Question

A farmer planned fence-posts at $6$ metre intervals along a straight side, posts at both ends. He bought posts but found he had $5$ less than needed for $6$ m spacing. However, with $8$ m spacing, he had exactly enough. What is the length of the side and how many posts did he buy?

• A. 100, 15
• B. 100, 16
• C. 120, 15
• D. 120, 16

Hint:

For fence-post problems, remember that number of posts = intervals + 1.

Answer 1

The Correct Option is D

Let $L$ = length of the side in metres, and $P$ = posts bought. Step 1: Posts needed for $6$ m spacing If spacing is $6$ m, number of intervals = $L/6$, number of posts = $(L/6) + 1$. He has $5$ less than this: \[ P = \left(\frac{L}{6} + 1\right) - 5 \] \[ P = \frac{L}{6} - 4 \quad (1) \] Step 2: Posts needed for $8$ m spacing If spacing is $8$ m, posts = $(L/8) + 1$. This equals his stock: \[ P = \frac{L}{8} + 1 \quad (2) \] Step 3: Equate (1) and (2) \[ \frac{L}{6} - 4 = \frac{L}{8} + 1 \] Multiply by 24: \[ 4L - 96 = 3L + 24 \] \[ L = 120 \] Step 4: Find $P$ From (2): \[ P = \frac{120}{8} + 1 = 15 + 1 = 16 \] \[ \boxed{L = 120\ \text{m},\ P = 16} \]