Higher November 2018 Paper 2 Q5
5 The diagram shows a floor in the shape of a trapezium.
John is going to paint the floor.
Each 5 litre tin of paint costs £16.99
1 litre of paint covers an area of 2 m2
John has £160 to spend on paint.
Has John got enough money to buy all the paint he needs?
You must show how you get your answer. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| No (supported) | P1 | calculates area of trapezium eg \(\frac{1}{2} \times 7 \times (10 + 16)\) (= 91) |
| P1 | for division by coverage eg \(\div 2\) or [area of trapezium] \(\div\, 2\) (= 45.5) or process to find coverage per tin eg \(5 \times 2\ (= 10)\) | |
| P1 | for division to find the number of tins eg \(\div 5\) or \(\text{``}45.5\text{''} \div 5\ (= 9.1)\) or [area of trapezium] \(\div \text{``}10\text{''}\ (= 9.1)\) | |
| P1 | (dep on at least P2) for a process to multiply a whole number of tins (rounded up) by 16.99 | |
| C1 | for ‘No’ supported by correct figures eg 169.9 or 90 and 91 |
Alternative
| Answer | Mark | Mark scheme |
|---|---|---|
| No (supported) | P1 | calculates area of trapezium eg \(\frac{1}{2} \times 7 \times (10 + 16)\) (= 91) |
| P1 | for process to find number of tins bought eg \(160 \div 16.99 = 9\) tins | |
| P1 | for using whole no. of tins to find total litres eg \(9 \times 5\ (= 45)\) | |
| P1 | (dep on at least P2) for a process to find the total coverage eg \(\text{``}45\text{''} \times 2\ (= 90)\) | |
| C1 | for ‘No’ supported by correct figures eg 169.9 or 90 and 91 |
Additional guidance
[area of trapezium] needs to be clearly stated if the process of finding the area is not clear
There must be a conclusion (“No” or equivalent wording) including the figure 169.9 and working showing processes followed.