Question

An oil tanker can be emptied by the main pump in 4 hours. An auxiliary pump can empty the tanker in 9 hours. If the main pump is started at 9 AM, when should the auxiliary pump be started so that the tanker is emptied by noon?

Short answer

The auxiliary pump should be started at 9:45 AM.

Step-by-step solution

- Determine the rates of the pumps

The main pump can empty the tanker in 4 hours. Therefore, the rate of the main pump is \(\frac{1}{4}\) of the tanker per hour. Similarly, the auxiliary pump can empty the tanker in 9 hours, so its rate is \(\frac{1}{9}\) of the tanker per hour.

- Set up the equation

Let \('t'\) be the time in hours after 9 AM when the auxiliary pump is started. Both pumps work together until noon, which is 3 hours after 9 AM. The main pump works for 3 hours, and the auxiliary pump works for \(3 - t\) hours.

- Write the combined work equation

Combine the work done by both pumps to equal 1 full tanker. The equation is: \[ \left( \frac{1}{4} \right) \times 3 + \left( \frac{1}{9} \right) \times (3 - t) = 1 \]

- Solve the equation for t

First, distribute and simplify the equation: \[ \frac{3}{4} + \frac{3 - t}{9} = 1 \] Combine the fractions by finding a common denominator: \[ \frac{3}{4} + \frac{3 - t}{9} = 1 \] \[ 27 + 4(3 - t) = 36 \] \[ 27 + 12 - 4t = 36 \] \[ 39 - 4t = 36 \] \[ -4t = -3 \] \[ t = \frac{3}{4} \]

- Convert t to a time

Since \( t = \frac{3}{4} \) hours, convert this to minutes by multiplying by 60 minutes per hour: \[ \frac{3}{4} \times 60 = 45 \text{ minutes} \] Thus, the auxiliary pump should be started 45 minutes after 9 AM.

Key concepts

Rate of Work

Understanding the 'rate of work' is crucial in work problems. The rate of work tells us how much of a task can be completed in a given time period. In this scenario, the oil tanker is emptied by two pumps. The main pump empties the tanker in 4 hours, giving it a rate of \(\frac{1}{4}\) of the tanker per hour. This means that every hour, the main pump completes \(\frac{1}{4}\) of the job. Similarly, the auxiliary pump, which can empty the tanker in 9 hours, operates at a rate of \(\frac{1}{9}\) of the tanker per hour. These rates of work help us set up equations to solve the problem.

Combined Work Formula

When multiple parties are working together on a single task, their combined work rates can be added to determine how quickly they can complete the task. This is known as the combined work formula. In our exercise, we need to find out when to start the auxiliary pump so that both pumps together can finish the tanker by noon. The formula we use adds the rates at which each pump empties the tanker. Here is the combined work equation:
\[ \frac{1}{4} \times 3 + \frac{1}{9} \times (3 - t) = 1 \]

The variable \('t'\) represents the time in hours after 9 AM when the auxiliary pump starts.

Algebraic Equations

Solving work problems often involves setting up and solving algebraic equations. Here we derive our equation from the combined work formula. Expanding and simplifying the equation step-by-step, we solve for 't', the time after 9 AM when the auxiliary pump should start:

- Expand the equation:
\[ \frac{3}{4} + \frac{3 - t}{9} = 1 \]

- Combine fractions by finding a common denominator:
\[ 27 + 4(3 - t) = 36 \]

- Simplify and solve for \('t'\):
\[ 39 - 4t = 36 \]
\[ -4t = -3 \]
\[ t = \frac{3}{4} \]
This approach ensures we find the precise moment when the auxiliary pump should start.

Time Conversion

In word problems involving time, converting units of work correctly is essential. In our exercise, we find that \( t = \frac{3}{4} \) hours. To make this information practical, we convert hours into minutes, since \( \frac{3}{4} \times 60 = 45 \) minutes. Therefore, the auxiliary pump should begin operation 45 minutes after 9 AM. This ensures that both the main and auxiliary pumps together empty the tanker by noon.