Subfield: Geometry


Question: Serge likes to paddle his raft down the Speed River from point $A$ to point $B$. The speed of the current in the river is always the same. When Serge paddles, he always paddles at the same constant speed. On days when he paddles with the current, it takes him 18 minutes to get from $A$ to $B$. When he does not paddle, the current carries him from $A$ to $B$ in 30 minutes. If there were no current, how long would it take him to paddle from $A$ to $B$ ?

Solution:
Suppose that the distance from point $A$ to point $B$ is $d \mathrm{~km}$.

Suppose also that $r_{c}$ is the speed at which Serge travels while not paddling (i.e. being carried by just the current), that $r_{p}$ is the speed at which Serge travels with no current (i.e. just from his paddling), and $r_{p+c}$ his speed when being moved by both his paddling and the current.

It takes Serge 18 minutes to travel from $A$ to $B$ while paddling with the current.

Thus, $r_{p+c}=\frac{d}{18} \mathrm{~km} / \mathrm{min}$.

It takes Serge 30 minutes to travel from $A$ to $B$ with just the current.

Thus, $r_{c}=\frac{d}{30} \mathrm{~km} / \mathrm{min}$.

But $r_{p}=r_{p+c}-r_{c}=\frac{d}{18}-\frac{d}{30}=\frac{5 d}{90}-\frac{3 d}{90}=\frac{2 d}{90}=\frac{d}{45} \mathrm{~km} / \mathrm{min}$.

Since Serge can paddle the $d \mathrm{~km}$ from $A$ to $B$ at a speed of $\frac{d}{45} \mathrm{~km} / \mathrm{min}$, then it takes him 45 minutes to paddle from $A$ to $B$ with no current.
Suppose that the distance from point $A$ to point $B$ is $d \mathrm{~km}$, the speed of the current of the river is $r \mathrm{~km} / \mathrm{h}$, and the speed that Serge can paddle is $s \mathrm{~km} / \mathrm{h}$.

Since the current can carry Serge from $A$ to $B$ in 30 minutes (or $\frac{1}{2} \mathrm{~h}$ ), then $\frac{d}{r}=\frac{1}{2}$.

When Serge paddles with the current, his speed equals his paddling speed plus the speed of the current, or $(s+r) \mathrm{km} / \mathrm{h}$.

Since Serge can paddle with the current from $A$ to $B$ in 18 minutes (or $\frac{3}{10} \mathrm{~h}$ ), then $\frac{d}{r+s}=\frac{3}{10}$.

The time to paddle from $A$ to $B$ with no current would be $\frac{d}{s} \mathrm{~h}$.



Since $\frac{d}{r}=\frac{1}{2}$, then $\frac{r}{d}=2$.

Since $\frac{d}{r+s}=\frac{3}{10}$, then $\frac{r+s}{d}=\frac{10}{3}$.

Therefore, $\frac{s}{d}=\frac{r+s}{d}-\frac{r}{d}=\frac{10}{3}-2=\frac{4}{3}$.

Thus, $\frac{d}{s}=\frac{3}{4}$, and so it would take Serge $\frac{3}{4}$ of an hour, or 45 minutes, to paddle from $A$ to $B$ with no current.
Suppose that the distance from point $A$ to point $B$ is $d \mathrm{~km}$, the speed of the current of the river is $r \mathrm{~km} / \mathrm{h}$, and the speed that Serge can paddle is $s \mathrm{~km} / \mathrm{h}$.

Since the current can carry Serge from $A$ to $B$ in 30 minutes (or $\frac{1}{2}$ h), then $\frac{d}{r}=\frac{1}{2}$ or $d=\frac{1}{2} r$.

When Serge paddles with the current, his speed equals his paddling speed plus the speed of the current, or $(s+r) \mathrm{km} / \mathrm{h}$.

Since Serge can paddle with the current from $A$ to $B$ in 18 minutes (or $\frac{3}{10} \mathrm{~h}$ ), then $\frac{d}{r+s}=\frac{3}{10}$ or $d=\frac{3}{10}(r+s)$.

Since $d=\frac{1}{2} r$ and $d=\frac{3}{10}(r+s)$, then $\frac{1}{2} r=\frac{3}{10}(r+s)$ or $5 r=3 r+3 s$ and so $s=\frac{2}{3} r$.

To travel from $A$ to $B$ with no current, the time in hours that it takes is $\frac{d}{s}=\frac{\frac{1}{2} r}{\frac{2}{3} r}=\frac{3}{4}$, or 45 minutes.

Final answer: 45 minute