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From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A1, the area of \(\bigtriangleup\)PXY = Area of \(\bigtriangleup\)YZR = A2 Area of \(\bigtriangleup\)PQR = A = A1 + 2A2 But from similarity of triangles \(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\) \(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\) A = 4A2 But, A = A1 + 2A A1 = 4A2 - 2A2 A1 = 2A2 \(\frac{A_1}{A_2}\) = 2 A1:A2 = 2:1 Area of XYZQ:Area of \(\bigtriangleup\)YZR = 2:1

x = 65o (An interior angle of a cyclic quadrilateral = opposite exterior angle).

Perimeter of \(\bigtriangleup\) PQR = PQ + QR + PR 24cm = 6cm + 8cm = PR 24 = 14 + PR PR = 24 - 14 = 10cm Area of \(\bigtriangleup\) PRS = \(\frac{1}{2} \times 10 \times 3cm^3\) = 15cm3

\(\bigtriangleup\) RST is equilateral triangle, hence o But o(Angle between a chord and a tangent at the point of contact = Angle in the alt. segment). From the diagram o + 60o = 120o

o(Angles in a triangle) 110 + 5x = 180o 5x = 180o - 110 = 70o x = \(\frac{70}{5}\) = 14o Also y + 50 + 2(14) = 180o y + 50 + 28 = 180o y = 180 - 78 y = 102o

2 x s = 280^\circ(Angle at centre = 2 x < at circum) S = \(\frac{280^o}{2}\) = 140^\circ < O = 360 - 280 = 80^\circ 60 + 80 + 140 + a = 360^\circ (< in a quad); 280 + a = 360 a = 360 - 280 a = 80^\circ

M = Mercedez; T = Toyota Number of staff = 0 + (20 - 5) + (15 - 5) + 5 0 + 15 + 10 + 5 = 30

Let the total score for the girls and the boys be A and B respectively. \(\frac{A}{20} = 52 \implies A = 52 \times 20\) A = 1040 \(\frac{B}{y} = 57 \implies B = 57y\) 1040 + 57y = 2750 57y = 2750 - 1040 57y = 1710 y = \(\frac{1710}{57}\) y = 30

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option D.

P\(\text{Usman fails}) = 1 - \frac{3}{5} = \frac{2}{5}\) P\(\text{Dele fails}) = 1 - \frac{1}{3} = \frac{2}{3}\) P\(\text{neither passes}) = \left(\frac{2}{5}\right) \times \left(\frac{2}{3}\right) = \frac{4}{15}\)

x = \(\frac{ Ky}{z}\) x = 5, y = 2 and z = 1 x = \(\frac{ Ky}{z}\) 5 = \(\frac{ K2}{1}\) K = \(\frac{5}{2}\) = 2.5 find x when y = 5 and z = 2 x = \(\frac{2.5 \times5}{2}\) x = \(\frac{12.5}{2}\) = 6.25.

No expanded working has been attached for this question yet. The verified answer is Option B.

No expanded working has been attached for this question yet. The verified answer is Option E.

The base of a cone is a circle, r = 7cm Base area of a cone = \(\pi\)r\(^2\) Curve surface area of a cone = \(\pi\)l \(\pi\)r\(^2\) : \(\pi\)l = 1 : 4 \(\frac{\pi r^2}{\pi l}\) = \(\frac{1}{4}\) \(\frac{r}{l} = \frac{1}{4}\) \(\frac{7}{l} = \frac{1}{4}\) cross multiplying l = 4 x 7 = 28cm.