Higher November 2021 Paper 1 Q6
6 Alison buys 5 apples and 3 pears for a total cost of $1.96
Greg buys 3 apples and 2 pears for a total cost of $1.22
Michael buys 10 apples and 10 pears.
Work out how much Michael pays for his 10 apples and 10 pears.
Show your working clearly.
(5)
| Scheme | Marks |
|---|---|
| \(5a + 3p = 1.96\) and \(3a + 2p = 1.22\) oe or \(5a + 3p = 196\) and \(3a + 2p = 122\) oe | M1 |
| E.g. \(15a + 9p = 5.88\) \(15a + 10p = 6.10\) Subtracting \((-p = -0.22)\) or E.g. \(10a + 6p = 3.92\) \(9a + 6p = 3.66\) Subtracting \((a = 0.26)\) or E.g. \(5a + 3p = 1.96\) and \(6a + 4p = 2.44\) oe Subtracting | M1 |
| E.g. \(5a + 3(\text{``}{0.22}\text{''}) = 1.96\) or \(3a + 2(\text{``}{0.22}\text{''}) = 1.22\) or E.g. \(5(\text{``}{0.26}\text{''}) + 3p = 1.96\) or \(3(\text{``}{0.26}\text{''}) + 2p = 1.22\) or E.g. \(a + p = 0.48\) oe | M1 |
\(10 \times \text{``}{0.26}\text{''} + 10 \times \text{``}{0.22}\text{''}\) or \((a + p =)\ 0.48 \times 10\) oe or \(k(a + p) = k(0.48) \times \dfrac{10}{k}\) | M1 |
| Working required Answer: 4.8(0) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for setting up both equations oe
allow apples and pears oe throughout
e.g. 5 apples + 3 pears = 1.96 and 3 apples + 2 pears = 1.22
M1: for a correct method to eliminate a or p: coefficients of a or p the same and correct operation to eliminate the selected variable (condone any one arithmetic error) or to find the cost of 1 apple and 1 pear
M1: (dep on M2) for substituting their value found (must be > 0) of one variable into one of the equations, or for repeating the above method to find the second variable, or for the third working column allow k(a + p) = k(0.48), or for a complete arithmetical method to find the other value
M1: (dep on M3) can be implied by 10(a + p) provided a and p must be > 0
A1: dep M2
M2 for an arithmetical method (must see the calculation to find 0.22 or 0.26 or 0.74 and 0.48 oe)
E.g. 6.1(0) – 5.88 (= 0.22) oe or 3.92 – 3.66 (= 0.26) oe or 1.96 – 1.22 (= 0.74) oe and 1.22 – “0.74” (= 0.48)
A complete arithmetical method to find the other value: e.g. 3 × 0.22 (= 0.66), 1.96 – “0.66” (= 1.3(0)), “1.3(0)” ÷ 5 (= 0.26) or 5 × 0.26 (= 1.3(0)), 1.96 – “1.3(0)” (= 0.66), “0.66” ÷ 3 (= 0.22) or Apple and pear is 0.48 oe
| Scheme | Marks |
|---|---|
| \(5a + 3p = 1.96\) and \(3a + 2p = 1.22\) oe or \(5a + 3p = 196\) and \(3a + 2p = 122\) oe | M1 |
E.g. \(3\left(\dfrac{1.96 - 3p}{5}\right) + 2p = 1.22\) or \(5\left(\dfrac{1.22 - 2p}{3}\right) + 3p = 1.96\) or \(3a + 2\left(\dfrac{1.96 - 5a}{3}\right) = 1.22\) or \(5a + 3\left(\dfrac{1.22 - 3a}{2}\right) = 1.96\) or \(p = 0.22\) or \(a = 0.26\) | M1 |
E.g. \((a =)\ \dfrac{1.96 - 3(0.22)}{5}\) or \((a =)\ \dfrac{1.22 - 2(0.22)}{3}\) or \((p =)\ \dfrac{1.96 - 5(0.26)}{3}\) or \((p =)\ \dfrac{1.22 - 3(0.26)}{2}\) | M1 |
| \(10 \times \text{``}{0.26}\text{''} + 10 \times \text{``}{0.22}\text{''}\) | M1 |
| Working required Answer: 4.8(0) | A1 |
Notes
M1: for setting up both equations oe
allow apples and pears oe throughout
e.g. 5 apples + 3 pears = 1.96 and 3 apples + 2 pears = 1.22
M1: for correctly writing a or p in terms of the other variable and correctly substituting (condone any one arithmetic error)
M1: (dep on M2) substituting their value found (must be > 0) of one variable into one of the equations or repeating the above method to find the second variable
M1: (dep on M3) can be implied by 10(a + p) provided a and p must be > 0
A1: dep M2