4MA0 Higher June 2015 Paper 4 Q6
6
Diagram NOT accurately drawn
\(LMN\) is parallel to \(PQR\).
\(QM = QR\).
Angle \(RMN = x^\circ\)
Angle \(MQR = y^\circ\)
(a) Write down an expression for \(y\) in terms of \(x\). (2)
Diagram NOT accurately drawn
\(ABCDEF\) is a hexagon.
(b) Work out the value of \(k\). (4)
| Scheme | Marks |
|---|---|
| angle \(MRQ\) (or \(RMQ\)) = \(x\) or \(\dfrac{180 - y}{2}\) | M1 |
| \(180 - 2x\) | A1 |
| (2) |
Notes
M1: could be marked on diagram or for a correct equation in \(x\) and \(y\)
A1: oe eg \(2(90 - x)\), \(2(180 - x) - 180\) etc
| Scheme | Marks |
|---|---|
| (6 − 2) × 180 oe (= 720) | M1 |
| "720" − (90 + 115 + 144 + 87), [720 – 436] or 284 | M1dep |
| "284" ÷ 2 | M1dep |
| 142 | A1 |
| (4) |
Notes
M1: or (180 − 360÷6) × 6
Alternative
| Scheme | Marks |
|---|---|
| 180−90 (= 90), 180−115 (= 65), 180−144 (= 36), 180−87 (= 93) | M1 |
| 360 – (“90” + ”65” + ”36” + ”93”) , [360−284] (= 76) | M1dep |
| 180 – (“76” ÷ 2) | M1dep |
| 142 | A1 |
Notes
M1: A correct method to find each of the exterior angles at A, C, D & E – angles could be seen on diagram.
M1dep: A correct method to find the total of the remaining exterior angles
M1dep: A correct method to find \(k\)