A farmer has decided to build a wire fence along one straight side of his property. For this, he planned to place several fence-posts at 6 m intervals, with posts fixed at both ends of the side. After he bought the posts and wire, he found that the number of posts he had bought was 5 less than required. However, he discovered that the number of posts he had bought would be just sufficient if he spaced them 8 m apart. What is the length of the side of his property and how many posts did he buy?

### Concept & Strategy This problem sets up a linear equivalence based on the fencepost principle: Number of posts = (Total Length / Interval) + 1. By equating the number of posts bought under two different spacing scenarios, we can solve for the total length. ### Step-by-Step Solution * **Given:** Scenario A (Planned): 6m intervals. Requires $P$ posts. Scenario B (Actual): Posts bought were $P - 5$. Scenario C (Adjusted): Spacing them 8m apart perfectly uses the bought posts. * Let $L$ be the total length of the side of the property. Let $N$ be the number of posts actually bought. * Create an equation for the *required* posts based on the 6m interval: $$\text{Required Posts} = \frac{L}{6} + 1$$ * Create an equation for the *bought* posts ($N$). We know he bought 5 less than required: $$N = \left(\frac{L}{6} + 1\right) - 5$$ $$N = \frac{L}{6} - 4$$ * Create an equation for the *bought* posts based on the 8m interval working perfectly: $$N = \frac{L}{8} + 1$$ * Set the two expressions for $N$ equal to each other to solve for $L$: $$\frac{L}{6} - 4 = \frac{L}{8} + 1$$ $$\frac{L}{6} - \frac{L}{8} = 5$$ * Find a common denominator (24) to solve: $$\frac{4L - 3L}{24} = 5$$ $$\frac{L}{24} = 5$$ $$L = 120 \text{ metres}$$ * Now substitute $L = 120$ back into either equation to find $N$ (posts bought): $$N = \frac{120}{8} + 1$$ $$N = 15 + 1 = 16$$ ### Exam Strategy & Shortcut Use Option Elimination based on divisibility. The length must be perfectly divisible by 8 (since spacing them 8m apart worked perfectly with no remainder). Both 100 and 120 are in the options. $100 / 8 = 12.5$ (invalid). $120 / 8 = 15$. This immediately isolates the 120m options. 15 gaps require 16 posts. Thus, 120m and 16 posts is the only logical answer. ### Common Pitfall Forgetting the "+1" rule for posts. If you use $\text{Posts} = \frac{L}{\text{Interval}}$ without adding 1 for the end post, the algebraic setup will fail, leading to an incorrect length or a negative number of posts. ### Final Answer Therefore, the correct answer is **120 m, 16**.