Higher November 2020 Unit 2 Q8
8. The lengths of the sides of a rectangle are given as 24 cm and 15 cm.
Each measurement is given correct to the nearest centimetre.
Calculate the difference between the greatest possible perimeter of the rectangle and the least possible perimeter of the rectangle. [3]
| Scheme | Marks | Comments |
|---|---|---|
| Sight of 24·5 AND 15·5 OR Sight of 23·5 AND 14·5 | B1 | Sight of (Greatest =) 80 OR (Least =) 76 implies B1 |
| 2(24·5 + 15·5) – 2(23·5 + 14·5) or equivalent | M1 | FT only for upper bounds of 24·4 AND 15·4 or 24·49 AND 15·49 (lower bounds must be 23·5 AND 14·5 else M0) |
| = 4(cm) | A1 | CAO If M0, award B1 and an SC1 for sight of (Greatest =) 80 AND (Least =) 76 |
Alternative method.
| Scheme | Marks | Comments |
|---|---|---|
| Difference between least and greatest length for each side = 1(cm) | B1 | |
| 4 × 1 | M1 | FT only for differences of 0·9 or 0·99 |
| = 4(cm) | A1 | CAO |