Mathematics · May / June 2023 · Paper 42 · Question 1

111 [3]

− 3

M2 for 180 –180 42 oe or 42 + 2 − 180 42 oe 2 −

or M1 for 180 42 oe 2

150 [3]

M1 for k ÷ (3 + 4 + 5) [×p] where p = 1, 3, 4 or 5 or 5 12 oe

B1 for 360 used

3 5 cao nfww [4]

B3 for 72 120

or B2 for [d = ] 72 or [h = ] 120

or M1 for 360 ÷ 5 oe isw or 180 – (360 ÷ 6) isw or for (6 – 2) × 180 [÷ 6]

x + 2x – 5 + x + 20 + 3x – 40 = 360 7x = 360 + 5 – 20 + 40 or better x = 55 55 and 125 or 105 and 75 Opposite angles sum to 180 oe [so PQRS is a cyclic quadrilateral ] [5]

M1 Accept equivalent equation e.g. 7x – 25 = 360 M1 FT their equation, accept e.g. 7x = 385 B1 B1dep Dep on M1M1B1 Accept 55 + 3 × 55 – 40 = 180 or 2 × 55 – 5 + 55 + 20 = 180 If B0 scored,

SC1 for 55, 75, 105 and 125 A1 Dep on M1M1B1B1

48.7 or 48.69 to 48.70… [3]

M2 for 360 − 50 ×× π ×oe 2 9 360

or M1 for 50 360 × × π ×oe 2 9