Foundation June 2023 Paper 2 Q23
23 A factory packs \(x\) boxes of teabags per hour.
Each box contains 80 teabags.
Show that the factory packs \(\dfrac{4x}{3}\) teabags per minute. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(80 \times x\) or \(80x\) or \(x \times 80\) or \(x80\) or \(x \div 60\) or \(\dfrac{x}{60}\) or \(\dfrac{1}{60}x\) or \(x\dfrac{1}{60}\) or \(80 \div 60\) or \(\dfrac{80}{60}\) | M1 | teabags per hour boxes per minute |
| \(\dfrac{80x}{60}\ \left(= \dfrac{4x}{3}\right)\) or \(80 \div 60 \times x\ \left(= \dfrac{4x}{3}\right)\) | A1 | oe showing 80 and 60 and \(x\) eg \(\dfrac{80 \times x}{60}\ \left(= \dfrac{4x}{3}\right)\) or \(x\dfrac{80}{60}\ \left(= \dfrac{4x}{3}\right)\) or \(\dfrac{80}{60} \times x\ \left(= \dfrac{4x}{3}\right)\) or \(80x \div 60\ \left(= \dfrac{4x}{3}\right)\) |
Additional guidance
| M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Do not allow M1 if only seen embedded in an incorrect expression or calculation eg \(80x \times 4 = 320x\) | M0 |
| \(60 \times \dfrac{4x}{3} = 80x\) (M1 allowed as \(80x\) is not embedded in an incorrect expression or calculation, A0 because using the given answer) | M1A0 |
| Condone \(x = 80 \div 60\) | M1A0 |
| \(\dfrac{80x}{60}\ \left(= \dfrac{4x}{3}\right)\) | M1A1 |
| \(\dfrac{80}{60} = \dfrac{4}{3}\) and \(\dfrac{4}{3} \times x\ \left(= \dfrac{4x}{3}\right)\) \(\dfrac{80}{60} = \dfrac{4}{3}\) and \(\dfrac{4x}{3}\) | M1A1 M1A0 |
| No equivalents allowed for M1 | |
| Ignore units | |
| Condone 1.33(…) for \(\dfrac{4}{3}\) | |
| Ignore non-contradictory working after M1A1 seen |