\begin{align*} y = {k \over x^2} \implies & \text{As } x \text{ increases, } y \text{ decreases} \\ \\ \therefore \text{Graph } & 3 \end{align*}

\begin{align*} \angle OBA & = 90^\circ \phantom{0} \text{ (Tangent perpendicular to radius)} \\ \\ \angle OAB & = 180^\circ - 90^\circ - 66^\circ \phantom{0} \text{ (Interior angles, } CB \phantom{.} // \phantom{.} OA) \\ & = 24^\circ \end{align*}

(a)

$$ \text{The vertical axis of the graph does not start from 0} $$

(b)

$$ \text{Reader may think that the newspaper circulation have fell to close to 0 (instead of 1.25 million)} $$

(a)

\begin{align*} 4^5 & = (2^2)^5 \\ & = 2^{10} \phantom{000000} [ (a^m)^n = a^{mn} ] \end{align*}

(b)

\begin{align*} {4a^2 \over 3b} \div {10ab \over 21} & = {4a^2 \over 3b} \times {21 \over 10ab} \\ & = {84a^2 \over 30a b^2} \\ & = {14a \over 5 b^2} \end{align*}

\begin{align*} \text{Let length of hexagon} & = x \\ \\ \text{Perimeter of hexagon} & = 6x \\ \\ \text{Perimeter of triangle} & = 6x \times 2 \\ & = 12x \\ \\ \text{Length of triangle} & = 12x \div 3 \\ & = 4x \\ \\ { \text{Length of 1 triangle in hexagon} \over \text{Length of triangle} } & = {x \over 4x} \\ & = {1 \over 4} \\ \\ { \text{Area of 1 triangle in hexagon} \over \text{Area of triangle} } & = \left(1 \over 4\right)^2 \phantom{000000} \left[ {A_1 \over A_2} = \left(l_1 \over l_2\right)^2 \right] \\ & = {1 \over 16} \\ \\ \therefore \text{Area of hexagon} & : \text{Area of triangle} \\ 6 & : 16 \\ 3 & : 8 \end{align*}

(a)

\begin{align*} \text{Median position} & = {23 + 1 \over 2} \\ & = 12 \\ \\ \text{Median} & = 83 \text{ minutes} \end{align*}

(b)

\begin{align*} \text{Range (original)} & = 102 - 75 \\ & = 27 \\ \\ \text{Let time for 24th man} & = x \end{align*} \begin{align*} x - 75 & = 33 && \text{ or } & 102 - x & = 33 \\ x & = 33 + 75 &&& -x & = 33 - 102 \\ x & = 108 &&& -x & = -69 \\ & &&& x & = 69 \end{align*}

(a)

\begin{align*} \text{P(Yellow)} & = 1- 0 .45 - 0.3 \\ & = 0.25 \end{align*}

(b)

\begin{align*} 0.45 - 0.3 & = 0.15 \\ & = {3 \over 20} \\ \\ \text{No. of counters} & = 9 \times {20 \over 3} \\ & = 60 \end{align*}

\begin{align*} {x \over 5} - {2x - 3 \over 4} & = {-3 \over 1} \\ {4x \over 20} - {5(2x - 3) \over 20} & = {-60 \over 20} \\ {4x - 5(2x - 3) \over 20} & = {-60 \over 20} \\ 4x - 5(2x - 3) & = -60 \\ 4x - 10x + 15 & = -60 \\ 4x - 10x & = -60 - 15 \\ -6x & = -75 \\ x & = {-75 \over -6} \\ x & = 12.5 \end{align*}

(a)

\begin{align*} x^2 + 9x - 4 & = x^2 + 9x + \left(9 \over 2\right)^2 - \left(9 \over 2\right)^2 - 4 \\ & = \left(x + {9 \over 2}\right)^2 - {81 \over 4} - 4 \\ & = \left(x + {9 \over 2}\right)^2 - {97 \over 4} \end{align*}

(b)

\begin{align*} y & = x^2 + 9x - 4 \\ y & = \left(x + {9 \over 2}\right)^2 - {97 \over 4} \\ \\ \text{Min. } & \text{point: } \left(-{9 \over 2}, -{97 \over 4}\right) \end{align*}

(a)

\begin{align*} T_n & = a + (n - 1)(d) \phantom{000000} [a \text{: first term, } d \text{: common difference}] \\ & = 11 + (n - 1)(9) \\ & = 11 + 9n - 9 \\ & = 9n + 2 \end{align*}

(b)

\begin{align*} T_n & = 335 \\ 9n + 2 & = 335 \\ 9n & = 335 - 2 \\ 9n & = 333 \\ n & = {333 \over 9} \\ n & = 37 \end{align*}

\begin{align*} \text{Kim's share} & = 285 \times {3 \over 2 + 3} \\ & = \$ 171 \\ \\ \text{Pat's and Xin's share} & = 285 - 171 \\ & = \$ 114 \\ \\ \text{Xin's share} & = {114 - 24 \over 2} \\ & = \$ 45 \\ \\ \text{Pat's share} & = 45 + 24 \\ & = \$ 69 \end{align*}

(a)

\begin{align*} y & = 16 - 4(-2)^2 \\ y & = 0 \end{align*}

(b)

\begin{align*} y & = 16 - 4x^2 \\ 4x^2 & = 16 - y \\ x^2 & = {16 - y \over 4} \\ x & = \pm \sqrt{16 - y \over 4} \end{align*}

\begin{align*} 60 \times {100 - 16 \over 100} & = 50.4 \% \phantom{000000} [50.4 \% \text{ of tank is filled after usage}] \\ \\ 378 l \times {100 \over 50.4} & = 750 \text{ litres} \end{align*}

\begin{align*} \text{Total amount} & = P \left(1 + {r \over 100}\right)^n \\ 1120 & = 850 \left(1 + {r \over 100}\right)^{12} \\ {1120 \over 850} & = \left(1 + {r \over 100}\right)^{12} \\ \pm \sqrt[12]{1120 \over 850} & = 1 + {r \over 100} \\ \\ {r \over 100} & = \sqrt[12]{1120 \over 850} - 1 \text{ or } - \sqrt[12]{1120 \over 850} - 1 \text{ (Not applicable since } r > 0) \\ {r \over 100} & = 0.023 \phantom{.} 253 \\ r & = 100 (0.023 \phantom{.} 253) \\ r & = 2.325 \phantom{.} 3 \\ r & \approx 2.33 \end{align*}

(a)(i)

\begin{align*} A \cap B & = \{ \text{s, a, r, e} \} \end{align*}

(a)(ii)

\begin{align*} A \cup B & = \{ \text{a, e, g, i, n, o, p, q, r, s, u} \} \\ \\ (A \cup B)' & = \{ \} \end{align*}

(b)

\begin{align*} ( P \cap Q )' \end{align*}

(a)

\begin{align*} 1 \text{ cm} & : 2 \phantom{.} 000 \phantom{.} 000 \text{ cm} \\ 1 \text{ cm} & : 20 \phantom{.} 000 \text{ m} \phantom{000000} [1 \text{ m} = 100 \text{ cm}] \\ 1 \text{ cm} & : 20 \text{ km} \phantom{00000000.} [1 \text{ km} = 1000 \text{ m}] \\ 47.5 \text{ cm} & : 950 \text{ km} \\ \\ \text{Distance on map} & = 47.5 \text{ cm} \end{align*}

(b)

\begin{align*} 1 \text{ cm} & : 20 \text{ km} \\ 1 \text{ cm}^2 & : 400 \text{ km}^2 \phantom{000000} [\text{Square both sides}] \\ 827. 007 \phantom{.} 5 \text{ cm}^2 & : 330 \phantom{.} 803 \text{ km}^2 \\ \\ \text{Area on map} & = 827.007 \phantom{.} 5 \text{ cm}^2 \end{align*}

\begin{align*} \tan 36^\circ & = {45 \over BC} \phantom{000000} \left[ {Opp \over Adj}\right] \\ BC \tan 36^\circ & = 45 \\ BC & = {45 \over \tan 36^\circ} \\ BC & = 61.937 \\ \\ \tan \angle ADB & = {45 \over 61.937 + 22} \\ \angle ADB & = \tan^{-1} \left( 45 \over 61.937 + 22 \right) \\ \angle ADB & = 28.196^\circ \\ \angle ADB & \approx 28.2^\circ \end{align*}

(a)

\begin{align*} 3(3x + 2y) - 5(x - 3y) & = 9x + 6y - 5x + 15y \\ & = 4x + 21y \end{align*}

(b)

\begin{align*} 12ab - 9ax - 8by + 6xy & = 3a(4b - 3x) - 2y(4b - 3x) \\ & = (4b - 3x)(3a - 2y) \end{align*}

(a)

\begin{align*} 1188 & = 2^2 \times 3^3 \times 11 \end{align*}

(b)(i)

\begin{align*} \text{LCM} & = 2^2 \times 3^3 \times 11 \\ \\ A & = 2 \times 3^{1 + 2} \times 11 \\ \\ B & = 2^2 \times 3^1 \times 11 \\ \\ \therefore p & = 2, q = 3, r = 1 \end{align*}

(b)(ii)

\begin{align*} A & = 2 \times 3^{1 + 2} \times 11 \\ \\ B & = 2^2 \times 3^1 \times 11 \\ \\ \text{HCF} & = 2 \times 3 \times 11 \\ & = 66 \end{align*}

(a)(i)

\begin{align*} \text{Angle } EDC = 102^\circ \text{ because } \angle AED + \angle ADC = 180^\circ \text{ (Interior angles, } AE \phantom{.} // \phantom{.} CD) \end{align*}

(a)(ii)

\begin{align*} & \text{Angle } BCD = 96^\circ \text{ because sum of interior angles} = (5 - 2) \times 180^\circ = 540^\circ \text{ and } \\ & \angle BCD = 540^\circ - 102^\circ - 78^\circ - 120^\circ - 144^\circ = 96^\circ \end{align*}

(b)

\begin{align*} \angle BAC & = \angle BCA \phantom{0} \text{ (Isosceles triangle)} \\ & = {180^\circ - 144^\circ \over 2} \\ & = 18^\circ \\ \\ \angle EAC & = 120^\circ - 18^\circ \\ & = 102^\circ \\ \\ \angle DEA + \angle EAC & = 78^\circ + 102^\circ \\ & = 180^\circ \\ \\ \text{By the converse} & \text{ of interior angles, } ED \phantom{.} // \phantom{.} AC \\ \\ \text{Since } ED \phantom{.} // \phantom{.} AC & \text{ and } AE \phantom{.} // \phantom{.} CD, ACDE \text{ is a parallelogram} \end{align*}

(a)

\begin{align*} \text{Triangles } BGD & \text{ and } BAG \text{ are similar} \\ \\ {BD \over BG} & = {BG \over BA} \\ {BD \over 36} & = {36 \over 24} \\ 24 BD & = 36 (36) \\ BD & = {36(36) \over 24} \\ BD & = 54 \text{ m (Shown)} \end{align*}

(b)

\begin{align*} \text{Area of triangle } BGD & = {1 \over 2} \times 36 \times 54 \\ & = 972 \text{ m}^2 \\ \\ \text{Area of triangle } BAG & = {1 \over 2} \times 24 \times 36 \\ & = 432 \text{ m}^2 \\ \\ DC & = 54 - 37 \\ & = 17 \text{ m} \\ \\ [ \text{Need to find } & \text{length of } FC ] \\ \\ \text{Area of trapezium } ACFE & = {1 \over 2} \times (30 + FC) \times (37 + 24) \\ 1586 & = {1 \over 2} \times (30 + FC) \times (61) \\ {1586 \over {1 \over 2} \times 61 } & = 30 + FC \\ 52 & = 30 + FC \\ 52 - 30 & = FC \\ 22 & = FC \\ \\ \text{Area of triangle } DCF & = {1 \over 2} \times 17 \times 22 \\ & = 187 \text{ m}^2 \\ \\ \text{Total area} & = 972 + 432 + 187 + 1586 \\ & = 3177 \text{ m}^2 \end{align*}

(a) Remember to use the mid-values (2.5, 7.5, 11.5, ...) for calculations!

\begin{align*} \text{Mean} & = { \sum fx \over \sum f} \\ & = { 1888 \over 140 } \\ & = 13 {17 \over 35} \text{ hours} \end{align*}

(b)(i)

\begin{align*} \text{No. of adults watch } \le n \text{ hours per week} & = 140 \times {100 - 40 \over 100} \\ & = 84 \\ \\ n & = 14.75 \end{align*}

(b)(ii)

\begin{align*} \text{Required probability} & = { 60 - 12 \over 140} \\ & = {12 \over 35} \end{align*}

(a)

\begin{align*} \overrightarrow{OP} & = \overrightarrow{OC} + \overrightarrow{CP} \\ & = \textbf{c} + m (\textbf{a} - \textbf{c}) \\ & = \textbf{c} + m \textbf{a} - m \textbf{c} \\ & = m \textbf{a} + (1 - m) \textbf{c} \end{align*}

(b)(i)

\begin{align*} \overrightarrow{OP} & = m \textbf{a} + (1 - m) \textbf{c} \\ & = (1 - m) \textbf{c} + m \textbf{a} \\ \\ \overrightarrow{OB} & = 4(1 - m) \textbf{c} + 4 m \textbf{a} \\ & = 4 [ (1 - m) \textbf{c} + m \textbf{a} ] \\ & = 4 \overrightarrow{OP} \\ \\ \therefore O, & P \text{ and } B \text{ lie on a straight line} \end{align*}

(b)(ii)

\begin{align*} \overrightarrow{CB} & = \overrightarrow{CO} + \overrightarrow{OB} \\ k \textbf{a} & = - \textbf{c} + 4(1 - m) \textbf{c} + 4 m \textbf{a} \\ k \textbf{a} & = - \textbf{c} + 4 \textbf{c} - 4m \textbf{c} + 4m \textbf{a} \\ k \textbf{a} - 3 \textbf{c} & = 4m \textbf{a} - 4m \textbf{c} \end{align*} \begin{align*} \text{Com} & \text{paring } \textbf{a}, &&& \text{Com} & \text{paring } \textbf{c}, \\ k & = 4m &&& 3 & = 4m \\ k & = 4 \left(3 \over 4\right) &&& {3 \over 4} & = m \\ k & = 3 \end{align*}

(b)(iii)

\begin{align*} \overrightarrow{CP} & = {3 \over 4} (\textbf{a} - \textbf{c}) \\ \\ \overrightarrow{CA} & = \overrightarrow{CO} + \overrightarrow{OA} \\ & = - \textbf{c} + \textbf{a} \\ & = \textbf{a} - \textbf{c} \\ \\ \therefore \overrightarrow{CP} & = {3 \over 4} \overrightarrow{CA} \\ \\ \therefore CP : CA & = 3 : 4 \end{align*}

(a)

\begin{align*} \text{No. of girls} & = 16 - 9 \\ & = 7 \end{align*}

(b)

\begin{align*} \textbf{F} & = \left( \begin{matrix} 30 \\ 26 \\ 24 \end{matrix} \right) \end{align*}

(c)

\begin{align*} \textbf{M} & = \textbf{HF} \\ & = \underset{2 \times 3}{\left( \begin{matrix} 10 & 12 & 16 \\ 14 & 16 & 20 \end{matrix} \right)} \underset{3 \times 1}{\left( \begin{matrix} 30 \\ 26 \\ 24 \end{matrix} \right)} \\ & = \underset{2 \times 1}{\left( \begin{matrix} (10)(30) + (12)(26) + (16)(24) \\ (14)(30) + (16)(26) + (20)(24) \end{matrix} \right)} \\ & = \left( \begin{matrix} 996 \\1316 \end{matrix} \right) \end{align*}

(d)

\begin{align*} & 996 \text{ is the total fees collected for the morning session while } 1316 \text{ is the total fees} \\ & \text{collected for the afternoon session.} \end{align*}

(e)

\begin{align*} \text{Total fees} & = 5(996 + 1316) \phantom{000000} [\text{'5 days each week'}] \\ & = \$ 11 \phantom{.} 560 \end{align*}

(f)

\begin{align*} \text{Let } \textbf{A} \text{ denote } & \text{the number of children in each group in week 2} \\ \\ \textbf{A} & = \left( \begin{matrix} 10 \times {150 \over 100} & 12 \times {150 \over 100} & 16 \times {75 \over 100} \\ 14 \times {150 \over 100} & 16 \times {150 \over 100} & 20 \times {75 \over 100} \end{matrix} \right) \\ & = \left( \begin{matrix} 15 & 18 & 12 \\ 21 & 24 & 15 \end{matrix} \right) \\ \\ \left( \begin{matrix} 15 & 18 & 12 \\ 21 & 24 & 15 \end{matrix} \right) \left( \begin{matrix} 30 \\ 26 \\ 24 \end{matrix} \right) & = \left( \begin{matrix} (15)(30) + (18)(26) + (12)(24) \\ (21)(30) + (24)(26) + (15)(24) \end{matrix} \right) \\ & = \left( \begin{matrix} 1206 \\ 1614 \end{matrix} \right) \\ \\ \text{Total fees in week 2} & = 5(1206 + 1614) \\ & = \$ 14 \phantom{.} 100 \\ \\ \text{Percentage change} & = { \text{Final} - \text{Initial} \over \text{Initial} } \times 100 \\ & = { 14 \phantom{.} 100 - 11 \phantom{.} 560 \over 11 \phantom{.} 560} \times 100 \\ & = 21 {281 \over 289} \% \end{align*}

(a)

\begin{align*} y & = 2(7) + {6 \over 7} - 9 \\ y & = 5.8571 \\ y & \approx 5.9 \end{align*}

(b)

(c)

\begin{align*} 2x + {6 \over x} & = 10 \\ 2x + {6 \over x} - 9 & = 10 - 9 \\ \underbrace{ 2x + {6 \over x} - 9 }_\text{Curve} & = 1 \\ \\ \therefore \text{Draw } & y = 1 \\ \\ \text{From graph, } & x = 0.7 \text{ or } 4.3 \end{align*}

(d)(i)

(d)(ii)

$$ x = 0.9 \text{ or } 5.1 $$

(d)(iii)

\begin{align*} \text{Curve: } & y = 2x + {6 \over x} - 9 \\ \text{Line: } & y = {2 \over 3}x - 1 \\ \\ 2x + {6 \over x} - 9 & = {2 \over 3}x - 1 \\ x \left( 2x + {6 \over x} - 9 \right) & = x \left({2 \over 3}x - 1\right) \\ 2x^2 + 6 - 9x & = {2 \over 3}x^2 - x \\ 2x^2 - {2 \over 3}x^2 - 9x + x + 6 & = 0 \\ {4 \over 3}x^2 - 8x + 6 & = 0 \\ 3 \left({4 \over 3}x^2 - 8x + 6\right) & = 3(0) \\ 4x^2 - 24x + 18 & = 0 \\ 2x^2 - 12x + 9 & = 0 \\ \\ \therefore A & = -12, B = 9 \end{align*}

(a)

\begin{align*} AC^2 & = AB^2 + BC^2 - 2(AB)(BC) \cos \angle ABC \phantom{000000} [\text{Cosine rule}] \\ & = 660^2 + 950^2 - 2(660)(950) \cos 80^\circ \\ AC & = \sqrt{ 660^2 + 950^2 - 2(660)(950) \cos 80^\circ } \\ & = 1 \phantom{.} 058.463 \\ & \approx 1060 \text{ m (Shown)} \end{align*}

(b)

\begin{align*} {\sin \angle ADC \over 1 \phantom{.} 058.463} & = {\sin 24^\circ \over 480} \phantom{00000000000.} [\text{Sine rule}] \\ 480 \sin \angle ADC & = 1 \phantom{.} 058.463 \sin 24^\circ \phantom{0000} [\text{Cross-multiply}] \\ \sin \angle ADC & = { 1 \phantom{.} 058.463 \sin 24^\circ \over 480} \\ \angle ADC & = 180^\circ - \sin^{-1} \left( { 1 \phantom{.} 058.463 \sin 24^\circ \over 480 } \right) \phantom{000000} [\text{Obtuse angle } ADC] \\ & = 116.245^\circ \\ \\ \angle NCD & = 180^\circ - 116.245^\circ \phantom{0} \text{ (Interior angles)} \\ & = 63.755^\circ \\ \\ \text{Bearing of } D \text{ from } C & = 360^\circ - 63.755^\circ \phantom{0} \text{ (Angles at a point)} \\ & = 296.245^\circ \\ & \approx 296.2^\circ \end{align*}

(c)

\begin{align*} \text{Total distance} & = 660 + 950 + 1 \phantom{.} 058.463 \\ & = 2 \phantom{.} 668. 463 \text{ m} \\ & = 2. 668 \phantom{.} 463 \text{ km} \\ \\ \text{Time} & = {\text{Distance} \over \text{Speed}} \\ & = {2.668 \phantom{.} 463 \text{ km} \over 9.5 \text{ km/h}} \\ & = 0.28089 \text{ hours} \\ & = 16.8534 \text{ minutes} \\ & = 16 \text{ minutes } (0.8534 \times 60) \text{ seconds} \\ & \approx 16 \text{ minutes 50 seconds (to nearest 10 seconds)} \end{align*}

(a)

\begin{align*} \text{By Py} & \text{thagoras theorem,} \\ \text{Slant length, } l & = \sqrt{r^2 + h^2} \\ & = \sqrt{ 5.5^2 + 8^2 } \\ & = 9.7082 \text{ cm} \\ \\ \text{Curved surface area} & = \pi r l \phantom{000000000000} [\text{Formula provided}] \\ & = \pi (5.5) (9.7082) \\ & = 167.745 \\ & \approx 168 \text{ cm}^2 \end{align*}

(b)(i)

\begin{align*} {V_1 \over V_2} = \left(l_1 \over l_2\right)^3 & = \left(8 - 2 \over 8\right)^3 = {27 \over 64} \phantom{000000} [\text{Water forms a cone that is similar to the glass cone}] \\ \\ {27 \over 64} \times 100 & = 42 {3 \over 16} \% \\ \\ \text{Amir is wrong } & \text{as glass is only } 42 {3 \over 16} \% \ \text{filled.} \end{align*}

(b)(ii)

\begin{align*} {27 \over 64} \times 100 & = 42 {3 \over 16} \% \end{align*}

(b)(iii)

\begin{align*} \text{Volume of glass cone} & = {1 \over 3} \pi r^2 h \phantom{000000} [\text{Formula provided}] \\ & = {1 \over 3} \pi (5.5)^2 (8) \\ & = 80{2 \over 3} \pi \text{ cm}^3 \\ \\ \text{Volume of water} & = 80{2 \over 3} \pi \times {27 \over 64} \\ & = 34{1 \over 32} \pi \text{ cm}^3 \\ \\ \text{Volume of cylinder} & = \pi r^2 h \\ 34{1 \over 32} \pi & = \pi r^2 (2.5) \\ { 34{1 \over 32} \pi \over 2.5 \pi } & = r^2 \\ 13{49 \over 80} & = r^2 \\ \\ r & = \sqrt{ 13{49 \over 80} } \\ r & = 3.6895 \\ r & \approx 3.69 \text{ cm} \end{align*}

(a)(i)

\begin{align*} \text{Since tan} & \text{gents at } B \text{ and } C \text{ meet at } F, BF = CF \phantom{0} [S] \\ \\ \angle BFD & = \angle CFE \phantom{0} \text{ (Vertically opposite angles)} [A] \\ \\ \text{Since tan} & \text{gents at } D \text{ and } E \text{ meet at } F, DF = EF \phantom{0} [S] \\ \\ \therefore \text{Triangles } & BDF \text{ and } CEF \text{ are congruent } (SAS \text{ test}) \end{align*}

(a)(ii)(a)

\begin{align*} \angle OBA & = x^\circ \phantom{0} \text{ (Isosceles triangle } OBA \text{ with } OA = OB) \\ \\ \angle ABC & = 90^\circ \phantom{0} \text{ (Right angle in a semicircle)} \\ \\ \angle OBC & = (90 - x)^\circ \end{align*}

(a)(ii)(b)

\begin{align*} \angle OBF & = \angle OCF = \angle PDF = \angle PEF = 90^\circ \phantom{0} \text{ (Tangent perpendicular to radius)} \\ \\ \angle CBF & = 90^\circ - (90 - x)^\circ \\ & = x^\circ \\ \\ \angle BFC & = 180^\circ - x^\circ - x^\circ \phantom{0} \text{(Isosceles triangle } FBC \text{ with } FB = FC) \\ & = (180 - 2x)^\circ \\ \\ \angle DFE & = (180 - 2x)^\circ \phantom{0} \text{ (Vertically opposite angles)} \\ \\ \angle DPE & = 360^\circ - 90^\circ - 90^\circ - (180 - 2x)^\circ \phantom{0} \text{ (Angle sum of quadrilateral)} \\ & = 180^\circ - 180^\circ + 2x^\circ \\ & = 2x^\circ \end{align*}

(b)(i)

\begin{align*} \text{Reflex } \angle LKM & = (2\pi - 1.8) \text{ radians} \phantom{000000} [360^\circ = 2\pi \text{ radians}] \\ \\ \text{Major arc } LM & = r \theta \phantom{0000000000000000000.} [\text{Formula provided}] \\ & = (12)(2 \pi - 1.8) \\ & = (24\pi - 21.6) \text{ cm} \end{align*}

(b)(ii)

\begin{align*} \text{Minor sector area } KLM & = {1 \over 2} r^2 \theta \phantom{0000000000000.} [\text{Formula provided}] \\ & = {1 \over 2} (12)^2 (1.8) \\ & = 129.6 \text{ cm}^2 \\ \\ \text{Area of triangle } KLM & = {1 \over 2} ab \sin C \phantom{0000000000.} [\text{Formula provided}] \\ & = {1 \over 2} (12)(12) (\sin 1.8) \phantom{000} [\text{Radian mode!}] \\ & = 70.117 \text{ cm}^2 \\ \\ \text{Shaded area (minor segment)} & = 129.6 - 70.117 \\ & = 59.483 \text{ cm}^2 \\ \\ \text{Area of circle} & = \pi r^2 \\ & = \pi (12)^2 \\ & = 144 \pi \text{ cm}^2 \\ \\ \text{Required percentage} & = { 59.483 \over 144 \pi} \times 100 \\ & = 13.148 \\ & \approx 13.1 \% \end{align*}

(a)

\begin{align*} \text{Total number of students (frequency)} & = p + 10 + 13 + 9 + 6 + q + 2 \\ 50 & = p + q + 40 \\ 50 - 40 & = p + q \\ 10 & = p + q \\ 10 - q & = p \phantom{0} \text{--- (1)} \\ \\ \text{Total movies watched} & = (0 \times p) + (1 \times 10) + (2 \times 13) + (3 \times 9) + (4 \times 6) + (5 \times q) + (6 \times 2) \\ 2.68 \times 50 & = 0 + 10 + 26 + 27 + 24 + 5q + 12 \\ 134 & = 5q + 99 \\ 134 - 99 & = 5q \\ 35 & = 5q \\ {35 \over 5} & = q \\ 7 & = q \\ \\ \text{Substitute } & q = 7 \text{ into (1),} \\ p & = 10 - 7 \\ p & = 3 \\ \\ \therefore p & = 3, q = 7 \end{align*}

(b) Since question is worth 1 mark, it is not necessary to show workings.

\begin{align*} \text{Standard deviation} & = \sqrt{ { \sum fx^2 \over \sum f} - \left(\sum fx \over \sum f\right)^2 } \\ & = \sqrt{ { 486 \over 50 } - \left( 134 \over 50 \right)^2 } \\ & = 1.5929 \\ & \approx 1.59 \end{align*}

(c)

\begin{align*} \text{1. } & \text{Students watched more movies on average as the mean number of movies watched is higher.} \\ \\ \text{2. } & \text{Number of movies watched by students are more consistent as the standard deviation of} \\ & \text{movies watched is lower.} \end{align*}

(a)

\begin{align*} \text{Length of } AB & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ & = \sqrt{ [ -4 - (-7)]^2 + [5 - (-1)]^2 } \\ & = 6.7082 \\ & \approx 6.71 \text{ units} \end{align*}

(b)

\begin{align*} \text{Since } ABCD \text{ is a } & \text{parallelogram, } AB \text{ is parallel to } CD \\ \\ \implies \text{Gradient of } CD & = \text{Gradient of } AB \\ & = {y_2 - y_1 \over x_2 - x_1} \\ & = {5 - (-1) \over -4 - (-7)} \\ & = 2 \\ \\ \overrightarrow{BC} & = {8 \choose -2} \phantom{000000} [\text{Moving from } B(-4, 5) \text{ to } C] \\ \\ x \text{-coordinate of } C & = -4 + 8 = 4 \\ y \text{-coordinate of } C & = 5 + (-2) = 3 \\ \\ \therefore & \phantom{.} C(4, 3) \\ \\ y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } & C(4, 3), \\ 3 & = 2(4) + c \\ 3 & = 8 + c \\ 3 - 8 & = c \\ -5 & = c \\ \\ \text{Eqn of } CD: & \phantom{.} y = 2x - 5 \end{align*}

(c)(i)

\begin{align*} \overrightarrow{AC} & = {4 + 7 \choose 3 + 1} \\ & = {11 \choose 4} \\ \\ \overrightarrow{XC} & = {1 \over 2} \overrightarrow{AC} \\ & = {1 \over 2} {11 \choose 4} \\ & = {5.5 \choose 2} \end{align*}

(c)(ii)

\begin{align*} \overrightarrow{XC} & = {5.5 \choose 2} \\ \\ \overrightarrow{CX} & = {-5.5 \choose -2} \\ \\ x \text{-coordinate of } X & = 4 - 5.5 = -1.5 \\ y \text{-coordinate of } X & = 3 - 2 = 1 \\ \\ \therefore X(-1.5, 1) & \text{ and } \overrightarrow{OX} = {-1.5 \choose 1} \end{align*}

(a)

\begin{align*} \text{No. of pink balls} & = n - 21 \end{align*}

(b)

\begin{align*} \text{P(Pink, Pink)} & = {n - 21 \over n} \times {n - 22 \over n - 1} \\ {1 \over 8} & = { (n - 21)(n - 22) \over n(n - 1)} \\ n(n - 1) & = 8(n - 21)(n - 22) \phantom{000000} [\text{Cross-multiply}] \\ n^2 - n & = 8(n^2 - 22n - 21n + 462) \\ n^2 - n & = 8(n^2 - 43n + 462) \\ n^2 - n & = 8n^2 - 344n + 3696 \\ 0 & = 8n^2 - n^2 - 344n + n + 3696 \\ 0 & = 7n^2 - 343n + 3696 \\ 0 & = n^2 - 49n + 528 \phantom{0} \text{ (Shown)} \end{align*}

(c)

\begin{align*} 0 & = n^2 - 49n + 528 \\ 0 & = (n - 16)(n - 33) \end{align*} \begin{align*} n - 16 & = 0 && \text{ or } & n - 33 & = 0 \\ n & = 16 &&& n & = 33 \end{align*}

(d)

\begin{align*} \text{Reject } n = 16 \text{ since } n \text{ must be larger than } 21 \text{ (no. of blue balls)} \end{align*}

(e)

\begin{align*} \text{Case 1: P(Blue, Pink)} & = {21 \over n} \times {n - 21 \over n - 1} \\ & = {21 \over 33} \times {33 - 21 \over 33 - 1} \\ & = {21 \over 88} \\ \\ \text{Case 2: P(Pink, Blue)} & = {n -21 \over n} \times {21 \over n - 1} \\ & = {33 - 21 \over 33} \times {21 \over 33 - 1} \\ & = {21 \over 88} \\ \\ \text{Required probability} & = {21 \over 88} + {21 \over 88} \\ & = {21 \over 44} \end{align*}

(a)(i)

\begin{align*} \$ 1.824 \end{align*}

(a)(ii)

\begin{align*} \text{Lowest cost} & = 217 \times 1.796 \\ & = \$ 389.732 \\ \\ \text{Highest cost} & = 217 \times 1.86 \\ & = \$ 403.62 \\ \\ \text{Difference} & = 403.62 - 389.732 \\ & = \$ 13.888 \\ & \approx \$ 13.89 \end{align*}

(b)

\begin{align*} \text{1 year} & = 52 \text{ weeks} \\ \\ \text{Annual accommodation cost} & = 217 \times 52 \\ & = \text{£} 11 \phantom{.} 284 \\ \\ \text{Annual travel pass cost} & = 56.90 \times 12 \\ & = \text{£} 682.80 \\ \\ \text{Annual living costs} & = (60 + 57 + 25 + 15) \times 52 \\ & = \text{£} 8164 \\ \\ \text{Total cost} & = 11 \phantom{.} 284 + 682.80 + 8164 \\ & = \text{£} 20 \phantom{.} 130.80 \\ \\ \text{Total cost + extra costs} & = 20 \phantom{.} 130.80 \times {115 \over 100} \\ & = \text{£} 23 \phantom{.} 150.42 \\ \\ \text{For past 12 months}, & \text{ exchange rate is 0.542 - 0.576 pounds per dollar} \\ \\ \text{Use exchange rate} & \text{ of 0.54 pounds per dollar (worst case scenario)} \\ \\ \text{£} 0.54 & \rightarrow \$ 1 \\ \text{£} 1 & \rightarrow \$ {1 \over 0.54} = \$ 1.8518 \\ \\ \text{Amount in dollars} & = 23 \phantom{.} 150.42 \times 1.8518 \\ & = 42 \phantom{.} 869.95 \\ & \approx \$ 42 \phantom{.} 900 \end{align*}