(i)

\begin{align} \text{Perimeter} & = 2 \times \text{Length} + 2 \times \text{Breadth} \\ 112 & = 2 \times \text{Length} + 2 \times x \\ 112 & = 2 \times \text{Length} + 2x \\ 112 - 2x & = 2 \times \text{Length} \\ \\ \text{Length} & = {1 \over 2}(112 - 2x) \\ & = (56 - x) \text{ cm} \end{align}

(ii)

\begin{align} \text{Area} & = \text{Length} \times \text{Breadth} \\ 597 & = (56 - x) \times x \\ 597 & = x(56 - x) \\ 597 & = 56x - x^2 \\ x^2 - 56x + 597 & = 0 \phantom{0000} \text{ (Shown)} \end{align}

(iii)

\begin{align} x^2 & - 56x + 597 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-56) \pm \sqrt{(-56)^2 - 4(1)(597)} \over 2(1)} \\ & = {56 \pm \sqrt{748} \over 2} \\ & = 41.674 \text{ or } 14.325 \\ & \approx 41.67 \text{ or } 14.33 \text{ (2 d.p.)} \end{align}

(iv) For this part, I'm assuming the posters are pasted along the breadth, x = 14.325 cm

\begin{align} 4 \text{ m} & = 400 \text{ cm} \\ \\ {400 \over 14.325} & = 27.913 \\ \\ \therefore \text{Max. posters} & = 27 \end{align}

(i)

\begin{align} x \text{ mins} & \rightarrow 60 \text{ pages} \\ 1 \text{ min} & \rightarrow {60 \over x} \text{ pages} \end{align}

(ii)

\begin{align} (x + 2) \text{ mins} & \rightarrow 60 \text{ pages} \\ 1 \text{ min} & \rightarrow {60 \over x+2} \text{ pages} \end{align}

(iii)

\begin{align} {60 \over x} + {60 \over x + 2} & = 144 \\ {60(x + 2) \over x(x + 2)} + {60x \over x(x + 2)} & = 144 \\ {60(x + 2) + 60x \over x(x + 2)} & = 144 \\ {60x + 120 + 60x \over x^2 + 2x} & = 144 \\ {120x + 120 \over x^2 + 2x} & = {144 \over 1} \\ 120x + 120 & = 144(x^2 + 2x) \\ 120x + 120 & = 144x^2 + 288x \\ 0 & = 144x^2 + 288x - 120x -120 \\ 0 & = 144x^2 + 168x - 120 \\ 0 & = 6x^2 + 7x - 5 \phantom{00} \text{ (Shown)} \phantom{000000} [\text{Divide by 24}] \end{align}

(iv)

\begin{align} 0 & = 6x^2 + 7x - 5 \\ 0 & = (2x - 1)(3x + 5) \end{align} \begin{align} 2x - 1 & = 0 && \text{ or } & 3x + 5 & = 0 \\ 2x & = 1 &&& 3x & = -5 \\ x & = {1 \over 2} &&& x & = -{5 \over 3} \end{align}

(v) I think there's a typo in the question - it should be 'time taken by Printer B to print 1440 pages'

\begin{align} \text{Pages by Printer B in 1 min} & = {600 \over x + 2} \\ & = {600 \over {1 \over 2} + 2} \\ & = 240 \\ \\ \text{Time taken} & = {1440 \over 240} \\ & = 6 \text{ mins} \end{align}

(i)

\begin{align} \$ x & \rightarrow 1 \text{ kg} \\ \$ 1 & \rightarrow {1 \over x} \text{ kg} \\ \$ 600 & \rightarrow {600 \over x} \text{ kg} \end{align}

(ii)

\begin{align} \$ (x - 0.4) & \rightarrow 1 \text{ kg} \\ \$ 1 & \rightarrow {1 \over x - 0.4} \text{ kg} \\ \$ 600 & \rightarrow {600 \over x - 0.4} \text{ kg} \end{align}

(iii) Since brand B is cheaper, the food catering company will obtain more rice from brand B

\begin{align} {600 \over x - 0.4} - {600 \over x} & = 45 \\ {600x \over x(x - 0.4)} - {600(x - 0.4) \over x(x - 0.4)} & = 45 \\ {600x - 600(x - 0.4) \over x(x - 0.4)} & = 45 \\ {600x - 600x + 240 \over x^2 - 0.4x} & = 45 \\ {240 \over x^2 - 0.4x} & = {45 \over 1} \\ 240 & = 45(x^2 - 0.4x) \\ 240 & = 45x^2 - 18x \\ 0 & = 45x^2 - 18x - 240 \\ 0 & = 15x^2 - 6x - 80 \phantom{00} \text{ (Shown)} \phantom{000000} [\text{Divide by 3}] \end{align}

(iv)

\begin{align} 0 & = 15x^2 - 6x - 80 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-6) \pm \sqrt{(-6)^2 - 4(15)(-80)} \over 2(15)} \\ & = {6 \pm \sqrt{4836} \over 30} \\ & = 2.518 \text{ or } -2.118 \\ \\ \\ \text{Price of 1 kg from B} & = x - 0.4 \\ & = 2.518 - 0.4 \\ & = 2.118 \\ & \approx \$ 2.12 \end{align}

\begin{align} h & = -t^2 + 5t + 10 \phantom{00000000000000000000000.} [\text{Maximum curve } \cap] \\ h & = -(t^2 - 5t) + 10 \\ h & = -\left[ t^2 - 5t + \left(5 \over 2\right)^2 - \left(5 \over 2\right)^2 \right] + 10 \phantom{000000} [\text{Complete the square}] \\ h & = - ( t^2 - 5t + 2.5^2 - 2.5^2 ) + 10 \\ h & = - [(t - 2.5)^2 - 6.25] + 10 \\ h & = -(t - 2.5)^2 + 6.25 + 10 \\ h & = -(t - 2.5)^2 + 16.25 \\ \\ \text{Max. point: } & (2.5, 16.25) \\ \\ \text{Max. height} & = 16.25 \text{ m} \\ \\ \\ \text{When } & t = 0, \\ h & = -(0 - 2.5)^2 + 16.25 \\ & = 10 \text{ m} \\ \\ \text{Initial height} & = 10 \text{ m} \\ \\ \\ \text{Height difference} & = 16.25 - 10 \\ & = 6.25 \text{ m} \end{align}

(i)

\begin{align} \text{Area of triangle} & = {1 \over 2} \times BC \times AP \\ & = {1 \over 2} \times 2x \times (x + 5) \\ & = x(x + 5) \\ & = x^2 + 5x \\ \\ \text{Area of base (square)} & = 2x \times 2x \\ & = 4x^2 \\ \\ \text{Total surface area} & = 4x^2 + 4(x^2 + 5x) \\ & = 4x^2 + 4x^2 + 20x \\ & = (8x^2 + 20x) \text{ cm}^2 \end{align}

(ii)

\begin{align} \text{Total surface area} & = 8x^2 + 20x \\ 50 & = 8x^2 + 20x \\ 0 & = 8x^2 + 20x - 50 \\ 0 & = 4x^2 + 10x - 25 \phantom{00} \text{ (Shown)} \phantom{000000} [\text{Divide by 2}] \end{align}

(iii)

\begin{align} 0 & = 4x^2 + 10x - 25 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-10 \pm \sqrt{(10)^2 - 4(4)(-25)} \over 2(4)} \\ & = {-10 \pm \sqrt{500} \over 8} \\ & = 1.545 \text{ or } -4.045 \\ & \approx 1.55 \text{ or } -4.05 \end{align}

(iv)

\begin{align} BC > 0, \text{ thus } x = -4.045 \text{ must be rejected} \end{align}

(v)

\begin{align} \text{Floor area of model} & = 4x^2 \phantom{00000000} [\text{Area of square base}] \\ & = 4(1.55)^2 \\ & = 9.61 \text{ cm}^2 \\ \\ 1 & : 100 \\ 1 \text{ cm} & : 100 \text{ cm} \\ 1 \text{ cm} & : 1 \text{ m} \\ 1^2 \text{ cm}^2 & : 1^2 \text{ m}^2 \\ 1 \text{ cm}^2 & : 1 \text{ m}^2 \\ 9.61 \text{ cm}^2 & : 9.61 \text{ m}^2 \\ \\ \text{Actual floor area} & = 9.61 \text{ m}^2 \end{align}

(i)

\begin{align} \text{Time} & = {\text{Distance} \over \text{Speed}} \\ \\ \text{Time taken by Weiming} & = {2 \over x} + {10 - 2 \over x + 1} \\ & = \left({2 \over x} + {8 \over x + 1}\right) \text{ hours} \end{align}

(ii)

\begin{align} \text{Time taken by Yi Hao} & = {10 \over x} \text{ hours} \\ \\ \text{30 mins} & = {1 \over 2} \text{ hour} \\ \\ {10 \over x} - \left({2 \over x} + {8 \over x + 1}\right) & = {1 \over 2} \\ {10 \over x} - {2 \over x} - {8 \over x + 1} & = {1 \over 2} \\ {8 \over x} - {8 \over x + 1} & = {1 \over 2} \\ {8(x + 1) \over x(x + 1)} - {8x \over x(x + 1)} & = {1 \over 2} \\ {8(x + 1) - 8x \over x(x + 1)} & = {1 \over 2} \\ {8x + 8 - 8x \over x^2 + x} & = {1 \over 2} \\ {8 \over x^2 + x} & = {1 \over 2} \\ 2(8) & = x^2 + x \\ 16 & = x^2 + x \\ 0 & = x^2 + x - 16 \phantom{00} \text{ (Shown)} \end{align}

(iii)

\begin{align} 0 & = x^2 + x - 16 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-1 \pm \sqrt{(1)^2 - 4(1)(-16)} \over 2(1)} \\ & = {-1 \pm \sqrt{65} \over 2} \\ & = 3.531 \text{ or } -4.531 \\ & \approx 3.53 \text{ or } -4.53 \\ \\ \\ \text{Since } & x > 0, \phantom{.} x = -4.53 \text{ must be rejected} \end{align}

(iv)

\begin{align} \text{Time taken by Weiming} & = {2 \over 3.531} + {8 \over 3.531 + 1} \\ & = 2.332 \text{ hours} \\ & = 2 \text{ hours } (0.332 \times 60) \text{ mins} \\ & \approx 2 \text{ hours } 20 \text{ mins} \\ \\ 8:00 + 2:20 & = 10:20 \\ \\ \\ \text{Weiming finished} & \text{ the hike at } 10:20 \text{ am} \end{align}

(i)

\begin{align} \text{Time} & = {\text{Distance} \over \text{Speed}} \\ \\ \text{Time taken} & = \underbrace{{700 \over x - 30}}_\text{By bus} + \underbrace{700 \over x}_\text{By car} \\ 20 & = {700x \over x(x - 30)} + {700(x - 30) \over x(x - 30)} \\ 20 & = {700x + 700(x - 30) \over x(x - 30)} \\ 20 & = {700x + 700x - 21 \phantom{.} 000 \over x^2 - 30x} \\ {20 \over 1} & = {1400x - 21 \phantom{.} 000 \over x^2 - 30x} \\ 20(x^2 - 30x) & = 1400x - 21 \phantom{.} 000 \\ 20x^2 - 600x & = 1400x - 21 \phantom{.} 000 \\ 20x^2 - 600x - 1400x + 21 \phantom{.} 000 & = 0 \\ 20x^2 - 2000x + 21 \phantom{.} 000 & = 0 \\ x^2 - 100x + 1050 & = 0 \phantom{00} \text{ (Shown)} \phantom{000000} [\text{Divide by 20}] \end{align}

(ii)

\begin{align} x^2 & - 100x + 1050 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-100) \pm \sqrt{(-100)^2 - 4(1)(1050)} \over 2(1)} \\ & = {100 \pm \sqrt{5800} \over 2} \\ & = 88.078 \text{ or } 11.921 \\ & \approx 88.08 \text{ or } 11.92 \text{ (2 d.p.)} \end{align}

(iii) For x = 11.921, the speed of the bus would be negative (11.921 - 30). Thus, we use x = 88.078

\begin{align} \text{Time taken for return journey by car} & = {700 \over x} \\ & = {700 \over 88.078} \\ & = 7.9475 \\ & \approx 7.95 \text{ hours} \end{align}

(i)

\begin{align} x \phantom{0} l & \rightarrow 1 \text{ min} \\ 1 \phantom{0} l & \rightarrow {1 \over x} \text{ min} \\ 1500 \phantom{0} l & \rightarrow {1500 \over x} \text{ mins} \end{align}

(ii)

\begin{align} (x + 50) \phantom{0} l & \rightarrow 1 \text{ min} \\ 1 \phantom{0} l & \rightarrow {1 \over x + 50} \text{ min} \\ 1500 \phantom{0} l & \rightarrow {1500 \over x + 50} \text{ mins} \end{align}

(iii)

\begin{align} 30 \text{ s} & = {1 \over 2} \text{ min} \\ \\ {1500 \over x} - {1500 \over x + 50} & = {1 \over 2} \\ {1500(x + 50) \over x(x + 50)} - {1500x \over x(x + 50)} & = {1 \over 2} \\ {1500(x + 50) - 1500x \over x(x + 50)} & = {1 \over 2} \\ {1500x + 75 \phantom{.} 000 - 1500x \over x^2 + 50x} & = {1 \over 2} \\ {75 \phantom{.} 000 \over x^2 + 50x} & = {1 \over 2} \\ 2(75 \phantom{.} 000) & = x^2 + 50x \\ 150 \phantom{.} 000 & = x^2 + 50x \\ 0 & = x^2 + 50x - 150 \phantom{.} 000 \phantom{00} \text{ (Shown)} \end{align}

(iv)

\begin{align} 0 & = x^2 + 50x - 150 \phantom{.} 000 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-50 \pm \sqrt{(50)^2 - 4(1)(-150 \phantom{.} 000)} \over 2(1)} \\ & = {-50 \pm \sqrt{602 \phantom{.} 500} \over 2} \\ & = 363.104 \text{ or } -413.104 \\ & \approx 363.10 \text{ or } -413.10 \text{ (2 d.p.)} \end{align}

(v)

\begin{align} \text{Time taken by Pump B} & = {1500 \over x + 50} \\ & = {1500 \over 363.104 + 50} \\ & = 3.6310 \text{ mins} \\ & = 3 \text{ mins } (0.6310 \times 60) \text{ s} \\ & \approx 3 \text{ mins } 38 \text{ s} \end{align}

(i)

\begin{align} \text{S\$}x & \rightarrow \text{US\$}1 \\ \text{S\$}1 & \rightarrow \text{US\$} {1 \over x} \\ \text{S\$}2000 & \rightarrow \text{US\$} {2000 \over x} \end{align}

(ii)

\begin{align} \text{S\$}(x + 0.05) & \rightarrow \text{US\$}1 \\ \text{S\$}1 & \rightarrow \text{US\$} {1 \over x + 0.05} \\ \text{S\$}1000 & \rightarrow \text{US\$} {1000 \over x + 0.05} \end{align}

(iii)

\begin{align} {2000 \over x} + {1000 \over x + 0.05} & = 2370 \\ {2000(x + 0.05) \over x(x + 0.05)} + {1000x \over x(x + 0.05)} & = 2370 \\ {2000(x + 0.05) + 1000x \over x(x + 0.05)} & = 2370 \\ {2000x + 100 + 1000x \over x^2 + 0.05x} & = 2370 \\ {3000x + 100 \over x^2 + 0.05x} & = {2370 \over 1} \\ 3000x + 100 & = 2370(x^2 + 0.05x) \\ 3000x + 100 & = 2370x^2 + 118.5x \\ 0 & = 2370x^2 + 118.5x - 3000x - 100 \\ 0 & = 2370x^2 - 2881.5x - 100 \\ 0 & = 237x^2 - 288.15x - 10 \phantom{0} \text{ (Shown)} \phantom{000000} [\text{Divide by 10}] \end{align}

(iv)

\begin{align} 0 & = 237x^2 - 288.15x - 10 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-288.15) \pm \sqrt{(-288.15)^2 - 4(237)(-10)} \over 2(237)} \\ & = {288.15 \pm \sqrt{92 \phantom{.} 510. 422 \phantom{.} 5 } \over 474} \\ & = 1.24958 \text{ or } -0.03376 \\ & \approx 1.2496 \text{ or } -0.0338 \text{ (4 d.p.)} \end{align}

(v)

\begin{align} x + 0.05 & = 1.2496 + 0.05 \\ & = 1.2996 \\ & \approx 1.30 \\ \\ \text{Exchange rate: } & \text{US\$}1 = \text{S\$}1.30 \end{align}

\begin{align} \text{Time} & = { \text{Distance} \over \text{Speed} } \\ \\ \text{Time taken (with wind)} & = { 450 \over 165 + x} \\ \\ \text{Time taken (against wind)} & = {450 \over 165 - x} \\ \\ 5 \text{ hours } 30 \text{ mins} & = 5{1 \over 2} \text{ hours} \\ & = {11 \over 2} \text{ hours} \\ \\ {450 \over 165 + x} + {450 \over 165 - x} & = {11 \over 2} \\ {450(165 - x) \over (165 + x)(165 - x)} + {450(165 + x) \over (165 + x)(165 - x)} & = {11 \over 2} \\ {450(165 - x) + 450(165 + x) \over (165 + x)(165 - x)} & = {11 \over 2} \\ {74 \phantom{.} 250 - 450x + 74 \phantom{.} 250 + 450x \over (165)^2 - (x)^2} & = {11 \over 2} \\ {148 \phantom{.} 500 \over 27 \phantom{.} 225 - x^2} & = {11 \over 2} \\ 2(148 \phantom{.} 500) & = 11(27 \phantom{.} 225 - x^2) \\ 297 \phantom{.} 000 & = 299 \phantom{.} 475 - 11x^2 \\ 11x^2 & = 299 \phantom{.} 475 - 297 \phantom{.} 000 \\ 11x^2 & = 2475 \\ x^2 & = {2475 \over 11} \\ x^2 & = 225 \\ x & = \pm \sqrt{225} \\ x & = \pm 15 \\ \\ \\ \therefore \text{Speed of wind} & = 15 \text{ km/h} \end{align}

(i)

(ii)(a)

\begin{align} \text{Max. height} & = 202.5 \text{ cm} \end{align}

(ii)(b)

\begin{align} \text{Horizontal distance} & = 5.6 \text{ m} \end{align}

(iii)

\begin{align} \text{From graph, } & \text{balloon reaches ground (}y = 0) \text{ when } x = 6.4 \\ \\ \therefore t & = 6.4 \end{align}