Higher November 2024 Paper 3 Q12
12 Rudi invests £4500 in a savings account.
He gets compound interest at a rate of
2.4% for the first year
1.8% for each extra year.
(a) Work out the value of Rudi’s investment at the end of 3 years. (3)
Bruna buys a car for £7500
The value of the car depreciates by \(x\)% each year.
At the end of 2 years the value of the car is £4107
(b) Work out the value of \(x\). (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 4775.38 | M1 | for a method to find the value after 1 year, eg \(4500 \times 1.024\ (= 4608)\) oe |
| M1 | for a complete method to find the value after 3 years, eg \(\text{``}4608\text{''} \times 1.018^2\) or \(\text{``}4608\text{''} \times 1.018\ (= 4690.944)\) and \(\text{``}4690.944\text{''} \times 1.018\) oe | |
| A1 | accept 4775.37 SCB1 for 4770 or 4824 if M0 scored |
Additional guidance
Award of this mark implies the first M1
May be seen in more than 1 calculation
M2A0 is implied by 275.37 or 275.38
Correct answer not rounded to 2dp gains M2A0
| Answer | Mark | Mark scheme |
|---|---|---|
| 26 | P1 | for a start to the process, eg \(4107 \div 7500\ (= 0.5476)\) |
| P1 | for a process to find percentage change eg \(\sqrt{\text{``}0.5476\text{''}} \times 100\ (= 74)\) or \(\left(\sqrt{\text{``}0.5476\text{''}} - 1\right) \times 100\ (= -26)\) or \(1 - \sqrt{\text{``}0.5476\text{''}}\ (= 0.26)\) | |
| A1 | cao |
Additional guidance
0.74 implies P1