Further examples for Higher tier
Example 1
A tiler is tiling the floor of a room. The measurements of the room are 7.3 m by 8.5 m to the nearest 10 cm.
The tiles measure 150 mm by 150 mm to the nearest cm.
The tiles are to be laid in a repeating pattern like this:
Calculate the maximum and minimum number of tiles needed to complete the job. Ignore the thickness of the tile grout between each tile.
Solution
The 7.3 m wall
The wall that is 7.3 m long to the nearest 10 cm (0.1 m) has a maximum length of 7.35 m and a minimum length of 7.25 m.
The tiles are 15 cm by 15 cm to the nearest cm so they have a maximum length of 15.5 cm and a minimum length of 14.5 cm.
Maximum number of tiles along this wall = maximum wall length ÷ minimum tile length = 735 ÷ 14.5 = 50.69 tiles (two decimal places).
Minimum number of tiles along this wall = minimum wall length ÷ maximum tile length = 725 ÷ 15.5 = 46.77 tiles (two decimal places).
For this wall:
- the maximum number of tiles = 51
- the minimum number of tiles = 47
The 8.5 m wall
The wall that is 8.5 m long to the nearest 10 cm (0.1 m) has a maximum length of 8.55 m and a minimum length of 8.45 m.
The tiles are 15 cm by 15 cm to the nearest cm so they have a maximum length of 15.5 cm and a minimum length of 14.5 cm.
Maximum number of tiles along this wall = maximum wall length ÷ minimum tile length = 855 ÷ 14.5 = 58.97 tiles (two decimal places).
Minimum number of tiles along this wall = minimum wall length ÷ maximum tile length = 845 ÷ 15.5 = 54.52 tiles (two decimal places).
So for this wall:
- maximum number of tiles = 59
- minimum number of tiles = 55
The whole room
The maximum number of tiles needed for the whole room = 51 × 59 = 3,009.
The minimum number of tiles needed for the whole room = 47 × 55 = 2,585.
Example 2
A furniture maker has two tables. One is rectangular and measures 55 cm by 37 cm to the nearest cm. The other is circular with a diameter of 59 cm to the nearest cm.
To finish the tables, she needs to fit a thin strip of wood called veneer around their edges. She uses these measurements to calculate the perimeter of the rectangular table to be 184 cm, and the circumference of the circular table to be 185.4 cm. She then assumes that she will need more veneer for the edge of the circular table than for the rectangular table.
Show that this is not always correct.
Solution
Maximum perimeter = 55.5 cm + 37.5 cm + 55.5 cm + 37.5 cm = 186 cm
Minimum perimeter = 54.5 cm + 36.5 cm + 54.5 cm + 36.5 cm = 182 cm
Circular table:
\({Circumference} = {\pi} \times {diameter}\)
Maximum circumference = \(π \times \text{59.5~cm}\) = 186.9247629 cm = 186.9 cm (one decimal place)
Minimum circumference = \(π \times \text{58.5~cm}\) = 183.7831702 cm = 183.8 cm (one decimal place)
Minimum circumference of circular table \(\textless\) maximum perimeter of rectangular table
Therefore, it could be that she will need more veneer for the rectangular table than for the circular table.
