\begin{align*} \text{Amount Sima receives} & = { 480 \over 16 } \times 9 \\ & = \$ 270 \\ \\ \text{Amount Ken receives} & = 480 - 270 \\ & = \$ 210 \end{align*}

(a)

\begin{align*} x^\circ & = \sin^{-1} (0.9301) && \text{ or } & x^\circ & = 180^\circ - \sin^{-1} (0.9301) \\ & = 68.45^\circ &&& & = 111.55^\circ \end{align*}
\begin{align*} x & \approx 68.5 \text{ or } 111.6 \end{align*}

(b)

\begin{align*} \pi \text{ radians} & = 180^\circ \\ 1 \text{ radian} & = \left(180 \over \pi \right)^\circ \\ 1.36 \text{ radians} & = {180 \over \pi} \times 1.36 \\ & = 77.922 \\ & \approx 77.9^\circ \end{align*}

(a)

\begin{align*} (2 c^3 d^2 )^3 & = (2^3) (c^3)^3 (d^2)^3 \phantom{000000} [ (ab)^m = a^m \times b^m ] \\ & = 8 c^9 d^6 \phantom{0000000000000} [ (a^m)^n = a^{mn} ] \end{align*}

(b)

\begin{align*} 16 \times 5^4 + 9 \times 25^2 & = 5^k \\ 16 \times 5^4 + 9 \times (5^2)^2 & = 5^k \\ 16 \times 5^4 + 9 \times 5^4 & = 5^k \phantom{000000} [ (a^m)^n = a^{mn} ] \\ 5^4 (16 + 9) & = 5^k \phantom{000000} [ \text{Factorise } 5^4 ] \\ 5^4 (25) & = 5^k \\ 5^4 (5^2) & = 5^k \\ 5^6 & = 5^k \phantom{000000} [ a^m \times a^n = a^{m + n} ] \\ \\ \therefore k & = 6 \end{align*}

(a)

\begin{align*} \text{Power of each factor is a multiple of 2} \end{align*}

(b)(i)

\begin{align*} S & = 2^8 \times 7^{10} \times 11^{15} \\ \\ N & = 2^8 \times 7^{10} \times 11^{15} \times (2 \times 7^2) \\ & = 2^{9} \times 7^{12} \times 11^{15} \phantom{000000000} [\text{Powers are multiples of 3}] \\ \\ \therefore k & = 2 \times 7^2 \end{align*}

(b)(ii)

\begin{align*} S & = 2^8 \times 7^{10} \times 11^{15} \\ \\ T & = 2^{20} \times 7^8 \times 11^{15} \\ \\ \text{LCM} & = 2^{20} \times 7^{10} \times 11^{15} \end{align*}

(a)

\begin{align*} & \text{Simple interest rate} = 0.25 \% \times 12 = 3 \% \\ \\ & \text{Since interest is compounded monthly, the first year's interest is more than 3% of \$4500} \end{align*}

(b)

\begin{align*} \text{Total amount} & = P \left(1 + {r \over 100}\right)^n \\ & = 4500 \left(1 + {0.25 \over 100} \right)^{12} \\ & = 4 \phantom{.} 636.871 \\ & \approx \$ 4 \phantom{.} 636.87 \end{align*}

(a)

\begin{align*} x^2 - 14x + b & = x^2 - 14x + \left(14 \over 2\right)^2 - \left(14 \over 2\right)^2 + b \\ & = \underbrace{x^2 - 14x + 7^2}_{a^2 - 2ab + b^2 = (a - b)^2} - 49 + b \\ & = (x - 7)^2 - 49 + b \\ \\ (x - 7)^2 - 49 + b & = (x + a)^2 - 25 \end{align*} \begin{align*} a & = -7 &&& -49 + b & = -25 \\ & &&& b & = -25 + 49 \\ & &&& b & = 24 \end{align*}

(b)

\begin{align*} y & = x^2 - 14x + b \\ y & = (x - 7)^2 - 25 \\ \\ \text{Line of } & \text{symmetry: } x = 7 \end{align*}

(a)

\begin{align*} \text{No. of stamps} & = {550 \over 100} \times 118 \phantom{000000} [100\% + 18\% = 118\%] \\ & = 649 \end{align*}

(b)

\begin{align*} 100\% - 28\% & = 72\% \phantom{0000000} [\text{This represents postcards from EU and Asia}] \\ \\ 72\% \times {100 - 55 \over 100} & = 32.4 \% \phantom{000000} [\text{This represents postcards from Asia}] \\ \\ \text{Total number of postcards} & = {81 \over 32.4} \times 100 \\ & = 250 \end{align*}

(a)(i)

\begin{align*} \frac{50}{100} \times 40 & = 20 \\ \\ \text{Median height} & = 114 \text{ cm} \end{align*}

(a)(ii)

\begin{align*} \frac{25}{100} \times 40 & = 10 \\ \\ \frac{75}{100} \times 40 & = 30 \\ \\ \text{Interquartile range} & = Q_3 - Q_1 \\ & = 125 - 106 \\ & = 19 \text{ cm} \end{align*}

(b)

\begin{align*} \text{Number of children not allowed} & = 40 - 15 \\ & = 25 \\ \\ h & = 121 \end{align*}

(c)(i) Can use calculator function directly without formula

\begin{align*} \text{Mean height} & = { \sum fx \over \sum f } \\ & = { 3 \times 95 + 13 \times 105 + 8 \times 115 + 10 \times 125 + 6 \times 135 \over 3 + 13 + 8 + 10 + 6 } \\ & = 115.75 \text{ cm} \end{align*}

(c)(ii) Can use calculator function directly without formula

\begin{align*} \text{SD} & = \sqrt{ { \sum fx^2 \over \sum f } - \left( \sum fx \over \sum f\right)^2 } \\ \\ \sum fx^2 & = 3 \times 95^2 + 13 \times 105^2 + 8 \times 115^2 + 10 \times 125^2 + 6 \times 135^2 \\ & = 541 \phantom{.} 800 \\ \\ SD & = \sqrt{ { 541 \phantom{.} 800 \over 3 + 13 + 8 + 10 + 6 } - (115.75)^2 } \\ & = 12.121 \\ & \approx 12.1 \text{ cm} \end{align*}

\begin{align*} \underline{ \text{France} } \phantom{00000000000000000} & \\ \\ \text{SGD } 1.5953 & = \text{EUR } 1 \\ \text{SGD } 1 & = \text{EUR } {1 \over 1.5953} \\ \\ \text{Amount received in France} & = {1 \over 1.5953} \times 780 \\ & = \text{EUR } 488.936 \\ \\ \\ \underline{ \text{SG} } \phantom{00000000000000000000} & \\ \\ \text{Amount received in SG} & = 488.936 + 1 \\ & = \text{EUR } 489.936 \\ \\ \text{SGD }780 & = \text{EUR } 489.936 \\ \text{SGD }1 & = \text{EUR } {489.936 \over 780} \\ & = \text{EUR } 0.62812 \\ & \approx \text{EUR } 0.628 \end{align*}

\begin{align*} \text{Gradient of } AB & = {y_2 - y_1 \over x_2 - x_1} \\ {1 \over 3} & = {q - 3 \over p - (-2)} \\ {1 \over 3} & = {q - 3 \over p + 2} \\ p + 2 & =3(q - 3) \phantom{000000} [\text{Cross-multiply}] \\ p + 2 & = 3q - 9 \\ p & = 3q - 9 - 2 \\ p & = 3q - 11 \end{align*}

\begin{align*} {3a \over 4a - 1} - {2 \over 3a + 1} & = {3a(3a + 1) \over (4a - 1)(3a + 1)} - {2(4a - 1) \over (4a - 1)(3a + 1)} \\ & = {3a(3a + 1) - 2(4a -1) \over (4a - 1)(3a + 1)} \\ & = {9a^2 + 3a - 8a + 2 \over (4a - 1)(3a + 1)} \\ & = {9a^2 - 5a + 2 \over (4a - 1)(3a + 1)} \end{align*}

(a)(i)

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

(a)(ii)

\begin{align*} 15cd - 10ce + 12d^2 - 8de & = 5c (3d - 2e) + 4d(3d - 2e) \\ & = (3d - 2e)(5c + 4d) \phantom{000000000} [\text{Factorise by grouping method}] \end{align*}

(b)

\begin{align*} (2x + 3a)(5x - 2a) & = 10x^2 - 4ax + 15ax - 6a^2 \\ & = 10x^2 + 11ax - 6a^2 \end{align*}

\begin{align*} \text{P(White marble)} & = {7 + n \over 12 + 7 + n + 6} \\ {3 \over 5} & = {7 + n \over 25 + n} \\ 3(25 + n) & = 5(7 + n) \phantom{000000} [\text{Cross-multiply}] \\ 75 + 3n & = 35 + 5n \\ 3n - 5n & = 35 - 75 \\ -2n & = -40 \\ n & = {-40 \over -2} \\ n & = 20 \end{align*}

\begin{align*} D \text{ equidistant } & \text{from } A, B \text{ and } C \implies AD = BD = CD \\ \\ \text{Let } \angle DBC & = x \\ \\ \angle BDC & = 180^\circ - 2x \phantom{0} (\text{Angle sum of isosceles triangle}) \\ \\ \angle BDA & = 180^\circ - (180^\circ - 2x) \phantom{0} (\text{Adjacent angles on a straight line}) \\ & = 180^\circ - 180^\circ + 2x \\ & = 2x \\ \\ \angle DBA & = {180^\circ - 2x \over 2} \phantom{0} (\text{Base angle of isosceles triangle}) \\ & = {180^\circ \over 2} - {2x \over 2} \\ & = 90^\circ - x \\ \\ \therefore \angle ABC & = 90^\circ- x + x \\ & = 90^\circ \end{align*}

\begin{align*} (2n - 1)^2 - 2(n - 5)(2n - 1) & = \underbrace{(2n)^2 - 2(2n)(1) + 1^2}_{(a - b)^2 = a^2 - 2ab + b^2} - 2(2n^2 - n - 10n + 5) \\ & = 4n^2 - 4n + 1 - 2(2n^2 - 11n + 5) \\ & = 4n^2 - 4n + 1 - 4n^2 + 22n - 10 \\ & = 18n - 9 \\ & = 3(6n - 3) \\ \\ \therefore \text{Expression is a multiple} & \text{ of 3 for all integer values of } n \end{align*}

\begin{align*} & {V_1 \over V_2} = \left(l_1 \over l_2\right)^3 = \left(1 \over 3\right)^3 = {1 \over 27} \\ \\ & \text{The chart implies that sales in 2020 are 27 times the sales in 2010} \end{align*}

(a)

(b)

\begin{align*} (C' \cap D' ) \cup (C \cap D) \end{align*}

(a)

\begin{align*} \text{By simi} & \text{lar triangles,} \\ {h \over 20} & = {r \over 10} \\ 10h & = 20r \phantom{000000} [\text{Cross-multiply}] \\ h & = {20r \over 10} \\ h & = 2r \\ \\ \text{Since } d = 20 - h, & \phantom{.} h = 20 -d \\ \\ \therefore 20 - d & = 2r \\ -d & = 2r - 20 \\ d & = 20 - 2r \phantom{0} \text{(Shown)} \end{align*}

(b)

\begin{align*} {V_1 \over V_2} & = \left(l_1 \over l_2\right)^3 \\ { \text{Volume of small cone} \over \text{Volume of large cone} } & = \left(r \over 10\right)^3 \\ {1 \over 2} & = {r^3 \over 10^3} \phantom{000000} [\text{Volume of large cone = Volume of water + small cone}] \\ {1 \over 2} & = {r^3 \over 1000} \\ 1000 & = 2r^3 \phantom{0000000} [\text{Cross-multiply}] \\ {1000 \over 2} & = r^3 \\ 500 & = r^3 \\ \sqrt[3]{500} & = r \\ 7.937 & = r \\ \\ d & = 20 - 2(7.937) \\ & = 4.126 \\ & \approx 4.13 \end{align*}

\begin{align*} \require{cancel} 3x^2 - 8x - 16 & = (3x + 4)(x - 4) \\ \\ a^2 - b^2 & = (a + b)(a - b) \\ (2x + 1)^2 - (x + 3)^2 & = [(2x + 1) + (x + 3)][(2x + 1)- (x + 3)] \\ & = (2x + 1 + x + 3)(2x + 1 - x - 3) \\ & = (3x + 4)(x - 2) \\ \\ \therefore {3x^2 - 8x - 16 \over (2x + 1)^2 - (x + 3)^2} & = { \cancel{(3x + 4)} (x - 4) \over \cancel{(3x + 4)} (x - 2)} \\ & = {x- 4 \over x - 2} \end{align*}

(a)

\begin{align*} \textbf{T} & = \mathop{ \left( \begin{matrix} 4 & 3 \\ 5 & 2 \end{matrix} \right) }_{2 \times 2} \mathop{ \left( \begin{matrix} 25.8 & 49.2 & 16.1 \\ x & 68.5 & 4.2 \end{matrix} \right) }_{2 \times 3} \\ & = \mathop{ \left( \begin{matrix} 4(25.8) + 3x & 4(49.2) + 3(68.5) & 4(16.1) + 3(4.2) \\ 5(25.8) + 2x & 5(49.2) + 2(68.5) & 5(16.1) + 2(4.2) \end{matrix} \right) }_{2 \times 3} \\ & = \left( \begin{matrix} 103.2 + 3x & 402.3 & 77 \\ 129 + 2x & 383 & 88.9 \end{matrix} \right) \end{align*}

(b)

\begin{align*} \textbf{T} & = \left( \begin{matrix} 103.2 + 3x & 402.3 & 77 \\ 129 + 2x & 383 & 88.9 \end{matrix} \right) \phantom{0} \begin{matrix} \text{Namita} \\ \text{Sian} \end{matrix} \\ \\ 103.2 + 3x & = 129 + 2x + 5.6 \\ 103.2 + 3x & = 134.6 + 2x \\ 3x - 2x & = 134.6 - 103.2 \\ x & = 31.4 \end{align*}

(c)

\begin{align*} \textbf{T} & = \mathop{ \left( \begin{matrix} 4 & 3 \\ 5 & 2 \end{matrix} \right) }_{2 \times 2} \mathop{ \left( \begin{matrix} 1.6 \\ 2.8 \end{matrix} \right) }_{2 \times 1} \\ \\ \textbf{D} & = \left( \begin{matrix} 1.6 \\ 2.8 \end{matrix} \right) \end{align*}

(a)

\begin{align*} \text{Area of } A & = 3x(4x - y) \\ & = 12x^2 - 3xy \\ \\ \text{Area of } B & = 2x(3x + 2y) \\ & = 6x^2 + 4xy \\ \\ 12x^2 - 3xy & = 6x^2 + 4xy \\ -3xy - 4xy & = 6x^2 - 12x^2 \\ -7xy & = -6x^2 \\ 7xy & = 6x^2 \\ y & = {6x^2 \over 7x} \\ y & = {6x \over 7} \\ y & = {6 \over 7}x \end{align*}

(b)

\begin{align*} \text{Perimeter of } A & = 2(3x) + 2(4x - y) \\ & = 6x + 8x - 2y \\ & = 14x - 2y \\ \\ \text{Perimeter of } B & = 2(2x) + 2(3x + 2y) \\ & = 4x + 6x + 4y \\ & = 10x + 4y \\ \\ 10x + 4y & = 14x - 2y + 16 \\ 10x - 14x + 4y + 2y & = 16 \\ -4x + 6y & = 16 \\ \\ \text{Since } & y = {6 \over 7}x \text{ from (a),} \\ -4x + 6 \left({6 \over 7}x\right) & = 16 \\ -4x + {36 \over 7}x & = 16 \\ {8 \over 7}x & = 16 \\ x & = 16 \div {8 \over 7} \\ x & = 14 \end{align*}

(a)

\begin{align*} \angle N1CB & = 180^\circ - 60^\circ - 64^\circ \phantom{0} (\text{Interior angles}) \\ & = 56^\circ \\ \\ \text{Bearing of } B \text{ from } C & = 360^\circ - 56^\circ \\ & = 304^\circ \end{align*}

(b)

\begin{align*} AC^2 & = AB^2 + BC^2 - 2(AB)(BC) \cos \angle ABC \phantom{000000} [\text{Cosine rule}] \\ & = 63^2 + 95^2 - 2(63)(95) \cos 64^\circ \\ AC & = \sqrt{ 63^2 + 95^2 - 2(63)(95) \cos 64^\circ } \\ & = 88.015 \\ & \approx 88.0 \text{ m} \end{align*}

(c)

\begin{align*} \angle N2AB & = 180^\circ - 60^\circ \phantom{0} (\text{Interior angles}) \\ & = 120^\circ \\ \\ {\sin \angle BAC \over BC} & = {\sin \angle ABC \over AC} \phantom{000000} [\text{Sine rule}] \\ {\sin \angle BAC \over 95} & = {\sin 64^\circ \over 88.015} \\ 88.015 \sin \angle BAC & = 95 \sin 64^\circ \phantom{00000000} [\text{Cross-multiply}] \\ \sin \angle BAC & = {95 \sin 64^\circ \over 88.015} \\ \angle BAC & = \sin^{-1} \left( {95 \sin 64^\circ \over 88.015} \right) \\ & = 75.959^\circ \\ \\ \text{Bearing of } C \text{ from } A & = 360^\circ - 120^\circ - 75.959^\circ \\ & = 164.041^\circ \\ & \approx 164.0^\circ \end{align*}

(a)

\begin{align*} I & \propto {1 \over D^2} \\ I & = {k \over D^2} \\ \\ \text{When } D & = 150, I = 1370, \phantom{000000} [\text{Earth's numbers}] \\ 1370 & = {k \over 150^2} \\ {1370 \over 1} & = {k \over 150^2} \\ 1370(150^2) & = k \\ 30 \phantom{.} 825 \phantom{.} 000 & = k \\ \\ I & = { 30 \phantom{.} 825 \phantom{.} 000 \over D^2} \\ \\ \text{When } & D = 228, \\ I & = { 30 \phantom{.} 825 \phantom{.} 000 \over 228^2} \\ & = 592.97 \\ & \approx 593 \text{ watts/m}^2 \end{align*}

(b)

\begin{align*} I & \propto {1 \over D^2} \\ \\ \text{When } D \text{ is doubled, } & I \text{ will decrease to } {1 \over 4} \text{ of initial value} \end{align*}

(a)

\begin{align*} \text{Sum of first } n \text{ terms, } S_n & = {1 \over 2}n(5n + 1) \\ \\ S_1 & = {1 \over 2}(1) [5(1) + 1] \\ & = 3 \\ \\ \implies T_1 & = 3 \\ \\ \\ S_2 & = {1 \over 2} (2)[5(2) + 1] \\ & = 11 \\ \\ T_2 & = 11 - 3 \\ & = 8 \\ \\ \\ S_3 & = {1 \over 2} (3)[5(3) + 1] \\ & = 24 \\ \\ T_3 & = 24 - 8 - 3 \\ & = 13 \\ \\ \\ \text{First 3 terms: } & 3, 8, 13 \end{align*}

(b)

\begin{align*} T_n & = T_1 + (n - 1)(d) \phantom{000000} [d \text{ is the common difference}] \\ & = 3 + (n - 1)(5) \\ & = 3 + 5n - 5 \\ & = 5n - 2 \end{align*}

To minimise the length outside the cuboid, the rod must be placed along AG (or along BH)

\begin{align*} \text{By Py} & \text{thagoras theorem,} \\ AH^2 & = AE^2 + EH^2 \\ & = 8^2 + 6^2 \\ AH & = \sqrt{8^2 + 6^2} \\ & = 10 \text{ cm} \\ \\ \text{Length of rod} & = 10 + 7 \\ & = 17 \text{ cm} \\ \\ \text{By Py} & \text{thagoras theorem,} \\ AF^2 & = AE^2 + EF^2 \\ & = 8^2 + 12^2 \\ AF & = \sqrt{8^2 + 12^2} \\ & = 14.422 \text{ cm} \\ \\ \text{By Py} & \text{thagoras theorem,} \\ AG^2 & = AF^2 + FG^2 \\ & = 14.422^2 + 6^2 \\ AG & = \sqrt{14.422^2 + 6^2} \\ & = 15.62 \text{ cm} \\ \\ \text{Shortest length outside cuboid} & = 17 - 15.62 \\ & = 1.38 \text{ cm} \end{align*}

(a)

\begin{align*} \overrightarrow{OP} & = \overrightarrow{OM} + \overrightarrow{MP} \\ & = \textbf{a} + {3 \over 5} \overrightarrow{MN} \\ & = \textbf{a} + {3 \over 5} \left( \overrightarrow{MO} + \overrightarrow{ON} \right) \\ & = \textbf{a} + {3 \over 5} ( - \textbf{a} + \textbf{b} ) \\ & = \textbf{a} - {3 \over 5} \textbf{a} + {3 \over 5} \textbf{b} \\ & = {2 \over 5} \textbf{a} + {3 \over 5} \textbf{b} \\ & = {1 \over 5} (2 \textbf{a} + 3 \textbf{b} ) \phantom{0} \text{ (Shown)} \end{align*}

(b)

\begin{align*} \overrightarrow{OQ} & = \overrightarrow{OM} + \overrightarrow{MQ} \\ & = \textbf{a} + {1 \over 5} (\textbf{a} + 9 \textbf{b}) \\ & = \textbf{a} + {1 \over 5}\textbf{a} + {9 \over 5} \textbf{b} \\ & = {6 \over 5} \textbf{a} + {9 \over 5} \textbf{b} \\ \\ \text{Let } \overrightarrow{OP} & = k \overrightarrow{OQ} \\ {2 \over 5} \textbf{a} + {3 \over 5} \textbf{b} & = k \left( {6 \over 5} \textbf{a} + {9 \over 5} \textbf{b} \right) \\ {2 \over 5} \textbf{a} + {3 \over 5} \textbf{b} & = {6 \over 5} k \textbf{a} + {9 \over 5} k \textbf{b} \end{align*} \begin{align*} \text{Comparing } & \textbf{a}, &&& \text{Comparing } & \textbf{b}, \\ {2 \over 5} & = {6 \over 5} k &&& {3 \over 5} & = {9 \over 5}k \\ {2 \over 5} \div {6 \over 5} & = k &&& {3 \over 5} \div {9 \over 5} & = k \\ {1 \over 3} & = k &&& {1 \over 3} & = k \end{align*}
\begin{align*} & \therefore \overrightarrow{OP} = {1 \over 3} \overrightarrow{OQ} \text{ and } O, P \text{ and } Q \text{ lie on a straight line} \end{align*}

(a)(i)

\begin{align*} 3 \phantom{.} 771 \phantom{.} 721 & \approx 3 \phantom{.} 770 \phantom{.} 000 \\ & = 3.77 \times 10^6 \end{align*}

(a)(ii)

\begin{align*} \text{Percentage change} & = { \text{Final} - \text{Initial} \over \text{Initial} } \times 100 \\ & = { 4 \phantom{.} 044 \phantom{.} 210 - 3 \phantom{.} 273 \phantom{.} 363 \over 3 \phantom{.} 273 \phantom{.} 363 } \times 100 \\ & = 23.549 \\ & \approx 23.5 \% \end{align*}

(a)(iii)

\begin{align*} \text{Percentage of residents} & = 100 \% - 18.7 \% \\ & = 81.3 \% \\ \\ \text{Total population} & = {3 \phantom{.} 273 \phantom{.} 363 \over 81.3} \times 100 \\ & = 4 \phantom{.} 026 \phantom{.} 276.753 \\ & \approx 4 \phantom{.} 030 \phantom{.} 000 \end{align*}

(b)

\begin{align*} 7810 \text{ pax} & \rightarrow 1 \text{ km}^2 \\ 1 \text{ pax} & \rightarrow {1 \over 7810} \text{ km}^2 \\ 5.69 \text{ million pax} & \rightarrow {1 \over 7810} \times 5 \phantom{.} 690 \phantom{.} 000 \\ \\ \text{Area of Singapore} & = {1 \over 7810} \times 5 \phantom{.} 690 \phantom{.} 000 \\ & = 728.553 \\ & \approx 729 \text{ km}^2 \end{align*}

(c)(i)

\begin{align*} 3.9 \text{ cm} & = 0.039 \text{ m} \phantom{000000} [1 \text{ m} = 100 \text{ cm}] \\ \\ \text{Drawing} & : \text{Actual} \\ 0.039 & : 7.8 \times 10^{-6} \\ {0.039 \over 7.8 \times 10^{-6}} & : 1 \\ 5000 & : 1 \end{align*}

(c)(ii)

\begin{align*} \text{Drawing} & : \text{Actual} \\ 5000 & : 1 \\ 1 & : {1 \over 5000} \\ 8 & : {1 \over 5000} \times 8 \\ \\ \text{Actual diameter} & = {1 \over 5000} \times 8 \\ & = 0.0016 \text{ mm} \\ & = 1.6 \times 10^{-3} \text{ mm} \end{align*}

(a)

\begin{align*} 6 - 7x & = 5(1- 2x) \\ 6 - 7x & = 5 - 10x \\ -7x + 10x & = 5 - 6 \\ 3x & = -1 \\ x & = -{1 \over 3} \end{align*}

(b)

\begin{align*} 5 - y & > 7 \\ -y & > 7 - 5 \\ -y & > 2 \\ y & < - 2 \end{align*}

(c)(i)

\begin{align*} c & = {1 \over b} + {a \over 3 - a} \\ & = {1 \over 4} + {2 \over 3 - 2} \\ & = 2{1 \over 4} \end{align*}

(c)(ii) Topic: Change subject of formula

\begin{align*} c & = {1 \over b} + {a \over 3 - a} \\ {c \over 1} - {1 \over b} & = {a \over 3 - a} \\ {bc \over b} - {1 \over b} & = {a \over 3 - a} \\ {bc - 1 \over b} & = {a \over 3 - a} \\ (3 - a)(bc - 1) & = ab \phantom{000000} [\text{Cross-multiply}] \\ 3bc - 3 - abc + a & = ab \\ a - abc - ab & = 3 - 3bc \\ a(1 - bc - b) & = 3 - 3bc \\ a & = {3 - 3bc \over 1 - bc - b} \end{align*}

(d) Topic: Solve fractional equation & usage of quadratic formula

\begin{align*} {x \over 2x - 1} - {6 \over 4 - x} & = 3 \\ {x(4 - x) \over (2x - 1)(4 - x)} - {6(2x - 1) \over (2x - 1)(4 - x)} & = 3 \\ {x(4 - x) - 6(2x - 1) \over (2x - 1)(4 - x)} & = 3 \\ {4x - x^2 - 12x + 6 \over 8x - 2x^2 - 4 + x} & = 3 \\ {-x^2 - 8x + 6 \over - 2x^2 + 9x - 4} & = {3 \over 1} \\ -x^2 - 8x + 6 & = 3(-2x^2 + 9x - 4) \phantom{000000} [\text{Cross-multiply}] \\ -x^2 - 8x + 6 & = -6x^2 + 27x - 12 \\ -x^2 + 6x^2 - 8x - 27x + 6 + 12 & = 0 \\ 5x^2 - 35x + 18 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {- (-35) \pm \sqrt{ (-35)^2 - 4(5)(18)} \over 2(5)} \\ & = {35 \pm \sqrt{865} \over 10} \\ & = 6.441 \text{ or } 0.5589 \\ & \approx 6.44 \text{ or } 0.56 \phantom{0} \text{(2 d.p)} \end{align*}

(a)

\begin{align*} \text{Area of circle} & = \pi r^2 \\ & = \pi (3y)^2 \\ & = \pi (9y^2) \\ & = 9 \pi y^2 \\ \\ \text{Curved surface area} & = 2 \pi r^2 \\ & = 2 \pi (3y)^2 \\ & = 2 \pi (9y^2) \\ & = 18 \pi y^2 \\ \\ \text{Total surface area} & = 9\pi y^2 + 18 \pi y^2 \\ & = 27 \pi y^2 \phantom{0} \text{ (Shown)} \end{align*}

(b)

\begin{align*} \text{Curved surface area of cylinder} & = 2\pi r h \\ & = 2\pi (2y) (h) \\ & = 4 \pi hy \\ \\ \text{Area of 2 circles} & = 2 \pi r^2 \\ & = 2 \pi (2y)^2 \\ & = 2 \pi (4y^2) \\ & = 8 \pi y^2 \\ \\ \text{Total SA of cylinder} & = \text{Total SA of hemisphere} \\ 4 \pi h y + 8 \pi y^2 & = 27 \pi y^2 \\ 4 \pi h y & = 27 \pi y^2 - 8 \pi y^2 \\ 4 \pi h y & = 19 \pi y^2 \\ h & = {19 \pi y^2 \over 4 \pi y} \\ h & = {19 y \over 4} \\ h & = {19 \over 4} y \end{align*}

(c)

\begin{align*} \text{Volume of hemisphere} & = {2 \over 3} \pi r^3 \\ & = {2 \over 3} \pi (3y)^3 \\ & = {2 \over 3} \pi (27 y^3) \\ & = 18 \pi y^3 \\ \\ 18 \pi y^3 & = 500 \\ y^3 & = {500 \over 18 \pi} \\ y & = \sqrt[3]{500 \over 18 \pi} \\ y & = 2.0678 \\ \\ \text{From (b), } h & = {19 \over 4}(2.0678) \\ & = 9.8221 \\ \\ \text{Volume of cylinder} & = \pi r^2 h \\ & = \pi (2y)^2 (h) \\ & = \pi [ 2(2.0678)]^2 (9.8221) \\ & = 527.75 \\ & \approx 528 \text{ cm}^3 \end{align*}

(a)

\begin{align*} y & = {2 \over x^2} + 3x - 1 \\ \\ \text{When } & x = -3, \\ y & = {2 \over (-3)^2} + 3(-3) - 1 \\ y & = -9.778 \\ y & \approx -9.8 \text{ (1 d.p)} \end{align*}

(b)

Note: For \(x = 0\), \(y = \frac{2}{0^2} + 3(0) - 1\) is undefined. Graphically, this means that the curve does not cut the \(y\)-axis. (A level concept: The curve has a vertical asymptote \(x = 0\).)

(c)(i)

(c)(ii)

\begin{align*} \text{From graph, } x & = -0.65 \text{ and } 0.7 \end{align*}

(c)(iii)

\begin{align*} y & = {2 \over x^2} + 3x - 1 \phantom{0} \text{--- (1)} \\ y & = {5 \over 2}x + {7 \over 2} \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ {2 \over x^2} + 3x - 1 & = {5 \over 2}x + {7 \over 2} \\ x^2 \left( {2 \over x^2} + 3x - 1 \right) & = x^2 \left({5 \over 2}x + {7 \over 2}\right) \\ 2 + 3x^3 - x^2 & = {5 \over 2}x^3 + {7 \over 2}x^2 \\ 3x^3 - {5 \over 2}x^3 - x^2 - {7 \over 2}x^2 + 2 & = 0 \\ {1 \over 2}x^3 - {9 \over 2}x^2 + 2 & = 0 \\ 2 \left( {1 \over 2}x^3 - {9 \over 2}x^2 + 2 \right) & = 2(0) \\ x^3 - 9x^2 + 4 & = 0 \\ \\ \therefore A & = -9, B = 4 \end{align*}

(a)

\begin{align*} \angle OBE & = 90^\circ \phantom{0} \text{(Tangent perpendicular to radius)} \\ \\ \angle BOE & = 180^\circ - 90^\circ - 36^\circ \phantom{0} \text{(Angle sum of triangle)} \\ & = 54^\circ \\ \\ \angle OCB & = {180^\circ - 54^\circ \over 2} \phantom{0} \text{(Base angles of isosceles triangle } OBC) \\ & = 63^\circ \\ \\ \angle BAD & = 180^\circ - \angle BCD \phantom{0} \text{(Angles in opposite segments)} \\ & = 180^\circ - (63^\circ + 43^\circ) \\ & = 74^\circ \end{align*}

(b)

\begin{align*} \text{Reflex } \angle BOC & = 360^\circ - 54^\circ \phantom{0} \text{(Angles at a point)} \\ & = 306^\circ \\ \\ \text{Arc length} & = {\theta \over 360^\circ} \times 2 \pi r \\ \text{Major arc length } BADC & = {306^\circ \over 360^\circ} \times 2 \pi (8) \\ & = 13.6 \pi \text{ cm} \\ \\ \tan \angle OEB & = {OB \over BE} \phantom{000000} \left[ {Opp \over Adj} \right] \\ \tan 36^\circ & = {8 \over BE} \\ {\tan 36^\circ \over 1} & = {8 \over BE} \\ BE \tan 36^\circ & = 8 \\ BE & = {8 \over \tan 36^\circ} \\ BE & = 11.011 \text{ cm} \\ \\ \text{By Py} & \text{thagoras theorem,} \\ OE^2 & = OB^2 + BE^2 \\ & = 8^2 + 11.011^2 \\ OE & = \sqrt{ 8^2 + 11.011^2 } \\ & = 13.61 \text{ cm} \\ \\ \text{Perimeter} & = 13.6 \pi + 11.011 + (13.61 - 8) \\ & = 59.346 \\ & \approx 59.3 \text{ cm} \end{align*}

(a)

\begin{align*} \overrightarrow{XY} & = \overrightarrow{XO} + \overrightarrow{OY} \\ & = - \overrightarrow{OX} + \overrightarrow{OY} \\ & = - {7 \choose 3} + {3 \choose 4} \\ & = {-4 \choose 1} \\ \\ \overrightarrow{XY} & = m \textbf{p} + n \textbf{q} \\ {-4 \choose 1} & = m {-4 \choose 5} + n {2 \choose -3} \\ {-4 \choose 1} & = {-4m \choose 5m} + {2n \choose -3n} \\ {-4 \choose 1} & = {-4m + 2n \choose 5m - 3n} \\ \\ -4 = -4m + 2n & \text{ --- (1)} \phantom{000000} 1 = 5m - 3n \text{ --- (2)} \\ \\ \text{From } & (1), \\ -4 & = -4m + 2n \\ -2n & = -4m + 4 \\ 2n & = 4m - 4 \\ n & = 2m - 2 \text{ --- (3)} \\ \\ \text{Substitute } & \text{(3) into (2),} \\ 1 & = 5m - 3(2m - 2) \\ 1 & = 5m - 6m + 6 \\ 1 - 6 & = -m \\ -5 & = -m \\ m & = 5 \\ \\ \text{Substitute } & m = 5 \text{ into (3),} \\ n & = 2(5) - 2 \\ n & = 8 \\ \\ \therefore m & = 5, n = 8 \end{align*}

(b)(i) Since D lies on line AB, gradient of AB = gradient of AD

\begin{align*} \text{Gradient of } AB & = {y_2 - y_1 \over x_2 - x_1} \\ & = {-3 - 6 \over -5 - 4} \\ & = 1 \\ \\ \text{Gradient of } AD & = {y - 6 \over -1 - 4} \\ 1 & = {y - 6 \over -5} \\ -5 & = y - 6 \\ -5 + 6 & = y \\ 1 & = y \end{align*}

(b)(ii)

\begin{align*} C & (8, - 4), D(-1, 1) \\ \\ CD & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ & = \sqrt{ (-1 - 8)^2 + [1 - (-4)]^2 } \\ & = 10.296 \\ & \approx 10.3 \text{ units} \end{align*}

(b)(iii)

\begin{align*} \overrightarrow{AC} & = \overrightarrow{AO} + \overrightarrow{OC} \\ & = -\overrightarrow{OA} + \overrightarrow{OC} \\ & = -{4 \choose 6} + {8 \choose -4} \\ & = {4 \choose -10} \\ \\ \overrightarrow{AE} & = {3 \over 2} \overrightarrow{AC} \\ & = {3 \over 2} {4 \choose -10} \\ & = {6 \choose -15} \\ \\ \overrightarrow{OE} & = \overrightarrow{OA} + \overrightarrow{AE} \\ & = {4 \choose 6} + {6 \choose -15} \\ & = {10 \choose -9} \\ \\ \therefore & \phantom{.} E(10, -9), B(-5, -3) \\ \\ \text{Gradient of } BE & = {-9 - (-3) \over 10 - (-5)} \\ & = - {2 \over 5} \\ \\ y & = mx + c \\ y & = -{2 \over 5}x + c \\ \\ \text{Using } & B(-5, -3), \\ -3 & = -{2 \over 5}(-5) + c \\ -3 & = 2 + c \\ -3 - 2 & = c \\ -5 & = c \\ \\ \text{Eqn of } BE: & \phantom{.} y = -{2 \over 5}x - 5 \end{align*}

(a)(i)

\begin{align*} \text{For teachers, } Q_3 & = 50 \text{ mins} \\ \\ \implies 25 \% \text{ of teachers} & \text{ took more than 50 mins} \\ \\ \text{No. of teachers} & = {18 \over 25} \times 100 \\ & = 72 \end{align*}

(a)(ii)

\begin{align*} & \text{1. Teachers took longer time to travel to school on average as the median time is } \\ & \text{36 mins while the median time for students is 28 mins} \\ \\ & \text{2. The time taken by the teachers and by the students have the same consistency as} \\ & \text{the interquartile range for both are 18 mins} \end{align*}

(b)(i)

\begin{align*} \text{Students who travelled by bus} & = 24 - 8 - 6 \\ & = 10 \\ \\ \text{Required probability} & = {10 \over 24} \\ & = {5 \over 12} \end{align*}

(b)(ii)

\begin{align*} \text{Case 1: P(Both by bus)} & = {10 \over 24} \times {9 \over 23} = {15 \over 92} \\ \\ \text{Case 2: P(Both by cycling)} & = {6 \over 24} \times {5 \over 23} = {5 \over 92} \\ \\ \text{Case 3: P(Both by walking)} & = {8 \over 24} \times {7 \over 23} = {7 \over 69} \\ \\ \text{Required probability} & = {15 \over 92} + {5 \over 92} + {7 \over 69} \\ & = {22 \over 69} \end{align*}

(b)(iii)

\begin{align*} \text{Case 1: P(Walk, walk, cycle/bus)} & = {8 \over 24} \times {7 \over 23} \times {16 \over 22} = {56 \over 759} \\ \\ \text{Case 2: P(Walk, cycle/bus, walk)} & = {8 \over 24} \times {16 \over 23} \times {7 \over 22} = {56 \over 759} \\ \\ \text{Case 3: P(Cycle/bus, walk, walk)} & = {16 \over 24} \times {8 \over 23} \times {7 \over 22} = {56 \over 759} \\ \\ \text{Required probability} & = {56 \over 759} + {56 \over 759} + {56 \over 759} \\ & = {56 \over 253} \end{align*}

(a)(i)

\begin{align*} \require{bbox} \text{General case: } & \\ & \bbox[5px, border: 2px solid black]{ \phantom{00} n \phantom{00.}} \\ \bbox[5px, border: 2px solid black]{n + 7} & \bbox[5px, border: 2px solid black]{n + 8\phantom{0}} \bbox[5px, border: 2px solid black]{n + 9} \\ & \bbox[5px, border: 2px solid black]{n + 16} \\ \\ n + n + 16 & = 204 \\ 2n & = 204 - 16 \\ 2n & = 188 \\ n & = {188 \over 2} \\ n & = 94 \\ \\ \\ & \bbox[5px, border: 2px solid black]{ \phantom{0} 94 \phantom{0}} \\ \bbox[5px, border: 2px solid black]{101} & \bbox[5px, border: 2px solid black]{ \phantom{.} 102 \phantom{.} } \bbox[5px, border: 2px solid black]{103 } \\ & \bbox[5px, border: 2px solid black]{\phantom{0} 110} \end{align*}

(a)(ii)

\begin{align*} \text{Product of left and right} & = (n + 7)(n + 9) \\ & = n^2 + 9n + 7n + 63 \\ & = n^2 + 16n + 63 \\ \\ \text{Product of top and bottom} & = n(n + 16) \\ & = n^2 + 16n \\ \\ \text{Difference} & = n^2 + 16n + 63 - (n^2 + 16n) \\ & = n^2 + 16n + 63 - n^2 - 16n \\ & = 63 \phantom{0} \text{ (Shown)} \end{align*}

(b)

\begin{align*} T_k & = {k^2 - 1 \over 3k - 1} \\ {15 \over 4} & = {k^2 - 1 \over 3k - 1} \\ 15(3k - 1) & = 4(k^2 - 1) \phantom{000000} [\text{Cross-multiply}] \\ 45k - 15 & = 4k^2 - 4 \\ 0 & = 4k^2 - 45k - 4 + 15 \\ 0 & = 4k^2 - 45k + 11 \\ 0 & = (4k - 1)(k - 11) \end{align*} \begin{align*} 4k - 1 & = 0 && \text{ or } & k - 11 & = 0 \\ 4k & = 1 &&& k & = 11 \\ k & = {1 \over 4} \text{ (Reject)} \end{align*}
\begin{align*} T_{k + 1} & = T_{11 + 1} \\ & = T_{12} \\ & = {12^2 - 1 \over 3(12) - 1} \\ & = {143 \over 35} \end{align*}

(a)

\begin{align*} 165 \text{ W} & = {165 \over 1000} \text{ kW} \\ & = 0.165 \text{ kW} \\ \\ \text{Energy used} & = 0.165 \text{ kW} \times 5 \text{ h} \\ & = 0.825 \text{ kWh} \end{align*}

(b) Use the data from the bar graph for this question

\begin{align*} \text{Average daily peak sunlight hours} & = { 4.7 + 4.9 + 5 + 4.7 + 4.3 + 4.2 + 4.3 + 4.4 + 4.5 + 4.4 + 3.9 + 4 \over 12} \\ & = 4{53 \over 120} \text{ hours} \end{align*}

(c)

\begin{align*} \text{Cumulative usage in year 4} & \approx 3100 \text{ kWh} \\ \\ \text{Daily usage} & = {3100 \over 365} \\ & = 8{36 \over 73} \text{ kWh} \\ & = 8493 {11 \over 73} \text{ Wh} \\ \\ \text{Solar panel output} & = \text{Solar panel watts} \times \text{Peak sunlight hours} \times 75 \% \\ 8493 {11 \over 73} \text{ Wh} & = \text{Solar panel watts} \times \underbrace{4{53 \over 120} \text{ hours} }_\text{from (b)} \times {75 \over 100} \\ \\ \text{Solar panel watts} & = { 8493 {11 \over 73} \over 4{53 \over 120} \times {75 \over 100}} \\ & = 2549.53 \text{ W} \\ \\ \text{No. of type A needed} & = {2549.53 \over 300} = 8.498 \\ \text{No. of type B needed} & = {2549.53 \over 370} = 6.89 \text{ (closest to 7)}\\ \text{No. of type C needed} & = {2549.53 \over 390} = 6.537 \\ \text{No. of type D needed} & = {2549.53 \over 400} = 6.37 \\ \\ \text{Since solar power is unreliable} & \text{ and electricity usage is likely to increase, choose 7 type B panels} \\ \\ \text{Total area} & = (1.96 \times 1.00) \times 7 \\ & = 13.72 \text{ m}^2 \end{align*}