\begin{align*} 3y - 7 - 5y + 4 + 4y - 2 & = 3y - 5y + 4y - 7 + 4 - 2 \\ & = 2y - 5 \end{align*}

$$ -1.5 < x \le 8 $$

(a)

\begin{align*} \text{Deposit} & = 2250 \times {18 \over 100} \\ & = \$ 405 \\ \\ \text{Total amount} & = 405 + (24 \times 92.75) \\ & = \$2631 \end{align*}

(b)

\begin{align*} \text{Extra cost} & = 2631 - 2250 \\ & = \$ 381 \\ \\ \text{Required percentage} & = { 381 \over 2250 } \times 100 \\ & = 16 {14 \over 15} \% \end{align*}

(a)

$$ 15.5 \text{ hours} $$

(b)

\begin{align*} \text{Upper quartile, } Q_3 & = 20 \text{ hours} \\ \\ \implies 25 \% \text{ of adults} & \text{ spend more than 20 hours} \\ \\ \text{Lower quartile, } Q_1 & = 12.5 \text{ hours} \\ \\ \implies 25 \% \text{ of adults} & \text{ speed less than 12.5 hours} \\ \\ \therefore \text{Rishi is wrong as same number} & \text{ of adults spend less than 12.5 hours and more than 20 hours} \end{align*}

\begin{align*} {7x \over 6} - {3(x + 1) \over 8} - {7x - 6 \over 24} & = { 7x \over 6} - {3x + 3 \over 8} - {7x - 6 \over 24} \\ & = { 28x \over 24} - { 9x + 9 \over 24 } - {7x - 6 \over 24} \\ & = { 28x - (9x + 9) - (7x - 6) \over 24} \\ & = { 28x - 9x - 9 - 7x + 6 \over 24 } \\ & = { 12x - 3 \over 24} \\ & = { 3(4x - 1) \over 24} \\ & = { 4x - 1 \over 8} \phantom{000000} [\text{Simplest form}] \end{align*}

(a)

\begin{align*} 1 \text{ cm} & : 200 \phantom{.} 000 \text{ cm} \\ 1 \text{ cm} & : 2000 \text{ m} \phantom{000000} [1 \text{ m} = 100 \text{ cm} ] \\ 1 \text{ cm} & : 2 \text{ km} \phantom{00000000} [1 \text{ km} = 1000 \text{ m}] \\ \\ \therefore n & = 2 \end{align*}

(b)

\begin{align*} 1 \text{ cm} & : 2 \text{ km} \\ 18.9 \text{ cm} & : 37.8 \text{ km} \phantom{000000} [\text{Multiply both sides by } 18.9] \\ \\ \text{Actual distance} & = 37.8 \text{ km} \end{align*}

(c)

\begin{align*} 1 \text{ cm} & : 2 \text{ km} \\ (1)^2 \text{ cm}^2 & : (2)^2 \text{ km}^2 \\ 1 \text{ cm}^2 & : 4 \text{ km}^2 \\ 182.15 \text{ cm}^2 & : 728.6 \text{ km}^2 \phantom{000000} [\text{Multiply both sides by } 182.15] \\ \\ \text{Area on map} & = 182.15 \text{ cm}^2 \end{align*}

(a)

\begin{align*} 18a - 24b + 15c & = 3(6a - 8b + 5c) \end{align*}

(b)

\begin{align*} 3 + 2m^2 xy - 2my - 3mx & = 3 - 3mx + 2m^2 xy - 2my \\ & = 3(1 - mx) + 2my (mx - 1) \\ & = 3(1 - mx) - 2my (1 - mx) \phantom{000000} [\text{Grouping method}] \\ & = (1 - mx)(3 - 2my) \end{align*}

\begin{align*} \text{Let common difference} & = d \\ \\ T_1 & = w \\ \\ T_2 & = w + d = 15 \implies w = 15 - d \\ \\ T_3 & = 15 + d \\ \\ T_4 & = 15 + d + d = 15 + 2d \\ \\ T_5 & = 15 + d + d + d = 15 + 3d \\ \\ \text{Sum of first five terms} & = w + 15 + x + y + z \\ 105 & = (15 - d) + 15 + (15 + d) + (15 + 2d) + (15 + 3d) \\ 105 & = 75 + 5d \\ -5d & = 75 - 105 \\ -5d & = -30 \\ d & = {-30 \over -5} \\ d & = 6 \\ \\ w & = 15 - d = 15 - 6 = 9 \\ x & = 15 + d = 15 + 6 = 21 \\ y & = 15 + 2d = 15 + 2(6) = 27 \\ z & = 15 + 3d = 15 + 3(6) = 33 \end{align*}

\begin{align*} \text{Sector area} & = { \theta \over 360^\circ } \times \pi r^2 \\ \\ \text{Shaded area} & = { 96^\circ \over 360^\circ} \times \pi (90)^2 - {96^\circ \over 360^\circ} \times \pi (24)^2 \\ & = 6303.3 \\ & \approx 6300 \text{ cm}^2 \end{align*}

(a)

\begin{align*} \sin 58^\circ & = {AC \over 36} \phantom{000000} \left[ {Opp \over Hyp} \right] \\ {\sin 58^\circ \over 1} & = {AC \over 36} \\ 36 \sin 58^\circ & = AC \phantom{000000} [\text{Cross-multiply}] \\ \\ AC & = 30.53 \\ & \approx 30.5 \text{ m} \end{align*}

(b)

\begin{align*} \text{By Py} & \text{thagoras theorem,} \\ AB^2 & = BC^2 - AC^2 \\ AB & = \sqrt{BC^2 - AC^2} \\ & = \sqrt{ (36)^2 - (30.53)^2 } \\ & = 19.076 \text{ m} \\ \\ \text{Perimeter} & = 36 + 30.53 + 19.076 \\ & = 85.606 \\ & \approx 85.6 \text{ m} \end{align*}

(c)

\begin{align*} \angle NDB & = 180^\circ - 116^\circ \phantom{0} \text{(Interior angles)} \\ & = 64^\circ \\ \\ \text{Bearing of } B \text{ from } D & = 360^\circ - 64^\circ \phantom{0} \text{(Angles at a point)} \\ & = 296^\circ \end{align*}

(a)

\begin{align*} 80 \times {25 \over 100} & = 20 \\ \\ \text{Lower quartile, } Q_1 & = 150 \text{ g} \\ \\ 80 \times {75 \over 100} & = 60 \\ \\ \text{Upper quartile, } Q_3 & = 235 \text{ g} \\ \\ \text{Interquartile range} & = 235 - 150 \\ & = 85 \text{ g} \end{align*}

(b)

\begin{align*} 80 \times {2 \over 5} & = 32 \\ \\ \implies 32 \text{ peoples' } & \text{estimates are greater than } k \text{ grams} \\ \\ 80 - 32 & = 48 \\ \\ \implies 48 \text{ peoples' } & \text{estimates are less than or equal to } k \text{ grams} \\ \\ \text{From curve, } k & = 220 \end{align*}

(a)

\begin{align*} D & = {98.31 \over 18.31} - (18.31)(0.361)^2 \\ & = 2.9830 \\ & \approx 3.0 \text{ (2 s.f.)} \end{align*}

(b)

\begin{align*} D & = {a \over b} - bc^2 \\ bc^2 & = {a \over b} - D \\ bc^2 & = {a \over b} - {D \over 1} \\ bc^2 & = {a \over b} - {bD \over b} \\ bc^2 & = {a - bD \over b} \\ c^2 & = {1 \over b} \left(a - bD \over b\right) \\ c^2 & = {a - bD \over b^2} \\ c & = \pm \sqrt{ a - bD \over b^2 } \end{align*}

(a)

\begin{align*} \angle ABC & = 90^\circ \phantom{0} \text{ (Right angle in a semicircle)} \\ \\ \angle OBA & = 28^\circ \phantom{0} \text{ (Base angle of isosceles triangle, } OB = OA) \\ \\ \angle OBC & = 90^\circ - 28^\circ \\ & = 62^\circ \end{align*}

(b)

\begin{align*} \angle DCA & = 180^\circ - 60^\circ - 28^\circ \phantom{0} \text{ (Angle sum of triangle)} \\ & = 92^\circ \\ \\ \text{Since } \angle DCA & \ne 90^\circ, \phantom{.} DC \text{ is not a tangent to the circle} \end{align*}

\begin{align*} 6x + 5y & = 2 \\ 5(6x + 5y) & = 5(2) \\ 30x + 25y & = 10 \phantom{0} \text{--- (1)} \\ \\ 10x - 4y & = 65 \\ 3(10x - 4y) & = 3(65) \\ 30x - 12y & = 195 \phantom{0} \text{--- (2)} \\ \\ (1) & - (2), \\ 30x + 25y - (30x - 12y) & = 10 - 195 \\ 30x + 25y - 30x + 12y & = -185 \\ 37y & = -185 \\ y & = {-185 \over 37} \\ y & = -5 \\ \\ \text{Substitute } & y = -5 \text{ into (1),} \\ 30x + 25(-5) & = 10 \\ 30x - 125 & = 10 \\ 30x & = 10 + 125 \\ 30x & = 135 \\ x & = {135 \over 30} \\ x & = 4.5 \\ \\ \therefore x & = 4.5, y = -5 \end{align*}

\begin{align*} \text{Let } \angle CBA & = 9x^\circ \\ \\ \angle BAD & = 3x^\circ \\ \\ \angle CBA + \angle BAD & = 180^\circ \phantom{0} \text{ (Interior angles, } BC \phantom{.} // \phantom{.} AD) \\ 9x + 3x & = 180 \\ 12x & = 180 \\ x & = {180 \over 12} \\ x & = 15 \\ \\ \angle ADC & = 2x^\circ \\ \\ \angle BCD & = 360^\circ - 9x^\circ - 3x^\circ - 2x^\circ \phantom{0} \text{ (Angle sum of quadrilateral)} \\ & = 360^\circ - 14x^\circ \\ & = 360^\circ - 14(15)^\circ \\ & = 150^\circ \end{align*}

\begin{align*} \text{Total amount} & = P\left(1 + {r \over 100}\right)^n \phantom{000000} [ \text{Formula provided}] \\ P + 2132.16 & = P \left(1 + {3.2 \over 100} \right)^5 \\ P - P \left(1 + {3.2 \over 100} \right)^5 & = - 2132.16 \\ P \left[ 1 - \left(1 + {3.2 \over 100} \right)^5 \right] & = -2132.16 \\ P & = { - 2132.16 \over 1 - \left(1 + {3.2 \over 100} \right)^5 } \\ P & = 12 \phantom{.} 499.988 \\ P & \approx \$ 12 \phantom{.} 499.99 \end{align*}

(a)

\begin{align*} n ( (A' \cup B) \cap (A \cup B') ) & = 3 + 9 = 12 \end{align*}

(b)

\begin{align*} n ( (A \cap B') \cup (A \cup B)' ) & = 7 + 9 = 12 \end{align*}

(a)

\begin{align*} 263 & = 2.63 \times 10^2 \end{align*}

(b)(i)

\begin{align*} 3.4 \times 10^{99} & = A \times 10^{100} \\ 3.4 & = {A \times 10^{100} \over 10^{99}} \\ 3.4 & = A \times 10^{100 - 99} \\ 3.4 & = A \times 10 \\ \\ A & = {3.4 \over 10} \\ A & = 0.34 \\ \\ \therefore 3.4 \times 10^{99} & = 0.34 \times 10^{100} \end{align*}

(b)(ii)

\begin{align*} (4.7 \times 10^{100}) + (3.4 \times 10^{99}) & = (4.7 \times 10^{100}) + (0.34 \times 10^{100}) \phantom{000000} [\text{Use result from (i)}] \\ & = (4.7 + 0.34) \times 10^{100} \\ & = 5.04 \times 10^{100} \end{align*}

\begin{align*} \text{HCF} & = 2^4 \times 5 \\ \text{LCM} & = 2^6 \times 3^2 \times 5^3 \\ \\ 720 & = 2^4 \times 3^2 \times 5 \\ N & = 2^6 \times 5^3 \phantom{000000} [N \text{ must have factors } 2^6 \text{ and } 5^3 \text{ from LCM since they are not factors of 720}] \\ & = 8000 \end{align*}

\begin{align*} \sin (2x^\circ) & = \sin (180^\circ - 2x^\circ) = 0.561 \end{align*} \begin{align*} \sin (2x^\circ) & = 0.561 &&& \sin (180^\circ - 2x^\circ) & = 0.561 \\ 2x^\circ & = \sin^{-1} 0.561 &&& 180^\circ - 2x^\circ & = \sin^{-1} (0.561) \\ 2x^\circ & = 34.13^\circ &&& 180^\circ - 2x^\circ & = 34.13^\circ \\ x^\circ & = 17.065^\circ &&& -2x^\circ & = -145.87^\circ \\ x^\circ & \approx 17.1^\circ &&& x^\circ & = 72.935^\circ \\ & &&& x^\circ & \approx 72.9^\circ \end{align*} $$ \therefore x = 17.1 \text{ or } 72.9 $$

(a)

\begin{align*} \text{Total messages} & = 17.5 \times 8 \\ & = 140 \\ \\ 12 + 24 + 8 + 21 + 28 + 17 + 2p + 4p^2 & = 140 \\ 110 + 2p + 4p^2 & = 140 \\ 4p^2 + 2p + 110 - 140 & = 0 \\ 4p^2 + 2p - 30 & = 0 \\ 2p^2 + p - 15 & = 0 \\ (2p - 5)(p + 3) & = 0 \end{align*} \begin{align*} 2p - 5 & = 0 && \text{ or } & p + 3 & = 0 \\ 2p & = 5 &&& p & = -3 \text{ (Reject since } p > 0) \\ p & = {5 \over 2} \\ p & = 2.5 \text{ (Shown)} \end{align*}

(b)

\begin{align*} 1. & \text{ Li's sister received more messages on average as the mean number of messages} \\ & \text{ for her is higher than that of Li.} \\ \\ 2. & \text{The number of messages Li receives each day is more consistent as the standard} \\ & \text{ deviation for Li's data is lower than that of Li's sister.} \end{align*}

\begin{align*} (2x - 3)(3x^2 + 2x - 5) & = 6x^3 + 4x^2 - 10x - 9x^2 - 6x + 15 \\ & = 6x^3 + 4x^2 - 9x^2 - 10x - 6x + 15 \\ & = 6x^3 - 5x^2 - 16x + 15 \end{align*}

(a)

\begin{align*} \text{By Py} & \text{thagoras theorem,} \\ (3x + 4)^2 & = (3x - 1)^2 + y^2 \\ \underbrace{ (3x)^2 + 2(3x)(4) + (4)^2 }_{(a + b)^2 = a^2 + 2ab + b^2} & = \underbrace{ (3x)^2 - 2(3x)(1) + (1)^2 }_{(a - b)^2 = a^2 - 2ab + b^2} + y^2 \\ 9x^2 + 24x + 16 & = 9x^2 - 6x + 1 + y^2 \\ -y^2 & = 9x^2 - 6x + 1 - 9x^2 - 24x - 16 \\ -y^2 & = -30x - 15 \\ y^2 & = 30x + 15 \\ y^2 & = 15(2x + 1) \\ \\ \text{Since } x \text{ is an integer, } & 2x \text{ is even, } 2x + 1 \text{ is odd and } 15(2x + 1) \text{ is odd} \\ \\ \text{Thus if } y^2 = y \times y \text{ is} & \text{ an odd } \text{integer, }y \text{ is an odd integer} \end{align*}

(b)

\begin{align*} y^2 & = 15(2x + 1) \\ y \times y & = 15 \times (2x + 1) \\ \\ y & = 15 \\ \\ y = 2x + 1 & = 15 \\ 2x & = 15 - 1 \\ 2x & = 14 \\ x & = 7 \\ \\ \therefore x & = 7, y = 15 \end{align*}

\begin{align*} x^4 - 16 & = (x^2)^2 - (4)^2 \\ & = (x^2 + 4)(x^2 - 4) \phantom{000000000} [ a^2 - b^2 = (a + b)(a - b) ] \\ & = (x^2 + 4)(x^2 - 2^2) \\ & = (x^2 + 4)(x + 2)(x - 2) \phantom{0000} [ a^2 - b^2 = (a + b)(a - b) ] \\ \\ \therefore { 3x^2 + 6x \over x^4 - 16} & = { 3x (x + 2) \over (x^2 + 4)(x + 2)(x - 2)} \\ & = {3x \over (x^2 + 4)(x - 2)} \end{align*}

\begin{align*} x^2 - 12x + 17 & = 0 \\ x^2 - 12x + \left(12 \over 2\right)^2 - \left(12 \over 2\right)^2 + 17 & = 0 \\ \underbrace{ x^2 - 12x + 6^2 }_{ a^2 - 2ab + b^2 = (a - b)^2 } - 36 + 17 & = 0 \\ (x - 6)^2 - 19 & = 0 \\ (x - 6)^2 & = 19 \\ x - 6 & = \pm \sqrt{19} \end{align*} \begin{align*} x - 6 & = \sqrt{19} && \text{ or } & x - 6 & = - \sqrt{19} \\ x & = \sqrt{19} + 6 &&& x & = - \sqrt{19} + 6 \\ x & \approx 10.36 &&& x & \approx -1.64 \end{align*}

(a)

\begin{align*} \text{Radius} & = {15 \over 2} = 7.5 \text{ cm} \\ \\ \text{Volume of cylinder} & = \text{Volume of cone} \\ \pi r^2 h & = \underbrace{ { 1 \over 3} \pi r^2 h }_\text{Provided} \\ \pi (7.5)^2 (20) & = {1 \over 3} \pi (7.5)^2 (h) \\ \\ h & = { \pi (7.5)^2 (20) \over {1 \over 3} \pi (7.5)^2 } \\ h & = 60 \end{align*}

(b)

\begin{align*} \text{Volume of cylinder} & = \pi r^2 h \\ & = \pi (10)^2 (25) \\ & = 2500 \pi \text{ cm}^3 \\ \\ \text{Volume of cone} & = {1 \over 3} \pi r^2 h \\ & = {1 \over 3} \pi (10)^2 (90) \\ & = 3000 \pi \text{ cm}^3 \\ \\ \text{Total volume of container} & = 2500\pi + 3000\pi \\ & = 5500 \pi \text{ cm}^3 \\ \\ \text{Volume of 'empty cone'} & = {5500 \pi \over 2} \\ & = 2750 \pi \text{ cm}^3 \\ \\ {V_1 \over V_2} & = \left(l_1 \over l_2\right)^3 \\ {\text{Volume of 'empty cone'} \over \text{Volume of cone}} & = \left(d \over 90\right)^3 \\ {2750\pi \over 3000\pi} & = \left(d \over 90\right)^3 \\ {11 \over 12} & = \left(d \over 90\right)^3 \\ \\ {d \over 90} & = \sqrt[3]{11 \over 12} \\ {d \over 90} & = {0.97141 \over 1} \\ d & = 90(0.97141) \phantom{000000} [\text{Cross-multiply}] \\ d & = 87.427 \\ d & \approx 87.4 \text{ cm} \end{align*}

(a)

\begin{align*} \overrightarrow{OC} & = \overrightarrow{OA} + \overrightarrow{AC} \\ & = 3 \textbf{a} + \underbrace{ \overrightarrow{AC} }_\text{Need to find this} \\ \\ \overrightarrow{AC} & = {1 \over 3} \overrightarrow{AB} \\ & = {1 \over 3} ( \overrightarrow{AO} + \overrightarrow{OB} ) \\ & = {1 \over 3} ( - 3 \textbf{a} + 6 \textbf{b} ) \\ & = - \textbf{a} + 2 \textbf{b} \\ \\ \therefore \overrightarrow{OC} & = 3 \textbf{a} + (- \textbf{a} + 2 \textbf{b}) \\ & = 3 \textbf{a} - \textbf{a} + 2 \textbf{b} \\ & = 2 \textbf{a} + 2 \textbf{b} \phantom{0} \text{ (Shown)} \end{align*}

(b)

\begin{align*} \overrightarrow{AQ} & = \overrightarrow{AO} + \overrightarrow{OQ} \\ & = - 3 \textbf{a} + n \overrightarrow{OB} \\ & = - 3 \textbf{a} + n (6 \textbf{b} ) \\ & = - 3 \textbf{a} + 6 n \textbf{b} \\ \\ \overrightarrow{AP} & = \overrightarrow{AO} + \overrightarrow{OP} \\ & = - 3 \textbf{a} + {1 \over 2} \overrightarrow{OC} \\ & = - 3 \textbf{a} + {1 \over 2} ( \underbrace{ 2 \textbf{a} + 2 \textbf{b} }_\text{Part (a)} ) \\ & = - 3 \textbf{a} + \textbf{a} + \textbf{b} \\ & = - 2 \textbf{a} + \textbf{b} \\ \\ \overrightarrow{AQ} & = m \overrightarrow{AP} \\ - 3 \textbf{a} + 6 n \textbf{b} & = m ( - 2 \textbf{a} + \textbf{b} ) \\ - 3 \textbf{a} + 6 n \textbf{b} & = - 2m \textbf{a} + m \textbf{b} \end{align*} \begin{align*} \text{Comparing } & \text{terms with } \textbf{a}, &&& \text{Comparing } & \text{terms with } \textbf{b}, \\ -2 m & = -3 &&& 6n & = m \\ m & = {-3 \over -2} &&& n & = {m \over 6} \\ m & = 1.5 &&& n & = {1.5 \over 6} \\ & &&& n & = {1 \over 4} \end{align*} \begin{align*} \therefore \overrightarrow{OQ} & = n \overrightarrow{OB} \\ \overrightarrow{OQ} & = {1 \over 4} \overrightarrow{OB} \\ \\ \implies OQ:QB & = 1 : 3 \end{align*}

(c)

\begin{align*} { \text{Area of triangle } OAC \over \text{Area of triangle } OBC } & = { {1 \over 2} \times AC \times h \over {1 \over 2} \times BC \times h} \\ & = {AC \over BC} \\ & = {1 \over 2} \\ \\ \text{Area of triangle } OAC & = 25 \times {1 \over 2} \\ & = 12.5 \text{ cm}^2 \end{align*}

(a)

\begin{align*} -4 & \le 3 + 2x &&& 3 + 2x & < 5 \\ -4 - 3 & \le 2x &&& 2x & < 5 - 3 \\ -7 & \le 2x &&& 2x & < 2 \\ -{7 \over 2} & \le x &&& x & < 1 \\ -3.5 & \le x \end{align*} \begin{align*} \therefore -3.5 & \le x < 1 \\ \\ \text{Integers: } & -3, -2, -1, 0 \end{align*}

(b)

\begin{align*} \require{cancel} {20 p^3 q \over 7 r^2 } \div {8p^2 r \over 21 q^4} & = {20 p^3 q \over 7 r^2} \times {21 q^4 \over 8p^2 r} \\ & = { 420 p^3 q^{1 + 4} \over 56 p^2 r^{2 + 1} } \\ & = { ^{15} \cancel{420} p^\cancel{3} q^5 \over ^2 \cancel{56} \cancel{p^2} r^3 } \\ & = { 15 p q^5 \over 2 r^3 } \end{align*}

(c)

\begin{align*} \left( a^9 \over 64 b^{12} \right)^{-{2 \over 3}} & = \left( 64 b^{12} \over a^9 \right)^{2 \over 3} \phantom{000000} \left[ \left(a \over b\right)^{-n} = \left(b \over a \right)^n \right] \\ & = { 64^{2 \over 3} (b^{12})^{2 \over 3} \over (a^9)^{2 \over 3} } \\ & = { 16 b^{ 12 \times {2 \over 3} } \over a^{9 \times {2 \over 3}} } \\ & = { 16 b^8 \over a^6} \end{align*}

(d)

\begin{align*} {x \over x + 2} + {3 \over 4 - x} - {1 \over 1} & = {x(4 - x) \over (x + 2)(4 - x)} + {3(x + 2) \over (x + 2)(4 - x)} - {(x + 2)(4 - x) \over (x + 2)(4 - x)} \\ & = {x(4 - x) + 3(x + 2) - (x + 2)(4 - x) \over (x + 2)(4 - x)} \\ & = {4x - x^2 + 3x + 6 - (4x - x^2 + 8 - 2x) \over (x + 2)(4 - x)} \\ & = { - x^2 + 7x + 6 - (-x^2 + 2x + 8) \over (x + 2)(4 - x)} \\ & = { - x^2 + 7x + 6 + x^2 - 2x - 8 \over (x + 2)(4 - x)} \\ & = { 5x - 2 \over (x + 2)(4 - x)} \end{align*}

(a)(i)

\begin{align*} \textbf{T} & = \left( \begin{matrix} 12 \\ 15 \end{matrix} \right) \end{align*}

(a)(ii)

\begin{align*} \textbf{M} & = \textbf{CT} \\ & = \underset{3 \times 2}{\left( \begin{matrix} 120 & 40 \\ 90 & 30 \\ 50 & 20 \end{matrix}\right)} \underset{2 \times 1}{\left( \begin{matrix} 12 \\ 15 \end{matrix}\right)} \\ & = \underset{3 \times 1}{ \left( \begin{matrix} (120)(12) + (40)(15) \\ (90)(12) + (30)(15) \\ (50)(12) + (20)(15) \end{matrix}\right) } \\ & = \left( \begin{matrix} 2040 \\ 1530 \\ 900 \end{matrix}\right) \end{align*}

(a)(iii)

\begin{align*} \textbf{M} & = \left( \begin{matrix} 2040 \\ 1530 \\ 900 \end{matrix}\right) \begin{matrix} \text{Hall 1} \\ \text{Hall 2} \\ \text{Hall 3} \end{matrix} \\ \\ \text{Difference} & = {9 \over 10} (2040 + 1530 + 900) - {3 \over 5} (2040 + 1530 + 900) \\ & = \$ 1341 \end{align*}

(b)

\begin{align*} \text{No. of tickets sold} & = 5075 \times {100 - 56 \over 100} \\ & = \$ 2233 \end{align*}

(c)

\begin{align*} \text{Afternoon} & : \text{Evening} : \text{Night} \\ 3 & : \phantom{000} 8 \phantom{000000000000} [\text{Multiply by } 3] \\ & \phantom{:000.} 6 \phantom{000} : 11 \phantom{00000.} [\text{Multiply by } 4] \\ 9 & : \phantom{000} 24 \phantom{00} : 44 \\ \\ \text{Required ratio} & = 9 : 44 \end{align*}

(a)

\begin{align*} \text{Line } L: 4y & = 3x - 2 \\ \\ \text{Using } & (-4, p), \\ 4p & = 3(-4) - 2 \\ 4p & = -14 \\ p & = {-14 \over 4} \\ p & = -3.5 \\ \\ \text{Line } M: 2y & = kx + 13 \\ \\ \text{Using } & (-4, p) = (-4, -3.5), \\ 2(-3.5) & = k(-4) + 13 \\ -7 & = -4k + 13 \\ 4k & = 13 + 7 \\ 4k & = 20 \\ k & = {20 \over 4} \\ k & = 5 \\ \\ \therefore k & = 5, p = -3.5 \end{align*}

(b)

\begin{align*} \overrightarrow{OX} & = {-3 \choose 7} \\ \\ \overrightarrow{OY} & = \overrightarrow{OX} + \overrightarrow{XY} \\ & = {-3 \choose 7} + {5 \choose -2} \\ & = {2 \choose 5} \\ \\ \therefore & \phantom{.} Y(2, 5) \\ \\ \text{Gradient of } XY & = {5 - 7 \over 2 - (-3)} \phantom{000000} \left[ m = {y_2 - y_1 \over x_2 - x_1} \right] \\ & = -0.4 \\ \\ y & = mx + c \\ y & = -0.4x + c \\ \\ \text{Using } & X(-3, 7), \\ 7 & = -0.4(-3) + c \\ 7 & = 1.2 + c \\ -c & = 1.2 - 7 \\ -c & = -5.8 \\ c & = 5.8 \\ \\ \text{Line } XY : & \phantom{.} y = -0.4x + 5.8 \end{align*}

(c)

\begin{align*} \text{Length of } AB & = \sqrt{ (8 - 15)^2 + (a - 11)^2 } \phantom{000000} \left[ \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \right] \\ 25 & = \sqrt{ 49 + (a - 11)^2 } \\ (25)^2 & = \left( \sqrt{ 49 + (a - 11)^2 } \right)^2 \\ 625 & = 49 + (a - 11)^2 \\ -(a - 11)^2 & = 49 - 625 \\ -(a - 11)^2 & = -576 \\ (a - 11)^2 & = 576 \\ a - 11 & = \pm \sqrt{576} \end{align*} \begin{align*} a - 11 & = \sqrt{576} && \text{ or } & a - 11 & = - \sqrt{576} \\ a & = \sqrt{576} + 11 &&& a & = - \sqrt{576} + 11 \\ a & = 35 \text{ (Reject, since } a < 0) &&& a & = -13 \end{align*}

(a)(i)

\begin{align*} y & = 2(-2)^2 - {(-2)^3 \over 2} - 4 \\ & = 8 \end{align*}

(a)(ii)

(a)(iii)

(a)(iv)

\begin{align*} \text{Equation of curve: } & y = 2x^2 - {x^3 \over 2} - 4 \\ \\ \underbrace{ 4x^2 - x^3 - 4 }_\text{Need to match to eqn of curve} & = 0 \\ {1 \over 2} (4x^2 - x^3 -4) & = {1 \over 2} (0) \\ 2x^2 - {1 \over 2}x^3 - 2 & = 0 \\ 2x^2 - {x^3 \over 2} - 2 - 2 & = 0 - 2 \\ \underbrace{ 2x^2 - {x^3 \over 2} - 4}_\text{Curve} & = -2 \\ \\ \therefore \text{Draw } & y = -2 \\ \\ \text{From graph, } x & = -0.9 \text{ or } 1.2 \text{ or } 3.7 \end{align*}

(b)

\begin{align*} y & = k a^x \\ \\ \text{Using } & A(0, 5), \\ 5 & = k a^0 \\ 5 & = k(1) \\ 5 & = k \\ \\ \implies y & = 5 a^x \\ \\ \text{Using } & B(3, 135), \\ 135 & = 5a^3 \\ {135 \over 5} & = a^3 \\ 27 & = a^3 \\ \\ a & = \sqrt[3]{27} \\ a & = 3 \\ \\ \implies y & = 5 (3^x) \\ \\ \text{Using } & C(4, p), \\ p & = 5(3^4) \\ p & = 405 \end{align*}

(a)

\begin{align*} BC^2 & = AB^2 + AC^2 - 2(AB)(AC) \cos \angle BAC \phantom{000000} [\text{Cosine rule}] \\ 405^2 & = 175^2 + 280^2 - \underbrace{ 2(175)(280) \cos \angle BAC}_\text{One term} \\ 2(175)(280) \cos \angle BAC & = 175^2 + 280^2 - 405^2 \\ \cos \angle BAC & = { 175^2 + 280^2 - 405^2 \over 2(175)(280)} \\ \angle BAC & = \cos^{-1} \left[ 175^2 + 280^2 - 405^2 \over 2(175)(280) \right] \\ & = 121.14^\circ \\ & \approx 121.1^\circ \end{align*}

(b)

\begin{align*} \angle \text{North} AB & = 48^\circ \phantom{000000} [\text{Bearing of } B \text{ from } A] \\ \\ \angle \text{North} AD & = 90^\circ \phantom{000000} [D \text{ is due west of } A ] \\ \\ \angle DAC & = 360^\circ - 90^\circ - 48^\circ - 124.14^\circ \phantom{0} \text{ (Angles at a point)} \\ & = 97.86^\circ \\ \\ \angle ADC & = 180^\circ - 97.86^\circ - 32^\circ \phantom{0} \text{ (Angle sum of triangle)} \\ & = 50.14^\circ \\ \\ {AD \over \sin 32^\circ} & = { 280 \over \sin 50.14^\circ} \phantom{000000} [\text{Sine rule}] \\ AD \sin 50.14^\circ & = 280 \sin 32^\circ \phantom{000000} [\text{Cross-multiply}] \\ AD & = {280 \sin 32^\circ \over \sin 50.14^\circ} \\ AD & = 193.30 \\ AD & \approx 193 \text{ m} \end{align*}

(c)

\begin{align*} \tan \angle 34^\circ & = { TA \over 280} \phantom{000000} \left[ {Opp \over Adj} \right] \\ {\tan \angle 34^\circ \over 1} & = {TA \over 280} \\ 280 \tan \angle 34^\circ & = TA \phantom{0000000.} [\text{Cross-multiply}] \\ \\ TA & = 188.86 \text{ m} \\ \\ \tan \angle TBA & = {188.86 \over 175} \\ \angle TBA & = \tan^{-1} \left(188.86 \over 175\right) \\ & = 47.18^\circ \\ & \approx 47.2^\circ \end{align*}

(a)

\begin{align*} \text{Time taken} & = {\text{Distance} \over \text{Speed}} \\ \\ \text{Time taken on expressway} & = {40 \over x} \text{ h} \\ \\ \text{Time taken on local roads} & = {20 \over x - 30} \text{ h} \\ \\ 50 \text{ minutes} & = {50 \over 60} = {5 \over 6} \text{ h} \\ \\ {40 \over x} + {20 \over x - 30} & = {5 \over 6} \\ {40(x - 30) \over x(x - 30)} + {20x \over x(x - 30)} & = {5 \over 6} \\ {40(x - 30) + 20x \over x(x - 30)} & = {5 \over 6} \\ {40x - 1200 + 20x \over x(x - 30)} & = {5 \over 6} \\ {60x - 1200 \over x(x - 30)} & = {5 \over 6} \\ 6(60x - 1200) & = 5x(x - 30) \\ 360x - 7200 & = 5x^2 - 150x \\ 0 & = 5x^2 - 150x - 360x + 7200 \\ 0 & = 5x^2 - 510x + 7200 \\ 0 & = x^2 - 102x + 1440 \phantom{0} \text{ (Shown)} \phantom{000000} [\text{Divide every term by 5}] \end{align*}

(b)

\begin{align*} 0 & = x^2 - 102x + 1440 \\ \\ x & = { -b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = { - (-102) \pm \sqrt{ (-102)^2 - 4(1)(1440)} \over 2(1)} \\ & = { 102 \pm \sqrt{4644} \over 2} \\ & = {102 + \sqrt{4644} \over 2} \text{ or } {102 - \sqrt{4644} \over 2} \\ & \approx 16.93 \text{ or } 85.07 \end{align*}

(c)

\begin{align*} & \text{If } x = 16.93, \text{ speed on local roads} = 16.93 - 30 = -13.07 \text{ km/h} \\ \\ & \text{Therefore, we need to reject } x = 16.93. \end{align*}

(d)

\begin{align*} \text{Time taken on expressway} & = {40 \over 85.07} \\ & = 0.4702 \text{ h} \\ \\ \text{Time taken on local roads} & = {20 \over 85.07 - 30} \\ & = 0.3632 \text{ h} \\ \\ \text{Time difference} & = 0.4702 - 0.3632 \\ & = 0.107 \text{ h} \\ & = (0.107 \times 60) \text{ mins} \\ & = 6.42 \text{ mins} \\ & = 6 \text{ mins } (0.42 \times 60) \text{ seconds} \\ & \approx 6 \text{ mins } 25 \text{ seconds} \end{align*}

(a)(i)

\begin{align*} \text{Median position} & = {120 +1 \over 2} = 60.5 \\ \\ \implies \text{Median between } & 60\text{th and 61st apples} \\ \\ \text{Interval: } & 120 < m \le 140 \end{align*}

(a)(ii)

\begin{align*} \text{Mean} & = {\sum fx \over \sum f} \phantom{000000} [\text{Formula provided}] \\ & = {14 \phantom{.} 100 \over 120} \phantom{00000} [\text{Use mid-values 70, 90, 110, 130, 150}] \\ & = 117.5 \text{ g} \end{align*}

(a)(iii)

\begin{align*} \text{SD} & = \sqrt{ {\sum fx^2 \over \sum f} - \left( \sum fx \over \sum f\right)^2 } \phantom{0000000.} [\text{Formula provided}] \\ & = \sqrt{ {1 \phantom{.} 718 \phantom{.} 400 \over 120} - \left( 14100 \over 120\right)^2 } \phantom{00000} [\text{Use mid-values 70, 90, 110, 130, 150}] \\ & = 22.666 \\ & \approx 22.7 \text{ g} \end{align*}

(a)(iv)

\begin{align*} & \text{Mean mass of original 120 apples} = 117.5 \text{ g} \\ \\ & \text{Since mean mass of 200 apples decreased to 110 g, the mean mass of the additional 80 apples is} \\ & \text{lower than 110 g.} \end{align*}

(b)(i)

\begin{align*} & \text{The mass of the heavist tomato may not be 180 g and the mass of the lightest} \\ & \text{tomato may not be 80 g.} \end{align*}

(b)(ii)

\begin{align*} \text{Required probability} & = {32 + 48 \over 150} \\ & = {8 \over 15} \end{align*}

(b)(iii)

\begin{align*} \text{P(< 120g, < 120g, > 160g)} & = {10 + 20 \over 150} \times {10 + 20 - 1 \over 150} \times {40 \over 148} \\ & = {58 \over 5513} \\ \\ \text{P(< 120g, > 160g, <120g)} & = {10 + 20 \over 150} \times {40 \over 149} \times {10 + 20 - 1 \over 148} \\ & = {58 \over 5513} \\ \\ \text{P(> 160g, < 120g, < 120g)} & = {40 \over 150} \times {10 + 20 \over 149} \times {10 + 20 - 1 \over 148} \\ & = {58 \over 5513} \\ \\ \text{Required probability} & = {58 \over 5513} + {58 \over 5513} + {58 \over 5513} \\ & = {174 \over 5513} \end{align*}

(a)(i)

\begin{align*} \text{Since chords } AB & \text{ and } CD \text{ have the same length, } \\ OM & = ON \phantom{0} [S] \\ \\ BM & = {1 \over 2} BA \phantom{0} (M \text{ is midpoint } BA) \\ & = {1 \over 2} CD \phantom{0} \text{(Given)} \\ & = CN \phantom{0} [S] \\ \\ OB & = OC \phantom{0} \text{(Radius of circle)} [S] \\ \\ \therefore \text{Triangles } BMO & \text{ and } CNO \text{ are congruent } (SSS) \end{align*}

(a)(ii)

\begin{align*} \angle AOB & = \angle COD = 1.8 \text{ radians} \\ \\ \text{Reflex } \angle AOB & = (2\pi - 1.8) \text{ radians} \phantom{0} \text{ (Angles at a point)} \\ \\ \text{Major arc length } AB & = r \theta \phantom{0000000000000000000000000} [\text{Formula provided}] \\ & = (4)(2\pi - 1.8) \\ & = (8 \pi - 7.2) \text{ cm} \approx 17.9 \text{ cm} \phantom{000000} [\text{Either one is fine}] \end{align*}

(a)

\begin{align*} \text{Volume of candle} & = 5 \times 5 \times 5 \\ & = 125 \text{ cm}^3 \\ \\ \text{Mass of candle} & = 0.765 \times 125 \\ & = 95.625 \text{ g} \end{align*}

(b)

\begin{align*} \text{Mass of wax} & = 150 \text{ g} \times {100 \over 100 + 8} \phantom{000000} [100\% \text{ fragrance oil, } 8\% \text{ wax}] \\ & = 138{8 \over 9} \text{ g} \\ & \approx 139 \text{ g} \end{align*}

(c)

\begin{align*} \underline{ \text{Small candle} } &\\ \text{Volume} & = \pi r^2 h \\ & = \pi (2)^2 (8) \\ & = 32 \pi \text{ cm}^3 \\ \\ \text{Mass} & = 0.765 \times 32\pi \\ & = 24.48 \pi \text{ g} \\ \\ \text{Mass of wax} & = 24.48 \pi \times {100 \over 108} \\ & = 22{2 \over 3} \pi \text{ g} \\ & = 71.21 \text{ g} \\ \\ \text{Cost of wax (using 500 g)} & = \$ 16 \times {71.21 \over 500} \\ & = \$ 2.28 \\ \text{Mass of oil} & = 24.48\pi - 22{2 \over 3} \pi \\ & = 1{61 \over 75} \pi \text{ g} \\ \\ \text{Volume of oil} & = 1{61 \over 75} \pi \text{ ml} \\ & = 5.697 \text{ ml} \\ \\ \text{Cost of oil (using 30 ml)} & = \$ 12 \times {5.697 \over 30} \\ & = \$ 2.28 \\ \\ \text{Length of wick} & = 8 + 0.5 \phantom{000000} [5 \text{ mm} = 0.5 \text{ cm}] \\ & = 8.5 \text{ cm} \\ \\ \text{Cost of wick} & = \$ 8 \div 10 \\ & = \$ 0.80 \phantom{000000} [\text{Have to use entire 15 cm}] \\ \\ \text{Total cost} & = \$ 2.28 + \$ 2.28 + \$ 0.80 \\ & = \$ 5.36 \\ \\ \text{Selling price} & = \$ 6 \\ \\ \\ \underline{ \text{Large candle} } &\\ \\ {V_1 \over V_2} = {m_1 \over m_2} & = \left(l_1 \over l_2\right)^3 \\ & = \left(6 \over 4\right)^3 \\ & = {27 \over 8} \phantom{000000} [\text{Small candle: 8 units, Large candle: 27 units}] \\ \\ \text{Mass of wax} & = 22{2 \over 3} \pi \times {27 \over 8} \\ & = 76.5 \pi \text{ g} \\ & = 240.33 \text{ g} \\ \\ \text{Cost of wax (using 500g)} & = \$ 16 \times {240.33 \over 500} \\ & = \$ 7.69 \\ \\ \text{Volume of oil} & = 1{61 \over 75} \pi \times {27 \over 8} \\ & = 6.12 \pi \text{ cm}^3 \\ & = 19.23 \text{ cm}^3 \\ \\ \text{Cost of oil (Using 30 ml)} & = \$ 12 \times {19.23 \over 30} \\ & = \$ 7.69 \\ \\ \text{Length of wick} & = 8 \times {6 \over 4} + 0.5 \\ & = 12.5 \text{ cm} \\ \\ \text{Cost of wick} & = \$ 0.80 \\ \\ \text{Total cost} & = \$ 7.69 + \$ 7.69 + \$ 0.80 \\ & = \$ 16.18 \\ \\ \text{Selling price} & = \$ 17 \end{align*} \begin{align*} & \text{Sell small candle at \$6 each and large candle at \$17 each.} \\ \\ & \text{Calculations assume the worse case scenario, where Wei purchases supplies of} \\ & \text{smaller sizes that increases the cost of goods sold.} \end{align*}