\begin{align*} \left({3 \over 4}\right)^2, {3 \over 5}, 0.65, {3 \over 4} \end{align*}

\begin{align*} \left( x^8 \over y^6 \right)^{-{3 \over 2}} & = \left(y^6 \over x^8\right)^{3 \over 2} \phantom{000000} \left[ \left(a \over b\right)^{-n} = \left(b \over a\right)^n \right] \\ & = { (y^6)^{3 \over 2} \over (x^8)^{3 \over 2} }\\ & = { y^9 \over x^{12} } \phantom{0000000000} [ (a^m)^n = a^{mn}] \end{align*}

(a)

\begin{align*} kx^2 + (k + 1)x - 4 & = 0 \\ \\ \text{When } & x = -2, \\ k(-2)^2 + (k + 1)(-2) - 4 & = 0 \\ k(4) - 2(k + 1) - 4 & = 0 \\ 4k - 2k - 2 - 4 & = 0 \\ 4k - 2k & = 2 + 4 \\ 2k & = 6 \\ k & = {6 \over 2} \\ k & = 3 \end{align*}

(b)

\begin{align*} kx^2 + (k + 1)x - 4 & = 0 \\ 3x^2 + (3 + 1)x - 4 & = 0 \\ 3x^2 + 4x - 4 & = 0 \\ (3x - 2)(x + 2) & = 0 \end{align*} \begin{align*} 3x - 2 & = 0 && \text{ or } & x + 2 & = 0 \\ 3x & = 2 &&& x & = -2 \\ x & = {2 \over 3} \end{align*}
$$ \text{Other value of } x = {2 \over 3} $$

\begin{align*} \angle DBC & = 180^\circ - 129^\circ \phantom{0} \text{ (Adjacent angles on a straight line)} \\ & = 51^\circ \\ \\ \angle CDB & = 180^\circ - 38^\circ - 51^\circ \phantom{0} \text{ (Angle sum of triangle)} \\ & = 91^\circ \\ \\ \text{Since } \angle CDB & \ne 90^\circ, BC \text{ is not a diameter of the circle} \end{align*}

\begin{align*} (3n - 1)^2 + 2 & = \underbrace{ (3n)^2 - 2(3n)(1) + (1)^2 }_{ (a - b)^2 = a^2 - 2ab + b^2 } + 2 \\ & = 9n^2 - 6n + 1 + 2 \\ & = 9n^2 - 6n + 3 \\ & = 3(3n^2 - 2n + 1) \\ \\ \therefore \text{Expression } & \text{is a multiple of } 3 \text{ for all integer values of } n \end{align*}

(a)

\begin{align*} \text{Males married later in 2014 compared with 2004.} \end{align*}

(b)

\begin{align*} \text{Yes because from age group 30-49, there are more marriages per 1000 males} \end{align*}

(a)

$$ A \cup B' $$

(b)

$$ (A \cap B') \cup (B \cap A') $$

\begin{align*} \text{Total instalments} & = 113 \times 12 \\ & = \$ 1356 \\ \\ \text{Deposit} & = 1651 - 1356 \\ & = \$ 295 \\ \\ \text{Required percentage} & = {295 \over 1475} \times 100 \\ & = 20 \% \end{align*}

(a) Join points O and B with a straight line - it's the radius of the circle (thus OA = OB = OC)

\begin{align*} \angle OBC & = 41^\circ \phantom{0} (\text{Isosceles triangle } OBC) \\ \\ \angle OBA & = 23^\circ \phantom{0} (\text{Isosceles triangle } OBA) \\ \\ \angle ABC & = 41^\circ + 23^\circ \\ & = 64^\circ \end{align*}

(b)

\begin{align*} \angle AOC & = 2 \times \angle ABC \phantom{0} \text{ (Angle at centre} = 2 \times \text{Angle at circumference}) \\ & = 2 \times 64^\circ \\ & = 128^\circ \end{align*}

(c)

\begin{align*} \angle CDA & = 180^\circ - \angle ABC \phantom{0} \text{ (Angles in opposite segment)} \\ & = 180^\circ - 64^\circ \\ & = 116^\circ \\ \\ \angle OAD & = 180^\circ - 116^\circ \phantom{0} \text{ (Interior angles, } CD \phantom{.} // \phantom{.} OA) \\ & = 64^\circ \end{align*}

\begin{align*} \text{Time taken for flight} & = { \text{Distance} \over \text{Speed} } \\ & = { 5408 \text{ km} \over 721 \text{ km/h} } \\ & = 7.500 \text{ hours} \\ & = 7 \text{ hours 30 minutes} \\ \\ \text{Time of arrival (Adelaide time)} & = 09 \phantom{.} 25 + 07 \phantom{.} 30 \\ & = 16 \phantom{.} 55 \\ \\ \text{Singapore is behind by } & 1.5 \text{ hours (1 hour 30 minutes)} \end{align*}

\begin{align*} { V_1 \over V_2 } & = \left(l_1 \over l_2\right)^3 \\ & = \left(1 \over 20\right)^3 \\ & = {1 \over 8000} \\ \\ \text{Volume of model} & = 184 \div 8000 \\ & = 0.023 \text{ m}^3 \end{align*}

(a)

\begin{align*} & 4 \text{ workers take } 17.5 \text{ hours} \\ & 1 \text{ worker take } 17.5 \times 4 = 70 \text{ hours} \\ & 5 \text{ workers take } {70 \over 5} = 14 \text{ hours} \end{align*}

(b)

\begin{align*} F & = {k \over d^2} \\ \\ \text{Let initial value of } d & = a \\ \\ F & = {k \over a^2} \\ \\ \text{New value of } d & = a \times {120 \over 100} \\ & = 1.2a \\ \\ F & = {k \over (1.2a)^2} \\ & = {k \over 1.44a^2} \\ \\ \text{Percentage change} & = { \text{Final} - \text{Initial} \over \text{Initial} } \times 100 \\ & = { {k \over 1.44a^2} - {k \over a^2} \over {k \over a^2} } \times 100 \\ & = { {k \over 1.44a^2} - {1.44k \over 1.44a^2} \over {k \over a^2}} \times 100 \\ & = { {-0.44 k \over 1.44a^2} \over {k \over a^2} } \times 100 \\ & = { {-0.44 \over 1.44} \left(k \over a^2\right) \over {k \over a^2} } \times 100 \\ & = -{0.44 \over 1.44} \times 100 \\ & = -30{5 \over 9} \% \\ \\ \text{Percentage reduction} & = 30 {5 \over 9} \% \end{align*}

(a)

\begin{align*} 8p^2 q - 6p q^3 & = 2pq (4p - 3q^2) \end{align*}

(b)

\begin{align*} 6x^2 y - 2xy + 3x - 1 & = 2xy (3x - 1) + (3x - 1) \phantom{000000} [\text{Grouping}] \\ & = (3x - 1)(2xy + 1) \end{align*}

\begin{align*} \text{Sum of interior angles of hexagon} & = (n - 2) \times 180 \\ & = (6 - 2) \times 180 \\ & = 720^\circ \\ \\ \text{Each interior angle in hexagon} & = {720 \over 6} \\ & = 120^\circ \\ \\ \text{Interior angle of polygon } B & = 360^\circ - 90^\circ - 120^\circ \phantom{0} \text{ (Angles at a point)} \\ & = 150^\circ \\ \\ \text{Let no. of sides of polygon } B & = n \\ \\ \text{Sum of interior angles} & = (n - 2) \times 180 \\ 150 \times n & = 180n - 360 \\ 150n & = 180n - 360 \\ 150n - 180n & = -360 \\ -30n & = -360 \\ n & = {-360 \over -30} \\ n & = 12 \end{align*}

\begin{align*} \text{Let no. of males} & = x \\ \\ \text{Total height of males} & = 183 \times x \\ & = 183x \\ \\ \text{No. of females} & = x + 4 \\ \\ \text{Total height of females} & = 162 \times (x + 4) \\ & = 162x + 648 \\ \\ \text{No. of students} & = x + x + 4 \\ & = 2x + 4 \\ \\ \text{Total height of students} & = 171 \times (2x + 4) \\ & = 342x + 684 \\ \\ \therefore 342x + 684 & = 183x + 162x + 648 \\ 342x - 183x - 162x & = 648 - 684 \\ -3x & = -36 \\ x & = {-36 \over -3} \\ x & = 12 \\ \\ \text{Total number of students} & = 2x + 4 \\ & = 2(12) + 4 \\ & = 28 \end{align*}

\begin{align*} \text{Arc length} & = r \theta \\ \\ \text{Perimeter of minor sector} & = r \theta + 2r \\ \\ \text{Perimeter of major sector} & = r(2\pi - \theta) + 2r \phantom{000000} [ 360^\circ = 2 \pi \text{ radians}] \\ & = 2 \pi r - r \theta + 2r \\ \\ \text{Perimeter of major sector} & = 2 \times \text{Perimeter of minor sector} \\ 2\pi r - r \theta + 2r & = 2 (r \theta + 2r) \\ 2 \pi r - r \theta + 2r & = 2 r \theta + 4r \\ -r \theta - 2r \theta & = 4r - 2r - 2\pi r \\ -3r \theta & = 2r - 2\pi r \\ \theta & = {2r - 2\pi r \over -3r} \\ & = {r (2 - 2\pi) \over -3r} \\ & = {2 - 2 \pi \over -3} \\ & = 1.4277 \\ & \approx 1.43 \end{align*}

\begin{align*} {y \over 1} & = {x^2 + 3 \over x^2 - a} \\ y(x^2 - a) & = x^2 + 3 \phantom{000000} [\text{Cross-multiply}] \\ x^2 y - ay & = x^2 + 3 \\ x^2 y - x^2 & = 3 + ay \\ x^2 (y - 1) & = 3 + ay \\ x^2 & = {3 + ay \over y - 1} \\ x & = \pm \sqrt{3 + ay \over y - 1} \end{align*}

(a)

\begin{align*} | \overrightarrow{PQ} | & = \sqrt{x^2 + y^2} \\ {5 \sqrt{10} \over 2} & = \sqrt{ a^2 + (3a)^2 } \\ {5 \sqrt{10} \over 2} & = \sqrt{ a^2 + 9a^2 } \\ {5 \sqrt{10} \over 2} & = \sqrt{10a^2} \\ \left( 5 \sqrt{10} \over 2 \right)^2 & = 10a^2 \\ 62.5 & = 10a^2 \\ {62.5 \over 10} & = a^2 \\ 6.25 & = a^2 \\ \pm \sqrt{6.25} & = a \\ \pm 2.5 & = a \\ \\ a & = 2.5 \text{ or } - 2.5 \end{align*}

(b)(i)

\begin{align*} \overrightarrow{AB} & = {43 - (-5) \choose 38 - 1} \\ & = {48 \choose 37} \end{align*}

(b)(ii)

\begin{align*} \overrightarrow{AC} & = {11 - (-5) \choose 13 - 1} \\ & = {16 \choose 12} \\ \\ \text{Let } \overrightarrow{AB} & = k \overrightarrow{AC} \\ {48 \choose 37} & = k {16 \choose 12} \\ {48 \choose 37} & = {16k \choose 12k} \end{align*} \begin{align*} 16k & = 48 &&& 12k & = 37 \\ k & = {48 \over 16} &&& k & = {37 \over 12} \\ k & = 3 &&& k & = 3{1 \over 12} \end{align*}
\begin{align*} \therefore AC \text{ is not parallel to } AB \text{ and } C \text{ does not lie on line} \end{align*}

(a)

\begin{align*} \text{Median position} & = {30 +1 \over 2} \\ & = 15.5 \\ \\ \text{Median mass} & = {127 + 129 \over 2} \phantom{000000} [\text{15th and 16th apple}] \\ & = 128 \text{ grams} \end{align*}

(b)

\begin{align*} \text{Let } 14x & = y \\ \\ y - 103 & = 44 \\ y & = 44 + 103 \\ y & = 147 \\ \\ \therefore x & = 7 \end{align*}

(c)

\begin{align*} \text{Median position for bananas} & = {25 + 1 \over 2} \\ & = 13 \\ \\ \text{Median mass of bananas} & = 125 \text{ g} \\ \\ \text{Range for apples} & = 159 - 92 \\ & = 67 \text{ g} \end{align*}
\begin{align*} & \text{1. The mass of apples is higher as a whole as the median mass is higher} \\ \\ & \text{2. The mass of bananas is more consistent as the range is lower} \end{align*}

(a)

\begin{align*} 2 & | \underline{126} \\ 3 & | \underline{63} \\ 3 & | \underline{21} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ \therefore 126 & = 2 \times 3^2 \times 7 \end{align*}

(b)

\begin{align*} 126k & = 2 \times 3^2 \times 7 \times k \\ & = 2 \times 3^2 \times 7 \times (2 \times 7) \\ & = 2^2 \times 3^2 \times 7^2 \\ \\ \therefore k & = 2 \times 7 = 14 \end{align*}

(c)

\begin{align*} 126 & = 2 \times 3^2 \times 7 \\ \\ \text{HCF} = 21 & = 3 \times 7 \phantom{000000} [x \text{ needs to have factors 3 and 7}] \\ \\ x & = 3 \times 7 \times 11 \\ & = 231 \end{align*}

\begin{align*} y & = ax^2 + bx + 3 \\ \\ \text{Using } & (-2, 11), \\ 11 & = a(-2)^2 + b(-2) + 3 \\ 11 & = a(4) - 2b + 3 \\ 11 & = 4a - 2b + 3 \\ 11 - 3 & = 4a - 2b \\ 8 & = 4a - 2b \\ 2b & = 4a - 8 \\ b & = 2a - 4 \phantom{0} \text{--- (1)} \\ \\ \text{Using } & (4, -1), \\ -1 & = a(4)^2 + b(4) + 3 \\ -1 & = a(16) + 4b + 3 \\ -1 -3 & = 16a + 4b \\ -4 & = 16a + 4b \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ -4 & = 16a + 4(2a - 4) \\ -4 & = 16a + 8a - 16 \\ -4 + 16 & = 24a \\ 12 & = 24a \\ {12 \over 24} & = a \\ 0.5 & = a \\ \\ \text{Substitute } & a = 0.5 \text{ into (1),} \\ b & = 2(0.5) - 4 \\ b & = -3 \\ \\ \therefore a & = 0.5, b = -3 \end{align*}

(a)

$$ x = 0.5 $$

(b)

\begin{align*} x - x^2 & = -4 \\ {1 \over 2}(x - x^2) & = {1 \over 2}(-4) \\ {1 \over 2}(x - x^2) & = -2 \\ \\ \text{From graph, } x & = -1.55 \text{ or } 2.55 \end{align*}

(c)(i)

(c)(ii)

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {-4 - 5 \over 4 - (-2)} \\ & = -1.5 \\ \\ y & = mx + c \\ y & = -1.5x + c \\ \\ \text{Using } & (4, - 4), \\ -4 & = -1.5(4) + c \\ -4 & = -6 + c \\ -4 + 6 & = c \\ 2 & = c \\ \\ \text{Eqn: } & y = -1.5x + 2 \end{align*}

(a)

\begin{align*} {\sin \angle DMS \over 4140} & = {\sin 29^\circ \over 2392} \phantom{000000.} [\text{Sine rule}] \\ 2392 \sin \angle DMS & = 4140 \sin 29^\circ \phantom{000} [\text{Cross-multiply}] \\ \sin \angle DMS & = {4140 \sin 29^\circ \over 2392} \\ \angle DMS & = \sin^{-1} \left( {4140 \sin 29^\circ \over 2392} \right) \\ & = 57.044^\circ \\ & \approx 57.0^\circ \end{align*}

(b)

\begin{align*} \angle DSM & = 180^\circ - 29^\circ - 57.044^\circ \phantom{0} \text{ (Angle sum of triangle)} \\ & = 93.956^\circ \\ \\ \angle \text{North}MD & = 360^\circ - 290^\circ \phantom{0} \text{ (Angles at a point)} \\ & = 70^\circ \\ \\ \angle NSM & = 180^\circ - 57.044^\circ - 70^\circ \phantom{0} \text{ (Interior angles)} \\ & = 52.956^\circ \\ \\ \angle NSD & = 93.956^\circ - 52.956^\circ \\ & = 41^\circ \\ \\ \text{Bearing of } D \text{ from } S & = 360^\circ - 41^\circ \phantom{0} \text{ (Angles at a point)} \\ & = 319^\circ \end{align*}

(a)(i)

\begin{align*} {5t^2 \over v} \div {25t \over v^3} & = {5t^2 \over v} \times {v^3 \over 25t} \\ & = {5 t^2 v^3 \over 25 tv} \\ & = {tv^2 \over 5} \end{align*}

(a)(ii)

\begin{align*} {4 \over 3 - 2y} - {5 \over y + 3} & = {4(y + 3) \over (3 - 2y)(y + 3)} - {5(3 - 2y) \over (3 - 2y)(y + 3)} \\ & = {4(y + 3) - 5(3 - 2y) \over (3 - 2y)(y + 3)} \\ & = {4y + 12 - 15 + 10y \over (3 - 2y)(y + 3)} \\ & = {14y - 3 \over (3 - 2y)(y + 3)} \end{align*}

(b)

\begin{align*} \require{cancel} 16x^2 - 9 & = (4x)^2 - (3)^2 \\ & = (4x + 3)(4x - 3) \phantom{000000} [ a^2 - b^2 = (a + b)(a - b) ] \\ \\ 4x^2 - 9x - 9 & = (x - 3)(4x + 3) \\ \\ \therefore {16x^2 - 9 \over 4x^2 - 9x - 9} & = { \cancel{(4x + 3)} (4x - 3) \over (x - 3) \cancel{(4x + 3)} } \\ & = {4x - 3 \over x - 3} \end{align*}

(c)

\begin{align*} {20 \over x + 1} & = {2x + 5 \over 1} \\ 20 & = (x + 1)(2x + 5) \phantom{0000000} [\text{Cross-multiply}] \\ 20 & = 2x^2 + 5x + 2x + 5 \\ 0 & = 2x^2 + 7x + 5 - 20 \\ 0 & = 2x^2 + 7x - 15 \\ 0 & = (2x - 3)(x + 5) \end{align*} \begin{align*} 2x - 3 & = 0 && \text{ or } & x + 5 & = 0 \\ 2x & = 3 &&& x & = -5 \\ x & = {3 \over 2} \end{align*}

(a)

\begin{align*} \text{Percentage change} & = { \text{Final} - \text{Initial} \over \text{Initial} } \times 100 \\ & = {(3900 \times 12) - 44 \phantom{.} 000 \over 44 \phantom{.} 000} \times 100 \\ & = 6{4 \over 11} \% \end{align*}

(b)

\begin{align*} \text{Total amount} & = P \left(1 + {r \over 100} \right)^n \phantom{000000} [\text{Formula provided}] \\ & = (6500) \left(1 + {2.35 \over 100}\right)^6 \\ & = \$ 7 \phantom{.} 472.06 \\ \\ \text{Interest} & = 7 \phantom{.} 472.06 - 6500 \\ & = \$ 972.06 \end{align*}

(c)(i)

\begin{align*} 1 \text{ pound} & = \$ 1.97 \\ \\ 0.66 \text{ euros} & = \$ 1 \\ 1 \text{ euros} & = \$ {1 \over 0.66} = \$ 1.5151 \\ \\ \therefore \text{London Hotel costs more } & \text{since it costs more SGD for 1 pound} \end{align*}

(c)(ii)

\begin{align*} \text{Total cost for London Hotel} & = (145 \times 3) \times 1.97 \\ & = \$ 856.95 \\ \\ \text{Total cost for Paris Hotel} & = (145 \times 4) \times 1.5151 \\ & = \$ 878.758 \\ \\ \text{Total cost (including fees)} & = (856.95 + 878.758) \times {100 + 1.8 \over 100} \\ & = 1 \phantom{.} 766.95 \\ & \approx \$ 1767 \end{align*}

(a)

\begin{align*} \textbf{P} & = 7 \textbf{M} \\ & = 7 \left( \begin{matrix} 70 & 40 & 30 \\ 50 & 30 & 30 \end{matrix} \right) \\ & = \left( \begin{matrix} 490 & 280 & 210 \\ 350 & 210 & 210 \end{matrix} \right) \end{align*}

(b)

\begin{align*} \textbf{N} & = \left( \begin{matrix} 0.75 \\ 1.50 \\ 2.25 \end{matrix} \right) \end{align*}

(c)

\begin{align*} \textbf{T} & = \textbf{PN} \\ & = \underset{2 \times 3}{ \left( \begin{matrix} 490 & 280 & 210 \\ 350 & 210 & 210 \end{matrix} \right) } \underset{3 \times 1}{ \left( \begin{matrix} 0.75 \\ 1.50 \\ 2.25 \end{matrix} \right) } \\ & = \underset{2 \times 1}{ \left( \begin{matrix} 490 \times 0.75 + 280 \times 1.5 + 210 \times 2.25 \\ 350 \times 0.75 + 210 \times 1.5 + 210 \times 2.25 \end{matrix} \right) } \\ & = \left( \begin{matrix} 1260 \\ 1050 \end{matrix} \right) \end{align*}

(d)

\begin{align*} & \text{Cost of making apple pies for 7 days is \$1260 while that of cherry pies is \$1050.} \end{align*}

(e)

\begin{align*} \text{No. of pies sold in the week} & = \left( \begin{matrix} 490 \times {6 \over 7} & 280 \times {6 \over 7} & 210 \times {6 \over 7} \\ 350 \times {4 \over 5} & 210 \times {4 \over 5} & 210 \times {4 \over 5} \end{matrix} \right) \\ & = \left( \begin{matrix} 420 & 240 & 180 \\ 280 & 168 & 168 \end{matrix} \right) \\ \\ \text{Selling price for each size} & = \left( \begin{matrix} 0.75 \times {140 \over 100} \\ 1.50 \times {140 \over 100} \\ 2.25 \times {140 \over 100} \end{matrix} \right) \\ & = \left( \begin{matrix} 1.05 \\ 2.1 \\ 3.15 \end{matrix} \right) \\ \\ \left( \begin{matrix} 420 & 240 & 180 \\ 280 & 168 & 168 \end{matrix} \right) \left( \begin{matrix} 1.05 \\ 2.1 \\ 3.15 \end{matrix} \right) & = \left( \begin{matrix} 420 \times 1.05 + 240 \times 2.1 + 180 \times 3.15 \\ 280 \times 1.05 + 168 \times 2.1 + 168 \times 3.15 \end{matrix} \right) \\ & = \left( \begin{matrix} 1512 \\ 1176 \end{matrix} \right) \\ \\ \text{Total profit} & = (1512 - 1260) + (1176 - 1050) \\ & = \$ 378 \end{align*}

(a)(i)

\begin{align*} T_n & = a + (n - 1)(d) \phantom{000000} [a \text{: first term, } d \text{: common difference}] \\ & = 11 + (n - 1)(6) \\ & = 11 + 6n - 6 \\ & = 6n + 5 \end{align*}

(a)(ii)

\begin{align*} T_n & = 6n + 5 \\ \\ {T_n \over 3} & = {6n + 5 \over 3} \\ & = {6n \over 3} + {5 \over 3} \\ & = 2n + {5 \over 3} \\ \\ \text{Since } T_n \text{ is not } & \text{exactly divisble by } 3, \text{ no term is a multiple of 3} \end{align*}

(b)(i)

\begin{align*} T_5 & = {4(5) - 1 \over 205 - 5(5)} \\ & = {19 \over 180} \end{align*}

(b)(ii)

\begin{align*} T_k & = {4k - 1 \over 205 - 5k} \\ {7 \over 32} & = {4k - 1 \over 205 - 5k} \\ 7(205 - 5k) & = 32(4k - 1) \phantom{000000} [\text{Cross-multiply}] \\ 1435 - 35k & = 128k - 32 \\ -35k - 128k & = -32 - 1435 \\ -163k & = -1467 \\ k & = {-1467 \over -163} \\ k & = 9 \end{align*}

(b)(iii)

\begin{align*} T_n & > 1 \\ {4n - 1 \over 205 - 5n} & > 1 \\ {4n - 1 \over 205 - 5n} & > {205 - 5n \over 205 - 5n} \\ 4n - 1 & > 205 - 5n \\ 4n + 5n & > 205 + 1 \\ 9n & > 206 \\ n & > {206 \over 9} \\ n & > 22{8 \over 9} \\ \\ \text{Least value of } n & = 23 \end{align*}

(a)

\begin{align*} \overrightarrow{BC} & = {4 \choose -3} \\ \overrightarrow{CB} & = {-4 \choose 3} \\ \\ x \text{-coordinate of } B & = 3 +(-4) = -1 \\ y \text{-coordinate of } B & = -1 + 3 = 2 \\ \\ \overrightarrow{OB} & = {-1 \choose 2} \end{align*}

(b)

\begin{align*} \overrightarrow{BA} & = {-6 \choose -2} \\ \\ x \text{-coordinate of } A & = -1 + (-6) = -7 \\ y \text{-coordinate of } A & = 2 + (-2) = 0 \\ \\ \therefore & \phantom{.} A(-7, 0) \end{align*}

\begin{align*} \overrightarrow{CD} & = \overrightarrow{BA} = {-6 \choose -2} \\ \\ x \text{-coordinate of } D & = 3 + (-6) = -3 \\ y \text{-coordinate of } D & = -1 + (-2) = -3 \\ \\ \therefore & \phantom{.} D(-3, -3) \end{align*}

(c)

\begin{align*} \tan \angle BAE & = {Opp \over Adj} \\ & = {BE \over AE} \\ & = {2 \over 6} \\ \angle BAE & = \tan^{-1} \left(2 \over 6\right) \\ & = 18.434^\circ \\ \\ \tan DAF & = {DF \over AF} \\ & = {3 \over 4} \\ \angle DAF & = \tan^{-1} \left(3 \over 4\right) \\ & = 36.87^\circ \\ \\ \angle BAD & = 18.434 + 36.87 \\ & = 55.304^\circ \\ & \approx 55.3^\circ \end{align*}

(d)

\begin{align*} BD & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ & = \sqrt{ [-3 - (-1)]^2 + (-3 - 2)^2 } \\ & = \sqrt{ 29 } \\ & \approx 5.39 \text{ units} \end{align*}

(a)

\begin{align*} {3500 \over x} \end{align*}

(b)

\begin{align*} {3500 \over x - 10} \end{align*}

(c)

\begin{align*} {3500 \over x - 10} - {3500 \over x} & = 21 \\ {3500x \over x(x - 10)} - {3500(x - 10) \over x(x - 10)} & = 21 \\ {3500x - 3500(x - 10) \over x(x - 10)} & = 21 \\ {3500x - 3500x + 35 \phantom{.} 000 \over x(x - 10)} & = 21 \\ {35 \phantom{.} 000 \over x(x - 10)} & = {21 \over 1} \\ 35 \phantom{.} 000 & = 21x(x - 10) \phantom{000000} [\text{Cross-multiply}] \\ 35 \phantom{.} 000 & = 21x^2 - 210x \\ 0 & = 21x^2 - 210x - 35 \phantom{.} 000 \\ 0 & = 3x^2 - 30x - 5000 \phantom{0} \text{ (Shown)} \end{align*}

(d)

\begin{align*} 0 & = 3x^2 - 30x - 5000 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-30) \pm \sqrt{(-30)^2 - 4(3)(-5000)} \over 2(3)} \\ & = {30 \pm \sqrt{60 \phantom{.} 900} \over 6} \\ & = 46.129 \text{ or } -36.129 \\ & \approx 46.13 \text{ or } -36.13 \text{ (to 2 d.p.)} \end{align*}

(e)

\begin{align*} \text{Rate of pumps A and B combined} & = x + x - 10 \\ & = (2x - 10) \text{ litres per minute} \\ \\ \text{Time taken} & = {3500 \over 2x - 10} \\ & = {3500 \over 2(46.129) - 10} \\ & = 42.549 \text{ minutes} \\ & = 42 \text{ minutes } (0.549 \times 60) \text{ seconds} \\ & \approx 42 \text{ minutes } 30 \text{ seconds (to nearest 10 seconds)} \end{align*}

(a)

\begin{align*} \angle AOC & = \angle BOD \phantom{0} \text{ (Vertically opposite angles)}[A] \\ \\ OC & = OD \phantom{0} \text{ (Radius of circle)}[S] \\ \\ \angle OCA & = \angle ODB \phantom{0} \text{ (Tangent perpendicular to radius)} [A] \\ \\ \therefore \text{Triangles } & OAC \text{ and } OBD \text{ are congruent } (ASA) \end{align*}

(b)(i)

\begin{align*} \sin \angle OAC & = {Opp \over Hyp} \\ \sin 30^\circ & = {OC \over OA} \\ {1 \over 2} & = {OC \over 7} \phantom{000000} [OA \text{ is radius of large circle}] \\ 7 & = 2OC \\ {7 \over 2} & = OC \\ 3.5 & = OC \\ \\ \angle AOC & = 180^\circ - 90^\circ - 30^\circ \phantom{0} \text{ (Angle sum of triangle)} \\ & = 60^\circ \\ \\ \text{Area of triangle } OAC & = {1 \over 2} ab \sin C \phantom{0000000} [\text{Formula provided}] \\ & = {1 \over 2} (AO) (OC) \sin \angle AOC \\ & = {1 \over 2} (7)(3.5) \sin 60^\circ \\ & = 10.608 \\ & \approx 10.6 \text{ cm}^2 \end{align*}

(b)(ii)

\begin{align*} \text{Area of large circle} & = \pi r^2 \\ & = \pi (7)^2 \\ & = 49 \pi \text{ cm}^2 \\ \\ \text{Sector area (in small circle)} & = {\theta \over 360^\circ} \times \pi r^2 \\ & = {180^\circ - 60^\circ \over 360^\circ} \times \pi (3.5)^2 \\ & = 4{1 \over 12} \pi \text{ cm}^2 \\ \\ \text{Shaded area} & = 49 \pi - 2 \left(4{1 \over 12} \pi \right) - 2 \underbrace{ (10.608) }_\text{Area of triangle} \\ & = 107.065 \\ & \approx 107 \text{ cm}^2 \end{align*}

(a)

\begin{align*} \text{By Py} & \text{thagoras theorem,} \\ AC^2 & = AB^2 + BC^2 \\ & = 15^2 + 15^2 \\ AC & = \sqrt{15^2 + 15^2} \\ & = 21.213 \text{ cm} \\ \\ AM & = {21.213 \over 2} \\ & = 10.606 \text{ cm} \\ \\ \text{By Py} & \text{thagoras theorem,} \\ AE^2 & = AM^2 + EM^2 \\ & = (10.606)^2 + 20^2 \\ AE & = \sqrt{ (10.606)^2 + 20^2 } \\ & = 22.638 \\ & \approx 22.6 \text{ cm (Shown)} \end{align*}

(b)

\begin{align*} AF & = {15 \over 2} \\ & = 7.5 \text{ cm} \\ \\ \cos \angle EAF & = {Adj \over Hyp} \\ & = {AF \over AE} \\ & = {7.5 \over 22.638} \\ \angle EAF & = \cos^{-1} \left( {7.5 \over 22.638} \right) \\ & = 70.652^\circ \\ \\ \therefore \angle BAE & \approx 70.7^\circ \end{align*}

(c)

\begin{align*} \text{Area of triangle } AEB & = {1 \over 2} ab \sin C \phantom{000000} [\text{Formula provided}] \\ & = {1 \over 2} (AE)(AB) \sin \angle BAE \\ & = {1 \over 2} (22.638) (15) \sin 70.652^\circ \\ & = 160.196 \text{ cm}^2 \\ \\ \text{Area of base} & = 15 \times 15 \\ & = 225 \text{ cm}^2 \\ \\ \text{Total surface area} & = 225 + 4(160.196) \\ & = 865.784 \\ & \approx 866 \text{ cm}^2 \end{align*}

(d)

\begin{align*} {A_1 \over A_2} & = \left(l_1 \over l_2\right)^2 \\ {\text{Area of base of pyramid } PQRSE \over \text{Area of base of pyramid } ABCDE } & = \left(16 \over 20\right)^2 \\ & = {16 \over 25} \\ \\ \text{Area of } PQRS & = 225 \times {16 \over 25} \phantom{000000} [\text{From (c), area of base } ABCD = 225 \text{ cm}^2] \\ & = 144 \text{ cm}^2 \end{align*}

(a)(i)

\begin{align*} 34 - 15 & = 19 \\ \\ 46 - 34 & = 12 \\ \\ 50 - 46 & = 4 \end{align*}

(a)(ii)

\begin{align*} \text{Mean} & = { \sum fx \over \sum f} \\ & = {163 \over 50} \\ & = 3.26 \text{ kg} \end{align*}

(a)(iii)

\begin{align*} \text{SD} & = \sqrt{ {\sum fx^2 \over \sum f} - \left( \sum fx \over \sum f\right)^2 } \\ & = \sqrt{ {545.125 \over 50} - \left(163 \over 50\right)^2 } \\ & = 0.52431 \\ & \approx 0.524 \text{ kg} \end{align*}

(a)(iv)

\begin{align*} \text{The mid-values of each interval, i.e. 2.25 2.75 3.25 3.75 4.25, are used in the calculations.} \end{align*}

(a)(v)

\begin{align*} 100 - 90 & = 10 \% \\ \\ 50 \times {10 \over 100} & = 5 \\ \\ \text{From graph, } 5 & \text{ babies have mass } \le 2.575 \text{ kg} \\ \\ \implies 90 \% \text{ of babies} & \text{ have mass greater than 2.575 kg (not 2.8 kg)} \\ \\ \therefore \text{Data from } & \text{hospital does not support the claim} \end{align*}

(b)(i)

\begin{align*} \text{P(Mother under 30)} & = {3 + 7 \over 3 + 7 + 9 + 6} \\ & = {2 \over 5} \end{align*}

(b)(ii)(a)

\begin{align*} \text{P(Baby girl, Baby girl)} & = { 7 + 6 \over 25} \times {7 + 6 - 1 \over 24} \\ & = {13 \over 50} \end{align*}

(b)(ii)(b)

\begin{align*} \text{Case 1: } \text{P(Aged 30 or over with baby boy, Aged 30 or over with baby girl)} & = {9 \over 25} \times {6 \over 24} \\ & = {9 \over 100} \\ \\ \text{Case 2: } \text{P(Aged 30 or over with baby girl, Aged 30 or over with baby boy)} \\ & = {6 \over 25} \times {9 \over 24} \\ & = {9 \over 100} \\ \\ \text{Required probability} & = {9 \over 100} + {9 \over 100} \\ & = {9 \over 50} \end{align*}

(a)

\begin{align*} \text{Area of cardboard (before cut)} & = 20 \times 30 \\ & = 600 \text{ cm}^2 \\ \\ \text{Area of square} & = 5 \times 5 \\ & = 25 \text{ cm}^2 \\ \\ \text{Area of remaining cardboard} & = 600 - 4(25) \\ & = 500 \text{ cm}^2 \\ & = 0.05 \text{ m}^2 \phantom{00000} [1 \text{ m = 100 cm, 1 m}^2 = 10000 \text{ cm}^2] \\ \\ \text{Mass of box} & = 210 \times 0.05 \\ & = 10.5 \text{ g} \end{align*}

(b)(i)

\begin{align*} \text{Length} & = (30 - 2x) \text{ cm} \\ \\ \text{Breadth} & = (20 - 2x) \text{ cm} \\ \\ \text{Height} & = x \text{ cm} \\ \\ \text{Volume} & = l \times b \times h \\ & = (30 - 2x)(20 - 2x)(x) \\ & = x(30 - 2x)(20 - 2x) \end{align*}

(b)(ii)

\begin{align*} \text{If } x \ge 10, \text{ then the breadth of the box } & (20 - 2x) \text{ will be zero or negative} \\ \\ \text{Since } x > 0, & \text{ thus } 0 < x < 10 \end{align*}

(c)

\begin{align*} \text{Volume, } V & = x(20 - 2x)(30 - 2x) \end{align*}

$x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$y$	$504$	$832$	$1008$	$1056$	$1000$	$864$	$672$	$448$	$216$

\begin{align*} \text{Area of square} & = x \times x = x^2 \text{ cm}^2 \\ \\ \text{Area of remaining cardboard} & = (600 - 4x^2) \text{ cm}^2 \\ \\ \text{To minimise mass of cardboard, } & \text{choose largest value of } x \text{ (but ensure } V \ge 900 ) \\ \\ \text{From graph, } x & = 5.8 \\ \\ \text{Area of remaining cardboard} & = 600 - 4(5.8)^2 \\ & = 465.44 \text{ cm}^2 \\ & = 0.046 \phantom{.} 544 \text{ m}^2 \\ \\ \text{Mass of cardboard} & = 210 \times 0.046 \phantom{.} 544 \\ & = 9.7742 \\ & \approx 9.77 \text{ g} \end{align*}