Foundation June 2022 Paper 3 Q25
25 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 \(18\,000 \times \dfrac{30}{100}\) oe AND M2 for \(\dfrac{\textit{their } 12600 - 9450}{\textit{their } 12600}\) \([\times 100]\) oe or M1 for \(\dfrac{9450}{\textit{their } 12600}\) \([\times 100]\) oe If 0 or M1 scored, instead award SC2 for answer 25[%] with no or insufficient working If 0 scored, award SC1 for 0.25 or 75[%] with no or insufficient working | “correct working” requires at least M2 or M1M1 the first M1 implied by B2 M0 for e.g. 70% of 18 000 M0 for e.g. 70% \(\times\) 18 000 Accept 3150 for numerator M2 may be \(\left(1 - \textit{their } \dfrac{9450}{\textit{their } 12600}\right)\) \([\times 100]\) M1 may be seen as \(\dfrac{9450}{18000} = 0.525\) and then followed by \(\dfrac{0.525}{0.7}\) Trials for second M marks M2 for \(12600 \times 0.25 = 3150\) or M1 for \(12600 \times 0.75 = 9450\) Equation method B2M2 \(\dfrac{p}{100} \times 12600 = 3150\); leading to \(p = 25\) (scores 5 marks) B2M1 \(\dfrac{p}{100} \times 12600 = 9450\); leading to \(p = 75\) B2M1 \(18000 \times 0.7 \times m = 9450\) leading to \(m = 0.75\), as 12600 implied within this |