4MA0 Higher January 2015 Paper 4 Q14
14 Kwo invests HK$ 40000 for 3 years at 2.5% per year compound interest.
Work out the value of the investment at the end of 3 years.
(3)
| Scheme | Marks |
|---|---|
| 0.025 × 40 000 (=1000) or 1.025 × 40 000 (=41000) or 3000 | M1 |
| “41000” × 0.025 (=1025) and “42025” × 0.025 (=1050.625) OR 3075.62 or 3075.625 or 3075.63 | M1 |
| 43075.63 | A1 |
| (3 marks) |
Notes
M1: (dep)
method to find interest for year 2 and year 3
A1: accept 43075.62 or 43075.625
NB. An answer of 3075.62 or 3075.625 or 3075.63 score M2A0
M2 for \(40000 \times 1.025^3\)