Jacod begins painting at 12:00 noon. At 12:30 P.M. he estimate that 13 gallons of paint are left, and at 3:30 he estimates that 4 gallons of paint remains. If the paint is being used at a constant rate, how many gallons of paint did Jacod have when he started the job?

Answer

To solve the problem step by step, we will define the variables and set up equations based on the information given.

### Step 1: Define the initial amount of paint
Let \( x \) be the initial amount of paint (in gallons) that Jacod had when he started the job at 12:00 noon.

### Step 2: Determine the amount of paint left after 30 minutes
At 12:30 PM, Jacod estimates that 13 gallons of paint are left. Therefore, the amount of paint used in the first 30 minutes is:
\[ \text{Paint used in 30 minutes} = x - 13 \]

### Step 3: Determine the amount of paint left after 3.5 hours
At 3:30 PM, which is 3.5 hours (or 210 minutes) after he started, Jacod estimates that 4 gallons of paint are left. Thus, the amount of paint used in 3.5 hours is:
\[ \text{Paint used in 3.5 hours} = x - 4 \]

### Step 4: Calculate the rate of paint usage
Since the paint is being used at a constant rate, we can set up a ratio of the paint used to the time taken. The time taken for the first observation is 0.5 hours, and for the second observation, it is 3.5 hours. Therefore, we can write:
\[ \frac{x - 13}{0.5} = \frac{x - 4}{3.5} \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ (x - 13) \cdot 3.5 = (x - 4) \cdot 0.5 \]

### Step 6: Expand both sides of the equation
Expanding both sides results in:
\[ 3.5x - 45.5 = 0.5x - 2 \]

### Step 7: Rearrange the equation to isolate \( x \)
Rearranging the equation gives:
\[ 3.5x - 0.5x = 45.5 - 2 \]
\[ 3x = 43.5 \]

### Step 8: Solve for \( x \)
Dividing both sides by 3 gives:
\[ x = \frac{43.5}{3} = 14.5 \]

### Conclusion
Thus, the initial amount of paint Jacod had when he started the job is:
\[ \boxed{14.5} \text{ gallons} \]