\begin{align} \sqrt[5]{12.5^2 - {6.8 \over 0.037}} & = -1.9407 \\ & \approx -1.94 \end{align}

(a)

\begin{align} 3y^5 \times 5y^3 & = (3 \times 5) (y^5 \times y^3) \\ & = 15 y^{8} \phantom{000000000000} [a^m \times a^n = a^{m + n}] \end{align}

(b)

\begin{align} 3(2x - 1) - 2 & = 6x - 3 - 2 \\ & = 6x - 5 \end{align}

(a)

\begin{align} 14, 16, 19, & \underbrace{25, 32}, 32, 32, 40 \\ \\ \text{Median position} & = {8 + 1 \over 2} \\ & = 4.5 \\ \\ \text{Median} & = {25 + 32 \over 2} \\ & = 28.5 \end{align}

(b)

\begin{align} \text{Range} & = \text{Maximum} - \text{Minimum} \\ & = 40 - 14 \\ & = 26 \end{align}

(a)

\begin{align} T_1 & = 5 \\ T_2 & = 5 + 3 = 8 \\ \\ T_8 & = 5 + 7(3) = 26 \end{align}

(b)

\begin{align} T_n & = a + (n - 1)(d) \phantom{000000} [a \text{: first term, } d \text{: common difference}] \\ \\ T_n & = 5 + (n - 1)(3) \\ & =5 + 3n - 3 \\ & = 2 + 3n \end{align}

(c)

\begin{align} \text{Let } T_n & = 157 \\ 2 + 3n & = 157 \\ 3n & = 157 - 2 \\ 3n & = 155 \\ n & = 51{2 \over 3} \\ \\ \text{Since } n \text{ is not an integer}& \text{, not possible to have diagram with 157 dots} \end{align}

(a)

\begin{align} \text{P(Pink counter}) & = 1 - {2 \over 5} - {2 \over 15} \phantom{000000} [\text{Probability of all outcomes} = 1] \\ & = {7 \over 15} \end{align}

(b)

\begin{align} \text{Let } x \text{ denote the } & \text{number of red counters added} \\ \\ \text{P(Green counter)} & = {16 \over 8 + 16 + x} \\ {1 \over 4} & = {16 \over 24 + x} \\ 24 + x & = 4(16) \phantom{000000} [\text{Cross-multiply}] \\ 24 + x & = 64 \\ x & = 64 -24 \\ x & = 40 \\ \\ \therefore 40 \text{ red } & \text{counters added} \end{align}

\begin{align} \text{Total mass of Ajay's pots} & = 4a + 2b \\ 119 & = 4a + 2b \phantom{00} \text{--- (1)} \\ \\ \text{Total mass of Bhanu's pots} & = 5a + 3b \\ 165 & = 5a + 3b \\ -5a & = 3b - 165 \\ 5a & = -3b + 165 \\ a & = -{3 \over 5}b + 33 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 119 & = 4 \left(-{3 \over 5}b + 33 \right) + 2b \\ 119 & = -{12 \over 5}b + 132 + 2b \\ {12 \over 5}b - 2b & = 132 - 119 \\ {2 \over 5}b & = 13 \\ b & = 13 \div {2 \over 5} \\ b & = 32.5 \\ \\ \text{Substitute } & b = 32.5 \text{ into (2),} \\ a & = -{3 \over 5}(32.5) + 33 \\ a & = 13.5 \\ \\ \\ \therefore \text{Mass of small pot} & = 13.5 \text{ kg, Mass of large pot} = 32.5 \text{ kg} \end{align}

(a)

\begin{align} \text{Sum of interior angles} & = (n - 2) \times 180^\circ \\ & = (8 - 2) \times 180^\circ \\ & = 1080^\circ \\ \\ \text{Angle } BCD & = {1080^\circ \over 8} \\ & = 135^\circ \end{align}

(b) Note: I think my explanation is too long! Let me know if you know of a shorter way to prove that CB = JD

\begin{align} \angle CBJ & = 135^\circ - 90^\circ = 45^\circ \\ \\ \angle BCD + \angle CBJ & = 135^\circ + 45^\circ = 180^\circ \\ \\ \text{By the converse of } & \text{interior angles, } CD \phantom{.} // \phantom{.} BJ \\ \\ CD & = AB = JB \phantom{0} [S] \\ \\ \angle CDB & = \angle JBD \phantom{0} (\text{Alternate angles, } CD \phantom{.} // \phantom{.} BJ) [A] \\ \\ BD & \text{ is common side } [S] \\ \\ \text{Triangles } & CDB \text{ and } JBD \text{ are congruent } (SAS) \\ \\ \implies CB & = JD \\ \\ \text{Since } BC = CD & = DJ = JB, \phantom{.} BCDJ \text{ is a rhombus} \end{align}

(c)

\begin{align} \angle BJD & = \angle BCD = 135^\circ \\ \\ \angle BJL & = 90^\circ \\ \\ \angle KJL & = 60^\circ \\ \angle KJD & = 360^\circ - 135^\circ - 90^\circ - 60^\circ \phantom{000000} [\text{Angles at a point}] \\ & = 75^\circ \end{align}

\begin{align} 84 & = 2^2 \times 3 \times 7 \\ \\ 60 & = 2^2 \times 3 \times 5 \\ \\ 36 & = 2^2 \times 3^2 \\ \\ \text{Highest common factor} & = 2^2 \times 3 = 12 \phantom{000000} [\text{Length of cube = 12 cm}] \\ \\ \text{Cubes required} & = {84 \over 12} \times {60 \over 12} \times {36 \over 12} \\ & = 105 \end{align}

Diagram for the entire question:

(i)

\begin{align} & \text{Step 1: Place compass at } A \text{ and draw an arc of 11 cm (this indicates possible positions of } L) \\ & \text{Step 2: From } A, \text{ measure 40} ^\circ \text{(clockwise from North) and draw a line to meet the arc draw in Step 1} \\ & \text{Step 3: The intersection point is the lighthouse, } L \end{align}

(ii) The shortest distance is the perpendicular distance from B to AL

\begin{align} & \text{Step 1: From } B, \text{ draw an arc that intersects } AL \text{ at two points} \\ & \text{Step 2: From the first point on } AL, \text{ draw an arc} \\ & \text{Step 3: From the second point on } AL, \text{ draw an arc to intersect the arc from Step 2} \\ & \text{Step 4: Join the point of intersection to point } B \\ & \text{Step 5: Measure the bearing } (\text{clockwise from point } B) \end{align}

\begin{align} \text{Area of circle} & = \pi r^2 \\ \\ \text{Area of innermost circle} & = \pi (5)^2 = 25\pi \\ \\ \text{Area of outermost circle} & = \pi (23)^2 = 529 \pi \\ \\ \text{Area of shaded region} & = 529\pi - 25\pi = 504 \pi \\ \\ \text{Area of one 'ring'} & = {1 \over 3} (504 \pi) = 168\pi \\ \\ \text{Area of larger dotted circe} & = \underbrace{25\pi + 2(168\pi)}_\text{Innermost circle + 2 'ring'} = 361\pi \\ \\ \pi r^2 & = 361 \pi \\ r^2 & = 361 \\ r & = \sqrt{361} \\ r & = 19 \text{ cm} \end{align}

(a)

\begin{align} 30 - 45a & = 15(2 - 3a) \end{align}

(b)

\begin{align} (5x - 4y)^2 & = (5x)^2 - 2(5x)(4y) + (4y)^2 \phantom{000000} [(a - b)^2 = a^2 - 2ab + b^2] \\ & = 25x^2 - 40xy + 16y^2 \end{align}

\begin{align} \text{Width (TV)} & : \text{Height(TV)} &&& \text{Width (Laptop)} & : \text{Height(Laptop)} \\ 4 &: 3 &&& 8 & : 5 \\ 20 &: 15 &&& 24 & : 15 \end{align}
\begin{align} \text{Fraction not covered} & = {\text{Area of shaded region} \over \text{Area of laptop screen}} \\ & = {(24 - 20) \times 15 \over 24 \times 15} \\ & = {1 \over 6} \end{align}

(a)

\begin{align} \text{'Punctual' time} & = (t - 4) \text{ mins} \\ \\ \text{Time taken on Tuesday} & = t - 4 - 6 \\ & = (t - 10) \text{ mins} \end{align}

(b)

\begin{align} \text{Distance travelled} & = \text{Speed} \times \text{Time} \\ \\ \text{Distance travelled on Monday} & = 18 \times t \\ & = 18t \\ \\ \text{Distance travelled on Tuesday} & = 24 \times (t - 10) \\ & = 24(t- 10) \\ \\ 18t & = 24 (t - 10) \\ 18t & = 24t - 240 \\ 18t - 24t & = -240 \\ -6t & = -240 \\ t & = {-240 \over -6} \\ t & = 40 \\ \\ \text{Time taken on Monday} & = 40 \text{ mins} \end{align}

(a)

\begin{align} (6x - 20) \times 2 - 30 & > 1000 \\ 2(6x - 20) - 30 & > 1000 \end{align}

(b)

\begin{align} 2(6x - 20) - 30 & > 1000 \\ 2(6x - 20) & > 1000 + 30 \\ 12x - 40 & > 1030 \\ 12x & > 1030 + 40 \\ 12x & > 1070 \\ x & > {1070 \over 12} \\ x & > 89{1 \over 6} \\ \\ \text{Smallest } & \text{value of } x = 90 \end{align}

\begin{align} \angle ACB & = \angle ECD \phantom{0} (\text{Vertically opposite angles}) [A] \\ \\ \angle ABC & = \angle EDC \phantom{0} (\text{Alternate angles, } AB \phantom{.} // \phantom{.} DE) [A] \\ \\ AB & = ED \phantom{0} (\text{Given}) [S] \\ \\ \text{Triangles } & ACB \text{ and } ECD \text{ are congruent } (AAS) \\ \\ \implies AC & = EC \\ \\ \therefore \text{Since } C \text{ is } & \text{the midpoint of } AE, BD \text{ bisects } AE \end{align}

\begin{align} \angle QOR & = 2 \times \angle QPR \phantom{0} (\text{Angle at centre} = 2 \times \text{Angle at circumference}) \\ & = 2 \times 35^\circ \\ & = 70^\circ \\ \\ \angle OXR & = 180^\circ - 30^\circ - 70^\circ \phantom{0} (\text{Angle sum of triangle}) \\ & = 80^\circ \\ \\ \angle PXQ & = 80^\circ \phantom{0} (\text{Vertically opposite angles}) \\ \\ \angle OQP & = \angle XQP \\ & = 180^\circ - 80^\circ - 35^\circ \phantom{0} (\text{Angle sum of triangle}) \\ & = 65^\circ \end{align}

(a)

\begin{align} \textbf{T} & = \textbf{PQ} \\ & = \mathop{ \left( \begin{matrix} 5 & 10 & 4 & 2 \\ x & x + 2 & 2 & 1 \end{matrix} \right)}_{2 \times 4} \mathop{\left( \begin{matrix} 140 & 150 \\ 105 & 100 \\ 13 & 12 \\ 9 & 8 \end{matrix} \right)}_{4 \times 2} \\ \\ & = \mathop{\left( \begin{matrix} 5(140) + 10(105) + 4(13) + 2(9) & 5(150) + 10(100) + 4(12) + 2(8) \\ 140x + 105(x + 2) + 2(13) + 9 & 150x + 100(x + 2) + 2(12) + 8 \end{matrix} \right)}_{2 \times 2} \\ \\ & = \left( \begin{matrix} 1820 & 1814 \\ 140x + 105x + 210 + 26 + 9 & 150x + 100x + 200 + 24 + 8 \end{matrix} \right) \\ \\ & = \left( \begin{matrix} 1820 & 1814 \\ 245x + 245 & 250x + 232 \end{matrix} \right) \end{align}

(b)

\begin{align} & \text{The total amount Cheng needs to pay for all the supplies from the store} \\ & \text{and from the online supplier respectively} \end{align}

(c)

\begin{align} \text{Xin's total price in store} + \$ 2 & = \text{Xin's total online price} \\ 245x + 245 + 2 & = 250x + 232 \\ 245x - 250x & = 232 - 245 - 2 \\ -5x & = -15 \\ x & = {-15 \over -5} \\ x & = 3 \end{align}

(d)

\begin{align} \text{Xin's total price in store} & = 245x + 245 \\ & = 245(3) + 245 \\ & = \$ 980 \end{align}

\begin{align} 5 \sin x & = 2 \\ \sin x & = {2 \over 5} \end{align}
\begin{align} x & = \sin^{-1} \left(2 \over 5\right) && \text{ or } & x & = 180^\circ - \sin^{-1} \left(2 \over 5\right) \\ & = 23.578^\circ &&& & = 156.42^\circ \\ & \approx 23.6^\circ &&& & \approx 156.4^\circ \end{align}

\begin{align} {3a + 2c \over 1} & = {5 - c \over 3b} \\ {3b(3a + 2c) \over 3b} & = {5 - c \over 3b} \\ 3b(3a + 2c) & = 5 - c \\ 9ab + 6bc & = 5 - c \\ 6bc + c & = 5 - 9ab \\ c(6b + 1) &= 5 - 9ab \\ c & = {5 - 9ab \over 6b + 1} \end{align}

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

\begin{align} \sin \angle DAB & = {DB \over AD} \phantom{000000} \left[ {Opp \over Hyp} \right] \\ \sin 38^\circ & = {DB \over 12.8} \\ \\ DB & = 12.8 \sin 38^\circ \\ & = 7.8804 \text{ cm} \\ \\ \\ \cos \angle CBD & = {DB \over BC} \phantom{000000} \left[ {Adj \over Hyp} \right] \\ \cos \angle CBD & = {7.8804 \over 10.3} \\ \angle CBD & = \cos^{-1} \left({7.8804 \over 10.3}\right) \\ & = 40.085 \\ & \approx 40.1^\circ \end{align}

\begin{align} 25^{2x} & = 125^7 \\ (5^2)^{2x} & = (5^3)^7 \\ 5^{4x} & = 5^{21} \phantom{000000} [ (a^m)^n = a^{mn}] \\ \\ 4x & = 21 \\ x & = {21 \over 4} \\ x & = 5{1 \over 4} \end{align}

(i) Besides applying formula, you can use the in-built function in calculator

\begin{align} \text{Mean monthly salary} & = { 3 \times 2450 + 5 \times 2550 + 14 \times 2650 + 8 \times 2750 \over 30} \\ & = \$ 2640 \end{align}

(ii) Besides applying formula, you can use the in-built function in calculator

\begin{align} \text{Standard deviation} & = \sqrt{ { \sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ \\ \sum fx^2 & = 3 \times 2450^2 + 5 \times 2550^2 + 14 \times 2650^2 + 8 \times 2750^2 \\ & = 209 \phantom{.} 335 \phantom{.} 000 \\ \\ \text{Standard deviation} & = \sqrt{ { 209 \phantom{.} 335 \phantom{.} 000 \over 30} - (2640)^2 } \\ & = 90.737 \\ & \approx 90.7 \end{align}

(iii) Standard deviation remains the same as the spread remains unchanged

\begin{align} & \text{Mean will increase by } \$ 60 \\ & \text{Standard deviation will remain the same} \end{align}

(a)

\begin{align} N & = m \times 2^{3t} \\ \\ \text{When } & t = 1 \text{ and } N = 2000, \\ 2000 & = m \times 2^{3(1)} \\ 2000 & = m \times 8 \\ \\ m & = {2000 \over 8} \\ m & = 250 \end{align}

(b)

\begin{align} N & = m \times 2^{3t} \\ N & = 250 \times 2^{3t} \\ N & = 250 \times (2^3)^t \phantom{000000} [ (a^m)^n = a^{mn}] \\ N & = 250 \times 8^t \\ \\ \text{When } & 8^t = k, \\ N & = 250 \times k \\ N & = 250k \end{align}

(c)

\begin{align} N & = 250 \times 2^{3t} \\ \\ \text{When } & t = 0, \\ N & = 250 \times 2^{3(0)} \\ N & = 250 \phantom{000000} [\text{This is the initial no. of bacteria}] \\ \\ \text{When } & t = 2, \\ N & = 250 \times 2^{3(2)} \\ N & = 16 \phantom{.} 000 \\ \\ \text{Percentage increase} & = {\text{Increase} \over \text{Original}} \times 100 \\ & = {16 \phantom{.} 000 - 250 \over 250} \times 100 \\ & = 6300 \% \end{align}

(d)

\begin{align} \text{Since } N = 250 \times 2^{3t}, \text{ as } t & \text{ increases, } N \text{ increases at an increasing rate} \phantom{000000} [\text{Diagram 2 or 3}] \\ \\ \text{From (c), when } & t = 0, N = 250 \\ \\ \therefore & \phantom{.} \text{Diagram 3} \end{align}

(a)

\begin{align} {x \over 7} + {x - 5 \over 3} & = 1 \\ {3x \over 21} + {7(x - 5) \over 21} & = {21 \over 21} \\ {3x + 7(x - 5) \over 21} & = {21 \over 21} \\ 3x + 7(x - 5) & = 21 \\ 3x + 7x - 35 & = 21 \\ 10x & = 21 + 35 \\ 10x & = 56 \\ x & = {56 \over 10} \\ x & = 5.6 \end{align}

(b)

\begin{align} {6a^3 b \over 5} \div {3a^2 \over 10b} & = {6 a^3 b \over 5} \times {10b \over 3a^2} \\ & = {60 a^3 b^2 \over 15 a^2 } \\ & = 4 a b^2 \end{align}

(c)

\begin{align} x^2 + 9x - 16 & = 0 \\ x^2 + 9x + \left(9 \over 2\right)^2 - \left(9 \over 2\right)^2 - 16 & = 0 \\ x^2 + 9x + (4.5)^2 - (4.5)^2 - 16 & = 0 \\ (x + 4.5)^2 - 20.25 - 16 & = 0 \\ (x + 4.5)^2 - 36.25 & = 0 \\ (x + 4.5)^2 & = 36.25 \\ x + 4.5 & = \pm \sqrt{36.25} \\ x + 4.5 & = \pm 6.0207 \end{align}
\begin{align} x + 4.5 & = 6.0207 && \text{ or } & x + 4.5 & = -6.0207 \\ x & = 6.0207 - 4.5 &&& x & = -6.0207 - 4.5 \\ x & = 1.5207 &&& x & = -10.5207 \\ x & \approx 1.52 \text{ (2 d.p.)} &&& x & \approx -10.52 \text{ (2 d.p.)} \end{align}

(d)

\begin{align} \require{cancel} {4x^2 - 8ax - 3x + 6a \over 3x^2 - 12a^2} & = { 4x (x - 2a) - 3(x - 2a) \over 3(x^2 - 4a^2)} \\ & = { (x - 2a)(4x - 3) \over 3[ x^2 - (2a)^2] } \phantom{0000000000} [\text{Numerator: Factorise by grouping}] \\ & = { \cancel{(x - 2a)}(4x - 3) \over 3 (x + 2a) \cancel{(x - 2a)}} \phantom{00000000} [ a^2 - b^2 = (a + b)(a - b)] \\ & = { 4x - 3 \over 3(x + 2a) } \end{align}

(a)

\begin{align} & \text{Account B is better since Cheryl will earn interest on interest earned in prior years} \end{align}

(b)

\begin{align} I & = {PRT \over 100} \\ 1385 - P & = {P (1.8)(6) \over 100} \\ 1385 - P & = {10.8P \over 100} \\ 1385 - P & = 0.108 P \\ -P - 0.108 P & = -1385 \\ -1.108 P & = -1385 \\ P & = {-1385 \over -1.108} \\ P & = 1250 \\ \\ \text{Marcus } & \text{invested } \$ 1250 \end{align}

(c)

\begin{align} \text{Total amount} & = P \left(1 + {r \over 100}\right)^n \\ 4 \phantom{.} 719.92 & = P \left(1 + {1.2 \over 100} \right)^4 \\ 4 \phantom{.} 719.92 & = P (1.012)^4 \\ \\ P & = {4 \phantom{.} 719.92 \over (1.012)^4} \\ P & = 4 \phantom{.} 500 \\ \\ \text{Interest earned} & = 4 \phantom{.} 719.92 - 4 \phantom{.} 500 \\ & = \$ 219.92 \end{align}

(d)

\begin{align} \text{Fuel used} & = {5.8 \over 100} \times 745 \\ & = 43.21 \text{ litres} \\ \\ \text{Cost of fuel} & = 43.21 \times 1.52 \\ & = \text{€} 65.679 \phantom{.} 2 \\ \\ \text{Total cost} & = \underbrace{{(134.50 + 65.679 \phantom{.} 2) \over 0.66}}_\text{Total cost in SGD} \times {100 + 1.5 \over 100} \\ & = 307.851 \\ & \approx \$ 307.85 \end{align}

(a)(i)

\begin{align} \text{Median age} & = 41 \text{ years old} \phantom{000000} [\text{Cumulative percentage} = 50\%] \end{align}

(a)(ii)

\begin{align} \% \text{ of UK population aged 60 and below} & = 76\% \\ \\ \% \text{ of UK population aged over 60} & = 100 - 76 \\ & = 24 \% \end{align}

(a)(iii)

\begin{align} \text{80th percentile for Singapore} & = 61 \text{ years old} \end{align}

(b)

\begin{align} \% \text{ of SG population under 25} & = 27 \% \\ \\ \text{No. of people under 25} & = 4.03 \text{ million} \times {27 \over 100} \\ & = 1.0881 \text{ million} \\ & \approx 1.09 \text{ million} \end{align}

(c)

\begin{align} 1. & \text{The median age for Singapore (41) is higher than the median age for UK (40),} \\ & \text{ thus on average Singaporeans are older} \\ \\ 2. & \text{The interquartile range for Singapore (58 - 24 = 34) is lower than the interquartile } \\ & \text{ range for UK (59 - 21 = 38), thus the ages of Singaporeans are more consistent} \end{align}

(a)

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

(b)

(c)

\begin{align} \underbrace{{x^3 \over 5} - 2x + 1}_\text{Curve} & = k \phantom{0} \text{ has two solutions} \\ \\ \therefore \text{Horizontal line } & y = k \text{ must meet curve only twice} \\ \\ \\ \text{From graph, } k & = 3.4 \text{ or } -1.4 \end{align}

(d)

(a)(i)

\begin{align} \text{Volume of cyinder} & = \pi r^2 h \\ & = \pi (3.5)^2 (16) \\ & = 196 \pi \text{ cm}^3 \\ \\ \text{Volume of solid} & = 196\pi \times {1 \over 2} \\ & = 98\pi \text{ cm}^3 \\ & \approx 308 \text{ cm}^3 \end{align}

(a)(ii)

\begin{align} \text{Curved surface area} & = \underbrace{ {1 \over 2} \times 2 \pi r }_\text{Circumference of semicircle} \times h \\ & = {1 \over 2} \times 2 \pi (3.5) \times 16 \\ & = 56 \pi \text{ cm}^2 \\ \\ \text{Area of semi-circle} & = {1 \over 2} \times \pi r^2 \\ & = {1 \over 2} \times \pi (3.5)^2 \\ & = 6.125 \pi \text{ cm}^2 \\ \\ \text{Area of rectangle} & = 16 \times 7 \\ & = 112 \text{ cm}^2 \\ \\ \text{Total surface area} & = 56\pi + 2(6.125\pi) + 112 \\ & = 326.41 \\ & \approx 326 \text{ cm}^2 \end{align}

(b)(i)

\begin{align} {V_A \over V_C} & = \left(l_A \over l_C\right)^3 \\ {1 \over 8} & = \left(l_A \over l_C\right)^3 \\ \\ {l_A \over l_C} & = \sqrt[3]{1 \over 8} \\ & = {1 \over 2} \\ \\ \therefore h_A : h_C & = 1 : 2 \end{align}

(b)(ii)

\begin{align} {A_B \over A_C} & = \left(l_B \over l_C\right)^2 \phantom{000000} \left[ \text{Need to find value of } {l_B \over l_C} \text{ first} \right] \\ \\ {V_B \over V_C} & = \left(l_B \over l_C\right)^3 \\ {3 \over 8} & = \left(l_B \over l_C\right)^3 \\ \\ {l_B \over l_C} & = \sqrt[3]{3 \over 8} \\ & = 0.72112 \\ \\ {A_B \over A_C} & = \left(l_B \over l_C\right)^2 \\ & = (0.72112)^2 \\ & = 0.52001 \\ \\ \text{Percentage} & = 0.52001 \times 100 \\ & = 52.001 \\ & \approx 52.0 \% \end{align}

(a)

\begin{align} \angle DPE & = 96^\circ \phantom{0} (\text{Vertically opposite angles}) \\ \\ \angle DEP & = {180^\circ - 96^\circ \over 2} \phantom{0} (\text{Isosceles triangle } PED \text{ with } PD = PE) \\ & = 42^\circ \end{align}

(b)

\begin{align} \angle ABC & = \angle FCD = 90^\circ \phantom{0} (\text{Right angle in semi-circle}) [A] \\ \\ \angle ACB & = \angle DCP \phantom{0} (\text{Vertically opposite angles}) \\ & = \angle CDP \phantom{0} (\text{Isosceles triangle } PCD \text{ with } PC = PD) \\ & = \angle FDC \phantom{0} [A] \\ \\ \therefore \text{Triangles } & ABC \text{ and } FCD \text{ are similar } (AA) \end{align}

(c)(i)

\begin{align} CF & = DE = 7.21 \text{ cm} \phantom{0} (\text{Triangles } PED \text{ and } PFC \text{ are congruent}) \\ \\ \text{By Py} & \text{thagoras theorem,} \\ DF^2 & = DC^2 + CF^2 \\ 9.7^2 & = DC^2 + 7.21^2 \\ \\ DC^2 & = 9.7^2 - 7.21^2 \\ DC & = \sqrt{9.7^2 - 7.21^2} \\ & = 6.4889 \text{ cm} \\ \\ BC & = 9.10 - 6.4889 \\ & = 2.6111 \text{ cm} \\ \\ \text{Triangles } & ABC \text{ and } FCD \text{ are similar } \\ \\ {AB \over FC} & = {BC \over CD} \\ {AB \over 7.21} & = {2.6111 \over 6.4889} \\ AB & = {2.6111 \over 6.4889} \times 7.21 \\ & = 2.9012 \\ & \approx 2.90 \text{ cm} \end{align}

(c)(ii)

\begin{align} \angle PFC & = \angle PED = 42^\circ \phantom{0} (\text{Triangles } PED \text{ and } PFC \text{ are congruent}) \\ \\ \text{Triangles } & ABC \text{ and } FCD \text{ are similar } \\ \\ \implies \angle CAB & = \angle DFC = 42^\circ \\ \\ \angle AOB & = 180^\circ - 42^\circ - 42^\circ \phantom{0} (\text{Isosceles triangle } OAB \text{ with } OA = OB) \\ & = 96^\circ \\ \\ \text{By Py} & \text{thagoras theorem,} \\ AC^2 & = AB^2 + BC^2 \\ & = 2.9012^2 + 2.6111^2 \\ AC & = \sqrt{ 2.9012^2 + 2.6111^2 } \\ & = 3.9031 \\ \\ \text{Arc length} & = {\theta \over 360^\circ} \times 2 \pi r \\ & = {96^\circ \over 360^\circ} \times 2 \pi \left(3.9031 \over 2\right) \\ & = 3.2698 \\ & \approx 3.27 \text{ cm} \end{align}

(a)

\begin{align} EX & = 24 \times {3 \over 8} \\ & = 9 \text{ cm} \\ \\ \text{Area of trapezium} & = {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} \\ & = {1 \over 2} \times (9 + 24) \times 8 \\ & = 132 \text{ cm}^2 \end{align}

(b)

\begin{align} XF & = 24 - 9 \\ & = 15 \text{ cm} \\ \\ \text{By Py} & \text{thagoras theorem,} \\ XG^2 & = XF^2 + FG^2 \\ & = 15^2 + 13^2 \\ XG & = \sqrt{ 15^2 + 13^2 } \\ & = 19.849 \\ & \approx 19.8 \text{ cm} \end{align}

(c)

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ BG^2 & = BC^2 + CG^2 \\ & = 13^2 + 8^2 \\ BG & = \sqrt{13^2 + 8^2} \\ & = 15.264 \text{ cm} \\ \\ \text{By Py} & \text{thagoras theorem,} \\ XB^2 & = 8^2 + 15^2 \\ XB & = \sqrt{8^2 + 15^2} \\ & = 17 \text{ cm} \\ \\ \cos A & = {b^2 + c^2 - a^2 \over 2 bc } \phantom{000000} [\text{Cosine rule to find angle}] \\ \cos \angle XBG & = {XB^2 + BG^2 - XG^2 \over 2(XB)(BG)} \\ & = {17^2 + 15.264^2 - 19.849^2 \over 2(17)(15.264)} \\ & = 0.24665 \\ \angle XBG & = \cos^{-1} (0.24665) \\ & = 75.72^\circ \\ \\ \text{Area of triangle } BGX & = {1 \over 2} ab \sin C \\ & = {1 \over 2} (XB)(BG) \sin \angle XBG \\ & = {1 \over 2} (17)(15.264) \sin 75.72^\circ \\ & = 125.73 \\ & \approx 126 \text{ cm}^2 \end{align}

(a)(i)

\begin{align} PQ & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ & = \sqrt{ [2 - (-3)]^2 + (11 - 5)^2 } \\ & = 7.8102 \\ & \approx 7.81 \text{ units} \end{align}

(a)(ii)

\begin{align} \overrightarrow{OP} & = \binom{-3}5 \\ \\ \overrightarrow{OR} & = \overrightarrow{OP} + \overrightarrow{PR} \\ & = \binom{-3}5 + \binom{8}{-2} \\ & = \binom{5}{3} \\ \\ \therefore & \phantom{.} R(5, 3) \end{align}

(a)(iii)

\begin{align} \text{Gradient of } QR & = {y_2 - y_1 \over x_2 - x_1} \\ & = {3 - 11 \over 5 - 2} \\ & = -{8 \over 3} \\ \\ y & = mx + c \\ y & = -{8 \over 3}x + c \\ \\ \text{Using } & Q(2, 11), \\ 11 & = -{8 \over 3}(2) + c \\ 11 & = -{16 \over 3} + c \\ 11 + {16 \over 3} & = c \\ {49 \over 3} & = c \\ \\ \text{Eqn of } QR : & \phantom{0} y = -{8 \over 3}x + {49 \over 3} \end{align}

(b)

\begin{align} \overrightarrow{OC} & = 3 \overrightarrow{OA} \\ & = 3 \textbf{a} \\ \\ \overrightarrow{OD} & = {5 \over 3} \overrightarrow{OB} \\ & = {5 \over 3} \textbf{b} \\ \\ \overrightarrow{DC} & = \overrightarrow{DO} + \overrightarrow{OC} \\ & = -{5 \over 3} \textbf{b} + 3 \textbf{a} \\ \\ \text{Since } X & \text{ is a point on } OC, \\ \overrightarrow{OX} & = k \overrightarrow{OC}, \text{ where } k \text{ is a constant} \\ & = k (3 \textbf{a} ) \\ & = 3k \textbf{a} \\ \\ \overrightarrow{BX} & = \overrightarrow{BO} + \overrightarrow{OX} \\ & = - \textbf{b} + 3k \textbf{a} \\ \\ \text{Since } BX & \text{ is parallel to } DC, \\ \overrightarrow{BX} & = j \overrightarrow{DC}, \text{ where } j \text{ is a constant} \\ - \textbf{b} + 3k \textbf{a} & = j \left( -{5 \over 3} \textbf{b} + 3 \textbf{a} \right) \\ - \textbf{b} + 3k \textbf{a} & = -{5 \over 3} j \textbf{b} + 3j \textbf{a} \end{align}
\begin{align} \text{Comparing } & \text{terms with } \textbf{b}, &&& \text{Comparing } & \text{terms with } \textbf{a}, \\ -1 & = -{5 \over 3} j &&& 3k & = 3j \\ -1 \div -{5 \over 3} & = j &&& k & = j \\ {3 \over 5} & = j &&& k & = {3 \over 5} \end{align}
\begin{align} \overrightarrow{OX} & = 3k \textbf{a} \\ & = 3 \left(3 \over 5\right) \textbf{a} \\ & = {9 \over 5} \textbf{a} \\ \\ \overrightarrow{XD} & = \overrightarrow{XO} + \overrightarrow{OD} \\ & = -{9 \over 5} \textbf{a} + {5 \over 3} \textbf{b} \end{align}

(a)(i)

\begin{align} \xi & = \{ 1, 2, ..., 14, 15 \} \\ \\ A & = \{ 2, 3, 5, 7, 11, 13 \} \\ \\ A' & = \{ 1, 4, 6, 8, 9, 10, 12, 14, 15 \} \end{align}

(a)(ii)

\begin{align} A & = \{ 2, 3, 5, 7, 11, 13 \} \\ \\ B & = \{ 1, 2, 3, 5, 6, 10, 15 \} \\ \\ A \cup B & = \{ 1, 2, 3, 5, 6, 7, 10, 11, 13, 15 \} \\ \\ (A \cup B)' & = \{ 4, 8, 9, 12, 14 \} \end{align}

(a)(iii) Note $p \notin C$ means the number chosen from $B \cup C$ is in $B$ but not in $C$

\begin{align} B & = \{ 1, 2, 3, 5, 6, 10, 15 \} \\ \\ C & = \{ 3, 6, 9, 12, 15 \} \\ \\ B \cup C & = \{ 1, 2, 3, 5, 6, 9, 10, 12, 15 \} \\ \\ & \text{P(} p \notin C) = {4 \over 9} \phantom{000000} [\text{4 numbers in } B \text{ but not in } C: 1, 2, 5, 10] \end{align}

(b)(i)

\begin{align} \text{P(French class student studies Spanish)} & = {8 \over 8 + 7} \\ & = {8 \over 15} \end{align}

(b)(ii)

\begin{align} \text{P(Both Spanish class students do not study French)} & = {12 \over 8 + 12} \times {12 - 1 \over 8 + 12 - 1} \\ & = {33 \over 95} \end{align}

(b)(iii)

\begin{align} [\text{Out of 30 students, } & \text{ 20 study Spanish and 10 does not} ] \\ \\ \\ \text{P(Only one out of three study Spanish)} & = \text{P(1st study, 2nd & 3rd don't)} + \text{P(1st don't, 2nd study, 3rd don't)} \\ & \phantom{0} + \text{P(1st & 2nd don't, 3rd study)} \\ & = {20 \over 30} \times {10 \over 29} \times {9 \over 28} + {10 \over 30} \times {20 \over 29} \times {9 \over 28} + {10 \over 30} \times {9 \over 29} \times {20 \over 28} \\ & = {45 \over 203} \end{align}

(a)

\begin{align} 1 \text{ hour} & = 60 \text{ mins} \\ & = 3600 \text{ seconds} \\ \\ \text{Time taken for one pair} & = {3600 \over 7} \\ & = 514.285 \\ & \approx 510 \text{ seconds (to nearest 10 seconds)} \end{align}

(b)

\begin{align} \text{Mei's basic pay} & = 9.80 \times 9 \\ & = \$ 88.20 \\ \\ \text{Mei's additional payment} & = 231.75 - 88.20 \\ & = \$143.55 \\ \\ \text{Total no. of bracelets made} & = {143.55 \over 1.45} \\ & = 99 \\ \\ \text{Average no. of bracelets made} & = {99 \over 9} \\ & = 11 \text{ bracelets per hour (shown)} \end{align}

(c) This problem can broken down into two main problems:

1. Find the no. of necklaces Chen and Zhu make respectively in one day
2. Calculate each person's daily income, then annual income

\begin{align} \text{Let } x \text{ denote the time taken (in seconds)} & \text{ for Zhu to make one necklace} \\ \\ \text{No. of necklaces made by Zhu in one day (8 hours)} & = {8 \times 60 \times 60 \over x} \\ & = {28 \phantom{.} 800 \over x} \\ \\ \text{No. of necklaces made by Chen in one day (8 hours)} & = {8 \times 60 \times 60 \over x + 80} \\ & = {28 \phantom{.} 800 \over x + 80} \\ \\ {28 \phantom{.} 800 \over x} + {28 \phantom{.} 000 \over x + 80} & = 132 \phantom{000000} [\text{Total 132 necklaces made in one day}] \\ {28 \phantom{.} 800 (x + 80) \over x(x + 80)} + {28 \phantom{.} 800x \over x(x + 80)} & = {132 \over 1} \\ {28 \phantom{.} 800 (x + 80) + 28 \phantom{.} 800x \over x(x + 80)} & = {132x(x + 80) \over x(x + 80)} \\ 28 \phantom{.} 800 (x + 80) + 28 \phantom{.} 800 x & = 132x(x + 80) \\ 28 \phantom{.} 800 x + 2 \phantom{.} 304 \phantom{.} 000 + 28 \phantom{.} 800x & = 132x^2 + 10 \phantom{.} 560x \\ 0 & = 132x^2 - 47 \phantom{.} 040x - 2 \phantom{.} 304 \phantom{.} 000 \\ 0 & = 11x^2 - 3920x - 192 \phantom{.} 000 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = { -(-3920) \pm \sqrt{(-3920)^2 - 4(11)(-192 \phantom{.} 000)} \over 2(11)} \\ & = {3920 \pm \sqrt{23 \phantom{.} 814 \phantom{.} 400} \over 22} \\ & = 400 \text{ or } - 43{7 \over 11} \text{ (Reject, since } x > 0) \\ \\ \\ \text{No. of necklaces made by Zhu in one day} & = {28 \phantom{.} 800 \over x} \\ & = {28 \phantom{.} 800 \over 400} \\ & = 72 \\ \\ \text{Zhu's daily earnings} & = 9.80 \times 8 + 72 \times 1.65 \\ & = \$ 197.20 \\ \\ \text{Asumming Zhu goes on holiday, no. of working days} & = \underbrace{52 \times 5}_\text{52 weeks in a year} - 18 \\ & = 242 \\ \\ \text{Zhu's annual earnings} & = 197.20 \times 242 + \underbrace{(9.80 \times 8 \times 18)}_\text{18 days paid holiday} \\ & = \$ 49 \phantom{.} 133.60 > \$48 \phantom{.} 000 \\ \\ \\ \text{No. of necklaces made by Chen in one day} & = 132 - 72 \\ & = 60 \\ \\ \text{Chen's daily earnings} & = 9.80 \times 8 + 60 \times 1.65 \\ & = \$177.40 \\ \\ \text{Chen's annual earnings (with holiday)} & = 177.40 \times 242 + (9.80 \times 8 \times 18) \\ & = \$44 \phantom{.} 342 < \$ 48 \phantom{.} 000 \\ \\ \text{Chen's annual earnings (without holiday)} & = 177.40 \times (52 \times 5) \\ & = \$46 \phantom{.} 124 < \$ 48 \phantom{.} 000 \\ \\ \\ \therefore \text{Zhu can expect } & \text{to earn advertised minimum annual income} \\ \text{Chen cannot expect } & \text{to earn advertised minimum annual income} \end{align}