\begin{align*} 7 - 8x & = 25 \\ -8x & = 25 - 7 \\ -8x & = 18 \\ x & = {18 \over -8} \\ x & = -2.25 \end{align*}

(a)

\begin{align*} 112 & = 2^4 \times 7 \\ \\ 168 & = 2^3 \times 3 \times 7 \\ \\ \text{LCM} & = 2^4 \times 3 \times 7 \\ & = 336 \end{align*}

(b)

\begin{align*} 112 & = 2^4 \times 7 \\ \\ 168 & = 2^3 \times 3 \times 7 \\ \\ \text{HCF} & = 2^3 \times 7 \\ & = 56 \end{align*}

(a)

\begin{align*} {302.6^2 \over 12.76 - 10.84} & = 47 \phantom{.} 691.02 \\ & \approx 47 \phantom{.} 690 \text{ (4 s.f.)} \end{align*}

(b)

\begin{align*} 47 \phantom{.} 690 & = 4.769 \times 10^4 \end{align*}

\begin{align*} (3x - y)(2x + 3y) & = 6x^2 + 9xy - 2xy - 3y^2 \\ & = 6x^2 + 7xy - 3y^2 \end{align*}

(a)

\begin{align*} \text{Required percentage} & = {144 \over 360} \times 100 \\ & = 40 \% \end{align*}

(b)

\begin{align*} \text{Not possible since the total number of adults that took part in the parachute jump is not provided} \end{align*}

(a)

\begin{align*} x^2 - 12x + 17 & = x^2 - 12x + \left(12 \over 2\right)^2 - \left(12 \over 2\right)^2 + 17 \\ & = x^2 - 12x + 6^2 - 36 + 17 \\ & = (x - 6)^2 - 19 \\ \\ \therefore n & = -19 \end{align*}

(b)

\begin{align*} \text{For all real values of } x, (x - 6)^2 & \ge 0 \\ (x - 6)^2 - 19 & \ge -19 \\ \\ \therefore \text{When } x = 6, \text{ minimum value} & = -19 \end{align*}

(a) & (b)

(c) Note: Your answer may differ from mine if the size of the diagram is different (AC is 11.2 cm for me)

$$ CP = 5 \text{ cm} $$

\begin{align*} {2 \over 1} - {5 \over x} & = x(x + 2) \\ {2x \over x} - {5 \over x} & = x(x + 2) \\ {2x - 5 \over x} & = {x(x + 2) \over 1} \\ 2x - 5 & = x[x(x + 2)] \phantom{000000} [\text{Cross-multiply}] \\ 2x - 5 & = x^2(x + 2) \\ 2x - 5 & = x^3 + 2x^2 \\ 0 & = x^3 + 2x^2 - 2x + 5 \\ \\ \therefore a & = 2, b = -2 \end{align*}

\begin{align*} \angle BAD & = 37^\circ \phantom{0} (\text{Base angle of isosceles triangle}) \\ \\ \angle BDC & = 180^\circ - 124^\circ \phantom{0} (\text{Adjacent angles on a straight line}) \\ & = 56^\circ \\ \\ \angle ABC + \angle ADC & = 37^\circ + 50^\circ + 37^\circ + 56^\circ \\ & = 180^\circ \\ \\ \implies \angle ABC & \text{ and } \angle ADC \text{ are angles in opposite segments} \\ \\ \\ \angle DAB & = 180^\circ - 2(37^\circ) \phantom{0} (\text{Angle sum of triangle}) \\ & = 106^\circ \\ \\ \angle BCD & = 180^\circ - 50^\circ - 56^\circ \phantom{0} (\text{Angle sum of triangle}) \\ & = 74^\circ \\ \\ \angle DAB + \angle BCD & = 106^\circ + 74^\circ \\ & = 180^\circ \\ \\ \implies \angle DAB & \text{ and } \angle BCD \text{ are angles in opposite segments} \\ \\ \\ \therefore A,B, C \text{ and } D & \text{ lies on the circumference of the circle} \end{align*}

Note: The total amount of money both person have remains the same (regardless of the transfer)

\begin{align*} \text{(Before) } \text{M} & : \text{K} : \text{Total} &&& \text{(After) } \text{M} & : \text{K} : \text{Total} \\ 5 & : 3 \phantom{.} : 8 &&& 3 & :4 \phantom{.} : 7 \\ 35 & : 21 : 56 &&& 24 & : 32 : 56 \end{align*}
\begin{align*} \text{Amount Min has now} & = {22 \over 11} \times 24 \phantom{000000} [ 35 - 24 = 11 \text{ units} \rightarrow \$22 ] \\ & = \$ 48 \\ \end{align*}

\begin{align*} \require{cancel} {4y^2 - 7y - 15 \over y^3 - 9y} & = {(4y + 5)(y - 3) \over y(y^2 - 9)} \\ & = {(4y + 5)(y - 3) \over y(y^2 - 3^2)} \\ & = {(4y + 5)\cancel{(y - 3)} \over y(y + 3)\cancel{(y - 3)}} \phantom{0000000} [a^2 - b^2 = (a + b)(a - b)] \\ & = {4y + 5 \over y(y + 3)} \end{align*}

\begin{align*} {x + 1 \over 2} - {x^2 + 1 \over 2x + 3} & = {9 \over 2} \\ {x + 1 \over 2} - {9 \over 2} & = {x^2 + 1 \over 2x + 3} \\ {x + 1 - 9 \over 2} & = {x^2 + 1 \over 2x + 3} \\ {x - 8 \over 2} & = {x^2 + 1 \over 2x + 3} \\ (2x + 3)(x - 8) & = 2(x^2 + 1) \phantom{000000} [\text{Cross-multiply}] \\ 2x^2 - 16x + 3x - 24 & = 2x^2 + 2 \\ 2x^2 - 13x - 24 & = 2x^2 + 2 \\ 2x^2 - 2x^2 - 13x & = 2 + 24 \\ -13x & = 26 \\ x & = {26 \over -13} \\ x & = -2 \end{align*}

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {-10 - 11 \over 5 - (-2)} \\ & = -3 \\ \\ y & = mx + c \\ y & = -3x + c \\ \\ \text{Using } & (-2, 11), \\ 11 & = -3(-2) + c \\ 11 & = 6 + c \\ 11 - 6 & = c \\ 5 & = c \\ \\ \text{Eqn of line: } & y = -3x + 5 \end{align*}

(a)(i)

\begin{align*} 9x^3 \times x^9 & = 9x^{12} \phantom{000000} [a^m \times a^n = a^{m + n}] \end{align*}

(a)(ii)

\begin{align*} (16x^8)^{3 \over 4} & = 16^{3 \over 4} (x^8)^{3 \over 4} \phantom{000000} [ (ab)^m = a^m \times b^m ] \\ & = 8x^6 \phantom{00000000000} [ (a^m)^n = a^{mn} ] \end{align*}

(b)

\begin{align*} {81^p \over 3^q} & = 27^r \\ {(3^4)^p \over 3^q} & = (3^3)^r \\ {3^{4p} \over 3^q} & = 3^{3r} \phantom{000000} [ (a^m)^n = a^{mn} ] \\ 3^{4p - q} & = 3^{3r} \phantom{00000.} \left[ {a^m \over a^n} = a^{m - n} \right] \\ \\ \therefore 4p - q & = 3r \\ -q & = 3r - 4p \\ q & = 4p - 3r \end{align*}

\begin{align*} \angle ADE & = \angle BEC \phantom{0} (\text{Corresponding angles, } AD \phantom{.} // \phantom{.} BE )[A] \\ \\ \angle AED & = \angle BCE \phantom{0} (\text{Corresponding angles, } AE \phantom{.} // \phantom{.} BC )[A] \\ \\ \text{Since } & ABCD \text{ is a trapezium, } AB \phantom{.} // \phantom{.} DC \\ \\ \text{Since } & AE \phantom{.} // \phantom{.} BC \text{ and } AB \phantom{.} // \phantom{.} EC, \phantom{.} ABCE \text{ is a parallelogram} \\ \\ \implies AE & = BC \phantom{0} [S] \\ \\ \\ \therefore \text{Triangles } & ADE \text{ and } BEC \text{ are congruent } (AAS) \end{align*}

(a) This method of factorisation is called Grouping

\begin{align*} 9wx + y - 3x - 3wy & = 9wx - 3x - 3wy + y \\ & = 3x(3w - 1) - y(3w - 1) \\ & = (3w - 1)(3x - y) \end{align*}

(b)

\begin{align*} 5x^4 - 80y^4 & = 5(x^4 - 16y^4) \\ & = 5[ (x^2)^2 - (4y^2)^2 ] \\ & = 5 (x^2 + 4y^2)(x^2 - 4y^2) \phantom{000000} [ a^2 - b^2 = (a + b)(a - b)] \\ & = 5 (x^2 + 4y^2)[ x^2 - (2y)^2 ] \phantom{00000} [ a^2 - b^2 = (a + b)(a - b)] \\ & = 5 (x^2 + 4y^2)(x + 2y)(x - 2y) \end{align*}

(a)

\begin{align*} \textbf{S} & = \left( \begin{matrix} 15 & 80 & 152 \\ 13 & 75 & x \end{matrix} \right) \end{align*}

(b)

\begin{align*} \textbf{T} & = \textbf{S} \left( \begin{matrix} 280 \\ 120 \\ 90 \end{matrix} \right) \\ & = \mathop{ \left( \begin{matrix} 15 & 80 & 152 \\ 13 & 75 & x \end{matrix} \right) }_{2 \times 3} \mathop{ \left( \begin{matrix} 280 \\ 120 \\ 90 \end{matrix} \right) }_{3 \times 1} \\ & = \mathop{ \left( \begin{matrix} 15(280) + 80(120) + (152)(90) \\ 13(280) + 75(120) + 90x \end{matrix} \right) }_{2 \times 1} \\ & = \left( \begin{matrix} 27 \phantom{.} 480 \\ 12 \phantom{.} 640 + 90x \end{matrix} \right) \end{align*}

(c)

\begin{align*} \textbf{T} = \left( \begin{matrix} 27 \phantom{.} 480 \\ 12 \phantom{.} 640 + 90x \end{matrix} \right) & \begin{matrix} \text{Outward total} \\ \text{Return total} \end{matrix} \\ \\ \\ 12 \phantom{.} 640 + 90x - 27 \phantom{.} 480 & = 1360 \\ 90x - 14 \phantom{.} 840 & = 1360 \\ 90x & = 1360 + 14 \phantom{.} 840 \\ 90x & = 16 \phantom{.} 200 \\ x & = {16 \phantom{.} 200 \over 90} \\ x & = 180 \end{align*}

\begin{align*} \angle OPT & = 90^\circ \phantom{0} (\text{Tangent perpendicular to radius}) \\ \\ \angle OPR & = (90 - x)^\circ \\ \\ \angle POR & = 180 - 2(90 - x) \phantom{0} (\text{Angle sum of isosceles triangle since } OP = OR) \\ & = 180 - 180 + 2x \\ & = 2x^\circ \\ \\ \text{Reflex } \angle POR & = (360 - 2x)^\circ \phantom{0} (\text{Angles at a point}) \\ \\ \angle PQR & = {360 - 2x \over 2} \phantom{0} (\text{Angle at centre} = 2 \times \text{Angle at circumference}) \\ & = {360 \over 2} - {2x \over 2} \\ & = (180 - x)^\circ \end{align*}

(a)

\begin{align*} \text{Arc length} & = r \theta \phantom{000000} [\text{Formula for angle in radians}] \\ & = (8.2)(2.2) \\ & = 18.04 \text{ cm} \\ \\ \text{Perimeter} & = 18.04 + 2(8.2) \\ & = 34.44 \text{ cm} \end{align*}

(b)

\begin{align*} \text{Circumference of base of cone} & = \text{Arc length of sector} \\ 2 \pi r & = 18.04 \\ r & = {18.04 \over 2 \pi} \\ r & = 2.8711 \\ r & \approx 2.87 \text{ cm} \end{align*}

(a)(i)

\begin{align*} A & = \{ 17, 19, 23 \} \phantom{000000} [ \text{Need to be in universal set } \xi ] \end{align*}

(a)(ii)

\begin{align*} B & = \{ 20, 25 \} \\ \\ A \cup B & = \{ 17, 19, 20, 23, 25 \} \\ \\ (A \cup B)' & = \{ 16, 18, 21, 22, 24 \} \end{align*}

(b)(i)

\begin{align*} & \underline{ 11 \notin P \cap Q } \end{align*}

(b)(ii)(a)

\begin{align*} P \cap Q & = \{ 2, 4, 6, 8, 12 \} \\ \\ (P \cap Q) \cup Q' & = \{ 2, 4, 5, 6, 7, 8, 9, 10, 11, 12 \} \\ \\ n [ (P \cap Q) \cup Q' ] & = 10 \end{align*}

(b)(ii)(b)

\begin{align*} P \cup Q & = \{ 1, 2, 3, 4, 6, 8, 10, 12 \} \\ \\ (P \cap Q)' & = \{ 1, 3, 5, 7, 9, 10, 11 \} \\ \\ (P \cup Q) \cap (P \cap Q)' &= \{ 1, 3, 10 \} \\ \\ n [ (P \cup Q) \cap (P \cap Q)' ] & = 3 \end{align*}

(a)(i)

\begin{align*} \text{Required probability} & = {20 + 5 + 1 \over 50} \\ & = 0.52 \end{align*}

(a)(ii)

\begin{align*} \text{Median position} & = {50 + 1 \over 2} \\ & = 25.5 \implies \text{Between 25th and 26th female} \\ \\ \text{Interval: } & 20 \le \text{Handspan (cm)} < 22 \end{align*}

(a)(iii) Since it's worth 1 mark, you can calculate using the calculator function without any workings

\begin{align*} [ \text{Mid-values} & \text{ of each interval: } 17, 19, 21, 23, 25] \\ \\ \text{Mean} & = { \sum fx \over \sum f} \\ & = { (17 \times 3) + (19 \times 21) + (21 \times 20) + (23 \times 5) + (25 \times 1) \over 50 } \\ & = 20.2 \text{ cm} \end{align*}

(b)

\begin{align*} & 1. \text{ On average, males have longer handspans since median falls in the interval } 22 \le x < 24 \\ & \phantom{1.} \text{ while median for females falls in the interval } 20 \le x < 22 \\ \\ & 2. \text{ Majority of handspans for females are between 18 cm and 22 cm. On the other hand, } \\ & \phantom{2. } \text{ majority of handspans for males are between 20 cm and 26 cm } \end{align*}

(a)

\begin{align*} {x \over 7.5} & = {8 \over 10} \\ 10x & = 7.5(8) \phantom{000000} [\text{Cross-multiply}] \\ 10x & = 60 \\ x & = {60 \over 10} \\ x & = 6 \end{align*}

(b)

\begin{align*} \text{Volume of pyramid} & = {1 \over 3} \times \text{Base area} \times \text{Height} \\ \\ \text{Volume of small pyramid} & = {1 \over 3} \times (6 \times 6) \times 8 \\ & = 96 \text{ cm}^3 \\ \\ \text{Volume of large pyramid} & = {1 \over 3} \times (7.5 \times 7.5) \times 10 \\ & = 187.5 \text{ cm}^3 \\ \\ \text{Volume of glass block} & = 187.5 - 96 \\ & = 91.5 \text{ cm}^3 \\ \\ \text{Mass of glass block} & = 91.5 \times 2.6 \\ & = 237.9 \text{ g} \end{align*}

(a)

\begin{align*} T_n & = T_1 + (n - 1)(d) \phantom{000000} [d \text{ is the common difference}] \\ & = 10 + (n - 1)(4) \\ & = 10 + 4n - 4 \\ & = 4n + 6 \end{align*}

(b)

\begin{align*} \text{Let } 4n + 6 & = 264 \\ 4n & = 264 - 6 \\ 4n & = 258 \\ n & = {258 \over 4} \\ n & = 64.5 \\ \\ \text{Since } n \text{ is not } & \text{an integer, 264 is not a term of the sequence} \end{align*}

(c)

\begin{align*} S_n & = 2n^2 + 8n \\ 384 & = 2n^2 + 8n \\ 0 & = 2n^2 + 8n - 384 \\ 0 & = n^2 + 4n - 192 \\ 0 & = (n - 12)(n + 16) \end{align*} \begin{align*} n - 12 & = 0 && \text{ or } & n + 16 & = 0 \\ n & = 12 &&& n & = -16 \text{ (Reject, since } n > 0) \end{align*}

(a)(i)

\begin{align*} a & = {3 - (-6) \over -6 + 4} + 2 \\ & = -2.5 \end{align*}

(a)(ii) Topic: Change subject of equation

\begin{align*} a & = {3 - b \over b + 4} + 2 \\ a - 2 & = {3 - b \over b + 4} \\ {a - 2 \over 1} & = {3 - b \over b + 4} \\ (b + 4)(a - 2) & = 3 - b \phantom{000000} [\text{Cross-multiply}] \\ ab - 2b + 4a - 8 & = 3 - b \\ ab - 2b + b & = 3 + 8 - 4a \\ ab - b & = 11 - 4a \\ b(a - 1) & = 11 - 4a \\ b & = {11 - 4a \over a - 1} \end{align*}

(b) Topic: Solve simultaneous equations

Elimination method:

\begin{align*} 8x + 5y & = 9 \phantom{0} \text{--- (1)} \\ 6x - 2y & = 47 \phantom{0} \text{--- (2)} \\ \\ (1) & \times 2, \\ 16x + 10y & = 18 \phantom{0} \text{--- (3)} \\ \\ (2) & \times 5, \\ 30x - 10y & = 235 \phantom{0} \text{--- (4)} \\ \\ (3) & + (4), \\ 16x + 10y + 30x - 10y & = 18 + 235 \\ 46x & = 253 \\ x & = {253 \over 46} \\ x & = 5.5 \\ \\ \text{Substitute } & x = 5.5 \text{ into (1),} \\ 8(5.5) + 5y & = 9 \\ 44 + 5y & = 9 \\ 5y & = 9 - 44 \\ 5y & = - 35 \\ y & = {-35 \over 5} \\ y & = -7 \\ \\ \therefore x & = 5.5, y = -7 \end{align*}

Substitution method:

\begin{align*} 8x + 5y & = 9 \phantom{0} \text{--- (1)} \\ 6x - 2y & = 47 \phantom{0} \text{--- (2)} \\ \\ \text{From } & (2), \\ -2y & = 47 - 6x \\ 2y & = 6x - 47 \\ y & = 3x - 23.5 \phantom{0} \text{--- (3)} \phantom{0000} [\text{Divide every term by 2}] \\ \\ \text{Substitute } & \text{(3) into (1),} \\ 8x + 5(3x - 23.5) & = 9 \\ 8x + 15x - 117.5 & = 9 \\ 23x & = 9 + 117.5 \\ 23x & = 126.5 \\ x & = {126.5 \over 23} \\ x & = 5.5 \\ \\ \text{Substitute } & x = 5.5 \text{ into (3),} \\ y & = 3(5.5) - 23.5 \\ y & = -7 \\ \\ \therefore x & = 5.5, y = -7 \end{align*}

(c) Topic: Subtract two algebraic fractions

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

(d) Topic: Solve fractional equation

\begin{align*} {11 \over x - 3} & = {3x - 1 \over 1} \\ 11 & = (x - 3)(3x - 1) \phantom{000000} [\text{Cross-multiply}] \\ 11 & = 3x^2 - x - 9x + 3 \\ 11 & = 3x^2 - 10x + 3 \\ 0 & = 3x^2 - 10x + 3 - 11 \\ 0 & = 3x^2 - 10x - 8 \\ 0 & = (x - 4)(3x + 2) \end{align*} \begin{align*} x - 4 & = 0 && \text{ or } & 3x + 2 & = 0 \\ x & = 4 &&& 3x & = -2 \\ & &&& x & = -{2 \over 3} \end{align*}

(a)

\begin{align*} 100\% - 8\% & = 92\% \\ \\ \text{Remainder (after pension)} & = {5625 \over 100} \times 92 \\ & = \$ 5175 \\ \\ 100\% - 5\% & = 95 \% \\ \\ \text{Amount left (after pension & savings account)} & = {5175 \over 100} \times 95 \\ & = \$ 4 \phantom{.} 916.25 \end{align*}

(b)

\begin{align*} \text{Deposit} & = \$ 900 \times {1 \over 5} \\ & = \$ 180 \\ \\ \text{Total amount paid} & = \$ 180 + (\$ 64 \times 12 ) \\ & = \$ 948 \end{align*}

(c)

\begin{align*} 100\% + 7.5\% & = 107.5 \% \\ \\ \text{Monthly rent last year} & = {2064 \over 107.5} \times 100 \\ & = \$ 1920 \end{align*}

(d)

\begin{align*} 100\% + 1.6\% & = 101.6\% \\ \\ \text{Total cost in euros} & = {230 \over 100} \times 101.6 \\ & = 233.68 \text{ euros} \\ \\ 0.65 \text{ euros} & = 1 \text{ SGD} \\ 1 \text{ euros} & = {1 \over 0.65} \\ & = 1.5385 \text{ SGD} \\ \\ \text{Total cost in SGD} & = 233.68 \times 1.5385 \\ & = 359.516 \\ & \approx \$ 359.52 \end{align*}

(a)

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

(b)

(c)

\begin{align*} & \text{The graph of } y = {x^3 \over 4} - 2x + 3 \text{ only meets the } x \text{-axis } (y = 0) \text{ once} \end{align*}

(d)(i)

\begin{align*} x^3 - 10x + 4 & = 0 \\ {1 \over 4}(x^3 - 10x + 4) & = {1 \over 4}(0) \phantom{000000} \left[ \text{Multiply by } {1 \over 4} \text{ on both sides} \right] \\ {1 \over 4}x^3 - {5 \over 2}x + 1 & = 0 \\ {1 \over 4}x^3 - {5 \over 2}x + 1 + {1 \over 2}x & = 0 + {1 \over 2}x \\ {1 \over 4}x^3 - 2x + 1 & = {1 \over 2}x \\ {1 \over 4}x^3 - 2x + 1 + 2 & = {1 \over 2}x + 2 \\ \underbrace{{1 \over 4}x^3 - 2x + 3}_\text{Curve} & = {1 \over 2}x + 2 \\ \\ \therefore \text{Draw } & y = {1 \over 2}x + 2 \\ \\ \therefore a & = {1 \over 2}, b = 2 \end{align*}

(d)(ii)

(a)(i)(a)

\begin{align*} 200 \times 50 \% & = 100 \\ \\ \text{Median mass (A)} & = 94 \text{ g} \end{align*}

(a)(i)(b)

\begin{align*} 200 \times 60 \% & = 120 \\ \\ \text{60th percentile (B)} & = 105 \text{ g} \end{align*}

(a)(i)(c)

\begin{align*} 200 \times 25 \% & = 50 \\ \\ 200 \times 75 \% & = 150 \\ \\ \text{IQR (B)} & = Q_3 - Q_1 \\ & = 113 - 87 \\ & = 26 \text{ g} \end{align*}

(a)(ii)

\begin{align*} & \text{Variety A because more apples from variety A have mass between 80 g and 120 g than that of variety B.} \\ & \text{Thus, George can sell more packs of apples.} \end{align*}

(b)(i)

\begin{align*} \text{Required probability} & = \text{P(Two A apples chosen)} + \text{P(Two B apples chosen)} \\ & = \left( {9 \over 16} \times {8 \over 15} \right) + \left( {7 \over 16} \times {6 \over 15} \right) \\ & = {19 \over 40} \end{align*}

(b)(ii)

\begin{align*} \text{Required probability} & = \text{P(C, C, not C)} + \text{P(C, not C, C)} + \text{P(not C, C, C)} \\ & = \left( {4 \over 20} \times {3 \over 19} \times {16 \over 18} \right) + \left( {4 \over 20} \times {16 \over 19} \times {3 \over 18} \right) + \left( {16 \over 20} \times {4 \over 19} \times {3 \over 18} \right) \\ & = {8 \over 95} \end{align*}

(a)

\begin{align*} \angle ABC & = \angle EDC = 90^\circ \phantom{0} (\text{Right angle in semi-circle}) [A] \\ \\ \angle ACB & = \angle ECD \phantom{0} (\text{Vertically opposite angles}) [A] \\ \\ \therefore \text{Triangles } & ABC \text{ and } EDC \text{ are similar (AA)} \end{align*}

(b)(i)

\begin{align*} {A_1 \over A_2} & = \left(l_1 \over l_2\right)^2 \\ {\text{Area of triangle } ABC \over \text{Area of triangle } EDC } & = \left(AC \over EC \right)^2 \\ & = \left( 5 \times 2 \over 8 \times 2\right)^2 \\ & = {25 \over 64} \\ \\ \text{Area of triangle } ABC & : \text{Area of triangle } EDC \\ 25 & : 64 \\ 1 & : {64 \over 25} \end{align*}

(b)(ii)

\begin{align*} \angle BPC & = 28^\circ \times 2 \phantom{0} (\text{Angle at centre} = 2 \times \text{Angle at circumference}) \\ & = 56^\circ \\ \\ \text{Sector area } PBC & = { \theta \over 360^\circ } \times \pi r^2 \\ & = {56^\circ \over 360^\circ} \times \pi (5)^2 \\ & = 3{8 \over 9} \pi \text{ cm}^2 \\ \\ \text{Area of triangle } PBC & = {1 \over 2} a b \sin C \\ & = {1 \over 2} (5)(5) \sin 56^\circ \\ & = 10.363 \text{ cm}^2 \\ \\ \text{Shaded area in small circle} & = 3{8 \over 9} \pi - 10.363 \\ & = 1.8543 \text{ cm}^2 \\ \\ \text{Shaded area in large circle} & = 1.8543 \times {64 \over 25} \phantom{000000} [\text{ratio from (b)(i) - triangles } ABC, EDC \text{ similar}]\\ & = 4.747 \text{ cm}^2 \phantom{00000000.} [\text{sectors } PBC, QDC \text{ similar and triangles } PBC, QDC \text{ similar}] \\ \\ \text{Total shaded area} & = 1.8543 + 4.747 \\ & = 6.6013 \\ & \approx 6.60 \text{ cm}^2 \end{align*}

(a)(i)

\begin{align*} \overrightarrow{QR} & = \overrightarrow{QO} + \overrightarrow{OR} \\ & = - \overrightarrow{OQ} + \overrightarrow{OR} \\ & = - {-7 \choose -3} + {5 \choose 1} \\ & = {7 \choose 3} + {5 \choose 1} \\ & = {12 \choose 4} \\ \\ | \overrightarrow{QR} | & = \sqrt{x^2 + y^2} \\ & = \sqrt{ 12^2 + 4^2 } \\ & = \sqrt{160} \\ & \approx 12.6 \text{ units} \end{align*}

(a)(ii)

\begin{align*} Q(-7, -3), & \phantom{.} R(5, 1) \\ \\ \text{Gradient of } QR & = {y_2 - y_1 \over x_2 - x_1} \\ & = {1 - (-3) \over 5 - (-7)} \\ & = {1 \over 3} \\ \\ \text{Gradient of } QS & = {-2 - (-3) \over k - (-7)} \\ {1 \over 3} & = {1 \over k + 7} \phantom{000000} [\text{Gradient of } QR = \text{Gradient } QS \text{ since } Q, R, S \text{ lie on a straight line}] \\ k + 7 & = 3 \phantom{0000000000.} [\text{Cross-multiply}] \\ k & = 3 - 7 \\ k & = -4 \\ \\ \therefore & \phantom{.} S(-4, -2) \\ \\ \therefore \overrightarrow{OS} & = {-4 \choose -2} \end{align*}

(b)(i)

\begin{align*} \overrightarrow{AB} & = \overrightarrow{AO} + \overrightarrow{OB} \\ & = - \textbf{a} + \textbf{b} \\ \\ \overrightarrow{AP} & = {2 \over 5} \overrightarrow{AB} \\ & = {2 \over 5} ( - \textbf{a} + \textbf{b} ) \\ & = -{2 \over 5} \textbf{a} + {2 \over 5} \textbf{b} \end{align*}

(b)(ii)

\begin{align*} \overrightarrow{OP} & = \overrightarrow{OA} + \overrightarrow{AP} \\ & = \textbf{a} + \left(-{2 \over 5} \textbf{a} + {2 \over 5} \textbf{b} \right) \\ & = \textbf{a} - {2 \over 5} \textbf{a} + {2 \over 5} \textbf{b} \\ & = {3 \over 5} \textbf{a} + {2 \over 5} \textbf{b} \\ \\ \overrightarrow{OM} & = {1 \over 2} \overrightarrow{OP} \\ & = {1 \over 2} \left( {3 \over 5} \textbf{a} + {2 \over 5} \textbf{b} \right) \\ & = {3 \over 10} \textbf{a} + {1 \over 5} \textbf{b} \end{align*}

(b)(iii)

\begin{align*} \overrightarrow{PB} & = {3 \over 5} \overrightarrow{AB} \\ & = {3 \over 5} ( - \textbf{a} + \textbf{b} ) \\ & = -{3 \over 5} \textbf{a} + {3 \over 5} \textbf{b} \\ \\ \overrightarrow{MN} & = \overrightarrow{MO} + \overrightarrow{ON} \\ & = - \overrightarrow{OM} + {1 \over 2} \overrightarrow{OB} \\ & = - \left( {3 \over 10} \textbf{a} + {1 \over 5} \textbf{b} \right) + {1 \over 2} \textbf{b} \\ & = - {3 \over 10} \textbf{a} - {1 \over 5} \textbf{b} + {1 \over 2} \textbf{b} \\ & = -{3 \over 10} \textbf{a} + {3 \over 10} \textbf{b} \\ \\ \text{Let } \overrightarrow{PB} & = k \overrightarrow{MN} \phantom{000000} [\text{Checking whether vectors are parallel}] \\ -{3 \over 5} \textbf{a} + {3 \over 5} \textbf{b} & = k \left( -{3 \over 10} \textbf{a} + {3 \over 10} \textbf{b} \right) \\ -{3 \over 5} \textbf{a} + {3 \over 5} \textbf{b} & = -{3 \over 10}k \textbf{a} + {3 \over 10}k \textbf{b} \end{align*} \begin{align*} \text{Comp} & \text{aring } \textbf{a}, &&& \text{Comp} & \text{aring }\textbf{b}, \\ -{3 \over 5} & = -{3 \over 10} k &&& -{3 \over 5} & = -{3 \over 10} k \\ -{3 \over 5} \div -{3 \over 10} & = k &&& -{3 \over 5} \div -{3 \over 10} & = k \\ 2 & = k &&& 2 & = k \end{align*}
\begin{align*} \therefore \overrightarrow{PB} & = 2 \overrightarrow{MN} \\ \\ \text{Trapezium because } & \text{sides } PB \text{ and } MN \text{ are parallel} \end{align*}

(a)

\begin{align*} \angle DAB & = 235^\circ - 155^\circ \\ & = 80^\circ \\ \\ \text{By Sine} & \text{ rule,} \\ {\sin \angle ABD \over AD} & = {\sin \angle DAB \over DB} \\ {\sin \angle ABD \over 168} & = {\sin 80^\circ \over 325} \\ 325 \sin \angle ABD & = 168 \sin 80^\circ \phantom{000000} [\text{Cross-multiply}] \\ \sin \angle ABD & = {168 \sin 80^\circ \over 325} \\ \angle ABD & = \sin^{-1} \left( {168 \sin 80^\circ \over 325} \right) \\ & = 30.60^\circ \\ \\ \angle N1BA & = 180^\circ - 155^\circ \phantom{0} (\text{Interior angles}) \\ & = 25^\circ \\ \\ \text{Bearing of } D \text{ from } B & = 360^\circ - 30.60^\circ - 25^\circ \\ & = 304.4^\circ \end{align*}

(b)

\begin{align*} \text{By } & \text{Cosine rule,} \\ CB^2 & = CD^2 + DB^2 - 2 (CD)(DB) \cos \angle CDB \\ & = 151^2 + 325^2 - 2(151)(325) \cos 48^\circ \\ CB & = \sqrt{ 151^2 + 325^2 - 2(151)(325) \cos 48^\circ } \\ & = 250.50 \text{ m} \\ \\ \text{Perimeter of triangle } BCD & = 151 + 325 + 250.50 \\ & = 726.5 \text{ m} \\ \\ \text{Total cost of fence} & = 726.5 \text{ m} \times \$ 85 \\ & = \$ 61 \phantom{.} 752.50 \\ & \approx \$ 61 \phantom{.} 800 \text{ (to nearest 100 dollars)} \end{align*}

(a)

\begin{align*} \text{Width} & = (x + 6) \text{ cm} \\ \\ \text{Length} & = 2(x + 6) \text{ cm} \\ & = (2x + 12) \text{ cm} \\ \\ \text{By Py} & \text{thagoras theorem,} \\ AF^2 & = AB^2 + BF^2 \\ & = (2x + 12)^2 + (x + 6)^2 \\ & = (2x)^2 + 2(2x)(12) + 12^2 + x^2 + 2(x)(6) + 6^2 \phantom{000000} [(a + b)^2 = a^2 + 2ab + b^2] \\ & = 4x^2 + 48x + 144 + x^2 + 12x + 36 \\ & = 5x^2 + 60x + 180 \\ \\ \text{By Py} & \text{thagoras theorem,} \\ AG^2 & = AF^2 + FG^2 \\ 24^2 & = 5x^2 + 60x + 180 + x^2 \\ 576 & = 6x^2 + 60x + 180 \\ 0 & = 6x^2 + 60x + 180 -\phantom{.}576 \\ 0 & = 6x^2 + 60x - 396 \\ 0 & = x^2 + 10x - 66 \text{ (Shown)} \phantom{000000} [\text{Divide every term by 6}] \end{align*}

(b)

\begin{align*} x^2 & + 10x - 66 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-10 \pm \sqrt{(10)^2 - 4(1)(-66)} \over 2(1)} \\ & = {-10 \pm \sqrt{364} \over 2} \\ & = 4.539 \text{ or } -14.539 \\ & \approx 4.54 \text{ or } -14.54 \text{ (to 2 d.p.)} \end{align*}

(c)

\begin{align*} \tan \angle GAF & = {GF \over AF} \phantom{000000} \left[ {Opp \over Adj} \right] \\ & = {x \over \sqrt{ 5x^2 + 60x + 180} } \\ & = {4.539 \over \sqrt{ 5(4.539)^2 + 60(4.539) + 180} } \\ & = 0.19261 \\ \\ \angle GAF & = \tan^{-1} (0.19261) \\ & = 10.902^\circ \\ & \approx 10.9^\circ \end{align*}

(a)(i) 'recommended health benefits' - at least 150 minutes of moderate-intensity aerobic activity

\begin{align*} \text{Walking duration for 1 walk} & = {150 \text{ mins} \over 4} \\ & = 37.5 \text{ mins} \\ & = (37.5 \div 60) \text{ h} \\ & = 0.625 \text{ h} \\ \\ \text{Distance} & = \text{Speed} \times \text{Time} \\ \text{Distance covered for 1 walk} & = 5 \text{ km/h} \times 0.625 \text{ h} \\ & = 3.125 \text{ km} \end{align*}

(a)(ii)

\begin{align*} \text{Calories used by Zhao} & = {150 \text{ mins} \over 30 \text{ mins}} \times 135 \\ & = 675 \\ \\ \text{Calories used by Mei} & = {150 \text{ mins} \over 30 \text{ mins}} \times 120 \\ & = 600 \\ \\ \text{Difference} & = 675 - 600 \\ & = 75 \end{align*}

(b)

\begin{align*} \underline{\text{Zhao}} \phantom{00000000000000000000000} & \\ \text{Time spent (moderate activity)} & = {4 \text{ km} \times 3 \over 5 \text{ km/h}} \\ & = 2.4 \text{ h} \\ & = 144 \text{ mins} \\ \\ \text{Time spent (vigorous activity)} & = {6 \text{ km} \times 2 \over 8 \text{ km/h}} \\ & = 1.5 \text{ h} \\ & = 90 \text{ mins} \implies 180 \text{ mins of moderate-intensity aerobic activity} \\ \\ \text{Total mins} & = 144 + 180 \\ & = 324 \text{ mins} \implies \text{Additional health benefits } \checkmark \\ \\ \text{Calories used in total} & = \underbrace{{144 \text{ mins} \over 30 \text{ mins}} \times 135}_\text{Walk} + \underbrace{{90 \text{ mins} \over 30 \text{ mins}} \times 315}_\text{Jog} + \underbrace{{45 \text{ mins} \times 2 \over 60 \text{ mins}} \times (3 \times 80 \text{ kg})}_\text{Yoga} \\ & = 1953 \text{ calories} \end{align*}
\begin{align*} \underline{\text{Mei}} \phantom{00000000000000000000000} & \\ \text{Time spent (vigorous activity)} & = {8 \text{ km} \times 2 \over 9.5 \text{ km/h}} \\ & = 1{13 \over 19} \text{ h} \\ & = 101.05 \text{ mins} \\ & \approx 101 \text{ mins} \implies \text{Recommended health benefits } \checkmark \\ \\ \text{Calories used in total} & = \underbrace{{101.05 \text{ mins} \over 30 \text{ mins}} \times 345}_\text{Jog} + \underbrace{1 \text{ h} \times (4 \times 70 \text{ kg}) }_\text{Yoga} \\ & = 1442.075 \text{ calories} \\ \\ \\ \text{Percentage of extra calories Zhao used} & = {1953 - 1442.075 \over 1442.075} \times 100 \\ & = 35.430 \% \\ \\ \\ \therefore & \phantom{.} \text{Zhao is not correct} \\ \\ \text{Both will meet respective targets } & \text{for exercise, however Zhao will use only 35.4% more calories} \end{align*}