4MA0 Higher June 2017 Paper 4R Q11
11 Mabintou invested $7500 for 3 years at 4% per year compound interest.
Calculate the value of her investment at the end of 3 years.
(3)
| Scheme | Marks |
|---|---|
| 7500 × 0.04 or 300 or 7500 × 1.04 or 7800 or \(7500 \times 1.04^n\) (\(n \gt 1\)) | M1 |
Eg \(7500 + \dfrac{4}{100} \times 7500 + \dfrac{4}{100} \times (7500 + \text{“}300\text{”}) + \dfrac{4}{100} \times (7500 + \text{“}300\text{”} + \text{“}312\text{”})\) or 7500 + “300” + “312” + “324.48” | M1 |
| 8436.48 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: For interest for first year or
for 7500 × 0.04 × 3 oe or 900
or
for 7500 + 7500 × 0.04 × 3 oe
or an answer of 8400
M1: For a complete method
A1: Accept answers in the range 8436 – 8437
NB: Answer in the range 936 -937 gets M2A0
or M2 for \(7500 \times 1.04^3\) oe