\begin{align} \text{Length of } n \text{ bricks} & = 5 \times n \\ & = 5n \text{ cm} \end{align} \begin{align} 5n & > 28 &&& 5n & < 40 \\ n & > {28 \over 5} &&& n & < {40 \over 5} \\ n & > 5.6 &&& n & < 8 \end{align}
\begin{align} 5.6 < n < 8 & \\ \\ \text{Possible values of } n & = 6, 7 \end{align}

\begin{align} \text{Volume of cylinder} & = \text{Base area} \times \text{Height} \\ & = 50 \times h \\ & = 50h \text{ cm}^3 \end{align} \begin{align} 50h & \ge 200 &&& 50h & \le 320 \\ h & \ge {200 \over 50} &&& h & \le {320 \over 50} \\ h & \ge 4 &&& h & \le 6.4 \end{align}
$$ 4 \le h \le 6.4 $$

(a)

\begin{align} \text{Temperature of water} & = 24 + 7 \times t \\ & = (24 + 7t)^\circ \text{C} \end{align}

(b)

\begin{align} 24 + 7t & \ge 90 &&& 24 + 7t & \le 98 \\ 7t & \ge 90 -24 &&& 7t & \le 98 - 24 \\ 7t & \ge 66 &&& 7t & \le 74 \\ t & \ge {66 \over 7} &&& t & \le {74 \over 7} \\ t & \ge 9{3 \over 7} &&& t & \le 10{4 \over 7} \end{align}
$$ 9{3 \over 7} \le t \le 10{4 \over 7} $$

(a)

\begin{align} \text{Total amount} & = 3 \times 50 + n \times 10 \\ & = \$ (150 + 10n) \end{align}

(b)(i)

\begin{align} & 260 \le 150 + 10n < 310 \end{align}

(b)(ii)

\begin{align} 260 & \le 150 + 10n &&& 150 + 10n & < 310 \\ -10n & \le 150 - 260 &&& 10n & < 310 - 150 \\ -10n & \le -110 &&& 10n & < 160 \\ n & \ge {-110 \over -10} &&& n & < {160 \over 10} \\ n & \ge 11 &&& n & < 16 \end{align}
\begin{align} 11 \le n < 16 & \\ \\ \text{Min. no of \$ 10 notes } (n) & = 11 \end{align}

(a)(i)

$$ 19.5 \le x < 20.5 $$

(a)(ii)

$$ 12.5 \le y < 13.5 $$

(b)(i)

\begin{align} \text{Perimeter (smallest possible)} & = 2(19.5) + 2(12.5) \\ & = 64 \text{ m} \\ \\ \text{Perimeter (largest possible)} & = 2(20.5) + 2(13.5) \\ & = 68 \text{ m} \\ \\ \\ 64 \le & \text{ Perimeter} < 68 \end{align}

(b)(ii)

\begin{align} \text{Area (smallest possible)} & = 19.5 \times 12.5 \\ & = 243.75 \text{ m}^2 \\ \\ \text{Area (largest possible)} & = 20.5 \times 13.5 \\ & = 276.75 \text{ m}^2 \\ \\ \\ 243.75 \le & \text{ Area} < 276.75 \end{align}

\begin{align} \text{Perimeter} & = 2(2x + 1) + 2(x) \\ & = 4x + 2 + 2x \\ & = (6x + 2) \text{ cm} \end{align} \begin{align} 6x + 2 & \ge 38 &&& 6x + 2 & \le 80 \\ 6x & \ge 38 - 2 &&& 6x & \le 80 - 2 \\ 6x & \ge 36 &&& 6x & \le 78 \\ x & \ge {36 \over 6} &&& x & \le {78 \over 6} \\ x & \ge 6 &&& x & \le 13 \end{align}
$$ 6 \le x \le 13 $$

\begin{align} \text{Mean PSI} & = {40 + 54 + 44 + 38 + x \over 5} \\ & = {176 + x \over 5} \end{align} \begin{align} {176 + x \over 5} & > 45 &&& {176 + x \over 5} & < 48 \\ {176 + x \over 5} & > {45 \over 1} &&& {176 + x \over 5} & < {48 \over 1} \\ {176 + x \over 5} & > {225 \over 5} &&& {176 + x \over 5} & < {240 \over 5} \\ 176 + x & > 225 &&& 176 + x & < 240 \\ x & > 225 - 176 &&& x & < 240 - 176 \\ x & > 49 &&& x & < 64 \end{align}
\begin{align} 49 < x < 64 & \\ \\ \text{Smallest integer: } & 50 \end{align}

(a)

\begin{align} \text{No. of matches won} & = 38 - x - 2x \\ & = 38 - 3x \\ \\ \text{Total points} & = 3(38 - 3x) + 2x \\ & = 114 - 9x + 2x \\ & = 114 - 7x \end{align}

(b)(i)

\begin{align} 114 - 7x & \ge 85 &&& 114 - 7x & < 94 \\ -7x & \ge 85 - 114 &&& -7x & < 94 - 114 \\ -7x & \ge -29 &&& -7x & < -20 \\ x & \le {-29 \over -7} &&& x & > {-20 \over 7} \\ x & \le 4{1 \over 7} &&& x & > 2{6 \over 7} \end{align}
\begin{align} 2 {6 \over 7} < x < 4{1 \over 7} & \\ \\ \text{Largest value of } x & = 4 \\ \\ \text{Matches drawn} & = 2x \\ & = 2(4) \\ & = 8 \end{align}

(b)(ii) To find the greatest number of matches won (38 - 3x), the value of x must be the smallest

\begin{align} 2 {6 \over 7} < & \phantom{.} x < 4{1 \over 7} \\ \\ \text{Smallest value of } x & = 3 \\ \\ \text{Matches won} & = 38 - 3(3) \\ & = 29 \end{align}

\begin{align} \text{Let } x \text{ represent the num} & \text{ber of hours} \\ \\ \text{Charges from company } A & = 25 \times x \\ & = \$ 25x \\ \\ \text{Charges from company } B & = 35 + 18 \times x \\ & = \$ (35 + 18x) \\ \\ 35 + 18x & < 25x \\ 18x - 25x & < - 35 \\ -7x & < -35 \\ x & > {-35 \over -7} \\ x & > 5 \\ \\ \\ \text{Minimum hours} & = \text{Smallest integer value of } x \\ & = 6 \text{ hours} \end{align}

\begin{align} \text{Let } x \text{ represent the num} & \text{ber senior citizens} \\ \\ \text{Total number of bottles} & = 5 \times x + 12 \\ & = 5x + 12 \\ \\ \text{No. of senior citizens with 6 bottles} & = x - 1 \\ \\ 5x + 12 - 6(x - 1) & < 4 \phantom{000000} [\text{Last senior citizen receive less than 4}] \\ 5x + 12 - 6x + 6 & < 4 \\ 5x - 6x & < 4 - 12 - 6 \\ -x & < -14 \\ x & > 14 \\ \\ \text{Smallest no. of senior citizens} & = 15 \end{align}

\begin{align} \text{Time} & = {\text{Distance} \over \text{Speed}} \\ \\ \text{Let } x \text{ represent } & \text{distance of route } X \\ \\ \text{Time taken (route } X) & = {x \over 60} \\ \\ \text{Time taken (route } Y) & = {x + 1 \over 50} \\ \\ 8 \text{ mins} & = {8 \over 60} \text{ h} \\ & = {2 \over 15} \text{ h} \\ \\ {x + 1 \over 50} - {x \over 60} & \ge {2 \over 15} \\ {6(x + 1) \over 300} - {5x \over 300} & \ge {40 \over 300} \\ {6(x + 1) - 5x \over 300} & \ge {40 \over 300} \\ 6(x + 1) - 5x & \ge 40 \\ 6x + 6 - 5x & \ge 40 \\ 6x - 5x & \ge 40 - 6 \\ x & \ge 34 \\ \\ \text{Shortest possible distance (route } X) & = 34 \text{ km} \end{align}

(a)(i)

\begin{align} 5.7 + 0.11 & = 5.81 \text{ million} \end{align}

(a)(ii)

\begin{align} 5.81 + 0.11 & = 5.92 \text{ million} \end{align}

(a)(iii)

\begin{align} (5.7 + 0.11x) \text{ million} \end{align}

(b)

\begin{align} 5.7 + 0.11x & \ge 6.5 \\ 0.11x & \ge 6.5 - 5.7 \\ 0.11x & \ge 0.8 \\ x & \ge {0.8 \over 0.11} \\ x & \ge 7{3 \over 11} \\ \\ \text{Population will exceed } & 6.5 \text{ million first 8 years after 2020} \\ \\ \therefore \text{End of } & 2028 \end{align}

(c)

\begin{align} & \text{No. Other factors such as lack of resources, pandemics that may affect the rate of increase} \end{align}

\begin{align} \text{Cost for tables} & = 20 \times x \\ & = \$ 20 x \\ \\ \text{Cost for chairs} & = 50 \times 3 \\ & = \$ 150 \\ \\ \text{Cost for fans} & = 1000 - 20x - 150 - \underbrace{40}_\text{Transportation} \\ & = \$ (810 - 20x) \\ \\ \text{No. of fans} & = {810 - 20x \over 70} \end{align}
\begin{align} \text{No. of fans} & \ge 3 &&& \text{Budget for tables & chairs} & \ge \$ 600 \\ {810 - 20x \over 70} & \ge 3 &&& 20x + 150 & \ge 600 \\ {810 - 20x \over 70} & \ge {210 \over 70} &&& 20x & \ge 600 - 150 \\ 810 - 20x & \ge 210 &&& 20x & \ge 450 \\ -20x & \ge 210 - 810 &&& x & \ge {450 \over 20} \\ -20x & \ge -600 &&& x & \ge 22.5 \\ x & \le {-600 \over -20} \\ x & \le 30 \end{align}
\begin{align} 22.5 \le & \phantom{.} x \le 30 \\ \\ \text{Max. no of fans} & = {810 - 20x \over 70} \\ & = {810 - 20(23) \over 70} \phantom{00000000} [\text{Choose smallest integer value of } x ] \\ & = 5 \end{align}