Higher November 2019 Paper 2 Q11
11 The value of a house is £120 000
The value is expected to increase by 5% each year.
Work out the expected value after 4 years.
Give your answer to 2 significant figures.
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(120\,000 \times 1.05\) or 126 000 | M1 | oe eg \(120\,000 + 0.05 \times 120\,000\) may be implied by eg 144 000 |
| \(120\,000 \times 1.05^4\) or \(\dfrac{583\,443}{4}\) | M1dep | oe eg their \(126\,000 \times 1.05\) or 132 300 and their \(132\,300 \times 1.05\) or 138 915 and their \(138\,915 \times 1.05\) |
| 145 860(.75) or 145 860.8(0) or 145 861 or 145 900 or 146 000 | A1 | if no value given implied by M2 seen and 150 000 |
| 150 000 | B1ft | ft any answer seen with > 2sf condone 150 000.00 |
Additional guidance
| \(126\,000 \times 1.05^3\) | M1M1 |
| Answer only 145 860(.75) or 145 860.8(0) or 145 861 or 145 900 or 146 000 | M1M1A1B0 |
| Answer only 150 000 | Zero |
| For year on year working allow rounding/truncation if method shown for up to M2A0B1ft eg \(126\,000 \times 1.05 = 132\,000\) and \(132\,000 \times 1.05 = 138\,000\) and \(138\,000 \times 1.05 = 144\,900\) Answer 140 000 | M1 M1A0B1ft |
| 120 000, 126 000, 132 000, 138 000, 144 000 with no method shown does not imply truncation, this is just adding on 6 000 each year | M1M0A0 |
| \(120\,000 + 4 \times 0.05 \times 120\,000\) or \(120\,000 + 0.2 \times 120\,000\) implies M1 | M1M0A0 |
| Misreads can score up to M2A0B1ft | |
| Treat calculating 5 years as a misread but otherwise the wrong number of years eg \(120\,000 \times 1.05^2\) will score a maximum of M1M0A0B1ft |