June 2022 Paper 2 Q14
14 \(PQRS\) is a cyclic quadrilateral.
Not drawn accurately
Angle \(PSR = 4(x + 15^\circ)\)
Angle \(PQR\) is \(40^\circ\) smaller than angle \(PSR\).
Work out the value of \(x\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4(x + 15) + 4(x + 15) - 40 = 180\) or \(8(x + 15) - 40 = 180\) or \(4(x + 15) = \dfrac{180 + 40}{2}\) or \(4(x + 15) - 40 = \dfrac{180 - 40}{2}\) or \(y + 4(x + 15) = 180\) and \(y = 4(x + 15) - 40\) | M1 | oe equation in \(x\) or pair of equations in \(x\) and \(y\) \(y\) may be any letter other than \(x\) eg \(180 - (4x + 60) + 40 = 4x + 60\) or \(4(x + 15) = 110\) or \(4(x + 15) - 40 = 70\) or \(y + 4x = 120\) and \(y = 4x + 20\) implied by \(y = 70\) |
| \(4x + 60 + 4x + 60 - 40 = 180\) or \(8x + 120 - 40 = 180\) or \(8x = 100\) or \(100 \div 8\) or \(4x = 50\) or \(50 \div 4\) | M1dep | oe equation or calculation equation with brackets expanded and fractions eliminated eg \(120 - 4x + 40 = 4x + 60\) or \(8x + 80 = 180\) or \(4x + 60 = 110\) or \(4x + 20 = 70\) |
| 12.5 or \(\dfrac{25}{2}\) or \(12\dfrac{1}{2}\) | A1 | oe eg \(\dfrac{100}{8}\) or \(\dfrac{50}{4}\) SC2 2.5 oe |
Additional guidance
| Ignore simplification or conversion if correct answer seen | |
| 2nd M1 Allow unnecessary brackets eg \((4x + 60) + (4x + 60) - 40 = 180\) | M1M1 |
| 1st M1 may be implied if expansion error seen eg \(4(x + 15) = 4x + 15\) (may be seen on diagram) \(4x + 15 + 4x + 15 - 40 = 180\) | M1M0 |
| Only \(4x + 15 + 4x + 15 - 40 = 180\) | M0 |
| SC2 is when they have angle \(PQR\) \(40^\circ\) larger than angle \(PSR\) |