Foundation January 2020 Paper 2 Q24
24 The table shows the cost, in euros, of Brigitte’s car insurance in each of the years 2016, 2017 and 2018
| Year | 2016 | 2017 | 2018 |
|---|---|---|---|
| Cost of insurance (euros) | 500 | 545 | 592 |
Brigitte says,
“The percentage increase in the cost of my car insurance from 2017 to 2018 is more than the percentage increase in the cost of my car insurance from 2016 to 2017”
You must show how you get your answer. (4)
Henri wants to insure his car.
He gets a discount of 15% off the normal price.
Henri pays 952 euros for his car insurance after the discount.
| Scheme | Marks |
|---|---|
| 545 – 500 (= 45) or 592 – 545 (= 47) | M1 |
| \(\dfrac{45}{500} \times 100\ (= 9)\) or \(\dfrac{47}{545} \times 100\ (= 8.6)\) | M1 |
| \(\dfrac{45}{500} \times 100\ (= 9)\) and \(\dfrac{47}{545} \times 100\ (= 8.6)\) | M1 |
| Working required Answer: No, 9(%) and 8.6(%) | A1 |
| (4) |
Notes
M1: may be seen as part of a calculation
M1: for one correct expression (allow 8 or 8.7 from a correct expression for 8.6 throughout)
M1: for both correct expressions or having found “9%” finds 109% of 545: 1.09 × 545 (= 594.05) or 9% of 545 (49.05) or having found “8.6%” finds 108.6% of 500: 1.086 × 500 (= 543) or 8.6% of 500 (43)
A1: for no oe, 9% and 8.6% seen or
no oe and 9% and 594.05 or 8.6% and 543 or
No, 49.05 > 45 or No 594.05 > 592 oe
Alternative
| Scheme | Marks |
|---|---|
| \(\dfrac{545}{500} \times 100\ (= 109)\) or \(\dfrac{545}{500}\ (= 1.09)\) or/and \(\dfrac{592}{545} \times 100\ (= 108.6)\) or \(\dfrac{592}{545}\ (= 1.086)\) | M3 |
| Answer: No, 109(%) and 108.6(%) | A1 |
M3 for both correct expressions which should lead to 109 or 1.09 and 108.6 or 1.086
(allow 108 or 108.7 from correct working for 108.6 or 1.08 or 1.087 from correct working for 1.086 throughout)
(if not M3 then award M2 for one of these expressions)
A1: oe eg no and 1.09 and 1.086
(corrected from the printed mark scheme, which heads this “Alternative mark scheme for 8(a)”)
| Scheme | Marks |
|---|---|
| 952 ÷ 85 × 100 oe (= 1120) | M1 |
| 0.15 × “1120” or “1120” – 952 oe | M1 |
| Working required Answer: 168 | A1 |
| (3) | |
| (7 marks) |
Notes
M1: for a method to find price before discount
M1: for a correct method to find discount
M2 for \(\dfrac{952}{85} \times 15\)