1. ## Algebra

A ship's hull was punctured and filled with water before being sealed off. A pump removed water at a steady rate of 5 gallons per minute, but after 75% of the water was removed, the pump slowed. If the water was removed at an average rate of 4 gallons per minute, at what rate was the remaining 25% of the water removed ?
Thank you.

2. Hello, sahip!

An interesting problem . . .
I had to baby-talk my way through it.

A ship's hull was punctured and filled with water before being sealed off.
A pump removed water at a steady rate of 5 gallons per minute,
but after 75% of the water was removed, the pump slowed.
If the water was removed at an average rate of 4 gallons per minute,
at what rate was the remaining 25% of the water removed ?
Let $V$ = orginal amount of water (gallons).

Let $R$ = slower rate of pumping (gallons per minute).

The first 75% of the water was removed at 5 gallons per minute.
That is, $\tfrac{3}{4}V$ gallons were removed at 5 gal/min.
. . This took: . $\frac{\frac{3}{4}V}{5}$ minutes.

The last 25% of the water was removed at $R$ gallons per minute.
That is, $\tfrac{1}{4}V$ gallons were removed at $R$ gal/min.
. . This took: . $\frac{\frac{1}{4}V}{R}$ minutes.
The total time to empty the water was: . $T \:=\: \frac{\frac{3}{4}V}{5} + \frac{\frac{1}{4}V}{R} \:=\:\frac{(3R+5)V}{20R}$ minutes.

Recap: They pumped $V$ gallons of water in $T$ minutes
. . . . . and the average rate was 4 gallons per minute.

There is our equation: . $\frac{V}{\frac{V(3R+5)}{20R}} \:=\:4 \quad\Rightarrow\quad V \:=\:4\cdot\frac{V(3R+5)}{20R}\quad\Rightarrow\qua d 20RV \:=\:4V(3r+5)$

Divide by $4V\!:\;\;5R \:=\:3R + 5 \quad\Rightarrow\quad 2R \:=\:5 \quad\Rightarrow\quad R \:=\:\frac{5}{2}$

The last 25% of the water was pumped out at $2\tfrac{1}{2}$ gallons per minute.