0
$\begingroup$

Between a city and a village lie two bike tracks: 35km of paved road, and 20km of dirt road. Because of the thick forest between them, a cyclist can choose to ride only one.

The speed on the paved road is at least 4km/h faster than that on the dirt road, and the time it takes to complete the dirt track is shorter by a half hour than that of the paved road.

  1. In what range of time is it possible to complete the dirt track?
  2. In what range of time would it take for the cyclist to complete both tracks?

I solved the first question, but the second one stumped me:

$$\text{Let }v \text{ be the speed on the dirt road}\\\text{Let }x \text{ be the added speed on the paved road}$$ $$\frac{35}{v+x}=\frac{20}{v} + 0.5$$ From here, you can define $x$ in terms of $v$, and use the inequality $x > 4$, to get the correct range of: $$\boxed{10km/h < v < 16km/h}$$

I don't know how to definitively solve the second question. The total time it takes to complete both tracks would be: $$t = \frac{35}{v+x} + \frac{20}{v}$$

If you substitute $x = 4$ and the range of $v$ from before, you get the correct answer of: $$\boxed{3h < t < 4.5h}$$

...But why is this? This was just a pure guess by my part, as my first intuition was to describe $t$ in terms of $x$ for every speed limit & repeat the same process I did in the first question, but that turns into hell. How do I solve this without making this assumption?

$\endgroup$
1
$\begingroup$

From the first equation, we can solve for $x$:

$$x=\frac{30v-v^2}{v+40}$$

Plug this into the expression for $t$:

$$t=\frac{35}{v+\frac{30v-v^2}{v+40}}+\frac{20}{v}$$

Simplify:

$$t=\frac{40}{v}+\frac 1 2$$

From here, use the fact that $10 < v < 16$ to solve for the possible range of values for $t$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.