+60[/tex]

Now we subtract 6x from both sides:

[tex]\begin{gathered} 26x-6x=6x-6x+60 \\ 20x=60 \end{gathered}[/tex]

Now we divide both sides by 20:

[tex]\begin{gathered} \frac{20x}{20}=\frac{60}{20} \\ x=3 \end{gathered}[/tex]

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:

[tex]t=\frac{d}{v}[/tex]

Plugging in the values we get:

[tex]t=\frac{35}{3}=11.7[/tex]

therefore, the time is 11.7 hours.