(a)(i)

\begin{align*} \text{HK} \$ 100 & \rightarrow \text{S} \$ 16 \\ \text{HK} \$ 1 & \rightarrow \text{S} \$ {16 \over 100} = \$ 0.16 \\ \text{HK} \$ 55 & \rightarrow \text{S} \$ 0.16 \times 55 = \$ 8.80 \end{align*}

(a)(ii)

\begin{align*} \text{SG} \$ 16 & \rightarrow \text{HK} \$ 100 \\ \text{SG} \$ 1 & \rightarrow \text{HK} \$ {100 \over 16} = \$ 6.25 \\ \text{SG} \$ 20 & \rightarrow \text{HK} \$ 6.25 \times 20 = \$ 125 \end{align*}

(b)

$x ( \text{S} \$ ) $	$16$	$20$	$24$
$y ( \text{HK} \$) $	$100$	$125$	$150$

\begin{align*} \text{Gradient, } m & = {y_2 - y_1 \over x_2 - x_1} \\ & = {150 - 125 \over 24 - 20} \\ & = {25 \over 4} \\ \\ y \text{-intercept, } c & = 0 \\ \\ y & = {25 \over 4}x + 0 \\ y & = {25 \over 4}x \end{align*}

(a)(i)

\begin{align*} f & = {9 \over 5}c + 32 \\ & = {9 \over 5}(50) + 32 \\ & = 122 \end{align*}

(a)(ii)

\begin{align*} f & = {9 \over 5}c + 32 \\ 98 & = {9 \over 5}c + 32 \\ -{9 \over 5}c & = 32 - 98 \\ -{9 \over 5}c & = -66 \\ c & = -66 \div -{9 \over 5} \\ c & = 36{2 \over 3} \end{align*}

(b)

$ c $	$ 0 $	$ 50 $	$ 100 $
$ f $	$ 32 $	$ 122 $	$ 212 $

(c)

\begin{align*} \text{Increase in } c & = 49 - 21 \\ & = 28 \end{align*}

(a)

(b)(i)

$$ F = 106 \text{ N} $$

(b)(ii)

$$ F = 240 \text{ N} $$

(c)

$$ \text{Initial force} = 50 \text{ N} $$

(a)(i)

$$ \$ 1.50 $$

(a)(ii)

$$ \$ 4.90 $$

(a)(iii)

$$ \text{Company P} $$

(b)

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ \\ \text{Dollar per minute for company P} & = {6 - 0 \over 60 - 20} \\ & = \$ 0.15 \text{ per minute} \\ \\ \text{Dollar per minute for company Q} & = {5.4 - 0 \over 60 - 10} \\ & = \$ 0.108 \text{ per minute} \\ \\ \therefore \text{Company Q offers a } & \text{lower rate} \end{align*}

(i)

\begin{align*} \text{Duration} & = 60 - 40 \\ & = 20 \text{ minutes} \end{align*}

(ii)

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ \\ \text{Rate of increase} & = {188 - 92 \over 10 - 0} \\ & = 9.6 \text{ mg/d}l \text{ per minute} \end{align*}

(iii)

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {124 - 188 \over 20 - 10} \\ & = -6.4 \\ \\ \text{Rate of decrease} & = 6.4 \text{ mg/d}l \text{ per minute} \end{align*}

(i)

\begin{align*} \text{Duration} & = 19 - 15 \\ & = 4 \text{ minutes} \end{align*}

(ii)

\begin{align*} \text{Rate of increase} & = \text{Gradient} \\ & = {y_2 - y_1 \over x_2 - x_1} \\ & = {14 - 2 \over 15 - 0} \\ & = 0.8^\circ \text{C per minute} \end{align*}

(iii)

$$ 74^\circ \text{C} $$

(iv)

\begin{align*} \text{Rate of increase} & = \text{Gradient} \\ & = {y_2 - y_1 \over x_2 - x_1} \\ & = {74 - 14 \over 19 - 15} \\ & = 15^\circ \text{C per minute} \end{align*}

(i)

\begin{align*} \text{Gradient for (a)} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {10 - 35 \over 300 - 0} \\ & = -{1 \over 12} \\ \\ \text{Gradient for (b)} & = {26 - 45 \over 600 - 300} \\ & = -{19 \over 300} \\ \\ \text{They represent the litres of petrol} & \text{ consumed per kilometre for (a) & (b) respectively} \end{align*}

(ii)

\begin{align*} & \text{He is ascending the mountain for (a) and is descending the mountain for (b).} \\ & \text{Thus, more petrol is used on the ascent.} \end{align*}

(iii)

\begin{align*} \text{Petrol topped up} & = 45 - 10 \\ & = 35 \text{ litres} \\ \\ \text{Cost} & = {6.75 \times 35 \over 4.88} \\ & = 48.411 \\ & \approx \$ 48.41 \end{align*}

(i)

\begin{align*} \text{Volume of prism} & = \text{Cross-sectional area} \times \text{Length} \\ \\ { \text{Volume of smaller prism formed by water} \over \text{Volume of prism} } & = { \text{Area of smaller triangle} \times l \over \text{Area of larger triangle} \times l} \\ & = { \text{Area of smaller triangle} \over \text{Area of larger triangle} } \\ & = \left(24 \over 32\right)^2 \phantom{0000000} \left[ \text{Similar figures: } {A_1 \over A_2} = \left(l_1 \over l_2\right)^2 \right] \\ & = {9 \over 16} \\ \\ \text{Time taken} & = 12 \times {9 \over 16} \\ & = 6.75 \text{ seconds} \end{align*}

(ii)

\begin{align*} & \text{As water raises, the container goes from narrow to wide.} \\ & \text{Thus, water level increases at a decreasing rate.} \end{align*}

(a)

\begin{align*} 184 \text{ cents} = \$1.84 \end{align*}

(b)

\begin{align*} 1 \text{ km} & \rightarrow 6 \text{ cents} \\ 10 \text{ km} & \rightarrow 60 \text{ cents} \end{align*}

(c)(i)

\begin{align*} \text{Optimus bus service is cheaper} \end{align*}

(c)(ii)

\begin{align*} \text{Difference in fares} & = 220 - 212 \\ & = 8 \text{ cents} \end{align*}

(i) I think there's a typo in the question. It should be $h \text{ (m)} $ instead of $v \text{ (m/s)}$.

(ii)

\begin{align*} \text{Gradient} & = {55 - 0 \over 23 - 5.5} \\ & = 3.1428 \\ & \approx 3.14 \\ \\ \text{At } t = 16, \text{ the height } & \text{is increasing at approximately } 3.14 \text{ m/s} \end{align*}

(i)

$$ 48 \text{ km} $$

(ii)

\begin{align*} \text{Duration} & = 30 \text{ mins} + 1 \text{ hour 30 mins} \\ & = 2 \text{ hours} \end{align*}

(iii)

\begin{align*} \text{Distance from Q} & = 48 - 20 \\ & = 28 \text{ km} \end{align*}

(iv)

\begin{align*} \text{0900 to 1000 hours} \phantom{000} [\text{Steepest part of graph}] \end{align*}

(v)

\begin{align*} \text{Average speed} & = { \text{Total distance} \over \text{Total time}} \\ & = { 48 \text{ km} \over 6 \text{ hours} }\\ & = 8 \text{ km/h} \end{align*}

(i)

\begin{align*} \text{Acceleration} & = \text{Gradient of graph} \\ & = {15 - 0 \over 20 - 0} \phantom{0000000000} \left[ {y_2 - y_1 \over x_2 - x_1}\right] \\ & = 0.75 \text{ m/s}^2 \end{align*}

(ii)

\begin{align*} \text{Distance} & = \text{Area of trapezium} \\ & = {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} \\ & = {1 \over 2} \times (50 + 100) \times 15 \\ & = 1125 \text{ m} \end{align*}

(iii)

\begin{align*} \text{Distance} & = 100 \times 10 \\ & = 1000 \text{ m} \end{align*}

(iv)

\begin{align*} \text{Distance travelled in first 20 seconds (Car)} & = {1 \over 2} \times 20 \times 15 \\ & = 150 \text{ m} \\ \\ \text{Distance travelled in first 20 seconds (Lorry)} & = 20 \times 10 \\ & = 200 \text{ m} \\ \\ \implies \text{Lorry is still ahead} & \text{ of car after 20 seconds} \\ \\ \text{Distance travelled in first 70 seconds (Car)} & = 150 + 50 \times 15 \\ & = 900 \text{ m} \\ \\ \text{Distance travelled in first 70 seconds (Lorry)} & = 70 \times 10 \\ & = 700 \text{ m} \\ \\ \implies \text{Car is ahead of } & \text{lorry after 70 seconds} \\ \\ \text{Let time when car overtakes} & = t \text{ s}, \text{ where } 20 < t < 70 \\ \\ \text{Distance travelled in first } t \text{ seconds (Car)} & = 150 + (t - 20) \times 15 \\ & = 150 + 15t - 300 \\ & = (15t - 150) \text{ m} \\ \\ \text{Distance travelled in first } t \text{ seconds (Lorry)} & = t \times 10 \\ & = 10t \text{ m} \\ \\ 15t - 150 & = 10t \\ 15t - 10t & = 150 \\ 5t & = 150 \\ t & = {150 \over 5} \\ t & = 30 \text{ seconds} \end{align*}

(i)

\begin{align*} \text{Area under graph} & = \underbrace{ {1 \over 2} \times (v + 24) \times 20 }_\text{Area of trapezium} + 30 \times 24 + {1 \over 2} \times 40 \times 24 \\ & = 10(v + 24) + 720 + 480 \\ & = 10v + 240 + 720 + 480 \\ & = 10v + 1440 \\ \\ 1.84 \text{ km} & = 1840 \text{ m} \\ \\ 10v + 1440 & = 1840 \\ 10v & = 1840 - 1440 \\ 10v & = 400 \\ v & = {400 \over 10} \\ v & = 40 \end{align*}

(ii)

\begin{align*} \text{Acceleration} & = \text{Gradient of graph} \\ \\ \text{Acceleration during first 10 seconds} & = \text{Acceleration during first 20 seconds} \\ & = {24 - 40 \over 20 - 0} \phantom{0000000000} \left[ {y_2 - y_1 \over x_2 - x_1}\right] \\ & = -0.8 \text{ m/s}^2 \end{align*}

(iii)

\begin{align*} \text{Acceleration during first 10 seconds} & = -0.8 \text{ m/s}^2 \\ \\ \text{Change in speed after 10 seconds} & = -0.8 \times 10 \\ & = -8 \text{ m/s} \\ \\ \text{Speed after 10 seconds} & = 40 + (-8) \\ & = 32 \text{ m/s} \\ \\ \text{Distance travelled} & = \underbrace{ {1 \over 2} \times (40 + 32) \times 10 }_\text{Area of trapezium} \\ & = 360 \text{ m} \end{align*}

(i)

\begin{align*} \text{Acceleration during first 5 seconds} & = \text{Gradient of graph} \\ & = {14 - 6 \over 5 - 0} \phantom{00000000} \left[ {y_2 - y_1 \over x_2 - x_1}\right] \\ & = 1.6 \text{ m/s}^2 \\ \\ \text{Increase in speed after 3 seconds} & = 1.6 \times 3 \\ & = 4.8 \text{ m/s} \\ \\ \text{Speed after 3 seconds} & = 6 + 4.8 \\ & = 10.8 \text {m/s} \end{align*}

(ii)

\begin{align*} \text{Total distance} & = \underbrace{ {1 \over 2} \times (6 + 14) \times 5 }_\text{Area of trapezium} + 7 \times 14 \\ & = 148 \text{ m} \\ \\ \text{Average speed} & = {\text{Total distance} \over \text{Total time} } \\ & = { 148 \text{ m} \over 12 \text{ s} } \\ & = 12 {1 \over 3} \text{ m/s} \end{align*}

(iii)

\begin{align*} \text{Acceleration} & = \text{Gradient of graph} \\ -3.5 & = {0 - 14 \over T - 12} \phantom{00000000} \left[ {y_2 - y_1 \over x_2 - x_1} \right] \\ -3.5(T - 12) & = 0 - 14 \phantom{0000000000} [\text{Cross-multiply}] \\ -3.5T + 42 & = -14 \\ -3.5T & = -14 - 42 \\ -3.5T & = -56 \\ T & = {-56 \over -3.5} \\ T & = 16 \end{align*}

(i)

\begin{align*} \text{Average speed} & = { \text{Total distance} \over \text{Total time} } \\ \\ \text{Average speed (A)} & = {120 \text{ km} \over 2 \text{ hours} } \\ & = 60 \text{ km/h} \\ \\ \text{Average speed (B)} & = {120 \text{ km} \over 1 \text{ hours 50 minutes} } \\ & = {120 \text{ km} \over 1{5 \over 6} \text{ hours} } \\ & = 65 {5 \over 11} \text{ km/h} \end{align*}

(ii)

\begin{align*} \text{Distance} & = 120 - 80 \\ & = 40 \text{ km} \\ \\ \text{Duration} & = 40 \text{ minutes} \phantom{000000} [\text{1030 to 1110 hours}] \end{align*}

(iii)

\begin{align*} \text{Distance} & = 64 \text{ km} \\ \\ \text{Time} & = 1118 \text{ hours} \end{align*}

(iv)

\begin{align*} \text{Distance} & = 80 - 40 \\ & = 40 \text{ km} \end{align*}

(v)

\begin{align*} \text{Speed} & = \text{Gradient of graph between 1100 and 1200} \\ & = {120 - 40 \over 1200 - 1100} \\ & = {80 \text{ km} \over 1 \text{ hour} } \\ & = 80 \text{ km/h} \end{align*}

(a)

(b)(i)

\begin{align*} \text{Acceleration} & = \text{Gradient} \\ & = { 20 - 0 \over 10 - 0 } \phantom{000000} \left[ {y_2 - y_1 \over x_2 - x_1} \right] \\ & = 2 \text{ m/s}^2 \end{align*}

(b)(ii)

\begin{align*} \text{Let the time when the car comes to rest} & = T \text{ seconds} \\ \\ \text{Acceleration for last } (T - 30) \text{ seconds} & = \text{Gradient} \\ -1.8 & = {0 - 20 \over T - 30} \\ -1.8 (T - 30) & = 0 - 20 \phantom{000000} [\text{Cross-multiply}] \\ -1.8T + 54 & = -20 \\ -1.8T & = -20 - 54 \\ -1.8T & = -74 \\ T & = {-74 \over -1.8} \\ T & = 41{1 \over 9} \text{ seconds} \end{align*}

(a)(i)

\begin{align*} OA: v & = 8{1 \over 3}t \\ v & = 8{1 \over 3}t + 0 \phantom{000000} [y = mx + c] \\ \\ \text{Gradient} & = 8{1 \over 3} \\ \text{Acceleration} & = 8{1 \over 3} \text{ m/s}^2 \end{align*}

(a)(ii)

\begin{align*} OA: v & = 8{1 \over 3}t \\ \\ \text{Let } & v = 40, \\ 40 & = 8{1 \over 3}t \\ {40 \over 8{1 \over 3}} & = t \\ 4.8 & = t \\ \\ \therefore & \phantom{.} A(4.8, 40) \\ \\ \\ BC: v & = -12.5t + 175 \\ \\ \text{Let } & v = 40, \\ 40 & = -12.5t + 175 \\ 12.5t & = 175 - 40 \\ 12.5t & = 135 \\ t & = {135 \over 12.5} \\ t & = 10.8 \\ \\ \therefore & \phantom{.} B(10.8, 40) \\ \\ \text{Let } & v = 0, \\ 0 & = -12.5t + 175 \\ 12.5t & = 175 \\ t & = {175 \over 12.5} \\ t & = 14 \\ \\ \therefore & \phantom{.} C(14, 0) \end{align*}

(a)(iii)

\begin{align*} BC: v & = -12.5t + 175 \phantom{000000} [y = mx + c] \\ \\ \text{Gradient} & = -12.5 \\ \\ \text{Deceleration} & = 12.5 \text{ m/s}^2 \end{align*}

(b)

\begin{align*} \text{Distance} & = \text{Area under graph} \\ & = \underbrace{ {1 \over 2} \times (14 + 10.8 - 4.8) \times 40 }_\text{Area of trapezium} \\ & = 400 \text{ m} \end{align*}

(a)

\begin{align*} v & = 2t^2 - 10t + 15 \\ \\ a & = 2(4)^2 - 10(4) + 15 \\ a & = 7 \\ \\ b & = 2(10)^2 - 10(10) + 15 \\ b & = 115 \end{align*}

(b)

Bottom part only:

(c)(i)

$$ t = 7.1 $$

(c)(ii)

\begin{align*} \text{Gradient} & = {85 - 0 \over 12 - 3.5} \phantom{000000} \left[ {y_2 - y_1 \over x_2 - x_1} \right] \\ & = 10 \\ \\ \text{At } t = 5, \text{ the speed of the } & \text{particle is increasing at the rate of } 10 \text{ m/s}^2 \end{align*}

(a)

\begin{align*} \text{Volume of A (up to height of 20 cm)} & = 10 \times 12 \times 20 \\ & = 2400 \text{ cm}^3 \\ \\ \text{Volume of A} & = 10 \times 12 \times 30 \\ & = 3600 \text{ cm}^3 \\ \\ \text{Time taken to fill A (up to height of 20 cm)} & = {2400 \over 3600} \times 20 \\ & = 13{1 \over 3} \text{ s} \\ \\ \text{Volume of B (up to height of 20 cm)} & = 12 \times 12 \times 20 \\ & = 2880 \text{ cm}^3 \\ \\ \text{Volume of B} & = 2880 + \underbrace{\left({1 \over 2} \times 12 \times 10\right)}_\text{Area of triangle} \times 12 \\ & = 3600 \text{ cm}^3 \\ \\ \text{Time taken to fill B (up to height of 20 cm)} & = {2880 \over 3600} \times 20 \\ & = 16 \text{ s} \end{align*}

(b)(i)

$$ 15 \text{ cm} $$

(b)(ii)

$$ 12.5 \text{ cm} $$