A man sold half number of oranges more than half his oranges. He then sold half oranges more than his remaining oranges. He sold the same the third time and then had 7 oranges left. How many is his original number of oranges? 

Let's break down the problem step by step. 

Let the original number of oranges be \( x \).

### First Sale:
- The man sold half of his oranges plus half an orange more.
- This means he sold:
  \[
  \text{Oranges sold} = \frac{x}{2} + \frac{1}{2}
  \]
- The number of oranges remaining after the first sale is:
  \[
  \text{Remaining oranges} = x - \left(\frac{x}{2} + \frac{1}{2}\right) = \frac{x}{2} - \frac{1}{2}
  \]

### Second Sale:
- He then sold half of the remaining oranges plus half an orange more.
- The remaining oranges after the first sale is \( \frac{x}{2} - \frac{1}{2} \).
- The number of oranges sold in the second sale is:
  \[
  \text{Oranges sold} = \frac{1}{2} \left(\frac{x}{2} - \frac{1}{2}\right) + \frac{1}{2} = \frac{x}{4} - \frac{1}{4} + \frac{1}{2} = \frac{x}{4} + \frac{1}{4}
  \]
- The number of oranges remaining after the second sale is:
  \[
  \text{Remaining oranges} = \left(\frac{x}{2} - \frac{1}{2}\right) - \left(\frac{x}{4} + \frac{1}{4}\right) = \frac{x}{4} - \frac{3}{4}
  \]

### Third Sale:
- He then sold half of the remaining oranges plus half an orange more.
- The remaining oranges after the second sale is \( \frac{x}{4} - \frac{3}{4} \).
- The number of oranges sold in the third sale is:
  \[
  \text{Oranges sold} = \frac{1}{2} \left(\frac{x}{4} - \frac{3}{4}\right) + \frac{1}{2} = \frac{x}{8} - \frac{3}{8} + \frac{1}{2} = \frac{x}{8} + \frac{1}{8}
  \]
- The number of oranges remaining after the third sale is:
  \[
  \text{Remaining oranges} = \left(\frac{x}{4} - \frac{3}{4}\right) - \left(\frac{x}{8} + \frac{1}{8}\right) = \frac{x}{8} - \frac{7}{8}
  \]
- After this third sale, the man has 7 oranges left:
  \[
  \frac{x}{8} - \frac{7}{8} = 7
  \]

### Solving the Equation:
\[
\frac{x}{8} - \frac{7}{8} = 7
\]
Multiply the entire equation by 8 to eliminate the fraction:
\[
x - 7 = 56
\]
Add 7 to both sides:
\[
x = 63
\]

Thus, the original number of oranges was \( \boxed{63} \).

Do you have any questions or would you like more details?

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Here are 5 related questions:
1. How would the original number change if he had 8 oranges left instead of 7?
2. What would be the equation if the man sold one-third of his oranges plus one more each time?
3. How would the solution change if the man sold one additional orange at each sale?
4. Can you find a general formula for the remaining oranges after each sale?
5. What if the man sold a different fraction, such as one-fourth, instead of one-half?

**Tip:** When solving problems involving sequential steps, carefully track each step's results to ensure accuracy.

Learn how to solve a problem involving the sequential sale of oranges with a detailed step-by-step solution. This problem covers arithmetic operations and fractions, suitable for students in Grades 6-8.

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