\begin{align*} & \text{The length and breadth of the cards in the respective years are different. It is not known if} \\ & \text{the total spendings should be determined from the length or the area of the card.} \end{align*}

(a)

\begin{align*} (x^6)^{2 \over 3} & = x^{6 \times {2 \over 3}} \phantom{000000} [ (a^m)^n] \\ & = x^4 \end{align*}

(b)

\begin{align*} 3^a & = 3^7 + 3^7 + 3^7 \\ 3^a & = 3(3^7) \\ 3^a & = (3^1)(3^7) \\ 3^a & = 3^{1 + 7} \phantom{000000} [a^m \times a^n = a^{m + n}] \\ 3^a & = 3^8 \\ \\ \therefore a & = 8 \end{align*}

(a)

\begin{align*} 2 \text{|}& \underline{360} \\ 2 \text{|}& \underline{180} \\ 2 \text{|}& \underline{\phantom{0} 90} \\ 3 \text{|}& \underline{\phantom{0} 45 } \\ 3 \text{|}& \underline{\phantom{0} 15 } \\ 5 \text{|}& \underline{\phantom{00} 5 } \\ \text{|}& \underline{\phantom{00} 1} \\ \\ 360 & = 2^3 \times 3^2 \times 5 \end{align*}

(b)

\begin{align*} 360 & = 2^3 \times 3^2 \times 5 \\ \\ {360p \over q} & = { 2^3 \times 3^2 \times 5 \times 3 \over 5} \\ & = 2^3 \times 3^3 \phantom{000000000} [\text{Perfect cube}] \\ \\ \therefore p & = 3, \phantom{0} q = 5 \end{align*}

\begin{align*} \text{No. of Bengal tigers in 2011} & = 3200 \times {100+r \over 100} \\ & = 3200 \left(100+r \over 100\right) \\ \\ \text{No. of Bengal tigers in 2012} & = 3200 \left(100+r \over 100\right) \times {100+r \over 100} \\ & = 3200 \left(100 + r \over 100\right)^2 \\ \\ \text{No. of Bengal tigers in 2013} & = 3200 \left(100+r \over 100\right)^2 \times {100+r \over 100} \\ & = 3200 \left(100 + r \over 100\right)^3 \\ & . \\ & . \\ & . \\ \\ \text{No. of Bengal tigers in 2016} & = 3200 \left(100 + r \over 100\right)^6 \\ 3890 & = 3200 \left(100 + r \over 100\right)^6 \\ {3890 \over 3200} & = \left(100 + r \over 100\right)^6 \\ \sqrt[6]{3890 \over 3200} & = {100 + r \over 100} \\ 1.033 \phantom{.} 078 & = {100 + r \over 100} \\ 100(1.033 \phantom{.} 078) & = 100 + r \\ 103.307 \phantom{.} 8 & = 100 + r \\ \\ r & = 103.307 \phantom{.} 8 - 100 \\ & = 3.307 \phantom{.} 8 \\ & \approx 3.31 \end{align*}

Alternative solution by formula:

\begin{align*} \text{Total amount} & = P \left(1 + {r \over 100}\right)^n \\ 3890 & = 3200 \left(1 + {r \over 100}\right)^6 \\ {3890 \over 3200} & = \left(1 + {r \over 100}\right)^6 \\ \sqrt[6]{3890 \over 3200} & = 1 + {r \over 100} \\ 1.033 \phantom{.} 078 & = 1 + {r \over 100} \\ 1.033 \phantom{.} 078 - 1 & = {r \over 100} \\ 0.033 \phantom{.} 078 & = {r \over 100} \\ \\ r & = 100(0.033 \phantom{.} 078) \\ & = 3.307 \phantom{.} 8 \\ & \approx 3.31 \end{align*}

\begin{align*} {1 \over 2x - 3} - {3 \over 3x - 1} & = {3x - 1 \over (2x - 3)(3x - 1)} - {3(2x - 3) \over (2x - 3)(3x - 1)} \\ & = {3x - 1 - 3(2x - 3) \over (2x - 3)(3x - 1)} \\ & = {3x - 1 - 6x + 9 \over (2x - 3)(3x - 1)} \\ & = {8 - 3x \over (2x - 3)(3x - 1)} \end{align*}

(a)

\begin{align*} \text{P(not yellow counter)} & = 1 - P\text{(yellow counter)} \\ & = 1 - {3 \over 8 + 9 + 3} \\ & = {17 \over 20} \end{align*}

(b)

\begin{align*} \text{P(red counter)} & = {8 \over (8 + 9 + 3) - x} \\ {2 \over 3} & = {8 \over 20 - x} \\ \\ 2(20 - x) & = 3(8) \phantom{00000} [\text{Cross-multiply}] \\ 40 - 2x & = 24 \\ -2x & = 24 - 40 \\ -2x & =-16 \\ x & = {-16 \over -2} \\ x & = 8 \end{align*}

\begin{align*} (p + 2)(3p - 5) & = 0 \\ \\ p + 2 = 0 \phantom{00} & \text{ or } \phantom{00} 3p - 5 = 0 \\ p = -2 \phantom{.} & \phantom{00000000} 3p = 5 \\ & \phantom{000000000} p = {5 \over 3} \end{align*}

(a)

\begin{align*} 33\% & = 0.33 \\ \\ {1 \over 3} & = 0.333333333... \\ & = 0.\dot{3} \\ \\ \therefore 33\% & < {1 \over 3} \end{align*}

(b)

\begin{array} {r l c r l } 6 & \le 2x + 11 & \phantom{0} \text{ or } \phantom{0} & 2x+ 11 & < 19 \\ -5 & \le 2x & & 2x & < 8 \\ -2.5 & \le x & & x & < 4 \\ \\ & & \therefore -2.5 \le x < 4 \end{array}

(a)

\begin{align*} P & = k Q^3 \\ \\ \text{When } & Q = {1 \over 3} \text{ and } P = 2, \\ 2 & = k \left(1 \over 3\right)^3 \\ 2 & = k \left(1 \over 27\right) \\ 2(27) & = k \\ 54 & = k \\ \\ P & = 54Q^3 \\ \\ \text{When } & Q = {1 \over 6}, \\ P & = 54 \left(1 \over 6\right)^3 \\ P & = {1 \over 4} \end{align*}

(b)

\begin{align*} \require{cancel} P & = kQ^3 \\ \\ k & = {P \over Q^3} \\ \\ \text{When } Q \text{ is multiplied by } & {1 \over 2}, \phantom{0} P \text{ changes by factor of } m \\ k & = {mP \over \left({1 \over 2}Q\right)^3} \\ & = {mP \over {1 \over 8}Q^3} \\ & = 8m \left(P \over Q^3\right) \\ \\ \text{Since } & k = {P \over Q^3}, \\ \cancel{ {P \over Q^3} } & = 8m \cancel{ \left(P \over Q^3\right) } \\ 1 & = 8m \\ {1 \over 8} & = m \end{align*}

(a)

\begin{align*} x^2 - y^4 & = (x)^2 - (y^2)^2 \\ & = (x + y^2)(x - y^2) \phantom{000000} [ a^2 - b^2 = (a + b)(a - b)] \end{align*}

(b)

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

(a)

\begin{align*} \overrightarrow{AB} & = {-3 \choose 7} \\ \\ \overrightarrow{BA} & = {3 \choose -7} \\ \\ x \text{-coordinate of } A & = 5 + 3 = 8 \\ y \text{-coordinate of } B & = -2 - 7 = -9 \\ \\ \therefore & \phantom{.} A(8, -9) \end{align*}

(b)

\begin{align*} \left| {-3 \choose 7} \right| & = \sqrt{ (-3)^2 + (7)^2} \\ & = \sqrt{58} \\ & = 7.6158 \\ & \approx 7.62 \text{ units} \end{align*}

(a)

\begin{align*} \angle NSD & = 360^\circ - 317^\circ \phantom{000} (\angle \text{s at a point}) \\ & = 43^\circ \\ \\ \angle NDS & = 180^\circ - 43^\circ \phantom{000} (\text{Interior } \angle \text{s}) \\ & = 137^\circ \\ \\ \text{Bearing of Singapore from Delhi} & = 137^\circ \end{align*}

(b)

\begin{align*} \angle NDT & = 137^\circ - 55^\circ \\ & = 82^\circ \\ \\ \angle NTD & = 180^\circ - 82^\circ \phantom{000} (\text{Interior } \angle \text{s}) \\ & = 98^\circ \\ \\ \text{Reflex } \angle NTD & = 360^\circ - 98^\circ \phantom{000} (\angle \text{s at a point}) \\ & = 262^\circ \\ \\ \text{Bearings of Delhi from Tokyo} & = 262^\circ \end{align*}

(a)

\begin{align*} \angle OAT & = \angle OBT = 90^\circ \phantom{000} (\text{Tangent perpendicular to radius) [R]} \\ \\ OT & \text{ is a common side [H]} \\ \\ OA & = OB \phantom{000} (\text{Radius of circle ) [S]} \\ \\ \therefore \text{Triangles } & OAT \text{ and } OBT \text{ are congruent (RHS)} \end{align*}

(b)

(a)

\begin{align*} \text{Nur's speed} & = \text{Gradient from 10 am to 11 am} \\ & = {3.6 - 0 \over 11 - 10} \\ & = 3.6 \text{ km/h} \end{align*}

(b)(i)

$$ 12.24 \text{ pm} $$

(b)(ii)

\begin{align*} \text{Distance from village B} & = 14.8 - 6.8 \\ & = 8 \text{ km} \end{align*}

(c)

\begin{align*} 8 \text{ km/h} & = {8 \text{ km} \over 1 \text{ hour} } \\ & = {8000 \text{ m} \over 3600 \text{ seconds}} \\ & = 2{2 \over 9} \text{ m/s} \end{align*}

(a)(i)

\begin{align*} \xi & = \{ 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 \} \\ \\ R & = \{ 5, 6, 10, 15 \} \end{align*}

(a)(ii)

\begin{align*} \text{P} & = \{ 5, 7, 11, 13 \} \\ \\ P \cap R & = \{ 5 \} \end{align*}

(b)

\begin{align*} \text{Q} & = \{ 8, 12 \} \\ \\ \text{Correct statements: } & R \cap Q = \emptyset, 16 \notin P \end{align*}

\begin{align*} \text{Let } x \text{ denote Ben's} & \text{ initial savings} \\ \\ \text{Ann's initial savings} & = {7 \over 5} x \\ & = \$ 1.4x \\ \\ \text{Ann's latest savings} & = \$ (1.4x - 60) \\ \\ \text{Ben's latest savings} & = \$ (x - 60) \\ \\ {x - 60 \over 1.4x - 60} & = {2 \over 3} \\ 3(x- 60) & = 2(1.4x - 60) \phantom{00000} [\text{Cross-multiply}] \\ 3x - 180 & = 2.8x - 120 \\ 3x - 2.8x & = -120 + 180 \\ 0.2x & = 60 \\ x & = {60 \over 0.2} \\ x & = 300 \\ \\ \text{Amount of money in Ann's account} & = 1.4(300) - 60 \\ & = \$ 360 \end{align*}

(a)

\begin{align*} \text{Scale factor} & = {4 \over 10} \\ & = {2 \over 5} \\ \\ {h_{small} \over h_{large} } & = {2 \over 5} \\ {h_{small} \over 30} & = {2 \over 5} \\ h_{small} & = {2 \over 5} (30) \\ h_{small} & = 12 \text{ cm} \end{align*}

(b)

\begin{align*} {V_{small} \over V_{large}} & = {125 \over 500} \\ & = {1 \over 4} \\ \\ \left( {l_{small} \over l_{large} } \right)^3 & = {V_{small} \over V_{large} } \\ {l_{small} \over l_{large} } & = \sqrt[3] {V_{small} \over V_{large} } \\ {4 \over l_{large} } & = \sqrt[3]{1 \over 4} \\ {4 \over l_{large} } & = 0.62996 \\ \\ 4 & = 0.62996 l_{large} \\ \\ l_{large} & = {4 \over 0.62996} \\ & = 6.3496 \\ & \approx 6.35 \text{ cm} \end{align*}

(a)

\begin{align*} \textbf{Q} & = \left( \begin{matrix} 24 & 14 \\ 29 & x \\ 20 & 15 \end{matrix} \right) \end{align*}

(b)

\begin{align*} \textbf{T} & = \textbf{QP} \\ & = \left( \begin{matrix} 24 & 14 \\ 29 & x \\ 20 & 15 \end{matrix} \right) \left( \begin{matrix} 10 & 8 \\ 40 & 45 \end{matrix} \right) \\ & = \left( \begin{matrix} 24 \times 10 + 14 \times 40 & 24 \times 8 + 14 \times 45 \\ 29 \times 10 + x \times 40 & 29 \times 8 + x \times 45 \\ 20 \times 10 + 15 \times 40 & 20 \times 8 + 15 \times 45 \end{matrix} \right) \\ & = \left( \begin{matrix} 800 & 822 \\ 290 + 40x & 232 + 45x \\ 800 & 835 \end{matrix} \right) \end{align*}

(c)

\begin{align*} 800 - (290 + 40x) & = 70 \\ 800 - 290 - 40x & = 70 \\ 510 - 40x & = 70 \\ -40x & = 70 - 510 \\ -40x & = -440 \\ x & = {-440 \over -40} \\ x & = 11 \end{align*}

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ \\ m_{AB} & = {1 - (-5) \over -2 - 7} \\ & = -{2 \over 3} \\ \\ m_{PQ} \times m_{AB} & = -1 \\ m_{PQ} & = -{1 \over m_{AB}} \\ & = -{1 \over -{2 \over 3}} \\ & = {3 \over 2} \\ \\ y & = mx + c \\ y & = {3 \over 2}x + c \\ \\ \text{Using } &P(5, 4), \\ 4 & = {3 \over 2}(5) + c \\ 4 & = {15 \over 2} + c \\ \\ c & = 4 - {15 \over 2} \\ c & = -{7 \over 2} \\ \\ \therefore \text{Equation of line } PQ: \phantom{0} y & = {3 \over 2}x - {7 \over 2} \end{align*}

(a) Since the diagram is drawn to scale and there are different copies of the question, my measurements may differ from yours.

\begin{align*} \text{Length (on plan)} & = 5.25 \text{ cm} \\ \\ \text{Breadth (on plan)} & = 4.35 \text{ cm} \\ \\ \text{Area (on plan)} & = 5.25 \times 4.35 \\ & = 22.8375 \text{ cm}^2 \\ \\ \text{Plan } & : \text{ Actual} \\ 22.8375 \text{ cm}^2 & : 19.2 \text{ m}^2 \\ 1 \text{ cm}^2 & : 0.84972 \text{ m}^2 \\ \sqrt{1} \text{ cm} & : \sqrt{0.84972} \text{ m} \\ 1 \text{ cm} & : 0.91691 \text{ m} \\ \\ n & \approx 0.917 \end{align*}

(b)

\begin{align*} & 1 \text{ litre of varnish covers } 16 \text{ m}^2 \\ & 500 \text{ ml of varnish covers } 8 \text{ m}^2 \\ & 250 \text{ ml of varnish covers } 4 \text{ m}^2 \\ \\ \text{Least amount of money} & = 24 + 2(42.50) \\ & = \$ 109 \end{align*}

(a)(i)

$$ 180 \le \text{Height} \le 200 $$

(a)(ii)

\begin{align*} \text{Mean} & = {4 \times 110 + 5 \times 130 + 8 \times 150 + 11 \times 170 + 14 \times 190 + 12 \times 210 + 6 \times 230 \over 4 + 5 + 8 + 11 + 14 + 12 + 6} \\ & = 178{2 \over 3} \end{align*}

(b)(i)

\begin{align*} 1.92 \text{ m} & = 192 \text{ cm} \\ \\ \text{No. of trees below 192 cm} & = 36 \\ \\ \text{No. of trees at least 192 cm} & = 60 - 36 \\ & = 24 \end{align*}

(b)(ii)

\begin{align*} \text{Lower quartile, } Q_1 & = 156 \text{ cm} \\ \\ \text{Upper quartile, } Q_3 & = 204 \text{ cm} \\ \\ \text{Interquartile range} & = Q_3 - Q_1 \\ & = 204 - 156 \\ & = 48 \end{align*}

(b)(c)

\begin{align*} \text{The height of second set of trees is more consistent as the interquartile range is lower} \end{align*}

(a)(i)

\begin{align*} \text{Sum of first } n \text{ terms, } S_n & = pn^2 + qn \\ \\ \text{Sum of first two terms, } S_2 & = p(2)^2 + q(2) \\ 8 + 11 & = 4p + 2q \\ \\ 4p + 2q & = 19 \text{ (Shown)} \end{align*}

(a)(ii)

\begin{align*} p + q & = 8 \phantom{000} \text{ --- (1)} \\ 4p + 2q & = 19 \phantom{000} \text{ --- (2)} \\ \\ (1) & \times 2, \\ 2p + 2q & = 16 \phantom{000} \text{ --- (3)} \\ \\ (2) & - (3), \\ 4p + 2q - (2p + 2q) & = 19 - 16 \\ 4p + 2q - 2p - 2q & = 3 \\ 2p & = 3 \\ p & = {3 \over 2} \\ p & = 1.5 \\ \\ \text{Substitute } & p = 1.5 \text{ into } (2), \\ 4(1.5) + 2q & = 19 \\ 6 + 2q & = 19 \\ 2q & = 19 - 6 \\ 2q & = 13 \\ q & = {13 \over 2} \\ q & = 6.5 \end{align*}

(b)

\begin{align*} \text{Sum of first } n \text{ terms, } S_n & = 3n^2 + 7n \\ \\ S_{10} & = 3(10)^2 + 7(10) \\ & = 370 \\ \\ S_9 & = 3(9)^2 + 7(9) \\ & = 306 \\ \\ T_{10} & = S_{10} - S_{9} \\ & = 370 - 306 \\ & = 64 \end{align*}

(a)

\begin{align*} {4p^2r \over 3} \div {2r^3 \over p} & = {4p^2 r \over 3} \times {p \over 2r^3} \\ & = {4p^3 r \over 6 r^3} \\ & = {2 p^3 \over 3 r^2} \end{align*}

(b)(i)

\begin{align*} a & = {3(6) + 4(-2) \over 5 - (6)} \\ & = -10 \end{align*}

(b)(ii)

\begin{align*} {a \over 1} & = {3b + 4c \over 5 - b} \\ a(5 - b) & = 3b + 4c \phantom{00000} [\text{Cross-multiply}] \\ 5a - ab & = 3b + 4c \\ -ab - 3b & = 4c - 5a \\ -b(a + 3) & = 4c - 5a \\ -b & = {4c - 5a \over a + 3} \\ b & = -{4c - 5a \over a + 3} = {5a - 4c \over a + 3} \phantom{00000} [\text{Either is fine}] \end{align*}

(c)(i)

\begin{align*} 9 - 7x + x^2 & = x^2 - 7x + 9 \\ & = x^2 - 7x + \left(7 \over 2\right)^2 - \left(7 \over 2\right)^2 + 9 \phantom{000000} [\text{Complete the square}] \\ & = \left( x - {7 \over 2}\right)^2 - {49 \over 4} + 9 \\ & = (x - 3.5)^2 - 3.25 \\ & = -3.25 + (-3.5 + x)^2 \end{align*}

(c)(ii)

\begin{align*} y & = 9 - 7x + x^2 \\ & = (x - 3.5)^2 - 3.25 \\ \\ \text{Turning} & \text{ point is } (3.5, -3.25) \end{align*}

(d)

\begin{align*} {1 \over x - 3} + {6 \over x - 1} & = 2 \\ {x - 1 \over (x - 3)(x - 1)} + {6(x - 3) \over (x - 3)(x - 1)} & = 2 \\ {x - 1 + 6(x - 3) \over (x - 3)(x - 1)} & = 2 \\ {x - 1 + 6x - 18 \over x^2 - x - 3x + 3} & = 2 \\ {7x - 19 \over x^2 - 4x + 3} & = {2 \over 1} \\ 7x - 19 & = 2(x^2 - 4x + 3) \phantom{000000} [\text{Cross-multiply}] \\ 7x - 19 & = 2x^2 - 8x + 6 \\ 0 & = 2x^2 - 8x - 7x + 6 + 19 \\ 0 & = 2x^2 - 15x + 25 \\ 0 & = (x - 5)(2x - 5) \\ \\ x - 5 & = 0 \phantom{0} \text{ or } \phantom{0} 2x - 5 = 0 \\ x & = 5\phantom{00000000} 2x = 5 \\ & \phantom{=5000000000..} x = {5 \over 2} \end{align*}

(a)

\begin{align*} \text{By Cosine rule, } CX^2 & = CB^2 + XB^2 - 2(CB)(XB)\cos \angle CBX \\ & = (8.3)^2 + (6.4)^2 - 2(8.3)(6.4) \cos 27^\circ \\ & = 15.18946687 \\ \\ CX & = \sqrt{15.18946687} \\ & = 3.8973 \\ & \approx 3.90 \text{ cm} \end{align*}

(b)

\begin{align*} \text{By } & \text{Sine rule,} \\ {XB \over \sin \angle XAB} & = {AB \over \sin \angle AXB} \\ {6.4 \over \sin \angle XAB} & = {7.5 \over \sin 112^\circ} \\ 6.4\sin 112^\circ & = 7.5 \sin \angle XAB \phantom{000000} [\text{Cross-multiply}] \\ \sin \angle XAB & = {6.4 \sin 112^\circ \over 7.5} \\ \angle XAB & = \sin^{-1} \left( 6.4 \sin 112^\circ \over 7.5 \right) \\ & = 52.297 \\ & \approx 52.3 \text{ (1 d.p.)} \end{align*}

(c)

\begin{align*} \angle XBA & = 180^\circ - 112^\circ - 52.297^\circ \phantom{0} (\angle \text{ sum of triangle}) \\ & = 15.703^\circ \\ \\ \angle ABC & = 27^\circ + 15.703^\circ \\ & = 42.703^\circ \\ \\ \text{Area of triangle ABC} & = {1 \over 2}ab \sin C \\ & = {1 \over 2} (AB)(BC) \sin \angle ABC \\ & = {1 \over 2} (7.5)(8.3) \sin 42.703^\circ \\ & = 21.108 \\ & \approx 21.1 \text{ cm}^2 \end{align*}

(a)

\begin{align*} \text{Price of smarphone in UK in \$} & = {389 \over 0.58} \\ & = \$ 670.68965 \\ \\ \$ 670.68965 - \$ 620 & = \$ 50.68965 \\ & \approx \$ 50.69 \\ \\ \text{Smartphone is cheaper by } & \$50.69 \text{ in Singapore than in the UK} \end{align*}

(b)

\begin{align*} \text{Price of tablet before sale} & = \$ 785 \times {100 \over 100 - 6} \\ & = \$ 835.106383 \\ & \approx \$ 835 \text{ (to nearest dollar)} \end{align*}

(c)(i)

\begin{align*} \text{Difference in mobile phone subscriptions} & = 8.21 \times 10^6 - 7.29 \times 10^6 \\ & = 0.92 \times 10^6 \\ & = 9.2 \times 10^5 \end{align*}

(c)(ii)

\begin{align*} \text{Percentage increase of broadband subscriptions} & = { \text{Increase} \over \text{Original} } \times 100 \\ & = { 1.20 \times 10^7 - 7.85 \times 10^6 \over 7.85 \times 10^6} \times 100 \\ & = 52{136 \over 157} \% \end{align*}

(c)(iii)

\begin{align*} \text{Mean no. of SMS messages sent per person} & = { \text{SMS messages} \over \text{No. of people} } \\ & = {1.14 \times 10^{10} \over 5.54 \times 10^6} \\ & = 2057.761733 \\ \\ \text{Mean no. of SMS messages sent per person per day} & = {2057.761733 \over 365} \\ & = 5.6377 \\ & \approx 5.64 \end{align*}

(a)

\begin{align*} \text{When } & x = -4, \\ y & = {(-4)^3 \over 5} - 2(-4) + 1 \\ & = -3.8 \end{align*}

(b)

(c)

$$ \text{From graph, } x > 3.75 $$

(d)(i)

(d)(ii)

\begin{align*} y = {x^3 \over 5} - 2x + 1 & \phantom{00} \text{ --- (1)} \\ \\ 5y + x = 10& \phantom{00} \text{ --- (2)} \\ \\ \text{Substitute } (1) & \text{ into } (2), \\ 5\left( {x^3 \over 5} - 2x + 1 \right) + x & = 10 \\ x^3 - 10x + 5 + x & = 10 \\ x^3 - 10x + x + 5 - 10 & = 0 \\ x^3 - 9 - 5 & = 0 \phantom{0} \text{ (Shown)} \end{align*}

(d)(iii)

\begin{align*} \text{From graph, } x = -2.7 \text{ or } -0.6 \text{ or } 3.25 \end{align*}

(a)(i)

\begin{align*} \text{Top number} & = t \\ \\ \text{Next number} & = t + 6 \\ \\ \text{Next number} & = t + 6 + 6 \\ & = t + 12 \\ \\ \text{Bottom number} & = t + 12 + 6 \\ & = t + 18 \\ \\ (t)(t + 18) & = t^2 + 18t \end{align*}

(a)(ii)

\begin{align*} \text{Difference} & = (t + 6)(t + 12) - (t^2 + 18t) \\ & = t^2 + 12t + 6t + 72 - t^2 - 18t \\ & = t^2 - t^2 + 12t + 6t - 18t + 72 \\ & = 72 \\ \\ \therefore \text{Difference } & \text{is always divisible by } 72 \end{align*}

(a)(iii)

\begin{align*} \text{Sum of four numbers} & = t + (t + 6) + (t + 12) + (t + 18) \\ & = 4t + 36 \\ \\ 360 & = 4t + 36 \\ 360 - 36 & = 4t \\ 324 & = 4t \\ \\ t & = {324 \over 4} \\ & = 81 \\ \\ \text{Largest number} & = 81 + 3(6) \\ & = 99 \end{align*}

(b)(i)

\begin{align*} 3\text{rd term, } T_3 & = 36 \\ \\ \text{Let common difference} & = d \\ \\ T_4 & = 36 + d \\ \\ T_5 & = 36 + 2d \\ \\ T_6 & = 36 + 3d \\ \\ 36 + 3d & = 60 \\ 3d & = 60 - 36 \\ 3d & = 24 \\ d & = 8 \\ \\ T_n & = a + (n - 1)(d) \phantom{000000} [a \text{: first term}] \\ & = (36 - 8 - 8) + (n - 1)(8) \\ & = 20 + 8n - 8 \\ & = 12 + 8n \end{align*}

(b)(ii)

\begin{align*} T_n & = 12 + 8n \\ & = 4(3 + 2n) \\ \\ \text{Since } 4 \text{ is a factor} & \text{ of } T_n, \text{ all terms are multiples of 4} \end{align*}

(a)(i)

\begin{align*} \angle AOD & = 180^\circ - 23^\circ - 23^\circ \phantom{000} (\angle \text{ sum of isosceles triangle, } OA = OD) \\ & = 134^\circ \\ \\ \angle ABD & = 134^\circ \div 2 \phantom{000} (\angle \text{ at centre = 2 } \times \angle \text{ at circumference}) \\ & = 67^\circ \end{align*}

(a)(ii)

\begin{align*} \angle AED & = 180^\circ - \angle ABD \phantom{000} (\angle \text{s in opposite segment}) \\ & = 180^\circ - 67^\circ \\ & = 113^\circ \\ \\ \angle EAD & = 180^\circ - 113^\circ - 42^\circ \phantom{000} (\angle \text{ sum of triangle}) \\ & = 25^\circ \end{align*}

(b)(i)

\begin{align*} \text{Let } \angle POQ & = \theta \\ \\ \text{Arc length } PQ & = r \theta \\ & = 4\theta \text{ cm} \\ \\ \text{Perimeter of minor sector } OPQ & = 4\theta + 4 + 4 \\ & = (4 \theta + 8 ) \text{ cm} \\ \\ 4\theta + 8 & = 15.2 \\ 4\theta & = 15.2 - 8 \\ 4\theta & = 7.2 \\ \theta & = {7.2 \over 4} \\ \theta & = 1.8 \text{ radians} \end{align*}

(b)(ii)

\begin{align*} \text{Area of sector} & = {1 \over 2} r^2 \theta \\ \\ \text{Area of minor sector } OPQ & = {1 \over 2} (4)^2 (1.8) \\ & = 14.4 \text{ cm}^2 \\ \\ \text{Reflex } \angle POQ & = 2\pi - 1.8 \phantom{000} (\angle \text{s at a point}) \\ & = 4.4832 \text{ radians} \\ \\ \text{Area of major sector } ORS & = {1 \over 2} (6)^2 (4.4832) \\ & = 80.6976 \text{ cm}^2 \\ \\ \text{Area of major sector } OPQ & = {1 \over 2} (4)^2 (4.4832) \\ & = 35.8656 \text{ cm}^2 \\ \\ \text{Total shaded area} & = 14.4 + (80.6976 - 35.8656) \\ & = 59.232 \\ & \approx 59.2 \text{ cm}^2 \end{align*}

(a)(i)

\begin{align*} \overrightarrow{PQ} & = {-5 -4 \choose 1 - (-3)} \\ & = {-9 \choose 4} \end{align*}

(a)(ii)

\begin{align*} \overrightarrow{OR} & = {h \choose 3} \\ \\ \overrightarrow{PR} & = \overrightarrow{PO} + \overrightarrow{OR} \\ & = -{4 \choose -3} + {h \choose 3} \\ & = {-4 + h \choose 6} \\ \\ \overrightarrow{PR} & = k\overrightarrow{PQ} \\ {-4 + h \choose 6} & = k{-9 \choose 4} \\ {-4 + h \choose 6} & = {-9k \choose 4k} \end{align*} \begin{array} {r l c r l } -4 + h & = -9k & \phantom{0} \text{ or } \phantom{0} & 6 & = 4k \\ h & = -9k + 4 & & {6 \over 4} & = k \\ h & = -9(1.5) + 4 & & 1.5 & = k \\ h & = -9.5 \end{array}

(b)(i)

\begin{align*} OB : & \phantom{.} BC : OC \\ 2 : & \phantom{0} 3 \phantom{0}: 5 \\ \\ \overrightarrow{OC} & = {5 \over 2} \overrightarrow{OB} \\ & = {5 \over 2} \textbf{b} \\ \\ \overrightarrow{AC} & = \overrightarrow{AO} + \overrightarrow{OC} \\ & = -\textbf{a} + {5 \over 2} \textbf{b} \end{align*}

(b)(ii)

\begin{align*} AX : & \phantom{.} XB : AB \\ 1 : & \phantom{0} 2 \phantom{0}: 3 \\ \\ \overrightarrow{AB} & = \overrightarrow{AO} + \overrightarrow{OB} \\ & = -\textbf{a} + \textbf{b} \\ \\ \overrightarrow{XB} & = {2 \over 3} \left( -\textbf{a} + \textbf{b} \right) \\ & = -{2 \over 3} \textbf{a} + {2 \over 3} \textbf{b} \end{align*}

(b)(iii)

\begin{align*} \text{Since } Y & \text{ is a point on } OC, \\ \overrightarrow{BY} & = k\overrightarrow{OB} \\ & = k \textbf{b}, \text{ where } k \text{ is a constant} \\ \\ \overrightarrow{XY} & = \overrightarrow{XB} + \overrightarrow{BY} \\ & = -{2 \over 3}\textbf{a} + {2 \over 3}\textbf{b} + k\textbf{b} \\ & = -{2 \over 3}\textbf{a} + \left({2 \over 3} + k \right)\textbf{b} \\ \\ \text{Since } \overrightarrow{XY} & \text{ is parallel to } \overrightarrow{AC}, \\ \overrightarrow{XY} & = \lambda \overrightarrow{AC} \\ -{2 \over 3}\textbf{a} + \left({2 \over 3} + k \right)\textbf{b} & = \lambda \left( -\textbf{a} + {5 \over 2} \textbf{b} \right) \\ -{2 \over 3}\textbf{a} + \left({2 \over 3} + k \right)\textbf{b} & = -\lambda \textbf{a} + {5 \over 2}\lambda \textbf{b} \end{align*} \begin{array} {r l c r l } -\lambda & = -{2 \over 3} & \phantom{0} \text{ or } \phantom{0} & {2 \over 3} + k & = {5 \over 2} \lambda \\ \lambda & = {2 \over 3} & & k & = {5 \over 2}\lambda - {2 \over 3} \\ & & & k & = {5 \over 2}\left(2 \over 3\right) - {2 \over 3} \\ & & & k & = 1 \end{array}
\begin{align*} \therefore \overrightarrow{XY} & = -{2 \over 3} \textbf{a} + \left({2 \over 3} + 1 \right) \textbf{b} \\ & =-{2 \over 3} \textbf{a} + {5 \over 3} \textbf{b} \end{align*}

(a)

\begin{align*} \text{Area of trapezium} & = {1 \over 2}(a + b)(h) \\ \\ \text{Cross-sectional area } ABCG & = {1 \over 2}(4 + 5)(6) \\ & = 27 \text{ m}^2 \\ \\ \text{Cross-sectional area } CDEF & = {1 \over 2} (5 + 1.5 + 4) (10) \\ & = 52.5 \text{ m}^2 \\ \\ \text{Volume of barn} & = \text{Total cross-sectional area} \times \text{Length} \\ & = (27 + 52.5) \times (25) \\ & = 1987.5 \text{ m}^3 \end{align*}

(b)

\begin{align*} \text{By Pythagoras theorem, } AG^2 & = 6^2 + (5 - 4)^2 \\ & = 37 \\ \\ AG & = \sqrt{37} \text{ m} \\ \\ \text{By Pythagoras theorem, } FE^2 & = 10^2 + (1.5 + 5 - 4)^2 \\ & = 106.25 \\ \\ FE & = \sqrt{106.25} \text{ m} \\ \\ \text{Total area of two sloping roofs} & = \sqrt{37} \times 25 + \sqrt{106.25} \times 25 \\ & = 409.763 \\ & \approx 410 \text{ m}^2 \end{align*}

(c)

\begin{align*} \text{Let } N \text{ denote the foot of } & \text{perpendicular of } P \\ \\ \text{By Pythagoras theorem, } PD^2 & = 25^2 + 10^2 \\ & = 725 \\ \\ PD & = \sqrt{725} \text{ m} \\ \\ \tan \theta & = {Opp \over Adj} \\ \tan \theta & = {1.5 + 5 \over \sqrt{725}} \\ \theta & = \tan^{-1} \left( 6.5 \over \sqrt{725} \right) \\ & = 13.571 \\ & \approx 13.6^\circ \\ \\ \therefore \text{Angle of elevation of } P \text{ from } D & = 13.6^\circ \end{align*}

(a)(i)

\begin{align*} \text{Median position} & = {24 + 1 \over 2} \\ & = 12.5 \\ \\ \text{Median score for group A} & = {71 + 72 \over 2} \phantom{000000} [\text{Average of 12th & 13th}] \\ & = 71.5 \end{align*}

(a)(ii)

\begin{align*} \text{Range for group B} & = \text{Maximum score} - \text{Minimum score} \\ & = 93 - 50 \\ & = 43 \end{align*}

(a)(iii)

\begin{align*} \text{No. of students in Group A who scored more than 70} & = 14 \\ \\ \text{Required percentage} & = {14 \over 24} \times 100 \\ & = 58{1 \over 3} \% \end{align*}

(a)(iv)

\begin{align*} & \text{A greater proportion of students were awarded merit in group B (65%) than in group A } \left(58{1 \over 3}\% \right) \\ \\ & \text{A greater proportion of students were awarded distinction in group A } \left(20{5 \over 6}\%\right) \text{ than in group B (20%)} \end{align*}

(b)(i)

\begin{align*} {9 \over 25}& \times \underbrace{9 \over 25}_\text{Wrong} = {81 \over 625} \\ \\ \text{He should use } {8 \over 24} & \text{ as the counters are drawn without replacement} \end{align*}

(b)(ii)

(b)(iii)

\begin{align*} \text{P(only one counter is red)} & = \text{P(red, blue)} + \text{P(blue, red)} \\ & = {16 \over 25} \times {3 \over 8} + {9 \over 25} \times {2 \over 3} \\ & = {12 \over 25} \end{align*}

(a)

\begin{align*} \text{Amount of fuel used} & = {6.3 \text{ litres} \over 100 \text{ km} } \times 92 \text{ km} \\ & = 5.796 \\ & \approx 5.8 \text{ litres} \text{ (Shown)} \end{align*}

(b)

\begin{align*} \text{Distance} & = \text{Speed} \times \text{Time} \\ & = 85 \text{ km/h} \times {45 \over 60} \text{ h} \\ & = 63.75 \text{ km} \\ \\ \text{Amount of fuel used} & = {4.2 \text{ litres} \over 100 \text{ km}} \times 63.75 \text{ km} \\ & = 2.6775 \text{ litres} \end{align*}

(c)

\begin{align*} \text{Let distance of first stage} & = x \text{ km} \\ \\ \text{Time taken for first stage} & = { \text{Distance} \over \text{Speed} } \\ & = {x \text{ km} \over 60 \text{ km/h}} \\ & = {x \over 60} \text{ h} \\ \\ \text{Distance of second stage} & = (x - 25) \text{ km} \\ \\ \text{Time taken for second stage} & = { x - 25 \over 75} \text{ h} \\ \\ 3 \text{ h 15 mins} & = 3{15 \over 60} \text{ h} \\ & = 3.25 \text{ h} \\ \\ {x \over 60} + {x - 25 \over 75} & = 3.25 \\ {75x \over 60(75)} + {60(x - 25) \over 60(75)} & = 3.25 \\ {75x + 60(x - 25) \over 4500} & = 3.25 \\ {75x + 60x - 1500 \over 4500} & = 3.25 \\ {135x - 1500 \over 4500} & = 3.25 \\ 135x - 1500 & = 4500(3.25) \\ 135x - 1500 & = 14625 \\ 135x & = 14625 + 1500 \\ 135x & = 16125 \\ x & = {16125 \over 135} \\ x & = 119{4 \over 9} \\ \\ \text{Total distance} & = 119{4 \over 9} + \left( 119{4 \over 9} - 25 \right) \\ & = 213 {8 \over 9} \text{ km} \\ \\ \text{Assuming type of driving} & \text{ is 'Combined'}, \\ \text{Amount of fuel used} & = {5.0 \text{ litres} \over 100 \text{ km}} \times 213{8 \over 9} \text{ km} \\ & = 10{25 \over 36} \text{ litres} \\ \\ \text{Price of fuel} & = 10{25 \over 36} \times \$ 2.07 \times {100 - 5 \over 100} \\ & = \$21.030625 \\ \\ \text{Amount Hamid has to pay} & = 21.030625 \div 2 \\ & = 10.5153125 \\ & \approx \$ 10.52 \end{align*}