(a)

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

(b)

\begin{align*} 6x + 18xy & = 6x(1 + 3y) \end{align*}

\begin{align*} 4ax + 12by - 16ay - 3bx & = 4ax - 3bx - 16ay + 12by \\ & = x(4a - 3b) - 4y(4a - 3b) \\ & = (4a - 3b)(x - 4y) \end{align*}

(a)

(b)

\begin{align*} (5n + 1)^2 - (5n - 1)^2 & = \underbrace{ (5n)^2 + 2(5n)(1) + (1)^2 }_{ (a + b)^2 = a^2 + 2ab + b^2 } - \underbrace{ [ (5n)^2 - 2(5n)(1) + (1)^2]}_{ (a - b)^2 = a^2 - 2ab + b^2 } \\ & = 25n^2 + 10n + 1 - (25n^2 - 10n + 1) \\ & = 25n^2 + 10n + 1 - 25n^2 + 10n - 1 \\ & = 20n \\ \\ \therefore \text{Expression } & \text{is a multiple of 20} \end{align*}

\begin{align*} \text{Volume} & = l \times b \times h \\ 144 & = 2^4 \times 3^2 \\ \\ \text{Perimeter} & = 22 \\ l + b & = {22 \over 2} = 11 \\ \\ \text{Using prime} & \text{ factors} , \\ l & = 2^3 = 8 \\ b & = 3 \\ \\ h & = {144 \over 8 \times 3} \\ h & = 6 \text{ cm} \end{align*}

\begin{align*} { V_1 \over V_2 } & = \left(l_1 \over l_2\right)^3 \\ { \text{Volume of cone formed by water} \over \text{Volume of cone} } & = \left(h \over 30\right)^3 \\ {1 \over 2} & = \left(h \over 30\right)^3 \\ \\ \sqrt[3]{1 \over 2} & = {h \over 30} \\ \\ h & = 30 \times \sqrt[3]{1 \over 2} \\ & = 23.811 \\ & \approx 23.8 \text{ cm} \end{align*}

\begin{align*} & \text{The vertical axis is not uniformly divided.} \\ \\ & \text{Reader are unable to deduce how much the annual gas bill has increased from 2012 to 2015} \end{align*}

\begin{align*} {3 \over (x - 4)^2} - {1 \over 4 - x} & = {3 \over (x - 4)^2 } - {1 \over -(x - 4)} \\ & = {3 \over (x - 4)^2 } + {1 \over x - 4} \\ & = {3 \over (x - 4)^2 } + {x - 4 \over (x - 4)^2} \\ & = {3 + x - 4 \over (x - 4)^2 } \\ & = {x - 1 \over (x - 4)^2 } \end{align*}

(a)

\begin{align*} \angle ADC & = \angle AOC \div 2 \phantom{0} \text{ (Angle at centre} = 2 \times \text{Angle at circumference}) \\ & = 3x \div 2 \\ & = 1.5 x \end{align*}

(b)

\begin{align*} \angle ADC + \angle ABC & = 180^\circ \phantom{0} \text{ (Angles in opposite segments)} \\ 1.5x + 2x + 5 & = 180 \\ 3.5x & = 180 - 5 \\ 3.5x & = 175 \\ x & = {175 \over 3.5} \\ x & = 50 \end{align*}

(a)

\begin{align*} 9 - 8x + x^2 & = x^2 - 8x + 9 \\ & = x^2 - 8x + \left(8 \over 2\right)^2 - \left(8 \over 2\right)^2 + 9 \phantom{000000} [\text{Complete the square}] \\ & = x^2 - 8x + 4^2 - 16 + 9 \\ & = (x - 4)(x - 4) - 7 \\ & = (x - 4)^2 - 7 \\ & = -7 + (x - 4)^2 \end{align*}

(b)

\begin{align*} y & = 9 - 8x + x^2 \\ y & = (x - 4)^2 - 7 \\ \\ \text{Minimum } & \text{point: } (4, -7) \end{align*}

\begin{align*} \text{USD}4.48 & = 4.48 \times 1.242 \\ & = \$ 5.564 \phantom{.} 16 \\ \\ 1 \text{ gallon} & = 3.785 \text{ litres} \\ \\ \text{Cost per litre (in Hawaii)} & = {5.564 \phantom{.} 16 \over 3.785} \\ & = \$ 1.47 \text{ per litre} \\ \\ \therefore \text{Diesel is } & \text{cheaper in Hawaii} \end{align*}

(a)

\begin{align*} T & = 21 - {h \over 120} \\ 0 & = 21 - {h \over 120} \\ {h \over 120} & = {21 \over 1} \\ h & = 120(21) \phantom{000000} [\text{Cross-multiply}] \\ h & = 2520 \text{ m} \end{align*}

(b)

\begin{align*} T & = 21 - {h \over 120} \\ \\ \text{Let height of first aircraft} & = x \\ T & = {21 - x \over 120} \\ \\ \text{Let height of 2nd aircraft} 7 = y \\ T & = {21 - y \over 100} \\ \\ \text{Difference in temperature} & = {21 - x \over 120} - {21 - y \over 120} \\ 25 & = {21 - x - 21 + y \over 120} \\ {25 \over 1} & = {y - x \over 120} \\ 120(25) & = y - x \phantom{000000} [\text{Cross-multiply}] \\ 3000 & = y - x \\ \\ \therefore \text{Difference in height} & = 3000 \text{ m} \end{align*}

(a)

\begin{align*} \angle ECD & = 180^\circ - 45^\circ - 38^\circ \phantom{0} \text{ (Angle sum of triangle)} \\ & = 97^\circ \\ \\ \angle BCA & = 180^\circ - 97^\circ - 38^\circ - 22^\circ \phantom{0} \text{ (Interior angles, } BC \phantom{.} // \phantom{.} AD) \\ & = 23^\circ \\ \\ \angle ABC & = 180^\circ - 34^\circ - 23^\circ \phantom{0} \text{ (Angle sum of triangle)} \\ & = 123^\circ \\ \\ \text{Reflex } \angle ABC & = 360^\circ - 123^\circ \phantom{0} \text{ (Angles at a point)} \\ & = 237^\circ \end{align*}

(b)

\begin{align*} \angle ACD = \angle ECD & = 97^\circ \\ \\ \text{Since } \angle ACD \ne 90^\circ, & \text{ semicircle doesn not pass through } C \end{align*}

(a)

\begin{align*} \text{By similar} & \text{ triangles, } \\ {d \over 10} & = {13 \over 6.5} \\ 6.5d & = 10(13) \phantom{000000} [\text{Cross-multiply}] \\ 6.5d & = 130 \\ d & = {130 \over 6.5} \\ d & = 20 \\ \\ \text{Circumference} & = \pi d \\ & = \pi (20) \\ & = 20 \pi \text{ cm} \end{align*}

(b)

\begin{align*} \text{By Py} & \text{thagoras theorem,} \\ x & = \sqrt{ 20^2 - 13^2 } \\ & = 15.198 \\ \\ \text{Area of rectangle} & = 15.198 \times 13 \\ & = 197.574 \\ & \approx 198 \text{ cm}^2 \end{align*}

(a)

\begin{align*} y & = k a^{-x} \\ \\ \text{Using } & B(0, 4), \\ 4 & = k a^{-0} \\ 4 & = k(1) \\ 4 & = k \\ \\ y & = 4 a^{-x} \\ \\ \text{Using } & A(-2, 100), \\ 100 & = 4 a^{-(-2)} \\ 100 & = 4a^2 \\ {100 \over 4} & = a^2 \\ 25 & = a^2 \\ \pm \sqrt{25} & = a \\ \pm 5 & = a \\ \\ \therefore k & = 4, a= 5 \end{align*}

(b)

\begin{align*} \text{Gradient of } AB & = {y_2 - y_1 \over x_2 - x_1} \\ & = {4 - 100 \over 0 - (-2)} \\ & = -48 \\ \\ y & = mx + c \\ y & = -48x + c \\ \\ \text{Using } & A(-2, 100), \\ 100 & = -48(-2) + c \\ 100 & = 96 + c \\ 100 - 96 & = c \\ 4 & = c \\ \\ \text{Eqn of } AB: & \phantom{0} y = -48x + 4 \end{align*}

(a)

\begin{align*} 2 & | \underline{784} \\ 2 & | \underline{392} \\ 2 & | \underline{196} \\ 2 & | \underline{98} \\ 7 & | \underline{49} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 784 & = 2^4 \times 7^2 \end{align*}

(b)

\begin{align*} 784 & = 2^4 \times 7^2 \\ \\ \text{Power of each} & \text{ factor is a multiple of } 2 \end{align*}

(c)

\begin{align*} 784 \times {m \over n} & = {784 \over 1} \times {m \over n} \\ & = {784 m \over n} \\ & = { 2^4 \times 7^2 \times m \over n} \\ & = { 2^4 \times 7^2 \times 7 \over 2} \\ & = 2^3 \times 7^3 \\ \\ \therefore m & = 7, n = 2 \end{align*}

(a)

\begin{align*} \text{100 km} & \rightarrow 6.7 \text{ litres} \\ \text{1 km} & \rightarrow {6.7 \over 100} = 0.067 \text{ litre} \\ \\ \text{Petrol used} & = 19 \phantom{.} 629 \times 0.067 \\ & = 1 \phantom{.} 315.143 \text{ litres} \\ \\ \text{Total cost} & = 1 \phantom{.} 315.143 \times 2.42 \\ & = 3 \phantom{.} 182.646 \\ & \approx \$ 3 \phantom{.} 182.65 \end{align*}

(b)

\begin{align*} \text{Distance drove in 2014} & = 19 \phantom{.} 629 \times {100 \over 100 + 35} \\ & = 14 \phantom{.} 540 \text{ km} \end{align*}

(a)

\begin{align*} 1 \text{ cm} & : 2 \phantom{.} 500 \phantom{.} 000 \text{ cm} \\ 1 \text{ cm} & : 25 \phantom{.} 000 \text{ m} \phantom{000000} [1 \text{ m} = 100 \text{ cm}] \\ 1 \text{ cm} & : 25 \text{ km} \phantom{00000000.} [1 \text{ km} = 1000 \text{ m}] \\ 49.3 \text{ cm} & : 1232.5 \text{ km} \\ \\ \text{Actual length} & = 1232.5 \text{ km} \end{align*}

(b)

\begin{align*} 1 \text{ cm} & : 25 \text{ km} \\ 1^2 \text{ cm}^2 & : 25^2 \text{ km}^2 \\ 1 \text{ cm}^2 & : 625 \text{ km}^2 \\ 66.056 \text{ cm}^2 & : 41 \phantom{.} 285 \text{ km}^2 \\ \\ \text{Area on map} & = 66.056 \text{ cm}^2 \end{align*}

(a)

\begin{align*} \text{P(Student overestimated)} & = { 200 - 84 \over 200 } \\ & = {29 \over 50} \end{align*}

(b)

\begin{align*} 300 \times {10 \over 100} & = 30 \\ \\ \text{No. of students who estimated } \le 270 \text{ g} & = 64 \\ \\ \text{No. of students who estimated } \le 330 \text{ g} & = 100 \\ \\ \text{No. of students} & = 100 - 64 \\ & = 36 \end{align*}

(a)

\begin{align*} \angle \text{North} AB & = 47^\circ \phantom{000000} [\text{Use protractor to measure}] \\ \\ \text{Bearing of } B \text{ from } A& = 360^\circ - 47^\circ \phantom{0} \text{ (Angles at a point)} \\ & = 313^\circ \end{align*}

(b)

\begin{align*} \text{Length of line in shaded area on paper} & = 5.1 \text{ cm} \\ \\ \text{Actual distance} & = 5.1 \times 10 \\ & = 51 \text{ km} \\ \\ \text{Time} & = { \text{Distance} \over \text{Speed} } \\ & = { 51 \text{ km} \over 35 \text{ km/h} } \\ & = 1 {16 \over 35} \text{ hours} \\ & = 1 \text{ hour } {16 \over 35} \times 60 \text{ minutes} \\ & \approx 1 \text{ hour } 27 \text{ minutes} \end{align*}

(a)

\begin{align*} | \overrightarrow{AB} | & = \sqrt{ x^2 + y^2 } \\ & = \sqrt{ (-8)^2 + (15)^2 } \\ & = 17 \text{ units} \end{align*}

(b)

\begin{align*} \overrightarrow{DC} & = 2 \overrightarrow{AB} \\ & = 2 {-8 \choose 15} \\ & = {-16 \choose 30} \\ \\ \overrightarrow{CD} & = {16 \choose -30} \\ \\ x \text{-coordinate of } D & = 4 + 16 = 20 \\ y \text{-coordinate of } D & = 20 - 30 = -10 \\ \\ \therefore & \phantom{.} D(20, -10) \end{align*}

(c)

\begin{align*} & \text{Trapezium (since } DC \text{ is parallel to } AB) \end{align*}

(a)(i)

\begin{align*} \overrightarrow{OP} & = \overrightarrow{OA} + \overrightarrow{AP} \\ & = 5 \textbf{a} + \underbrace{ \overrightarrow{AP} }_\text{Need to find this} \\ \\ \overrightarrow{AC} & = \overrightarrow{AO} + \overrightarrow{OC} \\ & = -5 \textbf{a} + 5 \textbf{c} \\ \\ \overrightarrow{AP} & = {4 \over 5} \overrightarrow{AC} \\ & = {4 \over 5} (- 5 \textbf{a} + 5 \textbf{c} ) \\ & = - 4 \textbf{a} + 4 \textbf{c} \\ \\ \therefore \overrightarrow{OP} & = 5 \textbf{a} + (-4 \textbf{a} + 4 \textbf{c} ) \\ & = 5 \textbf{a} - 4 \textbf{a} + 4 \textbf{c} \\ & = \textbf{a} + 4 \textbf{c} \end{align*}

(a)(ii)

\begin{align*} \overrightarrow{PQ} & = \overrightarrow{PA} + \overrightarrow{AQ} \\ & = - \overrightarrow{AP} + 4 \textbf{a} + 8 \textbf{c} \\ & = - (-4 \textbf{a} + 4 \textbf{c}) + 4 \textbf{a} + 8 \textbf{c} \\ & = 4 \textbf{a} - 4 \textbf{c} + 4 \textbf{a} + 8 \textbf{c} \\ & = 8 \textbf{a} + 4 \textbf{c} \end{align*}

(b)(i)

\begin{align*} \text{Let } \overrightarrow{PQ} & = k \overrightarrow{CB} \\ 8 \textbf{a} + 4 \textbf{c} & = k ( 10 \textbf{a} + 5 \textbf{c} ) \\ 8 \textbf{a} + 4 \textbf{c} & = 10k \textbf{a} + 5k \textbf{c} \end{align*} \begin{align*} \text{Comparing } & \textbf{a}, &&& \text{Comparing } & \textbf{c} \\ 10k & = 8 &&& 5k & = 4 \\ k & = {8 \over 10} &&& k & = {4 \over 5} \\ k & = {4 \over 5} \end{align*}
\begin{align*} \therefore \overrightarrow{PQ} & = {4 \over 5} \overrightarrow{CB} \text{ and } PQ \text{ is parallel to } CB \end{align*}

(b)(ii)

\begin{align*} CB : PQ & = 5 : 4 \end{align*}

(a)(i)

\begin{align*} a & = {4 (5) - 5(-2) \over 5 + (-2)} \\ & = 10 \end{align*}

(a)(ii)

\begin{align*} {a \over 1} & = {4b - 5c \over b + c} \\ a(b + c) & = 4b - 5c \phantom{000000} [\text{Cross-multiply}] \\ ab + ac & = 4b - 5c \\ ab - 4b & = -5c - ac \\ b(a - 4) & = -5c - ac \\ b & = {-5c - ac \over a - 4} \end{align*}

(b)

\begin{align*} {2x - 3 \over 4} + {x \over 3} & = {3 \over 1} \\ {3(2x - 3) \over 12} + {4x \over 12} & = {36 \over 12} \\ 3(2x - 3) + 4x & = 36 \\ 6x - 9 + 4x & = 36 \\ 10x & = 36 + 9 \\ 10x & = 45 \\ x & = {45 \over 10} \\ x & = 4.5 \end{align*}

(c)

\begin{align*} 4x - 3y & = 18 \\ 8x - 6y & = 36 \phantom{0} \text{--- (1)} \\ \\ 6x + 2y & = 1 \\ 18x + 6y & = 3 \phantom{0} \text{--- (2)} \\ \\ (1) & + (2), \\ 8x - 6y + 18x + 6y & = 36 + 3 \\ 26x & = 39 \\ x & = {39 \over 26} \\ x & = 1.5 \\ \\ \text{Substitute } & x = 1.5 \text{ into (1),} \\ 8(1.5) - 6y & = 36 \\ 12 - 6y & = 36 \\ -6y & = 36 - 12 \\ -6y & = 24 \\ y & = {24 \over -6} \\ y & = -4 \\ \\ \therefore x & = 1.5, y = -4 \end{align*}

(d)

\begin{align*} 9x^2 - 4 & = (3x)^2 - (2)^2 \\ & = (3x + 2)(3x - 2) \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \\ \\ 3x^2 - 10x - 8 & = (3x + 2)(x - 4) \\ \\ {9x^2 - 4 \over 3x^2 - 10x - 8} & = { (3x + 2)(3x - 2) \over (3x + 2)(x - 4)} \\ & = {3x - 2 \over x - 4} \end{align*}

(a)

\begin{align*} \textbf{M} & = 12 \textbf{L} \\ & = 12 \left( \begin{matrix} 8 & 5 \\ 2 & 4 \end{matrix} \right) \\ & = \left( \begin{matrix} 96 & 60 \\ 24 & 48 \end{matrix} \right) \end{align*}

(b)

\begin{align*} \textbf{N} & = \left( \begin{matrix} 40 \\ 65 \end{matrix} \right) \end{align*}

(c)

\begin{align*} \textbf{P} & = \textbf{MN} \\ & = \underset{2 \times 2}{\left( \begin{matrix} 96 & 60 \\ 24 & 48 \end{matrix} \right)} \underset{2 \times 1}{\left( \begin{matrix} 40 \\ 65 \end{matrix} \right)} \\ & = \underset{2 \times 1}{\left( \begin{matrix} 96 \times 40 + 60 \times 65 \\ 24 \times 40 + 48 \times 65 \end{matrix} \right)} \\ & = \left( \begin{matrix} 7740 \\ 4080 \end{matrix} \right) \end{align*}

(d)

\begin{align*} & \text{Total fees collected for 12-week block of weekday sessions is \$ 7740} \\ & \text{Total fees collected for 12-week block of weekend sessions is \$ 4080} \end{align*}

(e)

\begin{align*} \text{Fees for basic} & = 40 \times {95 \over 100} \\ & = \$ 38 \\ \\ \text{Fees for advanced} & = 65 \times {95 \over 100} \\ & = \$ 61.75 \\ \\ \text{Total amount earned} & = (12 + 7) \times 12 \times 38 + (6 + 3) \times 12 \times 61.75 \\ & = \$ 15 \phantom{.} 333 \end{align*}

(a)

\begin{align*} \text{Since polygon is a regular } & \text{polygon, each side is equal} \\ \\ \implies AB & = CD \phantom{0} [S] \\ \implies BC & = DE \phantom{0} [S] \\ \\ \text{Since polygon is a regular } & \text{polygon, each interior angle is equal} \\ \\ \implies \angle ABC & = \angle CDE \phantom{0} [A] \\ \\ \therefore \text{Triangles } ABC & \text{ and } CDE \text{ are congruent } (SAS) \end{align*}

(b)(i)

\begin{align*} \text{Sum of interior angles} & = (n - 2) \times 180 \\ 160 \times n & = 180n - 360 \\ 160n & = 180n - 360 \\ 160n - 180n & = -360 \\ -20n & = -360 \\ n & = {-360 \over -20} \\ n & = 18 \end{align*}

(b)(ii)

\begin{align*} \angle BCA & = {180^\circ - 160^\circ \over 2} \phantom{000000} [\text{Base angle of isosceles triangle}] \\ & = 10^\circ \end{align*}

(b)(iii)

\begin{align*} \text{Since triangles } & ABC \text{ and } CDE \text{ are congruent,} \\ \angle BAC & = \angle DCE = 10^\circ \\ \angle BCA & = \angle DEC = 10^\circ \\ AC & = CE \\ \\ \angle ACE & = 160^\circ - 10^\circ - 10^\circ \\ & = 140^\circ \\ \\ \angle CEA & = {180^\circ - 140^\circ \over 2} \phantom{0} (\text{Base angle of isosceles triangle } CAE, \text{ since } AC = CE) \\ & = 20^\circ \\ \\ \angle DEA & = 20^\circ + 10^\circ \\ & = 30^\circ \end{align*}

(a)

\begin{align*} T_5 & = 6^2 + (14 + 3) \\ & = 53 \end{align*}

(b)

\begin{align*} \text{The sum of an even number and an odd number is always odd} \end{align*}

(c)

\begin{align*} & 5, 8, 11, 14, ... \\ \\ n \text{th term} & = a + (n - 1) d \phantom{000000} [a: \text{ first term, } d: \text{ common difference}] \\ & = 5 + (n - 1)(3) \\ & = 5 + 3n - 3 \\ & = 3n + 2 \\ \\ T_n & = (n + 1)^2 + 3n + 2 \\ & = \underbrace{ (n)^2 + 2(n)(1) + (1)^2 }_{ (a + b)^2 = a^2 + 2ab + b^2 } + 3n + 2 \\ & = n^2 + 2n + 1 + 3n + 2 \\ & = n^2 + 5n + 3 \phantom{0} \text{ (Shown)} \end{align*}

(d)

\begin{align*} T_n & = n^2 + 5n + 3 \\ \\ T_p & = p^2 + 5p + 3 \\ \\ T_{p + 1} & = (p + 1)^2 + 5(p + 1) + 3 \\ & = \underbrace{ (p)^2 + 2(p)(1) + (1)^2 }_{ (a + b)^2 = a^2 + 2ab + b^2 } + 5p + 5 + 3 \\ & = p^2 + 2p + 1 + 5p + 8 \\ & = p^2 + 7p + 9 \\ \\ T_{p + 1} - T_p & = p^2 + 7p + 9 - (p^2 + 5p + 3) \\ & = p^2 + 7p + 9 - p^2 - 5p - 3 \\ & = 2p + 6 \end{align*}

(e)

\begin{align*} T_{p + 1} - T_p & = 2p + 6 \\ \\ \text{Let } 2p + 6 & = 4 \\ 2p & = 4 - 6 \\ 2p & = -2 \\ p & = -1 \\ \\ \text{Since } p > 0, \text{ not possible } & \text{for two consecutive terms to have difference of } 4 \end{align*}

(a)

\begin{align*} y & = {(-3)^3 \over 2} - 5(-3) - 2 \\ & = -0.5 \end{align*}

(b)

(c)

\begin{align*} {x^3 \over 2} - 5x & = 8 \\ {x^3 \over 2} - 5x - 2 & = 8 - 2 \\ \underbrace{ {x^3 \over 2} - 5x - 2 }_\text{Eqn of curve} & = 6 \\ \\ \text{Draw } y = 6 \text{ and from } & \text{graph, line meets curve only once} \end{align*}

(d)

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {4 - (-10) \over -2 - 2} \\ & = -3.5 \end{align*}

(e)(i))

(e)(ii)

$$ x = 2.85 $$

(e)(iii)

\begin{align*} y & = {x^3 \over 2} - 5x - 2 \phantom{0} \text{--- (1)} \\ y & = 4 - 3x \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 4 - 3x & = {x^3 \over 2} - 5x - 2 \\ 0 & = {x^3 \over 2} - 5x + 3x - 2 - 4 \\ 0 & = {x^3 \over 2} - 2x - 6 \\ 2(0) & = 2 \left( {x^3 \over 2} - 2x - 6 \right) \\ 0 & = x^3 - 4x - 12 \\ \\ \therefore A & = -4, B = -12 \end{align*}

(a)

\begin{align*} \angle ADO & = 90^\circ \phantom{0} \text{ (Tangent perpendicular to radius)} \\ \\ \angle ABC & = 90^\circ \phantom{0} \text{ (Right angle in semi-circle)} \\ \\ \therefore \angle ABC & = \angle ADO \phantom{0} [A] \\ \\ \angle BAC & = \angle DAO \phantom{0} \text{ (Common angle}) [A] \\ \\ \text{Triangles } & ABC \text{ and } ADO \text{ are similar } (AA) \end{align*}

(b)

\begin{align*} {\text{Area of triangle } ADO \over \text{Area of triangle } ABC} & = \left( AO \over AC \right)^2 \phantom{000000} \left[ {A_1 \over A_2} = \left(l_1 \over l_2\right)^2 \right] \\ & = \left(1 \over 2\right)^2 \\ & = {1 \over 4} \\ \\ {\text{Area of triangle } ADO \over \text{Quadrilatral } ODBC} & = {1 \over 4 - 1} = {1 \over 3} \\ \\ \text{Triangles } ADO \text{ and } & BDO \text{ are congruent } (RHS) \\ \\ \implies \text{Area of triangle } ADO & = \text{Area of triangle } BDO \\ \\ {\text{Area of triangle } ADO \over \text{Area of triangle } BOC } & = {1 \over 3 - 1} = {1 \over 2} \\ \\ \text{Required ratio} & = 2: 1 \end{align*}

(c)

\begin{align*} \angle AOD & = \angle ACB = 65^\circ \phantom{0} \text{ (Triangles } ABC \text{ and } ADO \text{ are similar}) \\ \\ \cos \angle AOD & = {OD \over OA} \phantom{000000} \left[ {Adj \over Hyp} \right] \\ \cos 65^\circ & = {3 \over OA} \\ OA \cos 65^\circ & = 3 \\ OA & = {3 \over \cos 65^\circ} \\ OA & = 7.0986 \\ \\ \text{Shaded area} & = \text{Area of large circle} - \text{Area of small circle} \\ & = \pi (7.0986)^2 - \pi (3)^2 \\ & = 130.03 \\ & \approx 130 \text{ cm}^2 \end{align*}

(a)

\begin{align*} \text{Total no. of mugs} & = 140 \times {3 + 7+ 5 \over 5 - 3} \\ & = 1050 \end{align*}

(b)(i)

\begin{align*} x \text{ s} & \rightarrow 1 \text{ mug} \\ 1 \text{ s} & \rightarrow {1 \over x} \text{ mug} \\ 3600 \text{ s} & \rightarrow {1 \over x} \times {3600 \over 1} = {3600 \over x} \text{ mugs} \end{align*}

(b)(ii)

\begin{align*} (x - 50) \text{ s} & \rightarrow 1 \text{ mug} \\ 1 \text{ s} & \rightarrow {1 \over x - 50} \text{ mug} \\ 3600 \text{ s} & \rightarrow {1 \over x - 50} \times {3600 \over 1} = {3600 \over x- 50} \text{ mugs} \end{align*}

(b)(iii)

\begin{align*} \text{Total no. of mugs painted in 1 hour} & = {68 \over 4} \\ & = 17 \\ \\ {3600 \over x} + {3600 \over x - 50} & = 17 \\ {3600(x - 50) \over x(x - 50)} + {3600x \over x(x - 50)} & = 17 \\ {3600(x - 50) + 3600x \over x(x - 50)} & = {17 \over 1} \\ 3600(x - 50) + 3600x & = 17x(x - 50) \phantom{000000} [\text{Cross-multiply}] \\ 3600x - 180 \phantom{.} 000 + 3600x & = 17x^2 - 850x \\ 7200x - 180 \phantom{.} 000 & = 17x^2 - 850x \\ 0 & = 17x^2 - 850x - 7200x + 180 \phantom{.} 000 \\ 0 & = 17x^2 - 8050x + 180 \phantom{.} 000 \text{ (Shown)} \end{align*}

(b)(iv)

\begin{align*} 0 & = 17x^2 - 8050x + 180 \phantom{.} 000 \\ \\ x & = { - b \pm \sqrt{ b^2 - 4ac } \over 2a} \\ & = { - (-8050) \pm \sqrt{ (-8050)^2 - 4(17)(180 \phantom{.} 000) } \over 2(17) } \\ & = { 8050 \pm \sqrt{ 52 \phantom{.} 562 \phantom{.} 500 } \over 34 } \\ & = 450 \text{ or } 23 {9 \over 17} \end{align*}

(b)(v)

\begin{align*} [ \text{Use } x = 450 \text{ since for } x = 23{9 \over 17}&, \text{ time taken by Maryam is negative}] \\ \\ \text{No. of mugs painted by Maryam} & = {3600 \over x - 50} \\ & = {3600 \over 450 - 50} \\ & = 9 \end{align*}

(a)

\begin{align*} \text{By Co}& \text{sine rule,} \\ CB^2 & = CA^2 + AB^2 - 2 (CA) (AB) \cos \angle BAC \\ 6^2 & = 3^2 + 7^2 - 2(3)(7) \cos \angle BAC \\ 36 & = 9 + 49 - 42 \cos \angle BAC \\ 42 \cos \angle BAC & = 9 + 49 - 36 \\ 42 \cos \angle BAC & = 22 \\ \cos \angle BAC & = {22 \over 42} \\ \angle BAC & = \cos^{-1} \left(22 \over 42\right) \\ & = 58.411 \\ & \approx 58.4^\circ \end{align*}

(b)

\begin{align*} \text{Area of triangle } CAB & = {1 \over 2} ab \sin C \phantom{000000} [\text{Formula provided}] \\ & = {1 \over 2} (3)(7) \sin 58.411^\circ \\ & = 8.9441 \text{ cm}^2 \\ \\ \text{Surface area of prism} & = (6)(10) + (7)(10) + (3)(10) + 2(8.9441) \\ & = 177.888 \\ & \approx 178 \text{ cm}^2 \end{align*}

(c)

\begin{align*} \text{Area of triangle } CAB & = {1 \over 2} \times AB \times h \\ 8.9441 & = {1 \over 2} \times 7 \times h \\ 8.9441 & = 3.5 h \\ {8.9441 \over 3.5} & = h \\ 2.5554 & = h \\ \\ \text{Vertical distance} & \approx 2.56 \text{ cm} \end{align*}

(d)

\begin{align*} \text{By Py} & \text{thagoras theorem,} \\ AH^2 & = AC^2 - CH^2 \\ & = 3^2 - 2.5554^2 \\ AH & = \sqrt{ 3^2 - 2.5554^2 } \\ & = 1.5716 \\ \\ GF & = 1.5716 \\ \\ \text{By Py} & \text{thagoras theorem,} \\ AF^2 & = AG^2 + GF^2 \\ & = 10^2 + 1.5716^2 \\ AF & = \sqrt{ 10^2 + 1.5716^2 } \\ & = 10.123 \\ \\ \tan \angle EAF & = { EF \over AF} \phantom{000000} \left[ {Opp \over Adj} \right] \\ & = { 2.5554 \over 10.123 } \\ \angle EAF & = \tan^{-1} \left( 2.5554 \over 10.123 \right) \\ & = 14.167 \\ & \approx 14.2^\circ \end{align*}

(a)(i)

\begin{align*} \text{Percentage of students} & = { 6 + 4 + 1 \over 20} \times 100 \\ & = 55 \% \end{align*}

(a)(ii)

\begin{align*} \text{Median position} & = {20 + 1 \over 2} = 10.5 \\ \\ \text{Median time} & = { 31 + 32 \over 2} \phantom{000000} [\text{Average of 10th and 11th student}] \\ & = 31.5 \text{ minutes} \end{align*}

(a)(iii)

\begin{align*} & \text{The student who took 69 minutes will cause the mean to increase and } \\ & \text{thus the mean is not representative of the entire group.} \end{align*}

(a)(iv)

\begin{align*} \text{Standard deviation} & = \sqrt{ {\sum fx^2 \over \sum f} - \left( \sum fx \over \sum f \right)^2 } \\ & = \sqrt{ {26239 \over 20 } - \left( 691 \over 20 \right)^2 } \\ & = 10.874 \\ & \approx 10.9 \text{ minutes} \end{align*}

(a)(v)

\begin{align*} & \text{The timings of students in the second ground is more consistent as the standard} \\ & \text{deviation is lower.} \end{align*}

(b)(i)

(b)(ii)(a)

\begin{align*} \text{Case 1: P(W, W)} & = {9 \over 16} \times {8 \over 15} = {3 \over 10} \\ \\ \text{Case 2: P(R, R)} & = {5 \over 16} \times {4 \over 15} = {1 \over 12} \\ \\ \text{Case 3: P(B, B)} & = {2 \over 16} \times {1 \over 15} = {1 \over 120} \\ \\ \text{Required probability} & = {3 \over 10} + {1 \over 12} + {1 \over 120} \\ & = {47 \over 120} \end{align*}

(b)(ii)(b)

\begin{align*} \text{Case 1: P(1st ball white)} & = {9 \over 16} \\ \\ \text{Case 2: P(R, W)} & = {5 \over 16} \times {9 \over 15} = {3 \over 16} \\ \\ \text{Case 3: P(B, W)} & = {2 \over 16} \times {9 \over 15} = {3 \over 40} \\ \\ \text{Required probability} & = {9 \over 16} + {3 \over 16} + {3 \over 40} \\ & = {33 \over 40} \end{align*}

(a)

\begin{align*} \text{Cost of posting 1 copy} & = \$ 0.37 \phantom{000000} [\text{More than 20 g, less than 40 g}] \end{align*}

(b)

\begin{align*} \text{Pages to be printed} & = (4 \times 2) \times 6 \times 150 \\ & = 7200 \\ \\ \text{Toner required} & = {7200 \over 1300} \\ & = 5 {7 \over 13} \\ & \approx 6 \end{align*}

(c)

\begin{align*} \text{A4 paper required} & = 4 \times 6 \times 150 \\ & = 3600 \\ \\ \text{Packets required} & = {3600 \over 500} \\ & = 7.2 \\ & \approx 8 \\ \\ \text{Cost of paper} & = 8 \times \$3.90 \\ & = \$ 31.20 \\ \\ \text{Cost of paper (w. GST)} & = 31.20 \times {107 \over 100} \\ & = \$ 33.384 \\ \\ \text{Cost of toners} & = 6 \times \$ 147 \phantom{000000} [6 \text{ toners from (b)}]\\ & = \$ 882 \\ \\ \text{Cost of toners (w. GST)} & = 882 \times {107 \over 100} \\ & = \$ 943.74 \\ \\ \text{Envelopes required} & = 6 \times 150 \\ & = 900 \\ \\ \text{Cost of envelopes} & = 2 \times \$ 34 \phantom{0000000} [\text{Cheaper than 1 pack of 500 and 40 packs of 10}] \\ & = \$68 \\ \\ \text{Cost of envelopes (w. GST)} & = 68 \times {107 \over 100} \\ & = \$ 72.76 \\ \\ \text{Cost of postage (GST included)} & = 150 \times 6 \times 0.37 \\ & = \$ 333 \\ \\ \text{Total cost} & = \$33.384 + \$943.74 + \$ 72.76 + \$333 \\ & = \$ 1 \phantom{.} 382.884 \\ \\ \text{Cost per subscriber} & = {1 \phantom{.} 382.884 \over 150} \\ & = 9.2192 \\ & \approx \$ 9.22 \\ \\ \text{He should charge } \$ 9.22 & \text{ plus any amount (per subscriber) he wishes to profit from} \end{align*}