Higher January 2020 Paper 1 Q11
11 Max invests $6000 in a savings account for 3 years.
The account pays compound interest at a rate of 1.5% per year for the first 2 years.
The compound interest rate changes for the third year.
At the end of 3 years, there is a total of $6311.16 in the account.
Work out the compound interest rate for the third year.
Give your answer correct to 1 decimal place.
(3)
| Scheme | Marks |
|---|---|
6000 × \(1.015^2\) (= 6181.35) or 6000 + (0.015 × 6000) + (0.015 × (6000 + ‘90’)) (= 6181.35) or \((1.015)^2\) (= 1.030225) or \(\dfrac{6311.16}{6000}\) (= 1.05186) | M1 |
6311.16 ÷ ‘6181.35’ (= 1.021) (×100) or \(\dfrac{6311.16 - \text{‘}6181.35\text{’}}{\text{‘}6181.35\text{’}}\) (= 1.021) (×100) or ‘1.05186’ ÷ ‘1.030225’ (= 1.021) (×100) | M1 |
| 2.1 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for working out the total amount after two years
or working out the compound interest multiplier after two years
or working out the compound interest multiplier after three years
M1: (dep on M1) for a complete method to find the compound interest multiplier (×100)
A1: awrt 2.1