An oil tanker can be emptied by the main pump in 5 hours.  An auxiliary pump can empty the tanker in 13 hours.  If the main pump is started at 6 PM​, when should the auxiliary pump be started so that the tanker is emptied by 10 PM​?

We are given the rates at which two pumps can empty an oil tanker:

- The main pump can empty the tanker in 5 hours, so its rate is \( \frac{1}{5} \) of the tanker per hour.
- The auxiliary pump can empty the tanker in 13 hours, so its rate is \( \frac{1}{13} \) of the tanker per hour.

We need to find when to start the auxiliary pump so that the tanker is completely emptied by 10 PM if the main pump starts at 6 PM.

### Step 1: Work done by the main pump
The main pump runs from 6 PM to 10 PM, which is 4 hours. The amount of the tanker the main pump empties in 4 hours is:
\[
\text{Work by main pump} = 4 \times \frac{1}{5} = \frac{4}{5}
\]
This means the main pump alone empties \( \frac{4}{5} \) of the tanker by 10 PM.

### Step 2: Remaining work for the auxiliary pump
By 10 PM, \( \frac{1}{5} \) of the tanker remains to be emptied. This needs to be handled by the auxiliary pump.

Let \( t \) be the time (in hours) that the auxiliary pump operates before 10 PM. The amount of the tanker the auxiliary pump can empty in \( t \) hours is:
\[
\text{Work by auxiliary pump} = t \times \frac{1}{13}
\]
This must equal \( \frac{1}{5} \), since \( \frac{1}{5} \) of the tanker remains to be emptied.

### Step 3: Set up the equation and solve for \( t \)
We now have the equation:
\[
t \times \frac{1}{13} = \frac{1}{5}
\]
Multiply both sides by 13 to isolate \( t \):
\[
t = \frac{13}{5} = 2.6 \text{ hours}
\]
Thus, the auxiliary pump needs to run for 2.6 hours.

### Step 4: Determine the start time
To have the auxiliary pump running for 2.6 hours by 10 PM, we subtract 2.6 hours from 10 PM:
\[
10 \text{ PM} - 2.6 \text{ hours} = 7:24 \text{ PM}
\]
So, the auxiliary pump should be started at 7:24 PM.

### Conclusion:
The auxiliary pump should be started at **7:24 PM** to ensure the tanker is emptied by 10 PM.

---

Would you like more details or have any questions about this?

Here are 5 related questions to expand your understanding:
1. What would happen if both pumps were started at the same time?
2. How long would it take to empty the tanker if only the auxiliary pump was used?
3. If the tanker capacity was doubled, how would the start time for the auxiliary pump change?
4. How can we calculate the time it would take with only the main pump running?
5. How would the scenario change if there were a third pump with a different rate?

**Tip**: Always check if the combined work of multiple agents is modeled as the sum of their individual rates when solving such problems!

An oil tanker can be emptied by the main pump in 5 hours. An auxiliary pump can empty the tanker in 13 hours. If the main pump is started at 6 PM​, when should the auxiliary pump be started so that the tanker is emptied by 10 PM​?

Solve a work rate problem involving two pumps emptying an oil tanker. Learn how to calculate the exact time to start the auxiliary pump so that the tanker is emptied by 10 PM. This solution covers algebraic concepts and time/work relationships, ideal for students in Grades 8-10.

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