# Algebra-Some Current/Wind Problems

• May 29th 2007, 12:24 PM
SwissArmelle
Algebra-Some Current/Wind Problems
So these are really easy problems, but I haven't done them since the beginning of the year, so I could use some help.

1. A swimmer can swim 4 km with the current in 24 minutes. The same distance would take 40 minutes against the current. Find the speed of the swimmer and the rate of the current

2. A plane flies the first half of a 5600 km flight into the wind in 3.5 hours. The return trip, with the same wind, takes 2.5 hours. Find the speed of the wind and the speed of the plane in still air.
• May 29th 2007, 02:08 PM
Soroban
Hello, SwissArmelle!

Both can be solved with the same set-up.

Here's #1 . . .

We will use: . $\text{Speed }\times\text{ Time} \:=\:\text{Distance}$

Quote:

1. A swimmer can swim 4 km with the current in 24 minutes.
The same distance would take 40 minutes against the current.
Find the speed of the swimmer and the rate of the current

Let $S$ = speed of the swimmer (in km/hour).
Let $C$ = speed of the current.

With the current, his speed is $S + C$.
He swims 4 km in 24 minutes = $\frac{2}{5}$ hour.
We have: . $(S + C)\cdot\frac{2}{5} \:=\:4\quad\Rightarrow\quad S + C\:=\:10$ .[1]

Against the current, his speed is $S - C$
He swims 4 km in 40 minutes = $\frac{2}{3}$ hour.
We have: . $(S - C)\cdot\frac{2}{3}\:=\:4\quad\Rightarrow\quad S - C \:=\:6$ .[2]

Add [1] and [2]: . $2S \:=\:16\quad\Rightarrow\quad S \:=\:8$

Substitute into [1]: . $8 + C \:=\:10\quad\Rightarrow\quad C \:=\:2$

Therefore, the swimmer's speed is 8 km/hr; the current's speed is 2 km/hr.