Hector used a tool called an auger to remove corn from a storage bin at a constant rate. The bin contained 24,000 bushels of corn when Hector began to use the auger. After 5 hours of using the auger, 19,350 bushels of corn remained in the bin. If the auger continues to remove corn at this rate, what is the total number of hours Hector will have been using the auger when 12,840 bushels corn remain in the bin?

We are given the following information:

- The initial amount of corn in the storage bin: 24,000 bushels.
- After 5 hours, 19,350 bushels of corn remain in the bin.
- We need to find the total number of hours Hector will have been using the auger when 12,840 bushels of corn remain in the bin.

### Step 1: Find the rate at which the auger removes corn.
In 5 hours, the amount of corn in the bin decreased from 24,000 bushels to 19,350 bushels. The amount of corn removed is:

\[
\text{Corn removed in 5 hours} = 24,000 - 19,350 = 4,650 \text{ bushels}.
\]

Since this amount was removed in 5 hours, the rate at which corn is being removed by the auger is:

\[
\text{Rate of removal} = \frac{4,650}{5} = 930 \text{ bushels per hour}.
\]

### Step 2: Set up the equation to find when 12,840 bushels remain.
We want to find the total time \( t \) when 12,840 bushels remain in the bin. The number of bushels removed after \( t \) hours is given by the expression \( 24,000 - 930t \), and we know that we want this to equal 12,840:

\[
24,000 - 930t = 12,840.
\]

### Step 3: Solve for \( t \).
First, subtract 12,840 from both sides:

\[
24,000 - 12,840 = 930t.
\]

This simplifies to:

\[
11,160 = 930t.
\]

Now, divide both sides by 930 to solve for \( t \):

\[
t = \frac{11,160}{930} = 12.
\]

### Step 4: Interpret the result.
Hector will have been using the auger for **12 hours** when 12,840 bushels remain in the bin.

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Would you like more details or have any questions?

Here are 5 related questions to explore:
1. What would be the total time if the auger removed corn at a different rate?
2. How much corn would remain in the bin after 8 hours of use?
3. What if the initial amount of corn was different, say 30,000 bushels?
4. How long would it take to remove all the corn from the bin?
5. How would the problem change if the rate of removal increased over time?

**Tip:** When dealing with rates and changes over time, organizing the information into known variables and setting up equations step-by-step often helps solve the problem efficiently.

Learn how to solve a real-world problem involving the removal of corn from a bin using an auger at a constant rate. This step-by-step solution covers key concepts like linear equations and rates of change, and is suitable for students in Grades 6-8. In this example, find out how long it will take to reduce the corn in the bin to 12,840 bushels.

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