Foundation November 2020 Paper 2 Q24
24. ABCD is a parallelogram.
Diagram not drawn to scale
Work out the value of \(x\) and the value of \(y\).
You must show all your working. [5]
\(x\) = ………… \(y\) = …………
| Scheme | Marks | Comments |
|---|---|---|
| \(5x + 40 = 6x + 20\) | M1 | Allow for 5 × 20 + 40 = 6 × 20 + 20 which may be seen in stages |
| \(x = 20\) | A1 | |
| \(5 \times 20 + 40 + y + 35 = 180\) OR \(6 \times 20 + 20 + y + 35 = 180\) OR \(5 \times 20 + 40 + 2(y + 35) + 6 \times 20 + 20 = 360\) | M2 | FT ‘their 20’ for possible M2 provided previous M1 awarded. May be seen in stages. M1 for a correct equation \(5x + 40 + y + 35 = 180\) or \(6x + 20 + y + 35 = 180\) or \(5x + 40 + y + 35 + 6x + 20 + y + 35 = 360\) |
| \(y = 5\) | A1 | CAO |
| (5) |
Alternative
Alternative method (using simultaneous equations)
| Scheme | Marks | Comments |
|---|---|---|
| Writes two correct equations in \(x\) and \(y\) \(5x + 40 + y + 35 = 180\) or \(6x + 20 + y + 35 = 180\) or \(5x + 40 + y + 35 + 6x + 20 + y + 35 = 360\) | M2 | M1 for each correct equation May be simplified |
| Method to eliminate variable, e.g. equal coefficients and method to find second variable | m1 | Allow one error in one term but not with equal coefficients |
| Finds the value of the first variable | A1 | CAO \(x = 20\) OR \(y = 5\) |
| Second variable | A1 | FT ‘their first variable’ |