Foundation November 2024 Paper 1 Q9
9. Adrian works in a factory.
His normal rate of pay is £14 per hour.
Adrian usually works 22 hours per week and overtime if needed.
Any overtime hours are paid at a rate of \(1\frac{1}{2}\) times his normal rate of pay.
(a)
One week Adrian worked for 28 hours.
Calculate Adrian’s pay for this week. [4]
(b)
The following week, Adrian’s normal rate of pay increased.
He worked for 22 hours and was paid £330.
Show that Adrian’s rate of pay increased by less than 10%. [3]
| Scheme | Marks | Comments |
|---|---|---|
| (Adrian’s pay =) 14 × 22 + (14 × 1.5) × (28 – 22) (= 308 + 21 × 6) | M3 | M2 for (Overtime pay =) (14 × 1.5) × (28 – 22) (= £126) M1 for (Overtime rate =) 14 × 1.5 (= £21) |
| (£)434 | A1 | CAO If no marks award SC1 for sight of (£)308 (from 14 × 22) or (£)392 (from 14 × 28). |
| Scheme | Marks | Comments |
|---|---|---|
| (New normal rate of pay =) 330 ÷ 22 | M1 | |
| £15 | A1 | An indication of £1 extra per hour implies M1A1. |
| £15 AND £15.40 OR £1 AND £1.40 OR (\(\dfrac{15 - 14}{14}\) × 100 =) 7(.142…)% (< 10%) | A1 | FT ‘their £15’ provided M1 awarded and < £15.40. FT ‘their £15’ - £14 provided M1 awarded and < £1.40. Beware of 7% coming from wrong working. |
| (7) |
Alternative
9(b) Alternative method 1
| Scheme | Marks | Comments |
|---|---|---|
| 14 × 22 × 1.1 or 14 × 22 + \(\dfrac{14 \times 22}{10}\) oe | M2 | FT ‘their 14 × 22’ from (a) M1 for \(\dfrac{14 \times 22}{10}\) oe |
| = (£)338.8(0) (> £330) | A1 |
9(b) Alternative method 2
| Scheme | Marks | Comments |
|---|---|---|
| 330 - 14 × 22 (= 330 - 308) | M1 | FT ‘their 14 × 22’ from (a) |
| (10% of 14 × 22) = \(\dfrac{14 \times 22}{10}\) oe | M1 | FT ‘their 14 × 22’ from (a) |
| Correct comparison of 22 AND 10% of 308 | A1 | 22 must come from a correct method. 10% of 308 may be rounded or truncated. |