The average of three consecutive odd numbers is 12 more than one third of the first of these numbers. What is the last of the three numbers?

Options

A. 15
B. 17
C. 19
D. Data inadequate

Answer

To solve the problem step by step, we will follow these instructions:

1. **Define the first odd number**: Let the first odd number be \( x \). The next two consecutive odd numbers will then be \( x + 2 \) and \( x + 4 \).

2. **Calculate the average of the three numbers**: The average of these three consecutive odd numbers can be calculated as follows:
\[ \text{Average} = \frac{x + (x + 2) + (x + 4)}{3} = \frac{3x + 6}{3} = x + 2 \]

3. **Find one third of the first number**: One third of the first number \( x \) is given by:
\[ \frac{x}{3} \]

4. **Set up the equation based on the problem statement**: According to the problem, the average of the three consecutive odd numbers is 12 more than one third of the first number. Therefore, we can set up the equation:
\[ x + 2 = \frac{x}{3} + 12 \]

5. **Clear the fraction by multiplying the entire equation by 3**:
\[ 3(x + 2) = x + 36 \]
Simplifying this gives:
\[ 3x + 6 = x + 36 \]

6. **Rearranging the equation**: Move \( x \) to the left side:
\[ 3x - x = 36 - 6 \]
This simplifies to:
\[ 2x = 30 \]

7. **Solve for \( x \)**:
\[ x = \frac{30}{2} = 15 \]

8. **Determine the last of the three numbers**: The last of the three consecutive odd numbers is:
\[ x + 4 = 15 + 4 = 19 \]

Thus, the last of the three numbers is **19**.