Higher June 2025 Paper 1 Q13
13 Charlotte bought an apartment for $750 000
In the first year, the value of the apartment decreased by 4%
In the second year, the value of the apartment decreased by 6.5%
In the third year, the value of the apartment increased by \(x\)%
At the end of the third year, the value of Charlotte’s apartment was $698 445
Work out the value of \(x\)
(3)
| Scheme | Marks |
|---|---|
| 750 000 × (1 – 0.04) (= 720 000) oe and “720 000” × (1 – 0.065) (= 673 200) or 750 000 × (1 – 0.04) × (1 – 0.065) (= 673 200) or 750 000 × 0.96 × 0.935 (= 673 200) | M1 |
eg \(\dfrac{698\,445 - \text{``}{673\,200}\text{''}}{\text{``}{673\,200}\text{''}}(\times 100)\) (= 0.0375) or \(\dfrac{698\,445}{\text{``}{673\,200}\text{''}} - 1\) (= 0.0375) or 1.0375 – 1 (= 0.0375) oe or \(\dfrac{698\,445}{\text{``}{673\,200}\text{''}} \times 100 \;(= 103.75)\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 3.75 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: (NB: accept \(\left(1 - \dfrac{4}{100}\right)\) for 0.96 but not (1 – 4%) and accept \(\left(1 - \dfrac{6.5}{100}\right)\) for 0.935 but not (1 – 6.5%))
Calculations may be seen as part of an equation eg \(750\,000 \times 0.96 \times 0.935 \times \left(1 + \dfrac{x}{100}\right) = 698\,445\)
1.0375 or 25245 imply M1
M1: for a method to reach value one step away from \(x\) ie a method leading to 0.0375 or 103.75
A1: oe eg \(3\dfrac{3}{4}\) or \(\dfrac{15}{4}\)