Higher June 2022 Paper 2 Q26
26 Two objects, \(J\) and \(K\), are applying pressure to areas of ground.
\(\text{pressure} = \dfrac{\text{force}}{\text{area}}\)
For \(J\), the force is 18.9 newtons and the area is 0.45 m2
\[\text{pressure for } J : \text{pressure for } K = 7 : 8\]\[\text{area for } J : \text{area for } K = 9 : 5\]Work out the force for \(K\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(18.9 \div 0.45\) or 42 | M1 | oe |
| their \(42 \div 7 \times 8\) or 48 | M1dep | oe eg \(\dfrac{8}{7} \times\) their 42 or [1.14, 1.143] \(\times\) their 42 |
| \(0.45 \div 9 \times 5\) or 0.25 | M1 | oe eg \(\dfrac{5}{9} \times 0.45\) or [0.55, 0.56] \(\times 0.45\) \(\dfrac{8}{7} \times \dfrac{5}{9} \times 18.9\) oe scores M3 |
| 12 | A1 |
Additional guidance
| Up to M3 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts | |
| Any single calculation or set of calculations that are a correct method and lead to 12 | M3 |
| Note that the single calculation \(\dfrac{8}{7} \times \dfrac{5}{9} \times 18.9\) does not use 0.45 | M3 |
| An oe for \(\dfrac{8}{7} \times \dfrac{5}{9} \times 18.9\) is \(\dfrac{8}{7} \times \dfrac{18.9}{0.45} \times \dfrac{5}{9} \times 0.45\) | M3 |
| Values may be seen in ratios eg 42 : 48 | M1M1 |