• October 29th 2009, 04:34 PM
sahip
Can you help with this ? Thanks.

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 ?
• October 29th 2009, 08:23 PM
ux0
2.5 per minute, try to work out the details, if you cant figure it out post again
• October 30th 2009, 04:33 AM
HallsofIvy
Quote:

Originally Posted by sahip
Can you help with this ? Thanks.

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 ?

Suppose the water was removed at 5 gallons per minute for $T_1$ minutes. Then $5T_1= .75V$ where V is the volume of the hull. Suppose further that the pump ran for $T_2$ minutes at r gallons per minute. Then $rT_2= .25V$.

Divding the first equation by the second, $\frac{5T_1}{rT_2}= \frac {.75}{.25}= 3$ so $T_2= \frac{5}{3r}T_2$.

Then entire volume, V, was emptied in time $T_1+ T_2= T_1+ \frac{5}{3r}T_1= \frac{5+ 3r}{3r}T_2$ so the rate was $\frac{V}{\frac{5+3r}{3r}T_1}= = \frac{3r}{5+ 3r}\frac{V}{T_1}= 4$.

But $5T_1= .75V$ so $\frac{V}{T_1}= \frac{5}{.75}= \frac{20}{3}$.

That is, $\frac{20}{3}\frac{3r}{5+3r}= \frac{20r}{5+ 3r}= 4$.

Solve that for r.