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(a) \(\log_{10} x + \log_{10} y = 4 ... (1)\) \(\log_{10} x + 2 \log_{10} y = 3 ...(2)\) \((1) \implies \log_{10} (xy) = 4 \therefore xy = 10^{4} ... (3)\) \((2) \implies \log_{10} (xy^{2}) = 3 \therefore xy^{2} = 10^{3} ... (4)\) Divide (4) by (3) : \(y = 10^{-1} = 0.1\) Put y in (3), we have \(\frac{x}{10} = 10^{4} \implies x = 10^{5}\) \((x, y) = (10^{5}, 10^{-1})\) (b) (i) \(t \propto \frac{V}{P}\) \(t = \frac{kV}{P}\) when t = 10, V = 20 and P = 5. \(10 = \frac{20k}{5} \implies 50 = 20k\) \(k = \frac{5}{2}\) \(\therefore t = \frac{5V}{2P}\) (ii) When V = 50, P = 2, we have \(t = \frac{5(50)}{2(2)} = \frac{250}{4}\) = \(62.5 minutes\) (iii) when V = 40, t = 20 minutes \(20 = \frac{5(40)}{2P} \implies 40P = 200\) \(P = \frac{200}{40} = 5 pumps\)

(a) Area of curved surface of cylinder = \(2\pi r h\) = \(2 \times \frac{22}{7} \times 7 \times 10cm^{2} = 440 cm^{2}\) Area of the base of cylinder = \(\pi r^{2}\) = \(\frac{22}{7} \times 7 \times 7 = 154 cm^{2}\) Surface area of hemisphere = \(\frac{4\pi r^{2}}{2} = 2 \pi r^{2}\) = \(2 \times \frac{22}{7} \times 7 \times 7 = 308 cm^{2}\) \(\therefore\) Total surface area = 440 + 154 + 308 = 902cm\(^{2}\) (b) Volume of cylinder = \(\pi r^{2} h\) = \(\frac{22}{7} \times 7 \times 7 \times 10 = 1540 cm^{3}\) Volume of hemisphere = \(\frac{1}{2}(\frac{4\pi r^{3}}{3})\) = \(\frac{4 \times 22 \times 7 \times 7 \times 7}{2 \times 7 \times 3}\) = \(718.67 cm^{3}\) Total volume = 1540 + 718.67 = 2258.67 cm\(^{3}\).

(a)(i) Scale : On x- axis, 2cm = 1 unit On y- axis, 2cm = 5 units. (ii) Equation of line of symmetry is x = 1.25. (iii) Roots of the equation \(ax^{2} + bx + c = 0\) are x = 0.25 and x = 2.25. (b) Coordinates are D(0, 1), E(1, -2) and G(3, 4). Substituting for y and x in \(ax^{2} + bx + c = y\) D(0, 1) : \(1 = a(0^{2}) + b(0) + c \implies c = 1\) E(1, -2) : \(-2 = a(1^{2}) + b(1) + c \implies -2 = a + b + c\) \(a + b = -2 - 1 = -3 ... (1)\) G(3, 4) : \(4 = a(3^{2}) + b(3) + c \implies 4 = 9a + 3b + c\) \(9a + 3b = 4 - 1 = 3 ... (2)\) \(\implies 3a + b = 1 ... (2a)\) (2a) - (1) : \(2a = 4 \implies a = 2\) \(a + b = -3 \implies 2 + b = -3\) \(b = -3 - 2 = -5\) \(\therefore\) The equation is \(y = 2x^{2} - 5x + 1\) (c) The greatest value of y = 33.5.

(a) From the figure < QRS = 180^\circ - 56^\circ = 124^\circ (opposite angles of a parallelogram) \(\therefore\) < RQS = < PQS = \(\frac{180^\circ - 124^\circ}{2} = 28^\circ\) (base angles of an isosceles triangle) \(\therefore\) < SQT = < RSQ = 28^\circ (alternate angles PS // QT) (ii) < PQT = 90^\circ - 28^\circ = 62^\circ (< PQS is in a semi-circle) (b) \(\tan < BDC = \frac{30}{40} = 0.75\) \(< BDC = \tan^{-1} (0.75) = 36.87^\circ\) \(\therefore < BDF = 90^\circ - 36.87^\circ = 53.13^\circ \approxeq 53^\circ\) \(\tan < ADF = \frac{14}{30} = 0.4667\) \(< ADF = \tan^{-1} (0.4667) = 25.02^\circ \approxeq 25^\circ\)

(a) Marks (in %) Freq Class boundaries Cum Freq 44 - 46 2 43.5 - 46.5 2 47 - 49 5 46.5 - 49.5 7 50 - 52 11 49.5 - 52.5 18 53 - 55 20 52.5 - 55.5 38 56 - 61 42 55.5 - 61.5 80 62 - 64 46 61.5 - 64.5 126 65 - 67 36 64.5 - 67.5 162 68 - 70 9 67.5 - 70.5 171 71 - 73 3 70.5 - 73.5 174

An Isosceles Right Triangle is a right triangle that consists of two equal length legs. i.e opp = adj hyp\(^2\) = adj\(^2\) + opp\(^2\) 2\(^2\) = x\(^2\) + x\(^2\) 4 = 2x\(^2\) 2 = x\(^2\) \sqrt2 = x

△PQR is similar to △PNM \(\frac{PQ}{PN}\) = \(\frac{PR}{PM}\) \(\frac{7 + 3 }{5}\) \(\frac{5 + NR}{3}\) cross multiply 10 x 3 = (5 x 5) + 5 x NR 5NR = 30 - 25 = 5 NR = \(\frac{5}{5}\) = 1 cm.

Area of sector = \(\frac{\theta}{360^\circ} \times \pi r^2\) = \(\frac{60}{360} \times \frac{22}{7} \times 3.5 \times 3.5\) = \(\frac{77}{12}\) = \(6\frac{5}{12}\) cm\(^2\)

a, ar, ar\(^2\), ar\(^3\), ar\(^4\), ar\(^5\) 6th term = ar\(^5\) = \(\frac{-2}{27}\) 1st term = a = 18 => ar\(^5\) = 18r\(^5\) =\(\frac{-2}{27}\) r\(^5\) = \(\frac{-2}{27 * 18}\) → \(\frac{-1}{27 * 9}\) r\(^5\) = \(\frac{-1}{243}\) r = \(^5\)\sqrt[\(\frac{-1}{243}\)] r = \(\frac{-1}{3}\)

2 89 2 44 r 1 2 22 r 0 2 11 r 0 2 5 r 1 2 2 r 1 2 1 r 0 0 r 1 89\(_{10}\) = 101100\(1_2\)

\((2\frac{1}{6} - 1\frac{2}{3}) \div 2\frac{2}{3}\) = \((\frac{13}{6} - \frac{5}{3}) \div \frac{8}{3}\) = \(\frac{3}{6} \times \frac{3}{8}\) = \(\frac{3}{16}\)

\(56x^{-4} \div 14x^{-8}\) = \(4x^{- 4 - (-8)}\) = \(4x^{4}\)

Log\(_{10}\)4 + Log\(_{10}\)25 = Log\(_{10}\)(4 + 25) = Log\(_{10}\)100 = Log\(_{10}\)10\(^2\) = 2Log\(_{10}\)10 = 2 x 1 = 2

Length = 59.6m Breath = 40.3m Perimeter = (59.6m + 40.3m) x 2 = 99.9 x 2 = 199.8m Actual measurement Perimeter = (60 + 40) x 2 = 100 x 2 = 200m Error = 200m - 199.8m = 0.2 Error =0.2/200 x 100/1 = 0.1%

18, 12, 6, 0, -6, 12, -18, 24, -30 9th term = -30