\begin{align*} W & = { \sqrt{1234} \over 5.67 + 8.9} \\ W & = 2.411 \\ W & \approx 2.41 \end{align*}

(a)

\begin{align*} {150 \text{ g} \over 60 \text{ g}} & = 2.5 \text{ servings} \\ \\ \text{No. of scones} & = 8 \times 2.5 \\ & = 20 \end{align*}

(b)(i)

\begin{align*} {700 \text{ ml} \over 150 \text{ ml}} & = 4{2 \over 3} \text{ servings (based on milk)} \\ \\ {900 \text{ g} \over 275 \text{ g}} & = 3{3 \over 11} \text{ servings (based on flour)} \\ \\ {900 \text{ g} \over 60 \text{ g}} & = 15 \text{ servings (based on butter)} \\ \\ {900 \text{ g} \over 30 \text{ g}} & = 30 \text{ servings (based on sugar)} \\ \\ \\ [\text{Make 3 servings since amount of} & \text{ flour, butter and sugar is non-exact if } 3{3 \over 11} \text{ servings are made}] \\ \\ \text{Max. no. of scones} & = 8 \times 3 \\ & = 24 \end{align*}

(b)(ii)

\begin{align*} \text{Amount of butter} & = 60 \times 3 \\ & = 180 \text{ g} \end{align*}

\begin{align*} I & = {PRT \over 100} \\ I & = {(5250)(2.4)(4) \over 100} \\ I & = \$ 504 \\ \\ \text{Total amount} & = 5250 + 504 \\ & = \$ 5754 \end{align*}

\begin{align*} & \text{Calculation } B \text{ because the amount of gold Chen is selling is rounded up to 20 g } \\ & \text{while the price per gram is rounded up to \$ 60.} \end{align*}

(a)

\begin{align*} 3c - 4 & = 9c + 17 \\ 3c - 9c & = 17 + 4 \\ -6c & = 21 \\ c & = {21 \over -6} \\ c & = -{7 \over 2} \end{align*}

(b)

\begin{align*} 8ab - 3 + 12a - 2b & = 8ab - 2b + 12a - 3 \\ & = 2b(4a - 1) + 3(4a - 1) \\ & = (4a - 1)(2b + 3) \end{align*}

(c)

\begin{align*} (4x + 3y)(3x - 4y) & = 12x^2 - 16xy + 9xy - 12y^2 \\ & = 12x^2 - 7xy - 12y^2 \end{align*}

(a)

\begin{align*} -43 & < 2(5x + 3) &&& 2(5x + 3) - 14 & \le 42 \\ -43 & < 10x + 6 - 14 &&& 10x + 6 - 14 & \le 42 \\ -43 - 6 + 14 & < 10x &&& 10x & \le 42 - 6 + 14 \\ -35 & < 10x &&& 10x & \le 50 \\ {-35 \over 10} & < x &&& x & \le {50 \over 10} \\ -3.5 & < x &&& x & \le 5 \end{align*} \begin{align*} -3.5 < x \le 5 \end{align*}

(b)

(a)

\begin{align*} \overrightarrow{OP} & = {4 \choose -1} \\ \\ \overrightarrow{OQ} & = \overrightarrow{OP} + \overrightarrow{PQ} \\ & = {4 \choose -1} + {-7 \choose 10} \\ & = {-3 \choose 9} \\ \\ \therefore & \phantom{.} Q(-3, 9) \end{align*}

(b)

$$ [ \text{Magnitude of vector} = \sqrt{x^2 + y^2} ] $$ \begin{align*} \\ \overrightarrow{PQ} & = \sqrt{(-7)^2 + 10^2} \\ & = 12.206 \\ & \approx 12.2 \text{ units} \end{align*}

\begin{align*} \text{Total surface area of cube} & = 6 l^2 \\ 125 & = 6l^2 \\ {125 \over 6} & = l^2 \\ \sqrt{125 \over 6} & = l \\ \\ \text{Volume of cube} & = l^3 \\ & = \left( \sqrt{125 \over 6} \right)^3 \\ & = 95.09 \\ & \approx 95.1 \text{ cm}^3 \end{align*}

(a)

\begin{align*} 2 & | \underline{5500} \\ 2 & | \underline{2750} \\ 5 & | \underline{1375} \\ 5 & | \underline{275} \\ 5 & | \underline{55} \\ 11 & | \underline{11} \\ & | \underline{1} \\ \\ 5500 & = 2^2 \times 5^3 \times 11 \\ \\ x & = 2, y = 3 \end{align*}

(b)

\begin{align*} 793 \phantom{.} 800 k & = (2^3 \times 3^4 \times 5^2 \times 7^2) \times (3^2 \times 5 \times 7) \\ & = \underbrace{2^3 \times 3^6 \times 5^3 \times 7^3}_\text{Powers of each factor are multiple of 3} \\ \\ k & = 3^2 \times 5 \times 7 \\ k & = 315 \end{align*}

(a)

\begin{align*} & \text{Since the cost was recorded only at the end of the May and at the end of June, the increase in cost} \\ & \text{from the end of May to the end of June may not be linear (i.e. may not follow a straight line).} \end{align*}

(b)

\begin{align*} & \text{The cost was recorded at the end of each month but the highest cost may not occur} \\ & \text{at the end of the month.} \end{align*}

(c)(i)

\begin{align*} & \text{The vertical axis does not start from 0.} \end{align*}

(c)(ii)

\begin{align*} & \text{The reader may think that the increase/decrease between each month is larger than it actually is.} \end{align*}

\begin{align*} & \phantom{000000000000} 8a - 4b + 7 = 6c \\ \\ & \text{If } a, b \text{ and } c \text{ are integers, } 8a, 4b \text{ and } 6c \text{ are even integers.} \\ \\ & \text{On the left hand side, } 8a - 4b + 7 \text{ will be equals to an odd integer and not be equal to the right hand side } (6c). \\ \\ & \text{Thus, the equation is not valid if } a, b \text{ and } c \text{ are integers.} \end{align*}

\begin{align*} \text{Let } \angle DAC & = x^\circ \\ \\ \angle BCA & = \angle BAC \phantom{0} \text{ (Base angle of isosceles triangle}) \\ & = (42 + x)^\circ \\ \\ \angle ACD & = \angle ADC \phantom{0} \text{ (Base angle of isosceles triangle}) \\ & = (42 + x)^\circ \\ \\ 2(42 + x) + x & = 180 \phantom{0} \text{ (Angle sum of triangle)} \\ 84 + 2x + x & = 180 \\ 3x & = 180 - 84 \\ 3x & = 96 \\ x & = {96 \over 3} \\ x & = 32 \\ \\ \angle ACD & = (42 + 32)^\circ \\ & = 74^\circ \end{align*}

\begin{align*} \text{Let GNI in 2018} & = x \\ \\ \text{GNI in 2021} & = x \times {100 + 9.24 \over 100} \\ & = 1.0924 x \\ \\ \\ \text{GNI in 2019} & = x \times {100 + 3.63 \over 100} \\ & = 1.0363 x \\ \\ \text{GNI in 2020} & = 1.0363 x \times {100 - 5.56 \over 100} \\ & = 0.97868x \\ \\ \text{Percentage increase from 2020 to 2021} & = {\text{Final} - \text{Initial} \over \text{Initial} } \times 100 \\ & = { 1.0924 x - 0.97868x \over 0.97868x} \times 100 \\ & = {0.11372x \over 0.97868x} \times 100 \\ & = {0.11372 \over 0.97868} \times 100 \\ & = 11.619 \\ & \approx 11.6 \% \end{align*}

(a) Fast method:

\begin{align*} x \text{-intercepts: } & (2, 0) \text{ and } (8, 0) \\ \\ \text{When } & y = 0, \end{align*} \begin{align*} x & = 2 && \text{ or } & x & = 8 \\ x - 2 & = 0 &&& x - 8 & = 0 \end{align*} \begin{align*} 0 & = (x - 2)(x - 8) \\ \\ \therefore y & = (x - 2)(x - 8) \\ y & = x^2 - 8x - 2x + 16 \\ y & = x^2 - 10x + 16 \\ \\ b & = -10, c = 16 \end{align*}

(a) Longer method:

\begin{align*} x & \text{-intercepts: } (2, 0) \text{ and } (8, 0) \\ \\ \text{Subtitute } & (2, 0) \text{ into eqn of curve,} \\ 0 & = (2)^2 + b(2) + c \\ 0 & = 4 + 2b + c \\ -c & = 4 + 2b \\ c & = -4 - 2b \phantom{0} \text{--- (1)} \\ \\ \text{Subtitute } & (8, 0) \text{ into eqn of curve,} \\ 0 & = (8)^2 + b(8) + c \\ 0 & = 64 + 8b + c \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 0 & = 64 + 8b + (-4 -2b) \\ 0 & = 64 + 8b - 4 - 2b \\ -8b + 2b & = 64 - 4 \\ -6b & = 60 \\ b & = {60 \over -6} \\ b & = -10 \\ \\ \text{Substitute } & b = -10 \text{ into (1),} \\ c & = -4 - 2(-10) \\ c & = 16 \\ \\ \therefore b & = -10, c = 16 \end{align*}

(b)

\begin{align*} \text{From } (a), y & = x^2 - 10x + 16 \text{ has } x \text{-intercepts 2 and 8} \\ \\ \text{When } & x = 0, \\ y & = (0)^2 - 10(0) + 16 \\ y & = 16 \phantom{000000} [y \text{-intercept is 16}] \\ \\ \text{For } y & \text{-intercept to be 32,} \\ y & = 2(x^2 - 10x + 16) \\ y & = 2x^2 - 20x + 32 \end{align*}

(a)

\begin{align*} & \text{The tangent to the circle, } TA \text{, is perpendicular to the radius of the circle, } OA. \end{align*}

(b)

\begin{align*} \angle AOB & = 180^\circ - 90^\circ - 42^\circ \phantom{0} \text{ (Angle sum of triangle)} \\ & = 48^\circ \\ \\ 2 \times \angle ACB & = \angle AOB \phantom{0} \text{ (Angle at centre} = 2 \times \text{Angle at circumference}) \\ \angle ACB & = {48^\circ \over 2} \\ \angle ACB & = 24^\circ \end{align*}

(c)

\begin{align*} \angle COB & = 180^\circ - 2(23^\circ + 24^\circ) \phantom{0} \text{ (Angle sum of isosceles triangle } OCB) \\ & = 86^\circ \\ \\ \text{Reflex angle } \angle COB & = 360^\circ - 48^\circ - 86^\circ \phantom{0} \text{ (Angles at a point)} \\ & = 226^\circ \\ \\ \text{Area of major sector } OAC & = {\theta \over 360^\circ} \times \pi r^2 \\ & = {226^\circ \over 360^\circ} \times \pi (8.7)^2 \\ & = 149.27 \\ & \approx 149 \text{ cm}^2 \end{align*}

\begin{align*} 343^4 & = 49^{-5y} \\ (7^3)^4 & = (7^2)^{-5y} \\ 7^{12} & = 7^{-10y} \phantom{000000} [ (a^m)^n = a^{mn}] \\ \\ \therefore -10y & = 12 \\ y & = {12 \over -10} \\ y & = -1.2 \end{align*}

\begin{align*} {a \over 2a - 3} & = {3c \over d} + 1 \\ {a \over 2a - 3} & = {3c \over d} + {d \over d} \\ {a \over 2a - 3} & = {3c + d \over d} \\ ad & = (2a - 3)(3c + d) \phantom{000000} [\text{Cross-multiply}] \\ ad & = 6ac + 2ad - 9c - 3d \\ ad - 6ac - 2ad & = - 9c - 3d \\ -6ac - ad & = -9c - 3d \\ 6ac + ad & = 9c + 3d \\ a(6c + d) & = 9c + 3d \\ a & = {9c + 3d \over 6c + d} \phantom{0000000} \left[a = {-9c - 3d \over -6c - d} \text{ is ok!}\right] \end{align*}

\begin{align*} y & = x^2 + 6x + 11 \\ y & = x^2 + 6x + \left(6 \over 2\right)^2 - \left(6 \over 2\right)^2 + 11 \\ y & = \underbrace{ x^2 + 6x + 3^2}_{a^2 + 2ab + b^2 = (a + b)^2} + 2 \\ y & = (x + 3)^2 + 2 \\ \\ \text{Minimum} & \text{ point: } ( -3, 2) \end{align*}

(a)

\begin{align*} \angle PAS & = \angle QBP = 90^\circ \phantom{0} \text{ (Interior angle of a square = 90}^\circ) \phantom{.} [A] \\ \\ \text{Let } \angle APS & = x \\ \\ \angle BPQ & = 180^\circ - 90^\circ - x \phantom{0} \text{ (Adjacent angles on a straight line)} \\ & = 90^\circ - x \\ \\ \angle BQP & = 180^\circ - 90^\circ - (90^\circ - x) \phantom{0} \text{ (Angle sum of triangle)} \\ & = 90^\circ - 90^\circ + x \\ & = x \\ \\ \therefore \angle APS & = \angle BQP \phantom{0} [A] \\ \\ PS & = QP \phantom{0} \text{ (All sides in a square have the same length) } [S] \\ \\ \\ \text{Triangles } & APS \text{ and } BQP \text{ are congruent (AAS congruency test)} \end{align*}

(b)

\begin{align*} AB & = {124 \over 4} = 31 \text{ cm} \\ \\ SP & = {100 \over 4} = 25 \text{ cm} \\ \\ \text{Let } AP & = y \text{ cm} \\ \\ PB & = (31 - y) \text{ cm} \\ \\ \text{By Py} & \text{thagoras theorem,} \\ SP^2 & = SA^2 + AP^2 \\ 25^2 & = SA^2 + y^2 \\ 625 & = SA^2 + y^2 \\ 625 - y^2 & = SA^2 \\ \\ \text{Since triangles } & APS \text{ and } BQP \text{ are congruent,} \\ SA & = PB \\ SA^2 & = PB^2 \\ 625 - y^2 & = (31 - y)^2 \\ 625 - y^2 & = \underbrace{ (31)^2 - 2(31)(y) + (y)^2}_{(a - b)^2 = a^2 - 2ab + b^2} \\ 625 - y^2 & = 961 - 62y + y^2 \\ 0 & = y^2 + y^2 - 62y + 961 - 625 \\ 0 & = 2y^2 - 62 y + 336 \\ 0 & = y^2 - 31y + 168 \\ 0 & = (y - 7)(y - 24) \end{align*} \begin{align*} y - 7 & = 0 && \text{ or } & y - 24 & = 0 \\ y & = 7 \text{ (NA, since } AP>PB) &&& y & = 24 \\ \\ & &&& \therefore AP & = 24 \text{ cm} \\ \\ & &&& PB &= 31 - 24 = 7 \text{ cm} \end{align*}

(a)

$$ \left[ \text{Volume of cylinder} = \pi r^2 h \right] $$ \begin{align*} \require{cancel} \\ { \text{Volume of water} \over \text{Volume of container}} & = { \cancel{\pi r^2} (78) \over \cancel{\pi r^2} h } \\ & = { 78 \over h} \\ \\ {78 \over h} & = {65 \over 100} \phantom{000000} \left[65\% = {65 \over 100}\right] \\ 100(78) & = 65h \phantom{0000000}[ \text{Cross-multiply}] \\ 7800 & = 65h \\ {7800 \over 65} & = h \\ 120 & = h \\ \\ \text{Height} & = 120 \text{ cm} \end{align*}

(b)

$$ \left[ \text{For similar solids, }{V_1 \over V_2} = \left( { l_1 \over l_2} \right)^3 \right] $$ \begin{align*} \\ { \text{Volume of water} \over \text{Volume of cone} } & = \left(78 \over h\right)^3 \\ {65 \over 100} & = \left(78 \over h\right)^3 \\ \sqrt[3]{65 \over 100} & = {78 \over h} \\ 0.86624 & = {78 \over h} \\ {0.86624 \over 1} & = {78 \over h} \\ 0.86624 h & = 78 \phantom{000000} [\text{Cross-multiply}] \\ h & = {78 \over 0.86624} \\ h & = 90.044 \\ h & \approx 90.0 \text{ cm} \end{align*}

\begin{align*} \text{Distance travelled by } A & = \underbrace{ {1 \over 2} \times (20 + v) \times (60)}_\text{Area of trapezium} + 40 \times v \\ & = \left(10 + {1 \over 2}v\right) \times 60 + 40v \\ & = 600 + 30v + 40v \\ & = (600 + 70v) \text{ m} \\ \\ \text{Distance travelled by } B & = 100 \times (v + 3) \\ & = (100v + 300) \text{ m} \\ \\ \text{Distance travelled by } B & = \text{Distance travelled by } A + 75 \\ 100v + 300 & = 600 + 70v + 75 \\ 100v - 70v & = 600 + 75 - 300 \\ 30v & = 375 \\ v & = {375 \over 30} \\ v & = 12.5 \end{align*}

\begin{align*} {2x \over x + 4} + {5 \over 3x - 5} & = 3 \\ {2x(3x - 5) \over (x + 4)(3x - 5)} + {5(x + 4) \over (x + 4)(3x - 5)} & = 3 \\ {2x(3x - 5) + 5(x + 4) \over (x + 4)(3x - 5)} & = 3 \\ {6x^2 - 10x + 5x + 20 \over (x + 4)(3x - 5)} & = 3 \\ {6x^2 - 5x + 20 \over (x + 4)(3x - 5)} & = {3 \over 1} \\ 6x^2 - 5x + 20 & = 3(x + 4)(3x - 5) \phantom{00000} [\text{Cross-multiply}] \\ 6x^2 - 5x + 20 & = 3(3x^2 - 5x + 12x - 20) \\ 6x^2 - 5x + 20 & = 3(3x^2 + 7x - 20) \\ 6x^2 - 5x + 20 & = 9x^2 + 21x - 60 \\ 0 & = 9x^2 - 6x^2 + 21x + 5x - 60 - 20 \\ 0 & = 3x^2 + 26x - 80 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ x & = {-26 \pm \sqrt{(26)^2 - 4(3)(-80)} \over 2(3)} \\ x & = {-26 \pm \sqrt{1636} \over 6} \\ x & = 2.4079 \text{ or } -11.074 \\ x & \approx 2.41 \text{ or } -11.07 \text{ (2 dp)} \end{align*}

\begin{align*} \left(16 \over 81 a^4 c^{12} \right)^{-{3 \over 4}} & = \left( 81 a^4 c^{12} \over 16\right)^{3 \over 4} \phantom{00000000} \left[ \left(a \over b\right)^{-n} = \left(b \over a\right)^n \right] \\ & = \left( 3^4 a^4 c^{12} \over 2^4 \right)^{3 \over 4} \\ & = { (3^4)^{3 \over 4} (a^4)^{3 \over 4} (c^{12})^{3 \over 4} \over (2^4)^{3 \over 4} } \\ & = { 3^3 a^3 c^9 \over 2^3 } \phantom{00000000000000} [ (a^m)^n = a^{mn} ] \\ & = { 27 a^3 c^9 \over 8} \\ & = {27 \over 8} a^3 c^9 \phantom{00009999999000} [p = {27 \over 8}, q = 3, r = 9] \end{align*}

(a)(i)

\begin{align*} \text{Mean} & = { \sum fx \over \sum f} \\ & = {(8 \times 72.5) + (8 \times 76) + (7 \times 78) + (5 \times 80) + (4 \times 83) \over 8 + 8 + 7 + 5 + 4} \\ & = 77.0625 \text{ g} \phantom{00000} [\text{Exact}] \end{align*}

(a)(ii)

\begin{align*} \text{Standard deviation} & = \sqrt{ {\sum fx^2 \over \sum f} - \left(\sum fx \over \sum f\right)^2} \\ & = \sqrt{ { 190 \phantom{.} 402 \over 8 + 8 + 7 + 5 + 4} - (77.0625)^2 } \\ & = 3.3813 \\ & \approx 3.38 \text{ g} \end{align*}

(b)

\begin{align*} & \text{1. On average, the mass of kiwi fruits are larger than the mass of plums } \\ & \text{as the mean mass of kiwi fruits is higher.} \\ \\ & \text{2. The mass of kiwi fruits are more consistent than the mass of plums } \\ & \text{as the standard deviation of mass of kiwi fruits is lower.} \end{align*}

(a)

\begin{align*} \text{P(} X \text{ even}) & = 1 - x \phantom{000000} [\text{Total probability = 1 & there's only two outcomes: Even or odd number}] \end{align*}

(b)

\begin{align*} \text{P(Both odd)} & = {3 \over 20} \\ \text{P(} X \text{ odd}) \times \text{P(} Y \text{ odd}) & = {3 \over 20} \\ (x) \times \text{P(} Y \text{ odd}) & = {3 \over 20} \\ \text{P(} Y \text{ odd}) & = {3 \over 20} \div x \phantom{000000} \left[ x = {x \over 1}\right] \\ \text{P(} Y \text{ odd}) & = {3 \over 20} \times {1 \over x} \\ \text{P(} Y \text{ odd}) & = {3 \over 20x} \\ \\ \text{P(} X \text{ even, } Y \text{ odd)} & = {9 \over 20} \\ \text{P(} X \text{ even)} \times \text{P(} Y \text{ odd}) & = {9 \over 20} \\ (1 - x) \times \text{P(} Y \text{ odd}) & = {9 \over 20} \\ \text{P(} Y \text{ odd}) & = {9 \over 20} \div (1 - x) \\ \text{P(} Y \text{ odd}) & = {9 \over 20} \times {1 \over 1 - x} \\ \text{P(} Y \text{ odd}) & = {9 \over 20 - 20x} \\ \\ \therefore {3 \over 20x} & = {9 \over 20 - 20x} \\ 3(20 - 20x) & = 20x(9) \phantom{000000} [\text{Cross-multiply}] \\ 60 -60x & = 180x \\ -60x - 180x & = -60 \\ -240x & = -60 \\ x & = {-60 \over -240} \\ x & = {1 \over 4} \\ \\ \text{P(Both even)} & = \text{P(}X \text{ even)} \times \text{P(} Y \text{ even)} \\ & = (1 - x) \times \left(1 - {3 \over 20x}\right) \\ & = \left(1 - {1 \over 4}\right) \times \left[1 - {3 \over 20 \left(1 \over 4\right)}\right] \\ & = {3 \over 10} \end{align*}

\begin{align*} {6 \over 1 - 2x} & = {5 \over 1} \\ 6 & = 5(1 - 2x) \phantom{000000} [\text{Cross-multiply}] \\ 6 & = 5 - 10x \\ 10x & = 5 - 6 \\ 10x & = -1 \\ x & = -{1 \over 10} \end{align*}

\begin{align*} 12a^2c - 3b^2c & = 3c (4a^2 - b^2) \\ & = 3c [ (2a)^2 - (b)^2 ] \\ & = 3c (2a + b)(2a - b) \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \end{align*}

\begin{align*} 3x + 4y & = 1 \phantom{0} \text{--- (1)} \\ 5x - 6y & = 8 \phantom{0} \text{--- (2)} \\ \\ (1) & \times 5, \\ 15x + 20y & = 5 \phantom{0} \text{--- (3)} \\ \\ (2) & \times 3, \\ 15x - 18y & = 24 \phantom{0} \text{--- (4)} \\ \\ (3) & - (4), \\ 15x + 20y - (15x - 18y) & = 5 - 24 \\ 15x + 20y - 15x + 18y & = -19 \\ 38y & = -19 \\ y & = {-19 \over 38} \\ y & = -0.5 \\ \\ \text{Substitute } & y = -0.5 \text{ into (1),} \\ 3x + 4(-0.5) & = 1 \\ 3x - 2 & = 1 \\ 3x & = 1 + 2 \\ 3x & = 3 \\ x & = 1 \\ \\ \therefore x & = 1, y = -0.5 \end{align*}

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