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.
- In what range of time is it possible to complete the dirt track?
- 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?
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 $t$.