Problem of the Week

Problem B and Solution

Farmer Mac’s Bales

Problem

Farmer Mac’s hay bales are in the shape of rectangular prisms. Each hay bale is 2 m long by 2 m wide by 1.5 m high. The hay bales lie in rows in a field with one of the square sides of each hay bale on the ground and the rectangular sides of the hay bales facing each other. Farmer Mac leaves a 50 cm space between each hay bale.

• Determine the length of row of 20 hay bales.

• What is the total area, in m$$^2$$, of the ground underneath the 20 hay bales?

Solution

• We will look at two ways to determine the length of a row of 20 hay bales.

The first way is by using a table.
The diagram below illustrates the first three bales, with 0.5 m between them (since 50 cm = 0.5 m).

The first bale is 2 m long, and each new bale after adds $$2 + 0.5 = 2.5$$ m to the length. We will show this in the following table.

bales length
1 2
2 4.5
3 7
4 9.5
5 12
6 14.5
7 17
8 19.5
9 22
10 24.5
11 27
12 29.5
13 32
14 34.5
15 37
16 39.5
17 42
18 44.5
19 47
20 49.5

Therefore, the length of a row with 20 hay bales is 49.5 m.

The second way we will find the length of a row with 20 bales is by setting up an algebraic expression.
If we let $$b$$ represent the number of bales and $$s$$ represent the number of spaces, then an algebraic expression for the length of a row of bales, in m, is $2 \times b + 0.5 \times s$

Now, when there are 20 bales and 19 spaces the length becomes: \begin{aligned} 2 \times 20 + 0.5 \times 19 &= 40 + 9.5 \\ &= 49.5 \end{aligned}

Therefore, the length of a row with 20 hay bales is 49.5 m.

• The base of each hay bale is 2 m by 2 m.
So the area under each hay bale is $$2 \times 2 = 4 \text{ m}^2$$.
There are 20 hay bales in the row, so the total area of the ground underneath the 20 hay bales is $$20 \times 4 = 80 \text{ m}^2$$.