Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Options

A. (1, 3) and (5, 7)
B. (3, 5) and (5, 7)
C. (3, 5) and (7, 9)
D. (5, 7) and (7, 9)

Explanation

To solve the problem of finding all pairs of consecutive odd positive integers both of which are smaller than 10 and whose sum is more than 11, we can follow these steps:

### Step 1: Define the integers
Let the first odd positive integer be \( x \). Since we are looking for consecutive odd integers, the next consecutive odd integer will be \( x + 2 \).

### Step 2: Set up the inequalities
We need to satisfy two conditions:
1. Both integers must be smaller than 10.
2. Their sum must be more than 11.

This gives us the following inequalities:
1. \( x < 10 \)
2. \( x + 2 < 10 \)
3. \( x + (x + 2) > 11 \)

### Step 3: Simplify the inequalities
From the second inequality \( x + 2 < 10 \):
\[ x < 10 - 2 \implies x < 8 \]

From the third inequality \( x + (x + 2) > 11 \):
\[ 2x + 2 > 11 \implies 2x > 11 - 2 \implies 2x > 9 \implies x > \frac{9}{2} \implies x > 4.5 \]

### Step 4: Determine the integer values for \( x \)
Since \( x \) must be a positive odd integer, the possible values of \( x \) that satisfy \( 4.5 < x < 8 \) are:
- \( x = 5 \)
- \( x = 7 \)

### Step 5: Find the pairs of integers
Now we can find the pairs of consecutive odd integers:
1. For \( x = 5 \):
- The consecutive odd integers are \( 5 \) and \( 7 \).
2. For \( x = 7 \):
- The consecutive odd integers are \( 7 \) and \( 9 \).

### Step 6: Check the conditions
Now we check if these pairs meet the conditions:
- For the pair \( (5, 7) \):
- Both are less than 10: Yes.
- Their sum: \( 5 + 7 = 12 > 11 \): Yes.

- For the pair \( (7, 9) \):
- Both are less than 10: Yes.
- Their sum: \( 7 + 9 = 16 > 11 \): Yes.

### Conclusion
The pairs of consecutive odd positive integers both smaller than 10 and whose sum is more than 11 are:
- \( (5, 7) \)
- \( (7, 9) \)