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(a) \(\frac{\log_28 + \log_216 - 4\log_22}{\log_416}\) = \(\frac{\log_22^3 + \log_2 2^4 - 4\log_22}{\log_44^2}\) = \(\frac{3 \log_2 2 + 4\log_22 - 4 \log_2^2}{2 \log 4^4}\) But log\(_2\) 2 = \(\log_44 = 1\) Then, \(\frac{\log_28+\log_216 - 4 \log_2^2}{\log_416}\) = \(\frac{3 + 4 - 4}{2}\) = 1\(\frac{1}{2}\) (b) Subtract 241\(_{five}\) to get 1101\(_{five}\). Thereafter, converting to base ten to find the value of x = (5\(^3\) x 1) + (5\(^2\) x 1 + (5\(^1\) x 0) + (5\(^o\) x 1) = 125 + 25 + 0 + 1 = 151 (b) They subtracted 241\(_{five}\) from 1342\(_{five}\) to get 1101\(_{five}\). Thereafter, converting to base ten to find the value of x = (5\(^3\) x 1) + (5\(^2\) x 1) + (5\(^1\) x 0) + (5^\circx 1) = 125 + 25 + + 1 = 151.

Let C be the cost and n the number of tourists. then C = x + ny. Where x and y are constants 740 = x + 20y...........(1) 960 = x + 30y.............(2) Solving (1) and (2) simultaneously will yield x = 300 and y = 22. Then, C = 300 + 22n. If we substitute 15 for n this implies that C = 300 + (22 x 15)= $630.00

(a) Selling price = \(\frac{90 \times 5600}{100}\) = $5,040.00. Thereafter, to obtain the amount invested = 5,040 - 1310 = $3,730.00 The interest amount invested is \(\frac{3730 \times 6 \times 3}{100}\) = $671.40 (b) 2\(^x\)(2\(^{-14}\)) = 2. This will be simplified and solved for x to get x = 15. Similarly, 3\(^{-x}\)(3\(^{4y}\)) = 3 and this will yield -x + 4y = 1. Substituting for x and solving for y yielded y = 4.

(a) Using length of chord = 2r sin (\(\frac{\theta}{2}\)) = 22.42 and simplifying sin(\(\frac{\theta}{2}\)) = \(\frac{22.42}{33.18}\) = 0.6757 Taking the sine inverse of 0.6757 to obtain \(\frac{\theta}{2}\) = 42.51\(^o\) and multiplying through to get \(\theta\) = 58.02\(^o\) = 885\(^o\) (b(i), the expected diagram is (b)(ii) since T is equidistant from P and Q, then ITPI = ITQI. Using sum of the angles in a triangle. 70^\circ + x + x = 180^\circ and solving for x. x = 55^\circ. Therefore. the bearing of Q from P = 270^\circ — 30^\circ — 55^\circ = 185^\circ.

(a) The range; 20 - 15 = 5. (b) They drew the frequency distribution table as \(\sum\) f = 11\(\sum\) fx = 191 \(\sum\)f\(^2\) = 3337 Age(x) Tally Freq (f) Com. Freq fx fx\(^2\) 15 16 17 18 19 20 I II III III I I 1 2 3 3 1 1 1 3 6 9 10 11 15 32 51 54 19 20 225 512 867 972 361 400 (c) Median age as 17. (d)(i) Mean = \(\frac{\sum fx}{\sum f}\) = \(\frac{191}{11}\) = 17.3636 (d)(ii). The standard deviation was \(\sqrt{\frac{2227}{11} - (\frac{191}{11})^2}\) = \(\sqrt{1.8677}\) = 1.37(correct to two decimal places)

Let \(\mu\) = 40 n(M) = 23, n(M \(\cup\) P) = 5, n(M \(\cup\) C) = 8, n(P \(\cup\) C) = 5, n(M \(\cup\) P \(\cup\) C) = 3, n(C) only = x and n(P) only = 2x (a)(i) We have 13 + 2 + 3 + 5 + x + 2 + 2x = 40 and solving for x will yield x = 5, so that the number of students that studied only Physics = 2(5) = 10 (a)(ii) The number of students that studied only one subject is 13 + 10 + 5 = 28 (b) The probability that a student selected at random studies exactly two subjects = \(\frac{5 + 2 + 2}{40} = \frac{9}{48}\) or 0.225

(a) it will take 1 boy 10m days and when 4 more boys are added, it will take 8(m + 4) days. Since the rate of consumption was the same, then 10m = 8m + 32 and when solves, m = 16 (b) Let x be the number. so that \(\frac{5}{6}\)x = 10 + \(\frac{1}{2}\)x and multiplying through by 6 to get 5x = 60 + 2x. This was solved to get x = 20 (c) Let the gradient be m = \(\frac{\frac{1}{2} - (\frac{1}{2})}{2 - (-1)} = \frac{2}{3}\). The equation of the lines is y - y\(_1\) = m(x - x\(_1\)). Sunstituting for x\(_1\) and y\(_1\) y - (\(\frac{-1}{2}\)) = \(\frac{1}{2}\)(x - (-1)) and this can be simplified to y = \(\frac{1}{3}x - \frac{1}{6}\) or 2x - 6y - 1 = 0

(a) The required diagram was (b)(i) we have sin 75\(^o\) = \(\frac{x}{11}\) and this will give x = 10.625m. Since the ladder has moved up by 0.2 , the new point where the ladder touches the wall is 0.2 + 10.625 = 10.825m. Then sin\(\theta\) = \(\frac{10.825}{11}\) Sin\(\theta\) = 0.9841 and \(\theta\) = sin\(^{-1}\)(0.9841) = 79.77\(^o\) = 80\(^o\) (b)(ii) Cos 75\(^o\) = \(\frac{y}{11}\) so that y = 2.2847m Similarity, cos 79.77 = \(\frac{q}{11}\) and q = 1.9536m. Therefore, the distance the ladder was dragged = 2.8470 - 1.9536 = 0.8934m = 1m

(a) They completed the table of values as shown x -2.0 -1.5 -1.0 0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 y 11 0 -9 -16 -21 -24 -25 -24 -21 -16 -9 0 11 (b)(i) The graph (b)(ii)(\(\alpha\), deduce that 4x\(^2\) - 8x - 21 = 3 - 21 so that y = -18. From the graph, the point of intersection of y = -18 and y = 4x\(^2\) - 8x - 21 was atx = {-0.35 \(\pm\) 0.1, 2.3 \(\pm\) 0.1} (b)(ii)(\(\beta\)), the equation 4x\(^2\) - 8x - 21 = 0 which can be written as 4x\(^2\) - x - 7x - 21 = -x. This implies that 4x\(^2\) - 8x - 21 = -x. But, y = 4x\(^2\) - 8x - 21 so that y = -x, Next was to draw the line y = -x and from the graph, the point of intersection with the curve y = 4x\(^2\) - 8x - 21 was at x = {3.3 \(\pm\) 0.1, -1.6 \(\pm\) 0.1}

(a) since the curved surface areas were the same, it was expected that they have 5 x 12\(pi\) = 6 \(\pi\)l and solving for the slant height to get l = 10 cm. Then by the use of Pythagoras theorem, the height of the second cone is h\(^2\) = 10\(^2\) - 6\(^2\) and when simplified will result in h = 8 cm. (b(i), they were to find BA = \(\begin{pmatrix} 3 & -2 \\4 & 1\end{pmatrix}\) \(\begin{pmatrix} 2 & 5 \\ -1 & -3\end{pmatrix}\) = \(\begin{pmatrix} 8 & 21 \\ 7 & 17 \end{pmatrix}\) (b)(ii), the determinant of BA = (8 x 17) - (21 x 7) = 136 - 147 = -11

(a)110\(_x\) - 4\(0_5\) 1 x \(x^2}\) + 1 x \(x^1\) + 0 x \(x^o\) x = 4 x 5\(^1\) + 0 x 5\(^o\) x\(^2\) + x = 4 x 5 + 0 x\(^2\) + x - 20 = 0 (x - 4)(x + 5) = 0 x - 4 = 0 or x + 5 = 0 x = 4 or x = -5 x = 4 (b) Simplify \(\frac{15}{\sqrt{75}}\) + \(\sqrt{108}\) + \(\sqrt{432}\) = \(\frac{15}{\sqrt{25 \times 3}}\) + \(\sqrt{36 \times 3}\) = \(\frac{15}{5 \sqrt{3}\) + 6\(\sqrt{3}\) + 12\(\sqrt{3}\) = \(\frac{15 \sqrt{3}}{15 \times \sqrt{3} \times \sqrt{3}}\) + 18\(\sqrt{3}\) = \(\frac{15\sqrt{3}}{15}\) + 18\(\sqrt{3}\) \(\sqrt{3}\) + 18\(\sqrt{3}\) = 19\(\sqrt{3}\)

(a). < SOT = 2 < SOT = 2 x 68^\circ =136. Observe that < OSR = < OTR = 90\(^o\) So that < OSR + x + < OTR + < SOT = 360\(^o\). Next is to substitute and simplified to get x + 316\(o\) = 360\(o\) so that x = 44\(o\) (b) If we let capacity of tank A = a and capacity of tank B = b, then a = 600 + b......(1) a - 100 = 3(b - 100) a = 3b - 200........(2) Substituting (2) into (1) yielded 3b - 200 = 600 + b and solving for b, b = 400. Therefore. Capacity of tank B = 400litres Substituting b in (1) yielded A = 600 + 400 = 1000 Capacity of A =1000.

0,00 3597 = 0.00360 to 3 s.f

(0.064)\(^{- \frac{1}{3}}\) = (\(\frac{64}{1000}\))\(^{-\frac{1}{3}}\) = 3\(\sqrt{\frac{1000}{64}}\) = \(\frac{10}{4}\) = \(\frac{5}{2}\)

\(\frac{y + 1}{2} - \frac{2y - 1}{3}\) = \(\frac{4}{1}\) - \(\frac{3(y + 1) - 2(2y - 1)}{6} = \frac{4}{1}\) 3y + 3 - 4y + 2 = 24 - y + 5 = 24 - y = 24 - 5 = 19 y = - 19