Higher January 2023 Paper 2R Q9
9 Teresa invests \(\$\)2000 for 3 years in a savings account.
She gets 4% each year compound interest.
Give your answer correct to the nearest dollar. (3)
Sam invested \(\$T\)
The value of his investment decreased by 9% each year.
At the end of the first year, the value of Sam’s investment was \(\$\)1365
| Scheme | Marks |
|---|---|
| for 0.04 × 2000 oe (= 80) or 1.04 × 2000 oe (= 2080) | M1 |
| 1.04 × “2080” oe (= 2163.2) 1.04 × “2163.2” oe or \(2000 \times 1.04^3\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 2250 | A1 |
| (3) |
Notes
M1: for finding 4% or 104% of 2000
M1: for completing method to find total amount in the account at the end of 3 years
A1: accept 2249 – 2250
OR M2 for \(2000 \times 1.04^3\) oe or \(2000 \times 1.04^4\) oe (= 2339.72)
SC: if no other marks gained award M1 for 0.12 × 2000 oe or 240 or 1.12 × 2000 oe or 2240
accept (1 + 0.04) as equivalent to 1.04 throughout
| Scheme | Marks |
|---|---|
| eg 1365 ÷ (1 − 0.09) or 1365 ÷ 0.91 | M2 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 1500 | A1 |
| (3) | |
| (6 marks) |
Notes
M2: for a complete method
(M1) for 1365 ÷ (100 − 9) (= 15) or (100 − 9)% = 1365 or 91% = 1365 or eg \((1 - 0.09)T = 1365\) or eg \(T - 0.09T = 1365\)