Foundation June 2022 Unit 2 Q15
15. The length of a rectangle is double its width.
The area of the rectangle is greater than 60 cm2.
The perimeter of the rectangle is less than 40 cm.
Give a possible width and length of the rectangle.
Calculate the area and the perimeter of this rectangle.
You must show all your working.
Use the answer spaces to clearly identify which is the area and which is the perimeter. [5]
Width = …………… cm Length = …………… cm
Area = …………… cm2 Perimeter = …………… cm
| Scheme | Marks | Comments | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| length = 2 × width | B1 | Answer lines take precedence Note: correct answer 5·47… (cm) < width < 6·66…. (cm) Must be in the correct order for B1. | ||||||||
| Area = width × length | M1 | M1 for using the correct method (not for stating the formula). FT ‘their width’ × ‘their length’ | ||||||||
| Area correctly evaluated AND > 60 (cm2) | A1 | |||||||||
| Perimeter = 2 × (width + length) or equivalent | M1 | M1 for using the correct method (not for stating the formula) FT 2 × (‘their width’ + ‘their length’) | ||||||||
| Perimeter correctly evaluated AND < 40 (cm) | A1 | If answer space is left blank:
Note: (W and L need not be whole numbers)
|
Corrected by Westie's Workshop
- The board's mark scheme printed “5·47… (cm) ≤ width ≤ 6·66…. (cm)”. Changed to “5·47… (cm) < width < 6·66…. (cm)” because the question says the area is greater than 60 cm² and the perimeter is less than 40 cm, so the end values themselves do not work (for example a width of \(\tfrac{20}{3}\) cm gives a perimeter of exactly 40 cm).
Original: see the board's mark scheme PDF (link above).