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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.

When the phrase 'a number of' is used in a sentence, it is followed by a plural verb, hence 'have been' is the correct answer.

The contracted word with a subject is 'He has'. Has is usually not contracted except it comes before 'been'

It is grammatically correct to say 'a dangerously low level'

The past perfect tense 'had been hoping' complete the sentences appropriately.