Higher June 2022 Paper 6 Q9
9 A garage is trying to sell a car.
The price of the car is normally £18 000.
In a sale, the price of the car is reduced by 30%.
As a special offer, the sale price is then reduced by \(r\)%.
The special offer price is £9450.
Find the value of \(r\).
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 25[%] with correct working | 5 | B2 for 12 600 or M1 for \(18\,000 \times \dfrac{70}{100}\) oe or for \(18\,000 \times \dfrac{30}{100}\) | “correct working” requires at least M2 or M1M1 with the first M1 implied by B2 M0 for e,g. 70% of 18 000 M0 for e.g. 70% × 18 000 |
| AND M2 for \(\dfrac{\textit{their } 12600 - 9450}{\textit{their } 12600}\) [× 100] oe or M1 for \(\dfrac{9450}{\textit{their } 12\,600}\) [× 100] oe | Accept 3150 for numerator M2 may be seen as \(\left(1 - \dfrac{\textit{their } 9450}{\textit{their } 12600}\right)\)[× 100] M1 may be seen as \(\dfrac{9450}{18000} = 0.525\) and then followed by \(\dfrac{0.525}{0.7}\) | ||
| If 0 or M1 scored, instead award SC2 for answer 25[%] with no or insufficient working If 0 scored, instead award SC1 for 0.25 or 0.75 or 75[%] with no or insufficient working | Trials for second M marks M2 for 12600 × 0.25 = 3150 or M1 for 12600 × 0.75 = 9450 Equation method B2M2 or B2M2 for \(\dfrac{p}{100} \times 12600 = 3150\) leading to \(p = 25\) (scores 5 marks) B2M1 for \(\dfrac{p}{100} \times 12600 = 9450\) leading to \(p = 75\) B2M1 for 18000 × 0.7 × \(m\) = 9450 leading to \(m = 0.75\), as 12600 implied within this | ||