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(a) (i) (ii) \(32 - x + x + 28 - x + 12 = 45\) \(72 - x = 45\) \(x = 72 - 45 = 27\) 27 students offered both Physics and Government. (iii) \(n(P \cup G) = 45 - 12 = 33\) (b) \(p = \frac{2u}{1 - u} ; q = \frac{1 + u}{1 - u}\) \(\frac{p + q}{p - q}\) \(p + q = \frac{2u}{1 - u} + \frac{1 + u}{1 - u}\) = \(\frac{2u + 1 + u}{1 - u}\) = \(3u + 1}{1 - u}\) \(p - q = \frac{2u - 1 - u}{1 - u}\) = \(\frac{u - 1}{1 - u}\) = \(\frac{-(1 - u)}{1 - u}\) = \(-1\) \(\frac{p + q}{p - q} = \frac{\frac{3u + 1}{1 - u}}{-1}\) = \(\frac{-(3u + 1)}{1 - u}\) = \(\frac{-(3u + 1)}{-(u - 1)}\) = \(\frac{3u + 1}{u - 1}\)

(a) 1 2 3 4 5 6 1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 4 4,1 4, 2 4, 3 4, 4 4, 5 4, 6 5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6 (b) (i) P(getting a total of 7) = \(\frac{6}{36}\) P(getting a total of 8) = \(\frac{5}{36}\) P(getting a total of 7 or 8) = \(\frac{6}{36} + \frac{5}{36}\) = \(\frac{11}{36}\) (ii) P(getting a total less than 4) = \(\frac{3}{36}\) = \(\frac{1}{12}\)

(a) (i) Length of sector = \(\frac{\theta}{360} \times 2\pi R\) Length of the base of the cone formed = \(2\pi r\) \(\frac{90}{360} \times 2 \times \pi \times 8 = 2\pi r\) \(4\pi = 2\pi r\) \(2r = 4 \implies r = 2 cm\) The base radius of the cone = 2 cm. (ii) Height of the cone \(h^{2} = 8^{2} - 2^{2}\) \(h^{2} = 64 - 4 = 60\) \(h = \sqrt{60}\) = \(7.746 cm\) (iii) volume of a cone = \(\frac{1}{3} \pi r^{2} l\) = \(\frac{1}{3} \times \frac{22}{7} \times 2 \times 2 \times 7.746\) = \(\frac{88 \times 7.746}{21}\) = \(32.46 cm^{3}\) (b)(i) \(100,000 cm = 1 km\) 1 cm represents 20,000 metres \(\therefore 4.5 cm = (4.5 \times 20,000) cm\) = 90,000 cm Converting to km, = \(\frac{90,000}{100,000}\) = 0.9km (ii) 1 metre represents 20,000 metres 16km = 16000m Distance on the map in metres = \(\frac{16000}{20000}\) = 0.8m (iii) Area of the forest on the map = \(\frac{85 \times (10^{5})^{2} cm^{2}}{(20000)^{2}}\) = \(\frac{85 \times 10^{10}}{4 \times 10^{8}}\) = \(2125 cm^{2}\)

(a)(i) \(4x < 2 + 3x ..... (1)\) \(4x - 3x < 2\) \(x < 2\) \(x - 8 < 3x ..... (2)\) \(-8 < 2x\) \(x > -4\) \(x < 2 ; x > -4\). (ii) (b) Let the digits of the number be x and y. \(x + y = 9 .... (1)\) \((10y + x) - (10x + y) = 45 \) \(10y - y + x - 10x = 45\) \(9y - 9x = 45 .... (2)\) From (1), y = 9 - x. Putting it into (2), we have \(9(9 - x) - 9x = 45\) \(81 - 9x - 9x = 45\) \(36 = 18x \implies x = 2\) \(\therefore y = 9 - x\) = \(9 - 2 = 7\) \(\therefore\) The bill = £27.

\(\stackrel\frown{AOB} = \stackrel\frown{BOC} = \stackrel\frown{COD} = \stackrel\frown{DOE} = 60^\circ\) In \(\Delta AOB\), Area of sector = \(\frac{60}{360} \times \frac{22}{7} \times 7^{2}\) = \(25.67 cm^{2}\) Area of triangle AOB = \(\frac{1}{2} \times \7 \times 7 \times \sin 60\) = \(24.5 \times 0.866\) = \(21.22 cm^{2}\) Area of the 4 sectors = \(4 \times 25.67 = 102.68 cm^{2}\) Area of the 4 triangles = \(4 \times 21.22 = 84.88 cm^{2}\) Area of the shaded portion = \((102.68 - 84.88)cm^{2}\) = \(17.80 cm^{2}\) \(\approxeq 18 cm^{2}\)

(a) (b) \(x - 5 + 2x + 2x + 2x + 7 + 2x + 2x - 1 = 144\) \(11x + 1 = 144\) \(11x = 143\) \(x = \frac{143}{11} = 13cm\)

(a) Scores (x) 1 2 3 4 5 7 8 9 10 Total No of pupils (f) 1 1 5 3 \(k^{2} + 1\) 6 2 3 4 \(k^{2} + 26\) fx 1 2 15 12 \(5k^{2} + 5\) 42 16 27 40 \(5k^{2} + 160\) \(\bar{x} = \frac{\sum fx}{\sum f}\) \(6 = \frac{5k^{2} + 160}{k^{2} + 26}\) \(6(k^{2} + 26) = 5k^{2} + 160\) \(6k^{2} + 156 = 5k^{2} + 160\) \(6k^{2} - 5k^{2} = 160 - 156\) \(k^{2} = 4\) \( k = \sqrt{4} = 2\) \(\therefore k = 2\) (ii) \(Median = \frac{5 + 7}{2}\) = \(\frac{12}{2} = 6\) (b) (c) Probability of scoring less than 6 = \(\frac{15}{30} \) = \(\frac{1}{2}\)

(a) From the pyramid, construct \(\Delta BFG\) by drawing a line from B to F. \(|BF|^{2} = 4^{2} + 3^{2}\) \(|BF| = 25 \implies |BF| = 5m\) In calculating the height, \(h^{2} = 5^{2} - (\frac{5}{2})^{2}\) \(h^{2} = \frac{75}{4}\) \(h = \sqrt{75}{4}\) = \(4.33 m\) Volume of a pyramid = \(\frac{1}{3} \times 4 \times 3 \times 4.33\) = \(17.32 m^{3}\) Volume of a cuboid = \(l \times b \times h\) = \(3 \times 4 \times 2\) = \(24 m^{3}\) Total volume of the shape = \(17.32 + 24\) = \(41.32 m^{3}\) (b) \(5 - (x - 2) = d\) \((x + 2) - 5 = d\) \(\therefore (x + 2) - 5 = 5 - (x - 2)\) \(x - 3 = 7 - x\) \(2x = 10\) \(x = 5\)

(a) \(y = 2\sin x - \cos 2x\) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ y -1.0 0.5 2.2 3.0 2.2 0.5 -1.0 (b)

(a) Note that triangles AXC and BXD are similar. \(\therefore \frac{6}{|XD|} = \frac{3}{4}\) \(|XD| = \frac{6 \times 4}{3}\) = \(8 cm\) (b)(i) From \(\Delta ZYX\), \(\stackrel\frown{YZX} = 90^\circ - 22^\circ = 68^\circ\) \(\stackrel\frown{YXZ} = 312^\circ - 270^\circ = 42^\circ\) \(\stackrel\frown{ZYX} = 180^\circ - (68^\circ + 42^\circ)\) = \(180^\circ - 110^\circ\) = \(70^\circ\) \(a^{2} = b^{2} + c^{2} - 2bc \cos A\) \(|XZ|^{2} = 50^{2} + 32^{2} - 2(50)(32) \cos 70\) = \(2500 +1024 - 3200 \times 0.342\) = \(3524 - 1094.46\) \(|XZ|^{2} = 2429.54\) \(|XZ| = \sqrt{2429.54} = 49.29 km\) \(\approxeq 49.3 km\) (to 1 d.p) (ii) \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) \(\frac{50}{\sin \theta} = \frac{49.3}{\sin 70}\) \(\sin \theta = \frac{50 \times \sin 70}{49.3}\) \(\sin \theta = 0.9530\) \(\theta = \sin^{-1} (0.9530)\) = \(72.37^\circ\) \(\therefore\) The bearing of Z from X = \(312^\circ - 72.37^\circ\) = \(239.63^\circ \approxeq 239.6^\circ\)

IV is correct.

If 13m + 24m = 41m 1 * m1 + 3 0 + 2 * m1 + 4 * m0 = 4 * m1 + 1 * m0 1 * m + 3 * 1 + 2 * m + 4 * 1 = 4 * m + 1 * 1 m + 3 + 2m + 4 = 4m + 1 3m + 7 = 4m + 1 4m -3m = 7 - 1 m = 6

\(\frac{321_4}{23_4}\) = \(\frac{(3\times4^{2})+(2\times4^{1})+(1\times4^{0})}{(2\times4^{0})+(3\times4^{0})}\) = \(\frac{3\times16+2\times4+1\times1}{2\times4+3\times1}\) = \(\frac{48+8+1}{8+3}\) =\(\frac{57}{11}\) = 5\(\hspace{1mm}remainder\hspace{1mm}2\) ∴ r = 2\(_{10}\) Now\(\hspace{1mm}convert\hspace{1mm}2_{10} \hspace{1mm}to\hspace{1mm}base\hspace{1mm}4\frac{4}{2}\) = 2\(\frac{4}{0}\) = 0\(\hspace{1mm}or\hspace{1mm}2\) ∴ r = 2

3\(\frac{1}{2}\) - (2\(\frac{1}{3}\) x 1\(\frac{1}{4}\)) + \(\frac{3}{5}\)) = \(\frac{7}{2}\) - (\(\frac{7}{3}\) x \(\frac{5}{4}\)) + \(\frac{3}{5}\)) = \(\frac{7}{2}\) - \(\frac{35}{12}\) + \(\frac{3}{5}\) = L.C.M = 60 = \(\frac{(210 - 175 + 36)}{60}\) = \(\frac{71}{60}\) = 1\(\frac{11}{60}\)

To find the interest on \( N250.00 \) for \( 6 \) months at the same rate, we follow these steps: Calculate the Interest Rate Given: P = N150.00, T = 2\(\frac{1}{2}\) years and I = N4.50 Using the formula for simple interest: \(I = P \times r \times T\) Rearranging to find the interest rate \( r \): \(r = \frac{I}{P \times T}\) Substituting the values: \(r = \frac{4.50}{150 \times 2.5}\) Calculating \( P \times T \): \(P \times T = 150 \times 2.5 = 375\) Now, substituting back to find \( r \): \(r = \frac{4.50}{375} = 0.012\) Calculate the Interest for N250.00 \) for 6 Months We need to calculate the interest for: P = N250.00, T = 6 months = 0.5yrs, r = 0.012 Using the interest formula again: \(I = P \times r \times T\) Substituting the values: \(I = 250 \times 0.012 \times 0.5\) \(I = 250 \times 0.012 = 3\) Then: \(I = 3 \times 0.5 = 1.50\) The interest on \( N250.00 \) for \( 6 \) months at the same rate is N1.5