\begin{align*} {1 \over 81} & = 3^k \\ {1 \over 3^4} & = 3^k \\ 3^{-4} & = 3^k \phantom{000000} \left[ a^{-n} = {1 \over a^n} \right] \\ \\ \therefore k & = -4 \end{align*}

$$ - 3 < x \le 6.5 $$

(a)

\begin{align*} \text{Range} & = \text{Maximum} - \text{Minimum} \\ & = 228 - 191 \\ & = 37 \text{ grams} \end{align*}

(b)

\begin{align*} \text{Total no. of apples} & = 20 \\ \\ \text{Median position} & = {20 + 1 \over 2} \\ & = 10.5 \\ \\ \text{Median mass} & = { 208 + 210 \over 2} \\ & = 209 \text{ grams} \end{align*}

(a)

\begin{align*} \text{Distance travelled} & = \text{Area under speed-time graph} \\ 693 & = {1 \over 2} \times 84 \times v \\ 693 & = 42 v \\ {693 \over 42} & = v \\ 16.5 & = v \\ \\ \text{Greatest speed} & = 16.5 \text{ m/s} \end{align*}

(b)

\begin{align*} \text{Acceleration} & = \text{Gradient of line} \\ & = {y_2 - y_1 \over x_2 - x_1} \\ & = {16.5 - 0 \over 34 - 0} \\ & = {33 \over 68} \text{ m/s}^2 \end{align*}

(a)

\begin{align*} \{ 2, 10 \} & \subset A \phantom{000000} [ \{2, 10 \} \text{ is a set}] \end{align*}

(b)

\begin{align*} 3 & \in B \end{align*}

(c)

\begin{align*} B \cap C & = \emptyset \end{align*}

\begin{align*} \text{Let no. of adults to join} & = x \\ \\ {58 + x \over 58 + 37 + x} & \ge {70 \over 100} \phantom{000000} \left[ 70 \% = {70 \over 100} \right] \\ {58 + x \over 95 + x} & \ge {7 \over 10} \\ {10 (58 + x) \over 10(95 + x)} & \ge {7(95 + x) \over 10(95 + x)} \\ 10(58 + x) & \ge 7(95 + x) \\ 580 + 10x & \ge 665 + 7x \\ 10x - 7x & \ge 665 - 580 \\ 3x & \ge 85 \\ x & \ge {85 \over 3} \\ x & \ge 28{1 \over 3} \\ \\ \text{Smallest number} & = 29 \end{align*}

\begin{align*} \angle CDE & = 108^\circ \\ \\ \angle EDF & = 360^\circ - 96^\circ - 108^\circ \phantom{0} \text{(Angles at a point)} \\ & = 156^\circ \\ \\ \text{Let no. of sides} & = n \\ \\ \text{Sum of interior angles} & = (n - 2) \times 180^\circ \\ 156^\circ \times n & = 180^\circ (n - 2) \\ 156n & = 180n - 360 \\ 156n - 180n & = -360 \\ -24n & = -360 \\ n & = {-360 \over -24} \\ n & = 15 \end{align*}

\begin{align*} \text{Volume of cylinder} & = \text{Volume of hemisphere} \\ \pi r^2 h & = {2 \over 3} \pi r^3 \\ h & = { {2 \over 3} \pi r^3 \over \pi r^2} \\ h & = {2 \over 3} r \\ \\ \text{Total surface area of cylinder} & = \underbrace{ 2 \pi r^2 }_\text{Area of circles} + \underbrace{ 2 \pi r h }_\text{Curved surface area} \\ & = 2 \pi r^2 + 2 \pi r \left({2 \over 3} r\right) \\ & = 2 \pi r^2 + {4 \over 3} \pi r^2 \\ & = {10 \over 3} \pi r^2 \text{ cm}^2 \end{align*}

\begin{align*} {3x - 4 \over 2} - {2x \over 3} & = {1 \over 1} \\ {3(3x - 4) \over 6} - {4x \over 6} & = {6 \over 6} \\ 3(3x - 4) - 4x & = 6 \\ 9x - 12 - 4x & = 6 \\ 9x - 4x & = 6 + 12 \\ 5x & = 18 \\ x & = {18 \over 5} \\ x & = 3.6 \end{align*}

\begin{align*} \overrightarrow{BC} & = \overrightarrow{BA} + \overrightarrow{AC} \\ & = {-5 \choose 1} + {-1 \choose 8} \\ & = {-6 \choose 9} \\ \\ | \overrightarrow{BC} | & = \sqrt{x^2 + y^2} \\ & = \sqrt{ (-6)^2 + 9^2 } \\ & \approx 10.8 \text{ units} \end{align*}

\begin{align*} \text{Total amount} & = P \left(1 + {r \over 100} \right)^n \\ P + 674.79 & = P \left(1 + {2.5 \over 100}\right)^4 \\ P + 674.79 & = P (1.1038) \\ P + 674.79 & = 1.1038P \\ P - 1.1038P & = -674.79 \\ -0.1038P & = -674.79 \\ P & = { -674.79 \over -0.1038} \\ P & \approx \$ 6 \phantom{.} 500. 87 \end{align*}

(a)

\begin{align*} 2x^2 - 5x - 12 & = (2x + 3)(x - 4) \end{align*}

(b)

\begin{align*} 2x^2 - 5x - 12 & = (2x + 3)(x - 4) \\ \\ \text{Let } x & = 2y - 3, \\ 2(y - 3)^2 - 5(2y - 3) - 12 & = [2(2y - 3) + 3](2y - 3 - 4) \\ & = (4y - 6 + 3)(2y - 7) \\ & = (4y - 3)(2y - 7) \end{align*}

\begin{align*} \angle ABC & = 90^\circ \phantom{0} \text{ (Right angle in semi-circle)} \\ \\ \tan \angle ACB & = {AB \over BC} \phantom{000000} \left[ {Opp \over Adj} \right] \\ & = {8.1 \over 13.8} \\ \angle ACB & = \tan^{-1} \left(8.1 \over 13.8\right) \\ & = 30.411^\circ \\ \\ x^\circ & = \angle ACB \phantom{0} \text{ (Angles in the same segment)} \\ & = 30.411^\circ \\ x & \approx 30.4 \end{align*}

(a)

\begin{align*} \text{Mean} & = { \sum fx \over \sum f } \\ & = { 7910 \over 200 } \\ & = 39.55 \text{ minutes} \end{align*}

(b)

\begin{align*} \text{SD} & = \sqrt{ {\sum fx^2 \over \sum f} - \left(\sum fx \over \sum f\right)^2 } \\ & = \sqrt{ { 330 \phantom{.} 600 \over 200 } - \left( 7910 \over 200 \right)^2 } \\ & = 9.4232 \\ & \approx 9.42 \text{ minutes} \end{align*}

(c)

\begin{align*} \text{Angle representing 21 to 30} & = 99^\circ \phantom{000000} [\text{Use protractor to measure}] \\ \\ \text{Angle representing 41 to 60} & = 72^\circ \\ \\ \text{No. of adults aged 41 to 60} & = 44 \times {72 \over 99} \\ & = 32 \end{align*}

\begin{align*} \angle OBA & = 2x^\circ \phantom{0} \text{(Isosceles triangle } OAB \text{ with } OA = OB ) \\ \\ \angle OBC & = 5x^\circ \phantom{0} \text{(Isosceles triangle } OAB \text{ with } OC = OB ) \\ \\ \angle BOC & = 2x^\circ \phantom{0} \text{(Alternate angles, } AB \phantom{.} // \phantom{.} OC) \\ \\ 180 & = 5x + 5x + 2x \\ 180 & = 12x \\ {180 \over 12} & = x \\ 15 & = x \end{align*}

(a)

\begin{align*} 28 & = 2^2 \times 7 \\ \\ 63 & = 3^2 \times 7 \\ \\ 28 \times 63 & = 2^2 \times 3^2 \times 7^2 \\ \\ 28 \times 63 \text{ is a perfect square} & \text{ since powers of each factor is a multiple of 2} \end{align*}

(b)

\begin{align*} 28k & = 2^2 \times 7 \times k \\ & = 2^2 \times 7 \times (2 \times 7^2) \\ & = 2^3 \times 7^3 \\ \\ \therefore k & = 2 \times 7^2 \\ & = 98 \end{align*}

(a)

\begin{align*} \textbf{P} & = \left( \begin{matrix} 28 & 170 & 95 \\ x & 176 & 89 \end{matrix} \right) \end{align*}

(b)

\begin{align*} \textbf{R} & = \textbf{P} \left( \begin{matrix} 3 \\ 5 \\ 4 \end{matrix} \right) \\ & = \underset{2 \times 3}{ \left( \begin{matrix} 28 & 170 & 95 \\ x & 176 & 89 \end{matrix} \right) } \underset{3 \times 1}{ \left( \begin{matrix} 3 \\ 5 \\ 4 \end{matrix} \right) } \\ & = \underset{2 \times 1}{ \left( \begin{matrix} (28)(3) + (170)(5) + (95)(4) \\ 3x + (176)(5) + (89)(4) \end{matrix} \right) } \\ & = \left( \begin{matrix} 1314 \\ 3x + 1236 \end{matrix} \right) \end{align*}

(c)

\begin{align*} \text{The total charge in the first week is \$ 1314 while that of the second week is \$} (3x + 1236). \end{align*}

(d)

\begin{align*} 1314 & = 3x + 1236 \\ 1314 - 1236 & = 3x \\ 78 & = 3x \\ {78 \over 3} & = x \\ 26 & = x \end{align*}

(a)

\begin{align*} \text{Let common difference} & = d \\ \\ 39 - d - d - d - d & = 11 \\ 39 - 4d & = 11 \\ -4d & = 11 - 39 \\ -4d & = -28 \\ d & = {-28 \over -4} \\ d & = 7 \\ \\ p & = 39 - 7 = 32 \\ q & = 32 - 7 = 25 \\ r & = 25 - 7 = 18 \end{align*}

(b)

\begin{align*} T_n & = \underbrace{a}_\text{First term} + (n - 1)(d) \\ & = 39 + (n - 1)(-7) \\ & = 39 - 7(n - 1) \\ & = 39 - 7n + 7 \\ & = 46 - 7n \end{align*}

(c)

\begin{align*} \text{Let } T_n & = -246 \\ 46 - 7n & = -246 \\ -7n & = -246 - 46 \\ -7n & = -292 \\ n & = {-292 \over -7} \\ n & = 41{5 \over 7} \\ \\ \text{Since } n \text{ is not an integer, } & -246 \text{ is not a term of the sequence} \end{align*}

(a)

\begin{align*} & \text{Since minimum point of graph is } (0.75, -8.25), \text{ no solutions for } k < -8.25 \end{align*}

(b)

\begin{align*} y & = 2x^2 - 3x - 7 \phantom{0} \text{--- (1)} \\ y & = 5 - 2x \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 5 -2x & = 2x^2 - 3x - 7 \\ 0 & = 2x^2 - 3x + 2x - 7 - 5 \\ 0 & = 2x^2 - x - 12 \end{align*}

(c)(i)

\begin{align*} 2x^2 - 6x - 3 & = 0 \\ 2x^2 - 6x - 3 + 3x & = 0 + 3x \\ 2x^2 - 3x - 3 & = 3x \\ 2x^2 - 3x - 3 - 4 & = 3x - 4 \\ \underbrace{2x^2 - 3x - 7}_\text{Eqn of curve} & = 3x - 4 \\ \\ \therefore \text{Draw } & y = 3x - 4 \end{align*}

(c)(ii)

$$ y = 3x - 4 $$

$x$	$1$	$2$	$3$
$y$	$-1$	$2$	$5$

\begin{align*} \text{From graph, } x & = -0.45 \text{ or } 3.45 \end{align*}

(a)

\begin{align*} A & = {b(c + 2) \over 5 - c} \\ & = {(12.17)(1.615 + 2) \over 5 - 1.615} \\ & = 12.996 \\ & \approx 13.00 \text{ (2 d.p.)} \end{align*}

(b)

\begin{align*} {A \over 1} & = {b(c + 2) \over 5 - c} \\ A(5 - c) & = b(c + 2) \phantom{000000} [\text{Cross-multiply}] \\ 5A - Ac & = bc + 2b \\ - Ac - bc & = 2b - 5A \\ Ac + bc & = 5A - 2b \\ c(A + b) & = 5A - 2b \\ c & = {5A - 2b \over A + b} \end{align*}

(a)

\begin{align*} \text{Let volume} & = y \\ \text{Let cost} & = x \\ \\ y & = kx \phantom{000000} [\text{Direct proportion}] \\ {y \over x} & = k \\ \\ \text{When } x = 30.8 & \text{ and } y = 400, \\ k & = {400 \over 30.8} \\ k & = 12.987 \\ \\ \text{When } x = 19.25 & \text{ and } y = 250, \\ k & = {250 \over 19.25} \\ k & = 12.987 \\ \\ \text{When } x = 5.95 & \text{ and } y = 75, \\ k & = {75 \over 5.95} \\ k & = 12.605 \\ \\ \text{Since value of } k & \text{ is not same, cost is not proportional to volume} \end{align*}

(b)

\begin{align*} {V_1 \over V_2} & = \left(l_1 \over l_2\right)^3 \\ {75 \over 250} & = \left(h_1 \over 18.4\right)^3 \\ \sqrt[3]{75 \over 250} & = {h_1 \over 18.4} \\ 0.66943 & = {h_1 \over 18.4} \\ \\ h_1 & = 18.4(0.66943) \\ & = 12.317 \\ & \approx 12.3 \text{ cm} \end{align*}

(a)

\begin{align*} {x + 4 \over 5} & \ge {2 - x \over 3} \\ {3(x + 4) \over 15} & \ge {5(2 - x) \over 15} \\ 3(x + 4) & \ge 5(2 - x) \\ 3x + 12 & \ge 10 - 5x \\ 3x + 5x & \ge 10 - 12 \\ 8x & \ge -2 \\ x & \ge {-2 \over 8} \\ x & \ge -{1 \over 4} \end{align*}

(b)

\begin{align*} {2y \over 5 - 2y} - {3y \over (5 - 2y)^2 } & = {2y(5 - 2y) \over (5 - 2y)^2} - {3y \over (5 - 2y)^2} \\ & = {2y(5 - 2y) - 3y \over (5 - 2y)^2} \\ & = {10y - 4y^2 - 3y \over (5 - 2y)^2} \\ & = {7y - 4y^2 \over (5 - 2y)^2} \end{align*}

(c)

\begin{align*} {18 h^3 j^3 \over 5 k^3 } \div {3 h^5 k \over 10 j^2 } & = {18 h^3 j^3 \over 5 k^3 } \times { 10 j^2 \over 3 h^5 k } \\ & = { 180 h^3 j^5 \over 15 h^5 k^4 } \phantom{0000000000} [ a^m \times a^n = a^{m + n} ] \\ & = {12 j^5 \over h^2 k^4 } \end{align*}

(d)

\begin{align*} \left( 16 t^8 \over v^{12} \right)^{-{1 \over 4}} & = \left( v^{12} \over 16 t^8 \right)^{1 \over 4} \phantom{000000} \left[ \left(a \over b\right)^{-n} = \left(b \over a\right)^n \right] \\ & = \left( v^{12} \over 2^4 t^8 \right)^{1 \over 4} \\ & = { (v^{12})^{1 \over 4} \over (2^4)^{1 \over 4} (t^8)^{1 \over 4} } \\ & = { v^3 \over 2 t^2 } \phantom{000000000000} [ (a^m)^n = a^{mn} ] \end{align*}

(e)

\begin{align*} {2 \over x - 2} + {3 \over x + 4} & = 1 \\ {2(x + 4) \over (x - 2)(x + 4)} + {3(x - 2) \over (x + 4)(x - 2)} & = 1 \\ {2(x + 4) + 3(x - 2) \over (x - 2)(x + 4)} & = 1 \\ {2x + 8 + 3x - 6 \over (x - 2)(x + 4)} & = 1 \\ {5x + 2 \over (x - 2)(x + 4)} & = {1 \over 1} \\ 5x + 2 & = (x - 2)(x + 4) \phantom{000000} [\text{Cross-multiply}] \\ 5x + 2 & = x^2 + 4x - 2x - 8 \\ 0 & = x^2 + 2x - 5x - 8 - 2 \\ 0 & = x^2 - 3x - 10 \\ 0 & = (x - 5)(x + 2) \end{align*} \begin{align*} x - 5 & = 0 && \text{ or } & x + 2 & = 0 \\ x & = 5 &&& x & = -2 \end{align*}

(a)(i)

\begin{align*} \text{Mean monthly passengers} & = { 5.41 \times 10^7 \over 12 } \phantom{000000} [1 \text{ year} = 12 \text{ months}] \\ & = 4 \phantom{.} 508 \phantom{.} 333 \\ & \approx 4 \phantom{.} 510 \phantom{.} 000 \\ & = 4.51 \text{ million} \phantom{000000} [1 \text{ million} = 10^6 = 1 \phantom{.} 000 \phantom{.} 000 ] \end{align*}

(a)(ii)

\begin{align*} \text{Required percentage} & = { 5.09 \times 10^6 \over 5.41 \times 10^7 } \times 100 \\ & = 9.4085 \\ & \approx 9.41 \% \end{align*}

(b)

\begin{align*} \text{No. in 2014} & = (1.591 \times 10^6) \times {100 + 1.6 \over 100} \\ & = 1 \phantom{.} 616 \phantom{.} 456 \\ & = 1.616 \phantom{.} 456 \times 10^6 \end{align*}

(c)

\begin{align*} \text{No. in 2013} & = (1.509 \times 10^7) \times {100 \over 100 - 3.1} \\ & = 15 \phantom{.} 572 \phantom{.} 755 \\ & \approx 15 \phantom{.} 600 \phantom{.} 000 \\ & = 1.56 \times 10^7 \end{align*}

(a)

\begin{align*} y & = 50 \times 2^t \\ \\ \text{Let } & t = 0, \\ y & = 50 \times 2^0 \\ y & = 50 \end{align*}

(b)

\begin{align*} y & = 50 \times 2^t \\ \\ \text{Let } & t = 0.5, \\ y & = 50 \times 2^{0.5} \\ y & = 70.71 \\ y & \approx 70.7 \end{align*}

(c)

(d)

$$ \text{From graph, } t = 3.325 \text{ hours} $$

(e)(i)

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {400 - 60 \over 3.45 - 1} \\ & = 138.775 \\ & \approx 139 \end{align*}

(e)(ii)

\begin{align*} & \text{At } t = 2 \text{ hours, the number of bacteria } (y) \text{ is increasing at approx. 139 per hour} \end{align*}

(a)

\begin{align*} AB & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ & = \sqrt{ [4 - (-4)]^2 + (-2 - 10)^2 } \\ & = \sqrt{ 208 } \\ & \approx 14.4 \text{ units} \end{align*}

(b)

\begin{align*} \text{Gradient of } AB & = {y_2 - y_1 \over x_2 - x_1} \\ & = {-2 - 10 \over 4 - (-4) } \\ & = -1.5 \\ \\ y & = mx + c \\ y & = -1.5x + c \\ \\ \text{Using } & A(-4, 10), \\ 10 & = -1.5(-4) + c \\ 10 & = 6 + c \\ 10 - 6 & = c \\ 4 & = c \\ \\ \text{Eqn of } AB: & \phantom{.} y = -1.5x + 4 \end{align*}

(c)(i)

\begin{align*} \text{Line } p : & \phantom{.} 3x + 2y = 5 \\ \\ 2y & = -3x + 5 \\ y & = {1 \over 2}(-3x + 5) \\ y & = -1.5x + 2.5 \phantom{000000} [y = mx + c] \\ \\ \text{Gradient of line } p & = -1.5 \\ \\ \text{Since lines } AB \text{ and } & p \text{ have the same gradient but different } y \text{-intercepts,} \\ \text{both lines are parall} & \text{el and will not intersect.} \end{align*}

(c)(ii)

\begin{align*} \text{Line } p : & \phantom{.} y = -1.5x + 2.5 \phantom{0} \text{--- (1)} \\ \\ \text{Line } q : & \phantom{.} 6y = 4x - 37 \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 6(-1.5x + 2.5) & = 4x - 37 \\ -9x + 15 & = 4x - 37 \\ -9x - 4x & = -37 - 15 \\ -13x & = -52 \\ x & = {-52 \over -13} \\ x & = 4 \\ \\ \text{When } & x = 4 \text{ into (1),} \\ y & = -1.5(4) + 2.5 \\ y & = -3.5 \\ \\ \therefore & \phantom{.} (4, -3.5) \end{align*}

(a)

\begin{align*} \angle ACB & = 122^\circ - 64^\circ \\ & = 58^\circ \\ \\ AB^2 & = AC^2 + BC^2 - 2(AC)(BC) \cos \angle ACB \phantom{000000} [\text{Cosine rule}] \\ & = 104^2 + 135^2 - 2(104)(135) \cos \angle 58^\circ \\ AB & = \sqrt{ 104^2 + 135^2 - 2(104)(135) \cos \angle 58^\circ } \\ & = 118.999 \\ & \approx 119 \text{ m} \end{align*}

(b)

\begin{align*} \angle NBC & = 180^\circ - 122^\circ \phantom{0} \text{(Interior angles)} \\ & = 58^\circ \\ \\ {\sin \angle ABC \over 104} & = { \sin 58^\circ \over 118.999} \phantom{000000} [\text{Sine rule}] \\ 118.999 \sin \angle ABC & = 104 \sin 58^\circ \\ \sin \angle ABC & = {104 \sin 58^\circ \over 118.999} \\ \angle ABC & = \sin^{-1} \left( 104 \sin 58^\circ \over 118.999 \right) \\ & = 47.83^\circ \\ \\ \angle NBA & = 58^\circ - 47.83^\circ \\ & = 10.17^\circ \\ \\ \text{Bearing of } A \text{ from } B & = 360^\circ - 10.17^\circ \phantom{0} \text{(Angles at a point)} \\ & = 349.83^\circ \\ & \approx 349.8^\circ \end{align*}

(c)

\begin{align*} \sin \angle ACT & = {AT \over AC} \phantom{000000} \left[ {Opp \over Hyp} \right] \\ \sin 58^\circ & = {AT \over 104} \\ 104 \sin 58^\circ & = AT \\ 88.197 & = AT \\ \\ AT & \approx 88.2 \text{ m} \end{align*}

(d)

\begin{align*} \tan \angle DBC & = {DC \over CB} \phantom{000000} \left[ {Opp \over Adj} \right] \\ \tan 27^\circ & = {DC \over 135} \\ 135 \tan 27^\circ & = DC \\ 68.785 & = DC \\ \\ \text{Height of cliff} & \approx 68.8 \text{ m} \end{align*}

(a)

\begin{align*} \overrightarrow{RC} & = \overrightarrow{RD} + \overrightarrow{DC} \\ & = {1 \over 3} \overrightarrow{AD} + 2 \textbf{p} \\ & = {1 \over 3} (3 \textbf{q}) + 2 \textbf{p} \\ & = \textbf{q} + 2 \textbf{p} \end{align*}

(b)(i)

\begin{align*} \text{Since } & \overrightarrow{RC} = 2\overrightarrow{ST}, \text{ lines } RC \text{ and } ST \text{ are parallel} \\ \\ \angle DRC & = \angle DST \phantom{0} \text{(Corresponding angles, } RC \phantom{.} // \phantom{.} ST) [A] \\ \\ \angle RDC & = \angle SDT \phantom{0} \text{(Common angle)} [A] \\ \\ \therefore \text{Triangles }& RCD \text{ and } STD \text{ are similar } (AA) \end{align*}

(b)(ii)

\begin{align*} { \text{Area of triangle } RCD \over \text{Area of parallelogram } ABCD } & = { {1 \over 2} \times b \times h \over b \times h } \\ & = { {1 \over 2} \times RD \times h \over AD \times h} \phantom{000000} [\text{Common height}] \\ & = { {1 \over 2} \times RD \over AD } \\ & = { {1 \over 2} \times {1 \over 3} AD \over AD} \\ & = {1 \over 6} \\ \\ { \text{Area of triangle } STD \over \text{Area of triangle RCD} } & = \left( ST \over RC \right)^2 \phantom{000000} \left[\text{Similar triangles: } {A_1 \over A_2} = \left(l_1 \over l_2 \right)^2 \right] \\ & = \left(1 \over 2\right)^2 \\ & = {1 \over 4} \\ \\ \triangle STD & : \triangle RCD : ABCD \\ 1 & : \phantom{000.} 4 \phantom{00} \\ & \phantom{00000} 1 \phantom{00} : 6 \\ 1 & : \phantom{000.} 4 \phantom{00} : 24 \\ \\ \therefore \text{Ratio} & = 1: 24 \end{align*}

(b)(iii)

\begin{align*} \overrightarrow{QR} & = \overrightarrow{QA} + \overrightarrow{AR} \\ \\ \overrightarrow{QA} & = \overrightarrow{QB} + \overrightarrow{BA} \\ & = - \textbf{q} - 2 \textbf{p} \phantom{000000} [RD = QB] \\ \\ \overrightarrow{AR} & = {2 \over 3} \overrightarrow{AD} \\ & = {2 \over 3} ( 3 \textbf{q} ) \\ & = 2 \textbf{q} \\ \\ \therefore \overrightarrow{QR} & = - \textbf{q} - 2 \textbf{p} + 2 \textbf{q} \\ & = \textbf{q} - 2 \textbf{p} \end{align*}

(a)

\begin{align*} \text{Height} & = (x - 5) \text{ cm} \\ \\ \text{Width} & = 2(x - 5) \text{ cm} \end{align*}

(b)

\begin{align*} \text{External surface area} & = lb + 2lh + 2bh \\ & = x[2(x - 5)] + 2x(x - 5) + 2[2(x - 5)](x - 5) \\ & = 2x(x - 5) + 2x^2 - 10x + 4(x - 5)^2 \\ & = 2x^2 - 10x + 2x^2 - 10x + 4[ \underbrace{ (x)^2 - 2(x)(5) + (5)^2 }_{ (a - b)^2 = a^2 - 2ab + b^2 } \\ & = 4x^2 - 20x + 4 (x^2 - 10x + 25) \\ & = 4x^2 - 20x + 4x^2 - 40x + 100 \\ & = 8x^2 - 60x + 100 \\ \\ 48 & = 8x^2 - 60x + 100 \\ 0 & = 8x^2 - 60x + 100 - 48 \\ 0 & = 8x^2 - 60x + 52 \\ 0 & = 2x^2 - 15x + 13 \phantom{0} \text{ (Shown)} \end{align*}

(c)

\begin{align*} 0 & = 2x^2 - 15x + 13 \\ 0 & = (2x - 13)(x - 1) \end{align*} \begin{align*} 2x - 13 & = 0 && \text{ or } & x - 1 & = 0 \\ 2x & = 13 &&& x & = 1 \\ x & = {13 \over 2} \\ x & = 6.5 \end{align*}

(d)

\begin{align*} & \text{Reject } x = 1 \text{. If } x = 1, \text{ the height and width of the box have negative values.} \end{align*}

(e)

\begin{align*} \text{Volume} & = l \times b \times h \\ & = x \times 2(x - 5) \times (x - 5) \\ & = 6.5 \times 2(6.5 - 5) \times (6.5 - 5) \\ & = 29.25 \text{ cm}^3 \end{align*}

(a)

\begin{align*} [ \text{Note: } \angle OMA & = 90^\circ, OA = OB = 30] \\ \\ \cos \angle AOM & = {OM \over OA} \phantom{000000} \left[ {Adj \over Hyp} \right] \\ & = {20 \over 30} \\ \angle AOM & = \cos^{-1} \left(20 \over 30\right) \\ & = 48.189^\circ \\ \\ \angle AOB & = 48.189^\circ \times 2 \\ & = 96.378^\circ \\ & \approx 96.4^\circ \phantom{0} \text{ (Shown)} \end{align*}

(b)

\begin{align*} \text{Area of sector } OAB & = {\theta \over 360^\circ} \times \pi r^2 \\ & = { 96.378^\circ \over 360^\circ } \times \pi (30)^2 \\ & = 756.95 \text{ cm}^2 \\ \\ \text{Area of triangle } OAMB & = {1 \over 2} a b \sin C \phantom{000000} [\text{Formula provided}] \\ & = {1 \over 2} (OA) (OB) \sin \angle OAB \\ & = {1 \over 2} (30) (30) \sin 96.378^\circ \\ & = 447.21 \text{ cm}^2 \\ \\ \text{Area of shaded region} & = 756.95 - 447.21 \\ & = 309.74 \\ & \approx 310 \text{ cm}^2 \end{align*}

(c)(i)

\begin{align*} \text{Volume of prism} & = \text{Cross-sectional area} \times \text{Length} \\ & = \underbrace{309.74}_\text{From (b)} \times \underbrace{150}_\text{1.5 m in cm} \\ & = 46 \phantom{.} 461 \\ & \approx 46 \phantom{.} 500 \text{ cm}^3 \end{align*}

(c)(ii)

\begin{align*} \text{Area of semi-circle} & = {1 \over 2} \pi r^2 \\ & = {1 \over 2} \pi (30)^2 \\ & = 450 \pi \text{ cm}^2 \\ \\ \text{Volume of trough} & = 450 \pi \times 150 \\ & = 67 \phantom{.} 500 \pi \text{ cm}^3 \\ \\ \text{Volume of water to be added} & = 67 \phantom{.} 500 \pi - 46 \phantom{.} 461 \\ & = 165 \phantom{.} 596 \text{ cm}^3 \\ & = 165.596 \text{ litres} \phantom{000000} [1 \text{litre} = 1000 \text{ cm}^3] \\ & \approx 170 \text{ litres (to nearest 10 litres)} \end{align*}

(a)(i)(a)

\begin{align*} 80 \times {50 \over 100} & = 40 \\ \\ \text{Median speed} & = 42 \text{ km/h} \end{align*}

(a)(i)(b)

\begin{align*} 80 \times {25 \over 100} & = 20 \\ \\ \text{Lower quartile} & = 37 \text{ km/h} \\ \\ 80 \times {75 \over 100} & = 60 \\ \\ \text{Upper quartile} & = 46 \text{ km/h} \\ \\ \text{Interquartile range} & = 46 - 37 \\ & = 9 \text{ km/h} \end{align*}

(a)(ii)

\begin{align*} \text{From graph, speed of } 73 \text{ cars} & \le \text{Speed limit of 50 km/h} \\ \\ \text{Percentage of cars that exceeded} & = {80 - 73 \over 80} \times 100 \\ & = 8.75 \% \end{align*}

(a)(iii)

\begin{align*} \text{Median (afternoon)} & = 40 \text{ km/h} \\ \\ \text{Cars in the morning were faster } & \text{than cars in afternoon as the median speed is higher} \\ \\ \\ \text{Interquartile range (afternoon)} & = 42 - 34 \\ & = 8 \text{ km/h} \\ \\ \text{Speed of cars in the afternoon } & \text{were more consistent as the interquartile range is lower} \end{align*}

(b)(i)

\begin{align*} \text{P(No passenger)} & = {34 \over 34 + 17 + 13 + 11 + 5} \\ & = {17 \over 40} \end{align*}

(b)(ii)(a)

\begin{align*} \text{P(4 passengers, 4 passengers)} & = {5 \over 80} \times {4 \over 79} \\ & = {1 \over 316} \end{align*}

(b)(ii)(b)

\begin{align*} \text{Case 1: P(More than 2 passengers, less than 2 passengers)} & = { 11 + 5 \over 80 } \times {34 + 17 \over 79} \\ & = {51 \over 395} \\ \\ \text{Case 2: P(Less than 2 passengers, more than 2 passengers)} & = { 34 + 17 \over 80 } \times { 11 + 5 \over 79 } \\ & = {51 \over 395} \\ \\ \text{Required probability} & = {51 \over 395} + {51 \over 395} \\ & = {102 \over 395} \end{align*}

(a) Use a ruler (or a string if you take Geography) to measure

\begin{align*} \text{Length on map} & = 27.5 \text{ cm} \\ \\ 1 \text{ cm} & : 20 \phantom{.} 000 \text{ cm} \\ 1 \text{ cm} & : 200 \text{ m} \phantom{000000} [1 \text{ m} = 100 \text{ cm} ] \\ 1 \text{ cm} & : 0.2 \text{ km} \phantom{00000.} [1 \text{ km} = 1000 \text{ m}] \\ 27.5 \text{ cm} & : 5.5 \text{ km} \\ \\ \text{Actual dsitance} & = 5.5 \text{ km} \end{align*}

(b)

\begin{align*} \text{Time taken} & = { \text{Distance} \over \text{Speed} } \\ & = {5.5 \text{ km} \over 6.5 \text{ km/h} } \\ & = {11 \over 13} \text{ hour} \\ & = 50.77 \text{ minutes} \\ \\ \text{Weekly moderate-intensity time} & = 50.77 \times 3 \\ & = 152.31 \text{ minutes} \\ \\ \text{Yes, Nurul will } & \text{meet time target} \end{align*}

(c)

\begin{align*} \text{Calories used (walking)} & = {152.31 \text{ minutes} \over 30 \text{ minutes} } \times 150 \\ & = 761.55 \\ \\ \\ \text{Time taken (jogging)} & = { \text{Distance} \over \text{Speed} } \\ & = {5.5 \text{ km} \over 9.5 \text{ km/h} } \\ & = {11 \over 19} \text{ hour} \\ & = 34.74 \text{ minutes} \\ \\ \text{Weekly vigorous-intensity time} & = 34.74 \times 3 \\ & = 104.22 \text{ minutes} > 75 \text{ minutes} \\ \\ \checkmark & \text{ Meet weekly time target} \\ \\ \text{Calories used (jogging)} & = {104.22 \over 30} \times 345 \\ & = 1198.53 \\ \\ { \text{Calories used (jogging)} \over \text{Calories used (walking)} } & = { 1198.53 \over 761.55 } \\ & = 1.5738 \\ \\ \times & \text{ Will not use more than double amount of calories} \end{align*}