Chapter A.8, Problem 47AYU

To determine

To find: When should the auxiliary pump be started so that the tanker is emptied by noon?

Answer to Problem 47AYU

Started the auxiliary pump at 9:45 AM.

Explanation of Solution

Given:

An oil tanker can be emptied by the pump in 4 hours. An auxiliary pump can empty the tanker in 9 hours. If the main pump is started at 9 am.

Calculation:

The main pump’s emptying rate is $\frac{1}{4}$ tanker per hour. It will be open the entire 3 hours, so it will empty $\frac{3}{4}$ of the tank.

The other $\frac{1}{4}$ of tank must be emptied by the auxiliary pump.

We need to know how long it will take the auxiliary pump to empty the remaining $\frac{1}{4}$ tank which the main pump cannot empty in the 3 hours. The auxiliary pump’s rate is $\frac{1}{9}$ tanker per hour. Let $X$ be the number of hours it takes the auxiliary pump to empty the $\frac{1}{4}$ tank which the main pump will not be able to empty in the 3 hours allotted.

(rate in tankers per hour)(time required) $=\frac{1}{4}$ tank

$\left(\frac{1}{9}\right)x=\frac{1}{4}$

Therefore $4x=9$.

$x=\frac{9}{4}$

$x=2\frac{1}{4}$

$x=$ 2 hours 15 minutes.

So the auxiliary tank must be started 2 hours 15 minutes before 12 noon, so it must started at 9:45 AM.