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Construction joins R to P such that SRP = straight line R = 180o - 107o < p = 180o - (107o - 34o) 108 - 141o = 39o Angle < S = 39o (corr. Ang.) But in \(\bigtriangleup\)SRT < S = < T = 39o SRT = 180 - (39o + 39o) = 180o - 78o = 102o

Since QRU= Xo RSU = Xo, But RSU = 180o - (50o + 35o) = 180o - 85o = 95o x = 95o

Using angle bisector theorem: line PS bisects angle QPR QS/QP = SR/PR 3/4 = SR/9 4SR = 27 SR = \(\frac{27}{4}\) = 6\(\frac{3}{4}\)cm

SQ2 + OS2 = OQ2 + 52 = 142 SQ2 = 142 - 52 196 - 25 = 171 ST2 + TQ2 = SQ2 ST2 + 22 = 171 ST2 = 171 - 4 = 167 ST = \(\sqrt{167}\) = 12.92 = 12.9cm

Since area of square PQRS = 100cm2 each lenght = 10cm Also TUYS : XQVU = 1 : 16 lengths are in ratio 1 : 4, hence TU : UV = 1: 4 Let TU = x UV = 1: 4 hence TV = x + 4x = 5x = 10cm x = 2cm TU = 2cm UV = 8cm But TU = SY and UV = YR YR = 8cm

Number f students that took the test = \(\sum f\). Where f is the frequencies = 2 + 5 + 0 + 3 + 4 + 0 + 0 + 2 + 3 + 1 + 2 = 20

(a) \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\) = \(\log_{10} (\frac{30}{16}) - \log_{10} (\frac{5}{9})^{2} + \log_{10} (\frac{400}{243})\) = \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \div \frac{25}{81})\) = \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \times \frac{81}{25})\) = \(\log_{10} (10) = 1\) (b) \(\sqrt{\frac{P}{Q}} = \sqrt{\frac{3.6 \times 10^{-3}}{2.25 \times 10^{6}}\) = \(\sqrt{\frac{36 \times 10^{-4}}{2.25 \times 10^{6}}\) = \(\frac{6 \times 10^{-2}}{1.5 \times 10^{3}}\) = \(4 \times 10^{-2 - 3} = 4 \times 10^{-5}\)

\(P = {..., -5, -4, -3, -2, -1}\) \(Q = {..., -5, -3, -1, 1, 3, 5, ...}\) \(R = {-2, -1, 0, 1, 2, 3, 4, 5, 6}\) (a) \(Q \cap R = {-1, 1, 3, 5}\) (b) \(R' = {..., -5, -4, -3, 7, 8, ...}\) (c) \(P' = {0, 1, 2, 3, ...}\) \(P' \cup R' = {-5, -4, -3, 0, 1, 2, 3, ...}\) (d) \(Q = {..., -5, -3, -1, 1, 3, 5,...}\) \(P \cap Q = {..., -7, -5, -3, -1}\) \((P \cap Q)' = {..., -8, -6, -4, -2, 0, 1, 2, 3, ...}\)

In \(\Delta\) BPY, \(\tan 35^\circ = \frac{h}{x}\) \(h = x \tan 35 = 0.7 x ... (1)\) In \(\Delta\) BPX, \(\frac{h}{x + 4000} = \tan 15\) \(\frac{h}{x + 4000} = 0.3\) \(h = 0.3 (x + 4000) = 0.3x + 1200 ... (2)\) Equating (1) and (2) as the values of h, we have \(0.7x = 0.3x + 1200 \implies 0.7x - 0.3x = 1200\) \(0.4x = 1200 \times x = \frac{1200}{0.4} = 3000m\) \(\therefore h = 0.7x = 0.7 (3000) = 2100m\) \(\text{The total height = } 5200m + 2100m = 7300m\)

(a) \(\frac{3}{m + 2n} - \frac{2}{m - 3n}\) = \(\frac{3(m - 3n) - 2(m + 2n)}{(m + 2n)(m - 3n)}\) = \(\frac{3m - 9n - 2m - 4n}{(m + 2n)(m - 3n)}\) = \(\frac{m - 13n}{(m + 2n)(m - 3n)}\) (b) Let the numbers in the digit be x and z. Hence the number is \(10x + z\). When it is interchanged, we have \(10z + x\). \(10z + x = 10x + z + 9\) \(10z - z + x - 10x = 9\) \(9z - 9x = 9 \implies z - x = 1 ... (1)\) \(x + z = 11 ... (2)\) From (1), \(z = 1 + x\) \(\therefore x + 1 + x = 11\) \(2x + 1 = 11 \implies 2x = 11 - 1 = 10\) \(\therefore x = 5\) \(z = 1 + x = 1 + 5 = 6\) \(\therefore\) The original number = 56.

n(R) = number of red balls = 12; n(W) = number of white balls = 16; n(B) = number of blue balls = 8 Total number of balls = 36 (a) P(three are red) = \(\frac{12}{36} \times \frac{11}{35} \times \frac{10}{34}\) = \(\frac{11}{357}\) (b) P(first is blue and other two red) = \(\frac{8}{36} \times \frac{12}{35} \times \frac{11}{34}\) = \(\frac{44}{1785}\) (c) P(two are white and one is blue) = \(\frac{16}{36} \times \frac{15}{35} \times \frac{8}{34}\) = \(\frac{16}{357}\)

(a)(i) \(\log_{10} 45 = \log_{10} (3 \times 3 \times 5)\) = \(\log_{10} (3^{2} \times 5)\) = \(\log_{10} 3^{2} + \log_{10} 5\) = \(2 \log_{10} 3 + \log_{10} 5\) = \(2(0.477) + 0.699\) = \(0.954 + 0.699 = 1.653\) (ii) \(x^{0.8265} = 45\) Taking the log of both sides, \(\log_{10} x^{0.8265} = \log_{10} 45\) \(0.8265 \log_{10} x = \log_{10} 45\) \(\log_{10} x = \frac{1.653}{0.8265}\) \(\log_{10} x = 2\) \(x = 10^{2} = 100\) (b) \(\sqrt{\frac{2.067}{0.0348 \times 0.538}}\) No Log 2.067 \(0.0348\) = 0.3513 - 0.0348 \(\bar{2}.5416 +\) 0.538 \(\bar{1}.7308\) \(\bar{2}.2724\) = \(\bar{2}.2724\) = \(2.0789 \div 2 = 1.0395\) Antilog - 10.95 \(\therefore \sqrt{\frac{2.067}{0.0348 \times 0.538}} = 10.95\)

(a) Draw RC produced to E parallel to DE. \(\therefore < BCR = 52^\circ\) (alternate angles BA//RC) also < RCD = 48^\circ (alternate angles RC//DE) \(x^\circ = 360^\circ - (52^\circ + 48^\circ) = 360^\circ - 100^\circ = 260^\circ\) (angles at a point = 360^\circ) (b) \(y - 2x < 5 ... (i)\) \(2y + 2x \geq 4 \implies y + x \geq 2 ... (ii)\) \(y + 2x \leq 10 ... (iii)\) From (i), if \(y - 2x = 5\), when x = 0, y = 5, when x = 2, y = 9. From (ii), if \(y + x = 2\). when x = 0, y = 2, when x = 3, y = -1 From (iii), if \(y + 2x = 10\), when x = 0, y = 10, when x = 2, y = 6.

(a) Given : Circle ABC, centre O. To prove: < AOB = < ACB Construction : Join CO produced to P. Proof: With the lettering in the figure, OA = OB (radii) \(x_{1} = x_{2}\) (base angles of isosceles triangle) \(\therefore < AOP = x_{1} + x_{2}\) (exterior angle of triangle AOC) \(\therefore < AOP = 2x_{2} (x_{1} = x_{2})\) (base angles of an isosceles triangle) Similarly, \(< BOP = 2y_{2}\) \(\therefore < AOB = 2x_{2} + 2y_{2} \) = \(2(x_{2} + y_{2})\) \(\implies < AOB = 2 \times < ACB\) (proved) (b) From the figure, \(< ACB = \frac{130^\circ}{2} = 65^\circ\) (angle subtended at the centre of the circle) \(< ACO = 26^\circ\) (Base angles of isosceles triangle ACO) \(\therefore < BCO = 65^\circ - 26^\circ = 39^\circ\) \(< OBC = 39^\circ\) (base angles of isosceles triangle OBC) \(\therefore < COB = 180^\circ - (39^\circ + 39^\circ) = 102^\circ\)

(a) 5, 9, 13, ... is an A.P with the first term = 5 and common difference = 9 - 5 = 13 - 9 = 4. \(T_{n} = a + (n - 1)d\) (terms of an A.P) \(T_{25} = 5 + (25 - 1) \times 4\) = \(5 + 24 \times 4\) = \( 5 + 96 = 101\) (b) \(\frac{8}{9}, \frac{-4}{3}, 2, ...\) is a G.P with first term = \(\frac{8}{9}\) \(r = \frac{T_{2}}{T_{1}} = -\frac{4}{3} \div \frac{8}{9} = \frac{-4}{3} \times \frac{9}{8} = -\frac{3}{2}\) \(T_{n} = ar^{n - 1}\) (terms of a G.P) \(T_{5} = (\frac{8}{9})(-\frac{3}{2})^{5 - 1}\) = \(\frac{8}{9} \times \frac{81}{16} = \frac{9}{2}\) (c) \(T_{n} = ar^{n - 1}\) (terms of a G.P) \(T_{3} = ar^{2} = 48 ... (1)\) \(T_{6} = ar^{5} = 14\frac{2}{9} ... (2)\) \((2) \div (1) : r^{3} = \frac{128}{9} \div 48 = \frac{128}{9} \times \frac{1}{48}\) \(r^{3} = \frac{8}{27}\) \(r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}\) From (1), \(ar^{2} = 48 \implies a \times (\frac{2}{3})^{2} = 48\) \(4a = 48 \times 9 \implies a = \frac{48 \times 9}{4} = 108\) \(T_{2} = 108 \times \frac{2}{3} = 72\) \(T_{4} = 48 \times \frac{2}{3} = 32\) \(\therefore\) The first four terms of the sequence are \(108, 72, 48, 32\).