Higher June 2018 Paper 2 Q20
20. The diagram below shows a composite shape made by joining two rectangles.
Diagram not drawn to scale
(a) The total area of the composite shape is 47 cm2.
Show that \(6x^2 - 3x - 47 = 0\). [3]
Show that \(6x^2 - 3x - 47 = 0\). [3]
(b) Use the quadratic formula to solve \(6x^2 - 3x - 47 = 0\).
Give both of your answers correct to 2 decimal places. [3]
Give both of your answers correct to 2 decimal places. [3]
(c) Calculate the perimeter of the composite shape.
You must give a reason for any decision that you make. [3]
You must give a reason for any decision that you make. [3]
| Scheme | Marks | Comments |
|---|---|---|
| \(2x(3x-4) + 5x\ (=47)\) or \(2x(3x+1) - 5x\ (=47)\) or equivalent | M1 | Allow intention |
| \(6x^2 - 8x + 5x = 47\) or \(6x^2 + 2x - 5x = 47\) | A1 | Must be from convincing working shown |
| \(6x^2 - 3x - 47 = 0\) | A1 | Must be from convincing working shown |
| Scheme | Marks | Comments |
|---|---|---|
| \((x =)\ \dfrac{3 \pm\sqrt{(-3)^2 - 4\times6\times-47}}{2\times6}\) | M1 | Allow 1 slip in substitution, but must be correct formula |
| \(= \dfrac{3 \pm\sqrt{1137}}{12}\) | A1 | |
| 3.06 and -2.56 | A1 | Both solutions given to 2dp |
| Scheme | Marks | Comments |
|---|---|---|
| 32.6 (cm) | B2 | FT use of ‘their positive value’ for B1 only provided previous M1 in (b) awarded B1 for sight of \(10x + 2\) or equivalent, OR \(10\times3.06 + 2\) |
| Decision, e.g. ‘that the negative solution in (b) was not valid’, ‘only used the positive solution’ AND Reason, e.g. ‘as lengths can only be positive’ | E1 | Accept if the decision and/or reason is written in (b) |
| (9) |
Corrected by Westie's Workshop
- The board's mark scheme printed “only used the position solution”. Changed to “only used the positive solution” because it means the positive solution of the quadratic, since a length cannot be negative.
Original: see the board's mark scheme PDF (link above).