Higher June 2023 Paper 1 Q18
18 7 kg of carrots and 5 kg of tomatoes cost a total of 480p
cost of 1 kg of carrots : cost of 1 kg of tomatoes = 5 : 9
Work out the cost of 1 kg of carrots and the cost of 1 kg of tomatoes. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| (\(c\)) 30 (\(t\)) 54 | P1 | for setting up an equation, eg \(7c + 5t = 480\) or \(c : t = 5 : 9\) or \(\dfrac{c}{t} = \dfrac{5}{9}\) or \(9c = 5t\) or for starting to work with ratio of total costs, eg \(7 \times 5\ (= 35)\) and \(5 \times 9\ (= 45)\) or \(7 \times \dfrac{5}{14}\) and \(5 \times \dfrac{9}{14}\) or 35 : 45 or 7 : 9 |
| P1 | for a process to eliminate \(c\) or \(t\) from correct equations, eg \(7c + 9c = 480\) or \(7 \times \dfrac{5t}{9} + 5t = 480\) or \(7c + \dfrac{9c}{5} \times 5 = 480\) or for \(480 \div (\text{``}35\text{''} + \text{``}45\text{''})\ (= 6)\) or for a process to find total cost of carrots or total cost of tomatoes, eg \(480 \div (\text{``}7\text{''} + \text{``}9\text{''}) \times 7\ (= 210)\) or \(480 \div (\text{``}7\text{''} + \text{``}9\text{''}) \times 9\ (= 270)\) | |
| P1 | for a process to isolate \(t\) or \(c\), eg \(16c = 480\) or \(80c = 2400\) oe or \(80t = 4320\) oe or for one value correct eg \(c = 30\) or \(t = 54\) or for a process to find cost of 1 kg of carrots or 1 kg of tomatoes, eg \(5 \times \text{``}6\text{''}\ (= 30)\) or \(9 \times \text{``}6\text{''}\ (= 54)\) or \(\text{``}210\text{''} \div 7\ (= 30)\) or \(\text{``}270\text{''} \div 5\ (= 54)\) | |
| A1 | cao |