Higher January 2022 Paper 2R Q2
2 Mary saves for a holiday each year.
In 2020 she saved a total of $720
In 2021, each month she saved $78
The total amount Mary saved in 2021 was \(P\)% more than the total she saved in 2020
Roberto is going to go on holiday.
He has two coupons that will save him money on his holiday.
| Coupon A | Coupon B |
|---|---|
| 18% off the cost of the accommodation | 12.5% off the total cost of the accommodation and the flights |
For Roberto’s holiday
the cost of the accommodation is $1600
the cost of the flights is $800
Roberto can only use one of the coupons.
He wants to save as much money as he can.
Show your working clearly. (3)
| Scheme | Marks |
|---|---|
| 720 ÷ 12 (= 60) or 78 × 12 (= 936) | M1 |
78 – ‘60’ (= 18) or ‘936’ – 720 (= 216) or \(\text{‘}{x}\text{’} \times 720 = 936\) or \(720\left(1 + \dfrac{P}{100}\right) = \text{‘}{936}\text{’}\) or \((\text{‘}{x}\text{’} =)\ \dfrac{\text{‘}{936}\text{’}}{720}\ (= 1.3)\) oe | M1 |
\(\dfrac{\text{‘}{18}\text{’}}{60} \times 100\) or \(\dfrac{\text{‘}{216}\text{’}}{720} \times 100\) or \(\text{‘}{1.3}\text{’} \times 100 - 100\) oe or \((1.3 - 1) \times 100\) | M1 |
| 30 | A1 |
| (4) |
Notes
M1: complete method to find \(P\)
A1: ignore extra % sign if given by candidate.
| Scheme | Marks |
|---|---|
| 0.18 × 1600 (= 288) oe or 0.82 × 1600 + 800 (= 2112) | M1 |
| 0.125 × (1600 + 800) (= 300) oe or (1600 + 800) × 0.875 (= 2100) | M1 |
| Coupon B and correct figures seen | A1 |
| (3) | |
| (7 marks) |
Notes
M1: if 1600 × 18% seen, must have further processing of the 18% or the value (288) given.
A1: for Coupon B and
288 and 300 or
18.75(%) and 18(%) or
12(%) and 12.5(%) or
2112 and 2100
M2 for 1.5 × 12.5 (= 18.75) or 18 ÷ 1.5 (= 12)