We are given that the perimeter of a rectangle is 320 inches. The perimeter $P$ of a rectangle is calculated as:

$P = 2 \times (L + W)$

where $L$ is the length and $W$ is the width. We are also given the following conditions:

1. The length $L$ is an odd integer.
2. The width $W$ is 5 times the next consecutive odd integer after the length.

Step 1: Express the width in terms of the length

Let the length $L = x$, where $x$ is an odd integer. The next consecutive odd integer is $x + 2$. Therefore, the width is:

$W = 5 \times (x + 2)$

Step 2: Substitute into the perimeter formula

Substitute $L = x$ and $W = 5 \times (x + 2)$ into the perimeter formula:

$320 = 2 \times \left( x + 5(x + 2) \right)$

Step 3: Simplify the equation

Simplify the equation inside the parentheses:

$320 = 2 \times \left( x + 5x + 10 \right)$ $320 = 2 \times (6x + 10)$ $320 = 12x + 20$

Step 4: Solve for $x$

Now, solve for $x$:

$320 - 20 = 12x$ $300 = 12x$ $x = \frac{300}{12} = 25$

So, the length $L = 25$.

Step 5: Find the width

Now, substitute $x = 25$ back into the expression for the width:

$W = 5 \times (25 + 2) = 5 \times 27 = 135$

Final Answer:

The length of the rectangle is 25 inches, and the width is 135 inches.

Would you like more details or have any questions?

Here are 5 related questions:

1. How do we confirm if the perimeter condition is satisfied?
2. Can the length and width be fractional values in this scenario?
3. What happens if the perimeter was increased to 500 inches with the same conditions?
4. How would the solution change if the width were 4 times the next consecutive even integer?
5. Can you express the relationship between the length and width as a function of the perimeter?

Tip: In problems involving integer constraints, it helps to work systematically with consecutive integers or functions of those integers.