(a)
(i) \((16 - 2x) + (6 + x) + (19 - 3x) + 5x + 7x + 8x + 4x + 4 = 125\)
\(45 + 20x = 125 \implies 20x = 125 - 45 = 80\)
\(x = \frac{80}{20} = 4\)
(ii) \(n(P \cup Q \cap R') = (16 - 2x) + 5x + (6 + x)\)
= \(22 + 4x\)
= \(22 + 4(4) = 22 + 16 = 38\)
(b) In the diagram above, < XYW = 90° (angle in a semi-circle; XW is a diameter)
< XWY + 90° + 50° = 180° (sum of angles in a triangle)
< XWY = 180° - 140° = 40°
(i) < WYZ = 40° (alternate angles; \(\overline{XW} || \overline{YZ}\))
(ii) < WOZ = \(2 \times 40° \) (angle at centre = twice angle at circumference)
= 80°
\(\therefore < WEO = 180° - (40° + 80°)\)
= \(180° - 120° \)
= \(60°\)
\(\therefore < YEZ = < WEO = 60°\) (vertically opposite angles).