June 2016 Paper 2 Q14

14 \(P\), \(Q\), \(R\) and \(S\) are points on the circumference of a circle.

\[\begin{aligned} &w : y = 7 : 5 \\ &PQ = PS \end{aligned}\]

Not drawn accurately

Work out the size of angle \(x\). [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(180 \div (7 + 5)\) or 15	M1	oe
\((w =)\) 7 × their 15 or 105 or \((y =)\) 5 × their 15 or 75	M1dep	oe May be seen on diagram M2 105 : 75
\(\dfrac{180 - \text{their } w}{2}\) or \(\dfrac{\text{their } y}{2}\)	M1dep	oe dep on M1 M1
37.5	A1	oe SC2 52.5
Alternative method 2
\(w + y = 180\) and \(5w = 7y\) or \(w + \dfrac{5}{7}w = 180\) or \(y + \dfrac{7}{5}y = 180\)	M1	oe
\((w =)\ \dfrac{180 \times 7}{12}\) or 105 or \((y =)\ \dfrac{180 \times 5}{12}\) or 75	M1dep	oe May be seen on diagram M2 105 : 75
\(\dfrac{180 - \text{their } w}{2}\) or \(\dfrac{\text{their } y}{2}\)	M1dep	oe dep on M1 M1
37.5	A1	oe SC2 52.5
Alternative method 3
\(\dfrac{w}{180 - w} = \dfrac{7}{5}\) or \(\dfrac{y}{180 - y} = \dfrac{5}{7}\)	M1	oe
\((w =)\ \dfrac{180 \times 7}{12}\) or 105 or \((y =)\ \dfrac{180 \times 5}{12}\) or 75	M1dep	oe May be seen on diagram M2 105 : 75
\(\dfrac{180 - \text{their } w}{2}\) or \(\dfrac{\text{their } y}{2}\)	M1dep	oe dep on M1 M1
37.5	A1	oe SC2 52.5
Alternative method 4
\(y = 2x\) and \(w = \dfrac{7}{5} \times 2x\)	M1	oe
\(2x + \dfrac{7}{5} \times 2x = 180\) or \(4.8x = 180\)	M1dep	oe
180 ÷ their 4.8	M1dep	oe dep on M1 M1
37.5	A1	oe SC2 52.5

75 : 105 implies M1 M0 M0 unless recovered (award SC2 if answer 52.5)