Problem of the Week
Problem A and Solution
Fencepost Problem

Problem

Elyas helps his parents install fence posts in their backyard. They install a row of \(7\) equally-spaced posts. The posts are all the same size. The distance between the middle of the 2^nd post and the middle of the 5^th post is four and a half meters.

What is the distance between the middle of the first post and the middle of the last post?

Hint: You might start by drawing a diagram of the fence posts and labelling it with the distance you know.

Solution

There are many ways to calculate the distance. We start with a diagram of the posts, labelled with the distance of \(4 \frac{1}{2}\) metres between the middle of the 2^nd and 5^th posts.

Solution 1

Notice that there are a total of three gaps between the 2^nd and 5^th posts. There are also a total of three gaps between the 1^st and 2^nd posts and the 5^th and 7^th posts. Since the distances between the middle of adjacent posts are all the same, then the total distance between the 1^st and 2^nd posts and the 5^th and 7^th posts must also be \(4 \frac{1}{2}\) m. Since there are six gaps in total, the distance between the middle of the first post and the middle of the last post is \(4 + \frac{1}{2} + 4 + \frac{1}{2} = 9\) m.

Solution 2

Another way to solve this problem is to draw a number line from \(0\) to \(4 \frac{1}{2}\) m and space the 2^nd, 3^rd, 4^th, and 5^th fenceposts evenly along this line.

From this, we observe that the distance from the middle of one post to the middle of an adjacent post is \(1 \frac{1}{2}\) m. Since there are six gaps in total between adjacent posts, the distance between the middle of the first post and the middle of the last post is: \[1 + \frac{1}{2} + 1 + \frac{1}{2} + 1 + \frac{1}{2} + 1 + \frac{1}{2} + 1 + \frac{1}{2} + 1 + \frac{1}{2} = 9~\text{m}\]

Solution 3

First we convert the distance to centimetres: \(4 \frac{1}{2}~\text{m} = 450\) cm. Then we use a tape diagram. We observe that there are \(3\) gaps between the 2^nd and 5^th posts. That means we need to divide \(450\) cm into three equal distances. To make it easier, since \(450 = 45 \times 10\), we let one dot represent \(10\) cm, and then distribute \(45\) dots into the three equal pieces of our tape diagram.

Each piece has a total of \(15\) dots. This means the distance between adjacent posts is \(15 \times 10 = 150\) cm. Since there are \(6\) gaps in total between the first post and the last post, the total distance is: \[150 + 150 + 150 + 150 + 150 + 150 = 900~\text{cm~or}~9~\text{m}\]