\begin{align*} I & = {PRT \over 100} \\ & = {(2500)(1.6)(3) \over 100} \\ & = \$ 120 \\ \\ \text{Total value} & = 2500 + 120 \\ & = \$ 2620 \end{align*}

\begin{align*} \text{By Py} & \text{thagoras theorem,} \\ AC^2 & = AB^2 + BC^2 \\ 20.5^2 & = AB^2 + 13.3^2 \\ \\ AB^2 & = 20.5^2 - 13.3^2 \\ AB & = \sqrt{ 20.5^2 - 13.3^2 } \\ & = 15.6 \text{ cm} \end{align*}

(a)

\begin{align*} 82.5 & \approx 80 \\ \\ 3.35 & \approx 3 \\ \\ \text{Estimated distance} & = \text{Speed} \times \text{Time} \\ & = 80 \times 3 \\ & = 240 \text{ km} \end{align*}

(b)

\begin{align*} & \text{Since both the average speed and the time taken were rounded down,} \\ & \text{the actual distance travelled is greater than the estimated distance.} \end{align*}

(a)

\begin{align*} \text{P(Yellow)} & = 1 - 0.44 - 0.24 \phantom{000000} [\text{Total probability} = 1] \\ & = 0.32 \end{align*}

(b)

\begin{align*} \text{P(White)} & = 0.24 \\ \\ \text{Let } x & = \text{Number of white marbles} \\ \\ {x \over 175} & = 0.24 \\ {x \over 175} & = {0.24 \over 1} \\ x & = 175 (0.24) \phantom{000000} [\text{Cross-multiply}] \\ x & = 42 \end{align*}

\begin{align*} {3x \over 4} - {2(x - 4) \over 3} & = {3x \over 4} - {2x - 8 \over 3} \\ & = {9x \over 12} - {8x - 32 \over 12} \\ & = {9x - (8x - 32) \over 12} \\ & = {9x - 8x + 32 \over 12} \\ & = {x + 32 \over 12} \end{align*}

\begin{align*} x, y, & \underbrace{15}_\text{Median}, \underbrace{21, 21}_\text{Most frequent} \\ \\ \text{Mean} & = 12 \\ \text{Total} & = 12 \times 5 = 60 \\ \\ x + y & = 60 - 15 - 21 - 21 \\ x + y & = 3 \\ \\ \therefore 1, 2, & 15, 21, 21 \end{align*}

\begin{align*} \text{Volume} & = l \times b \times h \\ \\ \text{Height} & = {1200 \over 150} \\ & = 8 \text{ cm} \\ \\ 150 & = 2 \times 3 \times 5^2 \\ & = 2 \times 75 \phantom{000000} [\text{Not possible since } 75 \times 8 > 150] \\ & = 6 \times 25 \phantom{000000} [\text{Not possible since } 25 \times 8 > 150] \\ & = 10 \times 15 \\ \\ \therefore 15 \text{ cm by } & 10 \text{ cm by } 8 \text{ cm} \end{align*}

\begin{align*} \text{Number in 2018} & = 7 \phantom{.} 690 \phantom{.} 000 \times {100 + 140.6 \over 100} \\ & = 18 \phantom{.} 502 \phantom{.} 140 \end{align*}

(a)

\begin{align*} 6x + 7 & = -8 \\ 6x & = -8 - 7 \\ 6x & = -15 \\ x & = {-15 \over 6} \\ x & = - 2.5 \end{align*}

(b)

\begin{align*} 8a - 5b - 3(2a + 3b) & = 8a - 5b - 6a - 9b \\ & = 2a - 14b \end{align*}

\begin{align*} \text{Sector angle (Walk)} & = {24 \over 24 + 21 + 9} \times 360^\circ = 160^\circ \\ \\ \text{Sector angle (Bus)} & = {21 \over 24 + 21 + 9} \times 360^\circ = 140^\circ \\ \\ \text{Sector angle (Cycle)} & = {9 \over 24 + 21 + 9} \times 360^\circ = 60^\circ \end{align*}

\begin{align*} \text{Interior angle + Exterior angle} & = 180^\circ \\ \\ \text{Exterior angle} & = {180^\circ \over 8} \\ & = 22.5^\circ \\ \\ \text{Sum of exterior angles} & = 360^\circ \\ \\ \text{No. of sides} & = {360^\circ \over 22.5^\circ} \\ & = 16 \end{align*}

(a)

\begin{align*} \text{The elements of set } A \text{ are factors of } 12 \end{align*}

(b)

\begin{align*} A' \cap B' & = \{ 7, 8, 9, 10, 11, 13, 14 \} \phantom{000000} [\text{Elements outside of } A \text{ and } B] \end{align*}

(c)

\begin{align*} A \cap B' & = \{ 2, 4, 6, 12 \} \\ \\ A' \cap B & = \{ 5, 15 \} \\ \\ (A \cap B') \cup (A' \cap B) & = \{ 2, 4, 5, 6, 12, 15 \} \\ \\ \therefore n [ (A \cap B') \cup (A' \cap B) ] & = 6 \end{align*}

(a)

$$ 466 \phantom{.} 000 \phantom{00000} [\text{Use ruler to measure - your answer may differ due to printing}] $$

(b)

\begin{align*} & \text{The bar for 0-9 age group is shorter for 2017 than the bar for 2007} \end{align*}

(c)

\begin{align*} & \text{In general, the number of seniors and adults increased from 2007 to 2017, while} \\ & \text{the number of teenagers and children decreased from 2007 to 2017. This implies } \\ & \text{that the mean age of resident population has increased} \end{align*}

(a)

\begin{align*} \angle ABC & = \angle ADE \phantom{0} (\text{Given}) [A] \\ \\ \angle BAC & = \angle DAE \phantom{0} (\text{Common angle}) [A] \\ \\ \therefore \text{Triangles } & ABC \text{ and } ADE \text{ are similar (AA)} \end{align*}

(b)

\begin{align*} {AD \over AB} & = {AE \over AC} \\ {AD \over 10} & = {9 \over 12} \\ 12 AD & = 9(10) \phantom{000000} [\text{Cross-multiply}] \\ 12 AD & = 90 \\ AD & = {90 \over 12} \\ AD & = 7.5 \\ \\ CD & = AC - AD \\ & = 12 - 7.5 \\ & = 4.5 \text{ cm} \phantom{0} \text{(Shown)} \end{align*}

(a)

\begin{align*} (2x + 3p)^2 & = \underbrace{ (2x)^2 + 2(2x)(3p) + (3p)^2 }_{(a + b)^2 = a^2 + 2ab + b^2} \\ & = 4x^2 + 12px + 9p^2 \end{align*}

(b)

\begin{align*} (2x + 3p)^2 & = 4x^2 - 36x + 81 \\ 4x^2 + 12px + 9p^2 & = 4x^2 - 36x + 81 \\ \\ \text{Comparing } & \text{coefficients of } x, \\ 12p & = -36 \\ p & = {-36 \over 12} \\ p & = -3 \end{align*}

(a)

\begin{align*} s & = 3.5 (t - 2)^2 \\ 14 & = 3.5 (t - 2)^2 \\ {14 \over 3.5} & = (t - 2)^2 \\ 4 & = (t - 2)^2 \\ \pm \sqrt{4} & = t - 2 \\ \pm 2 & = t - 2 \end{align*} \begin{align*} t - 2 & = 2 && \text{ or } & t - 2 & = -2 \\ t & = 4 &&& t & = 0 \text{ (Reject, since } t > 2) \end{align*}

(b)

\begin{align*} E & = kv^2 \\ \\ \text{Let initial } & \text{value of } E = a \\ \\ a & = kv^2 \\ {a \over k} & = v^2 \\ \sqrt{a \over k} & = v \\ \\ \text{New value of } E & = a \times {100 - 75 \over 100} \\ & = {1 \over 4} a \\ \\ {1 \over 4}a & = kv^2 \\ {1 \over 4} \left(a \over k\right) & = v^2 \\ \sqrt{ {1 \over 4} \left(a \over k\right) } & = v \\ \sqrt{1 \over 4} \sqrt{a \over k} & = v \\ {1 \over 2} \sqrt{a \over k} & = v \\ \\ \text{Percentage change} & = {\text{Final} - \text{Initial} \over \text{Final}} \times 100 \\ & = { {1 \over 2} \sqrt{a \over k} - \sqrt{a \over k} \over \sqrt{a \over k} } \times 100 \\ & = { - {1 \over 2} \sqrt{a \over k} \over \sqrt{a \over k} } \times 100 \\ & = -{1 \over 2} \times 100 \\ & = -50 \% \\ \\ \text{Percentage reduction} & = 50 \% \end{align*}

\begin{align*} \left( 64 x^6 \over y^3 \right)^{2 \over 3} & = \left( 4^3 x^6 \over y^3 \right)^{2 \over 3} \\ & = { (4^3)^{2 \over 3} (x^6)^{2 \over 3} \over (y^3)^{2 \over 3} } \\ & = { 4^2 x^4 \over y^2 } \phantom{000000} [ (a^m)^n = a^{mn} ] \\ & = {16 x^4 \over y^2} \end{align*}

(a)

\begin{align*} 0. 000 \phantom{.} 925 \text{ km} & \rightarrow 1 \text{ litre} \\ 1 \text{ km} & \rightarrow {1 \over 0.000 \phantom{.} 925} \text{ litres} = 1081{3 \over 37} \text{ litres} \end{align*}

(b)(i)

\begin{align*} 1 \text{ cm} & : 60 \text{ cm} \\ 1 \text{ cm} & : 0.6 \text{ m} \phantom{000000} [1 \text{ m} = 100 \text{ cm}] \\ 550 \text{ cm} & : 330 \text{ m} \\ \\ \text{Length of model} & = 550 \text{ cm} \\ & = 5.5 \text{ m} \end{align*}

(b)(ii)

\begin{align*} {V_1 \over V_2} & = \left(l_1 \over l_2\right)^3 \\ & = \left(1 \over 60\right)^3 \\ & = {1 \over 216 \phantom{.} 000 } \\ \\ \text{Capacity of fuel tank of model} & = {1.228 \times 10^6 \over 216 \phantom{.} 000} \\ & = 5{37 \over 54} \text{ litres} \end{align*}

\begin{align*} \angle ACB & = 180^\circ - 82^\circ - 56^\circ \phantom{0} (\text{Angle sum of triangle}) \\ & = 42^\circ \\ \\ { AB \over \sin 42^\circ } & = { 15.6 \over \sin 56^\circ } \phantom{000000} [\text{Sine rule}] \\ AB \sin 56^\circ & = 15.6 \sin 42^\circ \phantom{000} [\text{Cross-multiply}] \\ AB & = {15.6 \sin 42^\circ \over \sin 56^\circ} \\ AB & = 12.591 \\ \\ \text{Area of triangle} & = {1 \over 2} ab \sin C \\ & = {1 \over 2} (AB)(AC) \sin \angle BAC \\ & = {1 \over 2} (12.591)(15.6) \sin 82^\circ \\ & = 97.254 \\ & \approx 97.3 \text{ cm}^2 \end{align*}

(a)(i)

\begin{align*} 144 & = 2^4 \times 3^2 \end{align*}

(a)(ii)

\begin{align*} \text{HCF} & = 12 = 2^2 \times 3 \\ \\ \text{LCM} & = 144 = 2^4 \times 3^2 \\ \\ x & = 2^2 \times 3^2 = 36 \\ y & = 2^4 \times 3 = 48 \end{align*}

(b)

\begin{align*} 784 \div {p \over q} & = {784 \over 1} \times {q \over p} \\ & = {784 \times q \over p} \\ & = {2^4 \times 7^2 \times q \over p} \phantom{000} \left[ \text{Perfect cube: } {2^4 \times 7^2 \times 7 \over 2} = 2^3 \times 7^3 \right] \\ \\ \therefore q & = 7, p = 2 \end{align*}

(a)

\begin{align*} 6x^2 - y + 3xy - 2x & = 6x^2 - 2x + 3xy - y \\ & = 2x (3x - 1) + y(3x - 1) \\ & = (3x - 1)(2x + y) \end{align*}

(b)

\begin{align*} 8x^2 - 6x - 9 & = 0 \\ (2x - 3)(4x + 3) & = 0 \end{align*} \begin{align*} 2x - 3 & = 0 && \text{ or } & 4x + 3 & = 0 \\ 2x & = 3 &&& 4x & = -3 \\ x & = {3 \over 2} &&& x & = -{3 \over 4} \end{align*}

(a)

\begin{align*} \textbf{M} & = \left( \begin{matrix} 5 & 4 & 2 \\ 6 & 3 & y \end{matrix} \right) \end{align*}

(b)

\begin{align*} \textbf{P} & = \textbf{MN} \\ & = \mathop{ \left( \begin{matrix} 5 & 4 & 2 \\ 6 & 3 & y \end{matrix} \right) }_{2 \times 3} \mathop{ \left( \begin{matrix} 36 & x \\ 40 & 39 \\ 68 & 65 \end{matrix} \right) }_{3 \times 2} \\ & = \mathop{ \left( \begin{matrix} 5 \times 36 + 4 \times 40 + 2 \times 68 & 5x + 4 \times 39 + 2 \times 65 \\ 6 \times 36 + 3 \times 40 + 68y & 6x + 3 \times 39 + 65y \end{matrix} \right) }_{2 \times 2} \\ & = \left( \begin{matrix} 476 & 5x + 286 \\ 336 + 68y & 6x + 117 + 65y \end{matrix} \right) \end{align*}

(c)

$$ \text{It represents the cost of Tom's purchases from centre } A $$

(d)

\begin{align*} 476 & = 286 + 5x \\ 476 - 286 & = 5x \\ 190 & = 5x \\ {190 \over 5} & = x \\ 38 & = x \end{align*}

(e)

\begin{align*} \text{Total cost in } A - 3 & = \text{Total cost in } B \\ 336 + 68y - 3 & = 6x + 117 + 65y \\ 68y - 65y & = 6x + 117 - 336 + 3 \\ 3y & = 6x - 216 \\ 3y & = 6(38) - 216 \\ 3y & = 12 \\ y & = {12 \over 3} \\ y & = 4 \end{align*}

Graph:

(a)

\begin{align*} \text{Gradient} & = { y_2 - y_1 \over x_2 - x_1 } \\ & = { 5 - (-3) \over 3 - 1} \\ & = 4 \end{align*}

(b)(i)

\begin{align*} x^3 + x^2 - 6x & = 0 \\ {1 \over 2} (x^3 + x^2 - 6x) & = {1 \over 2}(0) \\ {1 \over 2}x^3 + {1 \over 2}x^2 - 3x & = 0 \\ {1 \over 2}x^3 + {1 \over 2}x^2 - 3x - x & = 0 - x \\ {1 \over 2}(x^3 + x^2) - 4x & = -x \\ \underbrace{ {1 \over 2}(x^3 + x^2) - 4x + 3}_\text{Graph} & = -x + 3 \\ \\ \therefore \text{Draw } y & = -x + 3 \end{align*}

(a)

(a)

\begin{align*} \text{Acceleration} & = \text{Gradient} \\ & = {10 - 5 \over 20 - 0} \\ & = 0.25 \text{ m/s}^2 \end{align*}

(b)(i)

\begin{align*} \text{Distance} & = 8 \times T \phantom{000000} [\text{Area of rectangle}] \\ & = 8T \end{align*}

(b)(ii)

\begin{align*} \text{Distance travelled by } A \text{ after } T \text{ s} & = \underbrace{{1 \over 2} \times (5 + 10) \times 20}_\text{Area of trapezium} + 10 \times (T - 20) \\ & = 150 + 10T - 200 \\ & = 10T - 50 \\ \\ \text{Distance travelled by } A & = \text{Distance travelled by } B + 90 \\ 10T - 50 & = 8T + 90 \\ 10T - 8T & = 90 + 50 \\ 2T & = 140 \\ T & = {140 \over 2} \\ T & = 70 \end{align*}

(a)

\begin{align*} {2x + 5 \over 2} & > {1 - 4x \over 1} \\ {2x + 5 \over 2} & > {2 - 8x \over 2} \\ 2x + 5 & > 2 - 8x \\ 2x + 8x & > 2 - 5 \\ 10x & > -3 \\ x & > -{3 \over 10} \end{align*}

(b)(i)

\begin{align*} a & = 2(0.5)[ 0.5 - 3(-1.2)] \\ & = 4.1 \end{align*}

(b)(ii)

\begin{align*} a & = 2n(n - 3k) \\ a & = 2n^2 - 6kn \\ 6kn & = 2n^2 - a \\ n & = {2n^2 - a \over 6n} \end{align*}

(c)

\begin{align*} {3x \over x - 4} + {5 \over 2x + 1} & = {1 \over 1} \\ {3x(2x + 1) \over (x - 4)(2x + 1)} + {5(x - 4) \over (x - 4)(2x + 1)} & = {(x - 4)(2x + 1) \over (x - 4)(2x + 1)} \\ {3x(2x + 1) + 5(x - 4) \over (x - 4)(2x + 1)} & = {(x - 4)(2x + 1) \over (x - 4)(2x + 1)} \\ 3x(2x + 1) + 5(x - 4) & = (x - 4)(2x + 1) \\ 6x^2 + 3x + 5x - 20 & = 2x^2 + x - 8x - 4 \\ 6x^2 + 8x - 20 & = 2x^2 - 7x - 4 \\ 6x^2 - 2x^2 + 8x + 7x - 20 + 4& = 0 \\ 4x^2 + 15x - 16 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {- 15 \pm \sqrt{(15)^2 -4 (4)(-16)} \over 2(4)} \\ & = {- 15 \pm \sqrt{481} \over 8} \\ & = 0.86646 \text{ or } -4.6164 \\ & \approx 0.87 \text{ or } -4.62 \text{ (2 d.p.)} \end{align*}

(a)

\begin{align*} 90 \phantom{.} 182 \times 10^6 & = (9.0182 \times 10^4) \times 10^6 \\ & = 9.0182 \times 10^{10} \phantom{000000} [a^m \times a^n = a^{m + n}] \\ & \approx 9.02 \times 10^{10} \end{align*}

(b)

\begin{align*} \text{Percentage change} & = { \text{Final} - \text{Initial} \over \text{Initial} } \times 100 \\ & = {8007 - 6230 \over 6230} \times 100 \\ & = 28.523 \\ & \approx 28.5 \% \end{align*}

(c)

\begin{align*} \text{Value in 2017} & = (2.3 \times 10^{11} ) \times {100 + 9.3 \over 100} \\ & = \$ 2.5139 \times 10^{11} \end{align*}

(d)

\begin{align*} \text{Value in 2012} & = 1.8 \times 10^{10} \times {100 \over 100 - 6.1} \\ & = 1.9169 \times 10^{10} \\ & \approx \$ 1.92 \times 10^{10} \end{align*}

(e)

\begin{align*} 80.4 \text{ Yen} & = \$ 1 \\ 1 \text{ Yen} & = \$ {1 \over 80.4} \\ 155 \phantom{.} 000 \text{ Yen} & = {1 \over 80.4} \times 155 \phantom{.} 000 \\ & = \$ 1 \phantom{.} 927.860 \phantom{.} 6 \\ \\ \text{Cost with GST} & = 1 \phantom{.} 927.860 \phantom{.} 6 \times {107 \over 100} \\ & = 2062.81 \\ & \approx \$ 2063 \text{ (to nearest dollar)} \end{align*}

(a)

\begin{align*} \text{Area of rectangle} & = AD \times AB \\ 80 & = x \times AB \\ \\ AB & = {80 \over x} \\ \\ AQ & = \left( {80 \over x} - 4 \right) \text{ cm} \end{align*}

(b)

\begin{align*} \text{Area of triangle } PAQ & = {1 \over 2} \times AP \times AQ \\ & = {1 \over 2} \times (x - 4) \times \left({80 \over x} - 4 \right) \\ & = \left({1 \over 2}x - 2 \right) \left({80 \over x} - 4 \right) \\ & = \left({1 \over 2}x\right)\left({80 \over x}\right) - 2x - {160 \over x} + 8 \\ & = 40 - 2x - {160 \over x} + 8 \\ & = 48 - 2x - {160 \over x} \\ \\ \text{Shaded area, } y & = 80 - 2 \left( 48 - 2x - {160 \over x} \right) \\ & = 80 - 96 + 4x + {320 \over x} \\ & = {320 \over x} + 4x - 16 \phantom{0} \text{ (Shown)} \end{align*}

(c)

\begin{align*} y & = {320 \over x} + 4x - 16 \\ & = {320 \over 18} + 4(18) - 16 \\ & = 73.777 \\ & \approx 73.8 \end{align*}

(d)

(e)

$$ x = 12.6 $$

(f)

\begin{align*} \text{From graph, the smallest possible shaded area is approximately } 55.5 \text {cm}^2 \end{align*}

(a)

\begin{align*} \text{Volume of cone} & = {1 \over 3} \pi r^2 h \phantom{000000} [\text{Provided}] \\ & = {1 \over 3} \pi (3x)^2 (y) \\ & = {1 \over 3} \pi (9x^2)(y) \phantom{000000} [(3x)^2 = 3x \times 3x = 9x^2] \\ & = 3 \pi x^2 y \\ \\ \text{Volume of hemisphere} & = {2 \over 3} \pi r^3 \\ & = {2 \over 3} \pi (3x)^3 \\ & = {2 \over 3} \pi (27x^3)\phantom{00000000}[ (3x)^3 = 3x \times 3x \times 3x = 27x^3] \\ & = 18 \pi x^3 \\ \\ \therefore 3 \pi x^2 y & = 2 \times 18 \pi x^3 \\ 3 \pi x^2 y & = 36 \pi x^3 \\ y & = {36 \pi x^3 \over 3 \pi x^2} \\ y & = 12x \phantom{0} \text{ (Shown)} \end{align*}

(b)

\begin{align*} \text{Curved surface area of cone} & = \pi r l \phantom{000000} [\text{Provided; need to find slant height } l] \\ \\ \text{By Py}& \text{thagoras theorem,} \\ l & = \sqrt{ (3x)^2 + y^2 } \\ & = \sqrt{ (3x)^2 + (12x)^2 } \phantom{000000} [\text{Use equation from (a)}] \\ & = \sqrt{ 9x^2 + 144x^2 } \\ & = \sqrt{ 153 x^2 } \\ & = \sqrt{153} \sqrt{x^2} \\ & = \sqrt{153} x \\ \\ \text{Curved surface area of cone} & = \pi (3x) ( \sqrt{153} x ) \\ & = 3 \sqrt{153} \pi x^2 \\ \\ \text{Curved surface area of cylinder} & = 2 \pi r h \\ & = 2 \pi (3x)(2x) \\ & = 2 \pi (6x^2) \\ & = 12 \pi x^2 \\ \\ \text{Curved surface area of hemisphere} & = 2 \pi r^2 \\ & = 2 \pi (3x)^2 \\ & = 2 \pi (9x^2) \\ & = 18 \pi x^2 \\ \\ \text{Total surface area} & = 3 \sqrt{153} \pi x^2 + 12 \pi x^2 + 18 \pi x^2 \\ 300 & = x^2 ( 3 \sqrt{153} \pi + 12 \pi + 18 \pi) \\ \\ x^2 & = {300 \over 3 \sqrt{153} \pi + 12 \pi + 18 \pi } \\ x & = \sqrt{300 \over 3 \sqrt{153} \pi + 12 \pi + 18 \pi } \\ x & = 1.1928 \\ \\ H & = y + 2x + 3x \phantom{000000} [\text{Radius of hemisphere is also height of hemisphere}] \\ & = 12x + 2x + 3x \\ & = 17x \\ & = 17(1.1928) \\ & = 20.277 \\ & \approx 20.3 \text{ cm} \end{align*}

(a)

\begin{align*} y & = mx + c \\ y & = -{2 \over 9}x + c \\ \\ \text{Using } & A(5, 5), \\ 5 & = -{2 \over 9}(5) + c \\ 5 & = -{10 \over 9} + c \\ 5 + {10 \over 9} & = c \\ {55 \over 9} & = c \\ \\ y & = -{2 \over 9}x + {55 \over 9} \\ 9y & = 9 \left( -{2 \over 9}x + {55 \over 9} \right) \\ 9y & = -2x + 55 \\ 9y + 2x & = 55 \phantom{0} \text{ (Shown)} \end{align*}

(b)

\begin{align*} \text{Line } p: y & = -{2 \over 9}x + {55 \over 9} \phantom{0} \text{--- (1)} \\ \\ \text{Line } q: 3y & = 4x - 26 \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 3 \left( -{2 \over 9}x + {55 \over 9} \right) & = 4x - 26 \\ -{2 \over 3}x + {55 \over 3} & = 4x - 26 \\ -{2 \over 3}x - 4x & = -26 - {55 \over 3} \\ -{14 \over 3}x & = -{133 \over 3} \\ -14x & = -133 \\ x & = {-133 \over -14} \\ x & = 9.5 \\ \\ \text{Substitute } & x = 9.5 \text{ into (1),} \\ y & = -{2 \over 9} (9.5) + {55 \over 9} \\ y & = 4 \\ \\ \therefore & \phantom{.} (9.5, 4) \end{align*}

(c)(i)

\begin{align*} \text{Line } p: y & = -{2 \over 9}x + {55 \over 9} \\ \\ \text{Let } & x = 0, \\ y & = -{2 \over 9}(0) + {55 \over 9} \\ y & = {55 \over 9} \\ \\ \therefore & \phantom{.} B \left(0, 6{1 \over 9}\right) \\ \\ \\ \text{Line } q: 3y & = 4x - 26 \\ \\ \text{Let } & x = 0, \\ 3y & = 4(0) - 26 \\ 3y & = -26 \\ y & = -{26 \over 3} \\ \\ \therefore & \phantom{.} C \left(0, -8{2 \over 3}\right) \\ \\ \text{Let } & x = 5, \\ 3y & = 4(5) - 26 \\ 3y & = -6 \\ y & = {-6 \over 3} \\ y & = -2 \\ \\ \therefore & \phantom{.} D(5, -2) \end{align*}

\begin{align*} \text{Since } BC \text{ and } AD \text{ are vertical lines, } ABCD \text{ is a trapezium with height of } 5 \text{ units} \end{align*}

(c)(ii)

\begin{align*} \text{Area of trapezium} & = {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} \\ & = {1 \over 2} \times \left( 6{1 \over 9} + 8{2 \over 3} + 5 + 2 \right) \times 5 \\ & = 54 {4 \over 9} \text{ units}^2 \end{align*}

(a)

\begin{align*} \angle AED & = 90^\circ \phantom{0} \text{ (Right-angle in semicircle)} \\ \\ \angle ADE & = 180^\circ - 90^\circ - 35^\circ \phantom{0} \text{ (Angle sum of triangle)} \\ & = 55^\circ \\ \\ \angle BED & = 180^\circ - 55^\circ - 48^\circ \phantom{0} \text{ (Interior angles, } BE \phantom{.} // \phantom{.} CD) \\ & = 77^\circ \end{align*}

(b)

\begin{align*} \angle BCD & = 180^\circ - 77^\circ \phantom{0} \text{ (Angles in opposite segments)} \\ & = 103^\circ \\ \\ \angle ECD & = \angle EAD \phantom{0} \text{ (Angles in the same segment)} \\ & = 35^\circ \\ \\ \angle BCE & = 103^\circ - 35^\circ \\ & = 68^\circ \end{align*}

(c)

\begin{align*} \angle DOC & = 180^\circ - 48^\circ - 48^\circ \phantom{000000} [\text{Isosceles triangle with } OC = OD] \\ & = 84^\circ \\ \\ \text{Reflex } \angle AOC & = 180^\circ + 84^\circ \\ & = 264^\circ \\ \\ \text{Area of sector} & = {\theta \over 360^\circ} \times \pi r^2 \\ & = { 264^\circ \over 360^\circ } \times \pi (5)^2 \\ & = 57.595 \\ & \approx 57.6 \text{ cm}^2 \end{align*}

(a)

\begin{align*} {13 + 4 \over 8 + 2} & = {17 \over 10} \end{align*}

(b)

\begin{align*} \text{Numerator: } & 1, 5, 9, 13, ... \\ \\ \text{General term} & = a + (n - 1)(d) \phantom{000000} [a \text{: first term, } d \text{: common difference}] \\ & = 1 + (n - 1)(4) \\ & = 1 + 4n - 4 \\ & = 4n - 3 \\ \\ \\ \text{Denominator: } & 2, 4, 6, 8, ... \\ \\ \text{General term} & = 2 + (n - 1)(2) \\ & = 2 +2n - 2 \\ & = 2n \\ \\ \therefore T_n & = {4n - 3 \over 2n} \end{align*}

(c)

\begin{align*} T_n & = {4n - 3 \over 2n} \\ \\ T_{n + 1} & = {4(n + 1) - 3 \over 2(n + 1)} \\ & = { 4n + 4 - 3 \over 2(n + 1)} \\ & = { 4n + 1 \over 2(n + 1) } \\ \\ T_{n + 1} - T_n & = { 4n + 1 \over 2(n + 1) } - {4n - 3 \over 2n} \\ & = {n(4n + 1) \over 2n(n + 1)} - {(4n - 3)(n + 1) \over 2n(n + 1)} \\ & = {n(4n + 1) - (4n - 3)(n + 1) \over 2n(n + 1)} \\ & = {4n^2 + n - (4n^2 + 4n - 3n - 3) \over 2n(n + 1)} \\ & = {4n^2 + n - (4n^2 + n - 3) \over 2n(n + 1)} \\ & = {4n^2 + n - 4n^2 - n + 3 \over 2n(n + 1)} \\ & = {3 \over 2n(n + 1)} \phantom{0} \text{ (Shown)} \end{align*}

(d)

\begin{align*} \text{For } n \ge 1, & \phantom{0} 2n \ge 2 \text{ and } n + 1 \ge 2 \\ \\ \implies & 2n(n + 1) \ge 2 \times 2 \\ & 2n(n + 1) \ge 4 \\ \\ \implies & {3 \over 2n(n + 1)} \le {3 \over 4} \\ \\ \therefore & \phantom{.} T_{n + 1} - T_n < 1 \end{align*}

(a)(i)(a)

\begin{align*} \overrightarrow{RS} & = \overrightarrow{RO} + \overrightarrow{OS} \\ & = -\overrightarrow{OR} + \overrightarrow{OS} \\ & = -{-4 \choose 5} + {3 \choose -1} \\ & = {4 \choose -5} + {3 \choose -1} \\ & = {7 \choose -6} \end{align*}

(a)(i)(b)

\begin{align*} | \overrightarrow{RS} | & = \sqrt{ 7^2 + (-6)^2 } \\ & = 9.2195 \\ & \approx 9.22 \text{ units} \end{align*}

(a)(ii)

\begin{align*} \overrightarrow{ST} & = \overrightarrow{SO} + \overrightarrow{OT} \\ & = -\overrightarrow{OS} + \overrightarrow{OT} \\ & = -{3 \choose -1} + {k \choose 7} \\ & = {-3 \choose 1} + {k \choose 7} \\ & = {k -3 \choose 8} \\ \\ | \overrightarrow{ST} | & = \sqrt{ (k - 3)^2 + 8^2 } \\ & = \sqrt{ \underbrace{ (k)^2 - 2(k)(3) + (3)^2 }_{ (a - b)^2 = a^2 - 2ab + b^2 } + 64 } \\ & = \sqrt{ k^2 - 6k + 9 + 64 } \\ & = \sqrt{ k^2 - 6k + 73 } \\ \\ 10 & = \sqrt{k^2 - 6k + 73} \\ 10^2 & = k^2 - 6k + 73 \\ 100 & = k^2 - 6k + 73 \\ 0 & = k^2 - 6k + 73 - 100 \\ 0 & = k^2 - 6k - 27 \\ 0 & = (k + 3)(k - 9) \end{align*} \begin{align*} k + 3 & = 0 && \text{ or } & k - 9 & = 0 \\ k & = -3 &&& k & = 9 \end{align*}

(b)(i)

\begin{align*} \overrightarrow{CD} & = \overrightarrow{CG} + \overrightarrow{GA} + \overrightarrow{AD} \\ & = {1 \over 4} \overrightarrow{CB} - \textbf{q} + \textbf{p} \\ & = -{1 \over 4} \textbf{p} - \textbf{q} + \textbf{p} \\ & = {3 \over 4} \textbf{p} - \textbf{q} \end{align*}

(b)(ii)

\begin{align*} \overrightarrow{GF} & = \overrightarrow{GB} + \overrightarrow{BF} \\ & = {3 \over 4} \overrightarrow{CB} + {1 \over 2} \overrightarrow{BA} \\ & = -{3 \over 4} \textbf{p} + {1 \over 2} \overrightarrow{CD} \\ & = -{3 \over 4} \textbf{p} + {1 \over 2} \left({3 \over 4} \textbf{p} - \textbf{q} \right) \\ & = -{3 \over 4} \textbf{p} + {3 \over 8} \textbf{p} - {1 \over 2} \textbf{q} \\ & = -{3 \over 8} \textbf{p} - {1 \over 2} \textbf{q} \\ \\ \overrightarrow{DH} & = -{3 \over 8} \textbf{p} - {1 \over 2} \textbf{q} \\ \\ \overrightarrow{AH} & = \overrightarrow{AD} + \overrightarrow{DH} \\ & = \textbf{p} + \left(-{3 \over 8} \textbf{p} - {1 \over 2} \textbf{q}\right) \\ & = \textbf{p} - {3 \over 8} \textbf{p} - {1 \over 2} \textbf{q} \\ & = {5 \over 8} \textbf{p} - {1 \over 2} \textbf{q} \end{align*}

(a)(i)(a)

\begin{align*} 500 \times {50 \over 100} & = 250 \\ \\ \text{Median} & = 41.6 \text{ hours} \end{align*}

(a)(i)(b)

\begin{align*} 500 \times {25 \over 100} & = 125 \\ \\ \text{Lower quartile} & = 39.6 \text{ hours} \\ \\ 500 \times {75 \over 100} & = 375 \\ \\ \text{Upper quartile} & = 44 \text{ hours} \\ \\ \text{Interquartile range} & = 44 - 39.6 \\ & = 4.4 \text{ hours} \end{align*}

(a)(ii)

\begin{align*} \text{No. of workers who worked less than 45.1 hours} & = 415 \\ \\ \text{Percentage} & = {500 - 415 \over 500} \times 100 \\ & = 17 \% \end{align*}

(a)(iii)

\begin{align*} \text{The curve will shift to the left by 3 units} \end{align*}

(b)

\begin{align*} \text{No. of men working part-time} & = 16 \times {1 \over 8} \\ & = 2 \\ \\ \text{No. of women} & = 16 - 2 - 5 \\ & = 9 \\ \\ \text{Let no. of women wokring full-time} & = x \\ \\ \text{P(Two women working full time chosen)} & = {x \over 16} \times {x - 1 \over 15} \\ {1 \over 8} & = {x(x - 1) \over 240} \\ 240 & = 8x(x - 1) \phantom{000000} [\text{Cross-multiply}] \\ 240 & = 8x^2 - 8x \\ 0 & = 8x^2 - 8x - 240 \\ 0 & = x^2 - x - 30 \\ 0 & = (x + 5)(x - 6) \end{align*} \begin{align*} x + 5 & = 0 && \text{ or } & x - 6 & = 0 \\ x & = -5 \text{ (N.A.)} &&& x & = 6 \end{align*}
\begin{align*} \text{No. of women working part-time} & = 9 - 6 \\ & = 3 \end{align*}

(a)

\begin{align*} \tan \theta & = {Opp \over Adj} \\ \tan x^\circ & = {1 \over 10} \\ x^\circ & = \tan^{-1} \left(1 \over 10\right) \\ x^\circ & = 5.7105 \\ x & \approx 5.7 \text{ (1 d.p.)} \end{align*}

(b)

\begin{align*} \angle OBC & = 90^\circ - 5.7105^\circ \\ & = 84.2895^\circ \\ \\ OC^2 & = OB^2 + BC^2 - 2(OB)(BC) \cos \angle OBC \phantom{000000} [\text{Cosine rule}] \\ OC & = \sqrt{ OB^2 + BC^2 - 2(OB)(BC) \cos \angle OBC } \\ & = \sqrt{ 35.0^2 + 20.0^2 - 2(35.0)(20.0) \cos 84.2895^\circ } \\ & = 38.544 \\ & \approx 38.5 \text{ m (Shown)} \end{align*}

(c)

\begin{align*} OD^2 & = OB^2 + BD^2 - 2(OB)(BD)(\cos \angle OBD) \\ OD & = \sqrt{ OB^2 + BD^2 - 2(OB)(BD)(\cos \angle OBD) } \\ & = \sqrt{ 35^2 + 40^2 - 2(35)(40)(\cos 84.2895^\circ) } \\ & = 50.461 \text{ m} \\ \\ \text{Mass of } OA & = 48.3 \text{ m} \times 23.2 \text{ kg/m} \phantom{000000} [\text{FLC 68}] \\ & = 1120.56 \text{ kg} \\ & = 1.120 \phantom{.} 56 \text{ tonnes} \\ \\ \text{Mass of } OC \text{ and } OD & = (38.544 + 50.461) \text{ m} \times 10.8 \text{ kg/m} \phantom{000000} [\text{FLC 44}] \\ & = 961.254 \text{ kg} \\ & = 0.961 \phantom{.} 254 \text{ tonnes} \\ \\ \text{Total mass} & = 1.120 \phantom{.} 56 + 0.961 \phantom{.} 254 \\ & = 2.081 \phantom{.} 814 \text{ tonnes} < 2.1 \text{ tonnes} \\ \\ \therefore \text{Cables } & \text{meet the engineer's specification} \end{align*}