(i)

\begin{align*} {68.1 \over 37 - 4.59^2} & = 4.274 \phantom{.} 443 \phantom{.} 098 \end{align*}

(ii)

\begin{align*} 4.274 \phantom{.} 443 \phantom{.} 098 & \approx 4.3 \end{align*}

(a)

\begin{align*} 121.4 & \approx 121 \\ \\ \text{Greatest mass} & = 121.4 \text{ g} \end{align*}

(b)

\begin{align*} 13.5 & \approx 14 \\ \\ 13.5 \text{ cm}^3 & \rightarrow 121.4 \text{ g} \\ 1 \text{ cm}^3 & \rightarrow {121.4 \over 13.5} \approx 8.9926 \approx 8.99 \text{ g} \end{align*}

(i)

\begin{align*} \text{Difference} & = 19.4 - (-89.2) \\ & = 108.6^\circ \text{C} \end{align*}

(ii)

\begin{align*} \text{Average} & = {-89.2 + 19.4 \over 2} \\ & = -34.9 ^\circ \text{C} \end{align*}

\begin{align*} 1 - {2 \over 3} & = {1 \over 3} \\ \\ {1 \over 3} \times \left(1 - {1 \over 4}\right) & = {1 \over 4} \\ \\ \implies {1 \over 4} \text{ of money } & \text{left} \\ \\ \text{Weekly allowance} & = 27 \times 4 \\ & = \$ 108 \end{align*}

(a)

\begin{align*} 2 & |\underline{792} \\ 2 & |\underline{396} \\ 2 & |\underline{198} \\ 3 & |\underline{99} \\ 3 & |\underline{33} \\ 11 & |\underline{11} \\ & |\underline{1} \end{align*} $$ 792 = 2^3 \times 3^2 \times 11 $$

(b)

\begin{align*} 4 & = 2^2 \\ \\ 15 & = 3 \times 5 \\ \\ \text{LCM} & = 2^2 \times 3 \times 5 \\ & = 60 \\ \\ \text{No. of factors of } 60 & = (2 + 1)(1 + 1)(1 + 1) \\ & = 12 \\ \\ \therefore p & = 60 \phantom{000000} [\text{Any larger multiple of 60 has more than 12 factors}] \end{align*}

(a)

\begin{align*} 3 & | \underline{525} \\ 5 & | \underline{175} \\ 5 & | \underline{35} \\ 7 & | \underline{7} \\ & | \underline{1} \end{align*} $$ 525 = 3 \times 5^2 \times 7 $$

(b)

\begin{align*} 15 & = 3 \times 5 \\ \\ 35 & = 5 \times 7 \\ \\ \text{LCM} = 525 & = 3 \times 5^2 \times 7 \\ \\ 5^2 \text{ is not a factor of 15 or 35} & \implies 5^2 \text{ is a factor of } x \\ \\ \text{Possible } x & = 25, \phantom{.} 75, \phantom{.} 175 \text{ or } 525 \\ \\ x_1 & = 5^2 = 25 \\ \\ x_2 & = 5^2 \times 3 = 75 \end{align*}

(c)

\begin{align*} 525 & = 3 \times 5^2 \times 7 \\ \\ 525k & = 3^2 \times 5^2 \times 7^2 \phantom{000000} [\text{Perfect square}] \\ \\ k & = 3 \times 7 = 21 \end{align*}

(a)

\begin{align*} \sqrt{a} & = \sqrt{ 2^4 \times 3^6 } \\ & = \sqrt{ (2^2 \times 2^2) \times (3^3 \times 3^3)} \\ & = 2^2 \times 3^3 \end{align*}

(b)

\begin{align*} \text{LCM} & = 2^4 \times 3^6 \times 5 \times 7 \end{align*}

(c)

\begin{align*} \text{HCF} & = 2 \times 3 \end{align*}

\begin{align*} 96 & = 2^5 \times 3 \\ \\ \text{Each side } & \geq 3 \text{ cm} \\ \\ 96 & = 3 \times 4 \times 8 \\ & = 4 \times 4 \times 6 \\ \\ \text{Dimensions: } & 3 \text{ cm by 4 cm by 8 cm} \\ & \text{or } 4 \text{ cm by 4 cm by 6 cm} \end{align*}

\begin{align*} 3 & \\ 4 & = 2^2 \\ 6 & = 2 \times 3 \\ \\ \text{LCM} & = 2^2 \times 3 = 12 \text{ hours} \\ \\ \text{Time: } & 2100 \text{ hours} \end{align*}

(a)

\begin{align*} 402 \approx 400 \\ \\ 70.35 \approx 70 \\ \\ \text{Total cost} & \approx 400 \times 70 \\ & = \$ 28 \phantom{.} 000 \end{align*}

(b)

\begin{align*} 2.5 \text{ megabytes} & = 2.5 \times 10^6 \text{ bytes} \\ \\ \text{Total file size} & = (2.5 \times 10^6) \times 680 \\ & = 1700 \times 10^6 \\ & = 1.7 \times 10^3 \times 10^6 \\ & = 1.7 \times 10^9 \text{ bytes} \phantom{0000000} [a^m \times a^n = a^{m + n}] \\ & = {1.7 \times 10^9 \over 10^9} \text{ gigabytes} \\ & = 1.7 \text{ gigabytes} \end{align*}

(i)

\begin{align*} \text{Original price} & = 139.50 \times {100 \over 100 - 22.5} \\ & = \$ 180 \end{align*}

(ii)

\begin{align*} \text{GST} & = 180 \times {7 \over 100 + 7} \\ & \approx 11.7757 \\ & \approx \$ 11.78 \end{align*}

(i)

\begin{align*} 1020 \text{ billion} & = 1020 \times 10^9 \\ & = 1.02 \times 10^3 \times 10^9 \\ & = 1.02 \times 10^{3 + 9} \phantom{0000000} [a^m \times a^n = a^{m + n}] \\ & = 1.02 \times 10^{12} \end{align*}

(ii)

\begin{align*} \text{Total trade in 2020} & = 1020 \times {95 \over 100} \\ & = 969 \\ & \approx \$ 970 \text{ billion (to 2 s.f.)} \end{align*}

\begin{align*} 50 \text{ pm} & = 50 \times 10^{-12} \text{ m} \\ & = 5.0 \times 10^1 \times 10^{-12} \\ & = 5.0 \times 10^{1 + (-12)} \phantom{000000} [a^m \times a^n = a^{m + n}] \\ & = 5.0 \times 10^{-11} \text{ m} \end{align*}

(a)

\begin{align*} 2P - Q & = 2(7.8 \times 10^5) - (3.9 \times 10^3) \\ & = 1 \phantom{.} 556 \phantom{.} 100 \\ & = 1.5561 \times 10^6 \\ & \approx 1.56 \times 10^6 \end{align*}

(b)

\begin{align*} 0.3 (P + 4Q) & = 0.3[ (7.8 \times 10^5) + 4(3.9 \times 10^3)] \\ & = 238 \phantom{.} 680 \\ & = 2.3868 \times 10^5 \\ & \approx 2.39 \times 10^5 \end{align*}

(c)

\begin{align*} {P \over Q} & = { 7.8 \times 10^5 \over 3.9 \times 10^3 } \\ & = 200 \\ & = 2 \times 10^2 \end{align*}

(d)

\begin{align*} (2 P Q)^{-{1 \over 3}} & = [ 2 (7.8 \times 10^5) (3.9 \times 10^3) ]^{-{1 \over 3}} \\ & = 5.4777 \times 10^{-4} \\ & \approx 5.48 \times 10^{-4} \end{align*}

(a)

\begin{align*} 11^7 \div 11^{-1} & = 11^{7-(-1)} \phantom{000000} [a^m \div a^n = a^{m - n}] \\ & = 11^8 \end{align*}

(b)

\begin{align*} {1 \over 121^3} & = {1 \over (11^2)^3} \\ & = {1 \over 11^6} \phantom{000000} [ (a^m)^n = a^{mn} ] \\ & = 11^{-6} \phantom{00000} \left[ {1 \over a^n} = a^{-n} \right] \end{align*}

(c)

\begin{align*} \sqrt[5]{11} & = 11^{1 \over 5} \phantom{000000} \left[ \sqrt[n]{a^m} = a^{m \over n} \right] \end{align*}

(a)

\begin{align*} 36^c \times 2 & = 12 \\ 36^c & = {12 \over 2} \\ 36^c & = 6 \\ (6^2)^c & = 6 \\ 6^{2c} & = 6^1 \phantom{000000} [ (a^m)^n = a^{mn}] \\ \\ 2c & = 1 \\ c & = {1 \over 2} \end{align*}

(b)

\begin{align*} 1 \div 2x^{-5} & = 1 \div {2 \over x^5} \phantom{000000} \left[ a^{-n} = {1 \over a^n} \right] \\ & = {1 \over 1} \times {x^5 \over 2} \\ & = {x^5 \over 2} \end{align*}

(c)

\begin{align*} 8^2 \div 16^{3 \over 4} & = {1 \over 2^m} \\ (2^3)^2 \div (2^4)^{3 \over 4} & = 2^{-m} \phantom{000000} \left[ {1 \over a^n} = a^{-n} \right] \\ 2^6 \div 2^3 & = 2^{-m} \phantom{000000.} [ (a^m)^n = a^{mn} ] \\ 2^{6 - 3} & = 2^{-m} \phantom{000000.} [a^m \div a^n = a^{m - n} ] \\ 2^3 & = 2^{-m} \\ \\ -m & = 3 \\ m & = -3 \end{align*}

(d)

\begin{align*} 10^n & = {10^5 \times 10^1 \over (10^2)^3} \\ 10^n & = {10^6 \over 10^6} \phantom{000000} [ a^m \times a^n = a^{m + n} \text{ and } (a^m)^n = a^{mn} ] \\ 10^n & = 1 \\ 10^n & = 10^0 \\ n & = 0 \end{align*}

\begin{align*} \text{Total amount} & = P \left(1 + {r \over 100}\right)^n \\ 9462.54 & = 8400 \left(1 + {r \over 100}\right)^{4 \times 2} \\ {9462.54 \over 8400} & = \left(1 + {r \over 100}\right)^8 \\ \sqrt[8]{9462.54 \over 8400} & = 1 + {r \over 100} \\ 1.015 & \approx 1 + {r \over 100} \\ 1.015 - 1 & = {r \over 100} \\ 0.015 & = {r \over 100} \\ 100(0.015) & = r \\ 1.5 & = r \\ \\ \text{IR per annum} & = 1.5 \times 2 \phantom{000000} [\text{Interest compounded half-yearly}] \\ & \approx 3.00 \% \end{align*}

\begin{align*} n & = 6 \times 4 = 24 \\ \\ r & = 4.2 \div 4 = 1.05 \\ \\ \text{Total amount} & = x \left(1 + {r \over 100}\right)^n \\ 50 \phantom{.} 000 & = x \left(1 + {1.05 \over 100}\right)^{24} \\ { 50 \phantom{.} 000 \over \left(1 + {1.05 \over 100}\right)^{24} } & = x \\ 38 \phantom{.} 913.328 & \approx x \\ \\ \therefore x & \approx 38 \phantom{.} 913.33 \end{align*}

(a)

\begin{align*} \text{Value in 2022} & = 7400 \times {104 \over 100} \times {104 \over 100} \\ & = \$ 8003.84 \end{align*}

(b)

\begin{align*} A & = 7400 \times \left( \underbrace{{104 \over 100} \times {104 \over 100} \times ...}_{ t \text{ times} } \right) \\ & = 7400 \times (1.04)^t \\ & = 7400 (1.04)^t \end{align*}

(c)

\begin{align*} A & = 7400 (1.04)^t \\ \\ \text{Let } & t = 8, \\ A & = 7400 (1.04)^8 \\ & \approx \$ 10 \phantom{.} 127.41 \\ \\ \therefore \text{Value } & \text{will be at least } \$10 \phantom{.} 000 \end{align*}

\begin{align*} I & = {PRT \over 100} \\ I & = {P(4)(2) \over 100} \\ I & = 0.08P \\ \\ \text{Total amount (simple interest)} & = P + 0.08P \\ & = 1.08P \\ \\ \text{Total amount (compound interest)} & = P \left(1 + {4 \over 100}\right)^2 \\ & = P(1.04)^2 \\ & = 1.0816 P \\ \\ 1.0816 P - 1.08 P & = 10 \\ 0.0016 P & = 10 \\ P & = {10 \over 0.0016} \\ P & = \$ 6250 \end{align*}

\begin{align*} \text{Total amount} & = P \left(1 + {r \over 100}\right)^n \\ 115 \phantom{.} 762.5 & = 100 \phantom{.} 000 \left(1 + {5 \over 100}\right)^n \\ {115 \phantom{.} 762.5 \over 100 \phantom{.} 000} & = (1.05)^n \\ 1.157 \phantom{.} 625 & = (1.05)^n \\ \\ \text{By guess-and-check, } n & = 3 \end{align*}

(i)

\begin{align*} 1 \text{ m} & = 100 \text{ cm} \\ 1 \text{ m}^3 & = 100^3 \text{ cm}^3 = 1 \phantom{.} 000 \phantom{.} 000 \text{ cm}^3 \\ \\ 4 \phantom{.} 200 \phantom{.} 000 \text{ m}^3 & = 4 \phantom{.} 200 \phantom{.} 000 \times 1 \phantom{.} 000 \phantom{.} 000 \text{ cm}^3 \\ & = 4.2 \times 10^{12} \text{ cm}^3 \end{align*}

(ii)

\begin{align*} \text{Required percentage} & = {12.8 \times 10^6 \over 4 \phantom{.} 200 \phantom{.} 000} \times 100 \\ & \approx 304.762 \\ & \approx 305 \% \end{align*}

(a)

\begin{align*} k & = 2.74 \times 10^8 \div 10^6 \\ & = 2.74 \times 10^2 \phantom{000000} [a^m \div a^n = a^{m - n}] \\ & = 274 \end{align*}

(b)(i)

\begin{align*} \text{No. of times} & = { 97.3 \times 10^6 \over 3.81 \times 10^7} \\ & = 2.5538 \\ & \approx 2.55 \end{align*}

(b)(ii)

\begin{align*} \text{Average no. of phones per person} & = { 386 \times 10^6 \over 2.74 \times 10^8} \\ & \approx 1.4088 \\ & \approx 1.41 \end{align*}

(a)

\begin{align*} \text{Let initial cost} & = x \text{ dollars per kg} \\ \\ \text{Let initial consumption} & = y \text{ kg} \\ \\ \text{Total cost (initial)} & = \$ xy \\ \\ \text{New cost} & = x \times {125 \over 100} \\ & = 1.25x \text{ dollars per kg} \\ \\ \text{New consumption} & = {xy \over 1.25x} \\ & = {y \over 1.25} \\ & = {4 \over 5}y \\ \\ \text{Percentage change} & = { {4 \over 5}y - y \over y} \times 100 \\ & = { -{1 \over 5} y \over y } \times 100 \\ & = -{1 \over 5} \times 100 \\ & = -20 \% \\ \\ \text{Percentage reduction} & = 20 \% \end{align*}

(b)

\begin{align*} \text{Let initial length} & = x \\ \\ \text{Let initial width} & = y \\ \\ \text{Initial area} & = xy \\ \\ \text{New length} & = x \times {120 \over 100} = 1.2x \\ \\ \text{New width} & = y \times {80 \over 100} = 0.8y \\ \\ \text{New area} & = (1.2x)(0.8y) \\ & = 0.96xy \\ \\ \text{Percentage change} & = {0.96xy - xy \over xy} \times 100 \\ & = {-0.04xy \over xy} \times 100 \\ & = -0.04 \times 100 \\ & = -4 \% \\ \\ \text{Percentage decrease in area} & = 4 \% \end{align*}

\begin{align*} \text{Commission (Chan)} & = 200 \phantom{.} 000 \times {5 \over 100} + (2 \phantom{.} 800 \phantom{.} 000 - 200 \phantom{.} 000) \times {1.5 \over 100} \\ & = \$ 49 \phantom{.} 000 \\ \\ \text{Commission (Lim)} & = 200 \phantom{.} 000 \times {3.5 \over 100} + (2 \phantom{.} 800 \phantom{.} 000 - 200 \phantom{.} 000) \times {2.5 \over 100} \\ & = \$ 72 \phantom{.} 000 \\ \\ \text{Engage Chan and }& \text{Partners since the commission is lower} \end{align*}

\begin{align*} & 30 \text{ cows last } 20 \text{ days} \\ & 10 \text{ cows last } 20 \times 3 = 60 \text{ days } \phantom{00000} [\text{Fewer cows, food lasts longer}] \\ & 40 \text{ cows last } {60 \over 4} = 15 \text{ days} \end{align*}

\begin{align*} & 4 \text{ workers manufacture } 80 \text{ chairs in } 12 \text{ days} \\ & 24 \text{ workers manufacture } 80 \text{ chairs in } 12 \div 6 = 2 \text{ days} \phantom{000000} [\text{More workers, less time needed}] \\ & 24 \text{ workers manufacture } 300 \text{ chairs in } 2 \times 3.75 = 7.5 \text{ days} \phantom{00} [\text{More days to manufacture more chairs; } 80 \times 3.75 = 300] \end{align*}

\begin{align*} I & = {PRT \over 100} \\ \\ \text{Interest for \$800} & = {(800)(1.2)(2) \over 100} \\ & = \$ 19.20 \\ \\ \text{Interest for \$700} & = {(700)(1.2)(1) \over 100} \\ & = \$ 8.40 \\ \\ \text{Total amount} & = 800 + 700 + 19.20 + 8.40 \\ & = \$ 1 \phantom{.} 527.60 \end{align*}

(i)

\begin{align*} \text{Amount in NZ\$} & = 1800 \times 1.03 \\ & = \text{NZ\$} 1854 \end{align*}

(ii)

\begin{align*} \text{NZ\$}1.05 & = \text{S\$} 1 \\ \text{NZ\$}1 & = \text{S\$} {1 \over 1.05} = \text{S\$} {20 \over 21} \\ \\ \text{Amount received} & = 350 \times {20 \over 21} \\ & \approx \text{S\$} 333.33 \end{align*}

\begin{align*} \text{Taxable income} & = 240 \phantom{.} 000 - 1000 - 3(4000) - 2(5500) - 48 \phantom{.} 000 \\ & = \$ 168 \phantom{.} 000 \\ \\ \text{Income tax} & = 13 \phantom{.} 950 + (168 \phantom{.} 000 - 160 \phantom{.} 000) \times {18 \over 100} \\ & = \$ 15 \phantom{.} 390 \end{align*}

(i)

\begin{align*} \text{Deposit} & = 1400 \times {1 \over 4} \\ & = \$ 350 \\ \\ \text{Balance (before interest)} & = 1400 - 350 \\ & = \$ 1050 \\ \\ \text{Interest, } I & = {PRT \over 100} \\ & = { (1050)(6)(2) \over 100} \\ & = \$ 126 \\ \\ \text{Balance (after interest)} & = 1050 + 126 \\ & = \$ 1176 \\ \\ \text{Monthly instalment} & = {1176 \over 24} \phantom{000000} [2 \text{ years} = 24 \text{ months}] \\ & = \$ 49 \end{align*}

(ii)

\begin{align*} \text{Total paid (hire purchase)} & = 350 + 1176 \\ & = \$ 1526 \\ \\ \text{Extra amount paid} & = 1526 - 1400 \\ & = \$ 126 \phantom{000000} [\text{Equal to the interest}] \end{align*}

\begin{align*} \text{Cost per gram (A)} & = {1.65 \over 3 \times 80} \\ & = \$ 0.006 \phantom{.} 875 \text{ per gram} \\ \\ \text{Cost per gram (B)} & = {2 \times 2 - 1 \over 5 \times 60 \times 2} \\ & = \$ 0.005 \text{ per gram} \\ \\ \text{Cost per gram (C)} & = {1.80 \over 5 \times 60} \\ & = \$ 0.006 \text{ per gram} \\ \\ \text{Cost per gram (D)} & = {2 \times 2 - 1.10 \over 5 \times 60 \times 2} \\ & \approx \$ 0.004 \phantom{.} 833 \text{ per gram} \\ \\ \therefore \text{Ali } & \text{should buy brand D} \end{align*}

(i)

\begin{align*} \text{Volume of sphere, } V & = {4 \over 3} \pi r^3 \phantom{000000} [\text{Formula provided}] \\ \\ \implies y & = {4 \over 3} \pi x^3 \\ y & \propto x^3 \\ \\ \therefore n & = 3 \end{align*}

(ii)

\begin{align*} \text{Distance} & = \text{Speed} \times \text{Time} \\ \\ \text{Since } & \text{distance is fixed,} \\ k & = y \times x \\ {k \over x} & = y \\ k x^{-1} & = y \phantom{000000} \left[ a^{-n} = {1 \over a^n} \right] \\ \\ \implies y & \propto x^{-1} \\ \\ n & = -1 \end{align*}

\begin{align*} y & = {k \over x} \\ \\ \text{When } & x = 8 \text{ and } y = 2.5, \\ 2.5 & = {k \over 8} \\ {5 \over 2} & = {k \over 8} \\ 8(5) & = 2k \phantom{000000} [\text{Cross-multiply}] \\ 40 & = 2k \\ {40 \over 2} & = k \\ 20 & = k \\ \\ y & = {20 \over x} \\ \\ \text{When } & x = 10, \\ y & = {20 \over 10} \\ y & = 2 \\ \\ \text{When } & y = 0.8, \\ 0.8 & = {20 \over x} \\ 0.8x & = 20 \\ x & = {20 \over 0.8} \\ x & = 25 \\ \\ \text{Completed table: } & \begin{array}{|c|c|c|c|} \hline x & 8 & 10 & 25 \\ \hline y & 2.5 & 2 & 0.8 \\ \hline \end{array} \end{align*}

(a)(i)

\begin{align*} V & = k r^2 \\ \\ \text{When } & r = 5 \text{ and } V = 240, \\ 240 & = k(5)^2 \\ 240 & = k(25) \\ {240 \over 25} & = k \\ 9.6 & = k \\ \\ \therefore V & = 9.6 r^2 \end{align*}

(a)(ii)

\begin{align*} V & = 9.6 r^2 \\ \\ \text{When } & r = 8, \\ V & = 9.6 (8)^2 \\ & = 614.4 \\ & \approx 614 \text{ cm}^3 \text{ (to nearest whole number)} \end{align*}

(b)

\begin{align*} V & = 9.6 r^2 \\ \\ \text{Let radius of Q} & = x \\ \\ \text{Volume of } Q & = 9.6 x^2 \\ \\ \text{Radius of } P & = x \times {125 \over 100} \\ & = 1.25x \\ \\ \text{Volume of } P & = 9.6 (1.25x)^2 \\ & = 9.6(1.5625x^2) \\ & = 15x^2 \\ \\ { \text{Volume of } P \over \text{Volume of } Q } & = {15x^2 \over 9.6 x^2} \\ & = {15 \over 9.6} \\ & = {25 \over 16} \\ \\ \therefore & \phantom{.} 25: 16 \end{align*}

\begin{align*} y & = kx^2 \end{align*} \begin{align*} \text{When } & x = 2, &&& \text{When } & x = 5, \\ y_1 & = k(2)^2 &&& y_2 & = k(5)^2 \\ y_1 & = k(4) &&& y_2 & = k(25) \\ y_1 & = 4k &&& y_2 & = 25k \end{align*}
\begin{align*} y_2 - y_1 & = 32 \phantom{000000} [\text{Taking } y_2 > y_1] \\ 25k -4k & = 32 \\ 21k & = 32 \\ k & = {32 \over 21} \\ \\ \therefore y & = {32 \over 21}x^2 \\ \\ \text{When } & x = 3, \\ y & = {32 \over 21}(3)^2 \\ y & = 13{5 \over 7} \end{align*}

(a)

\begin{align*} F & = {k \over d^2} \\ \\ \text{When } & d = 8 \text{ and } F = 10, \\ 10 & = {k \over 8^2} \\ 10 & = {k \over 64} \\ 64(10) & = k \\ 640 & = k \\ \\ F & = {640 \over d^2} \end{align*}

(b)

\begin{align*} F & = {640 \over d^2} \\ \\ \text{When } & F = 25, \\ 25 & = {640 \over d^2} \\ 25d^2 & = 640 \\ d^2 & = {640 \over 25} \\ d & = \sqrt{640 \over 25} \\ d & \approx 5.0596 \\ d & \approx 5.06 \text{ cm} \end{align*}

(c)

\begin{align*} F & = {640 \over d^2} \\ \\ \text{When } d \text{ is doubled, } & F \text{ is } {1 \over 2} \times {1 \over 2} = {1 \over 4} \text{ of initial value} \\ \\ \text{New value of } F & = 12 \times {1 \over 4} \\ & = 3 \text{ N} \end{align*}

(a)

\begin{align*} y & = k \sqrt{x} \phantom{000000} [\text{Test } y = k \sqrt{x} \text{ for all pairs}] \\ \\ \text{When } & x = 4 \text{ and } y = 6, \\ 6 & = k \sqrt{4} \\ 6 & = 2k \\ 3 & = k \\ \\ \text{When } & x = 9 \text{ and } y = 9, \\ 9 & = k \sqrt{9} \\ 9 & = 3k \\ 3 & = k \\ \\ \text{When } & x = 16 \text{ and } y = 12, \\ 12 & = k \sqrt{16} \\ 12 & = 4k \\ 3 & = k \\ \\ \text{When } & x = 25 \text{ and } y = 15, \\ 15 & = k \sqrt{25} \\ 15 & = 5k \\ 3 & = k \\ \\ \therefore y & = 3 \sqrt{x} \end{align*}

(b)

\begin{align*} y & = 3 \sqrt{x} \\ \\ y & \propto \sqrt{x} \end{align*}

(c)

\begin{align*} y & = 3 \sqrt{x} \\ \\ \text{When } & y = 24, \\ 24 & = 3 \sqrt{x} \\ {24 \over 3} & = \sqrt{x} \\ 8 & = \sqrt{x} \\ 8^2 & = x \\ 64 & = x \end{align*}

(a)

\begin{align*} 1 \text{ cm} & : 20 \phantom{.} 000 \text{ cm} \\ 1 \text{ cm} & : 200 \text{ m} \phantom{000000} [1 \text{ m} = 100 \text{ cm} ] \\ 1 \text{ cm} & : 0.2 \text{ km} \phantom{00000.} [1 \text{ km} = 1000 \text{ m}] \\ \\ n & = 0.2 \end{align*}

(b)

\begin{align*} 1 \text{ cm} & : 0.2 \text{ km} \\ 18 \text{ cm} & : 3.6 \text{ km} \\ \\ \text{Actual distance} & = 3.6 \text{ km} \end{align*}

(c)

\begin{align*} 1 \text{ cm} & : 0.2 \text{ km} \\ 1^2 \text{ cm}^2 & : 0.2^2 \text{ km}^2 \\ 1 \text{ cm}^2 & : 0.04 \text{ km}^2 \\ 80 \text{ cm}^2 & : 3.2 \text{ km}^2 \\ \\ \text{Area on map} & = 80 \text{ cm}^2 \end{align*}

(a)

\begin{align*} 1 \text{ cm} & : 4 \text{ km} \\ 6 \text{ cm} & : 24 \text{ km} \\ \\ \text{Actual distance} & = 24 \text{ km} \end{align*}

(b)

\begin{align*} 1 \text{ cm} & : 4 \text{ km} \phantom{000000} [\text{First map}] \\ 1^2 \text{ cm}^2 & : 4^2 \text{ km}^2 \\ 1 \text{ cm}^2 & : 16 \text{ km}^2 \\ 40 \text{ cm}^2 & : 640 \text{ km}^2 \\ \\ \text{Actual area} & = 640 \text{ km}^2 \\ \\ 0.4 \text{ cm}^2 & : 640 \text{ km}^2 \phantom{000000} [\text{Second map}] \\ 1 \text{ cm}^2 & : 1600 \text{ km}^2 \\ \sqrt{1} \text{ cm} & : \sqrt{1600} \text{ km} \\ 1 \text{ cm} & : 40 \text{ km} \\ 1 \text{ cm} & : 40 \phantom{.} 000 \text{ m} \phantom{00000.} [1 \text{ km} = 1000 \text{ m}] \\ 1 \text{ cm} & : 4 \phantom{.} 000 \phantom{.} 000 \text{ cm} \phantom{00} [1 \text{ m} = 100 \text{ cm} ] \\ \\ \therefore n & = 4 \phantom{.} 000 \phantom{.} 000 \end{align*}

(a)

\begin{align*} 1 \text{ cm} & : 20 \text{ cm} \\ 1 \text{ cm} & : 0.2 \text{ m} \phantom{000000} [1 \text{ m} = 100 \text{ cm}] \\ 24 \text{ cm} & : 4.8 \text{ m} \\ \\ \text{Length of lorry} & = 4.8 \text{ m} \end{align*}

(b)

\begin{align*} 1 \text{ cm} & : 0.2 \text{ m} \\ 1^2 \text{ cm}^2 & : 0.2^2 \text{ m}^2 \phantom{000000} \left[ {A_1 \over A_2} = \left(l_1 \over l_2\right)^2 \right] \\ 1 \text{ cm}^2 & : 0.04 \text{ m}^2 \\ 250 \text{ cm}^2 & : 10 \text{ m}^2 \\ \\ \text{Area of platform of model} & = 250 \text{ cm}^2 \end{align*}

(c)

\begin{align*} 1 \text{ cm} & : 0.2 \text{ m} \\ 1^3 \text{ cm}^3 & : 0.2^3 \text{ m}^3 \phantom{000000} \left[ {V_1 \over V_2} = \left(l_1 \over l_2\right)^3 \right] \\ 1 \text{ cm}^3 & : 0.008 \text{ m}^3 \\ 30 \text{ cm}^3 & : 0.24 \text{ m}^3 \\ \\ \text{Volume of fuel} & = 0.24 \text{ m}^3 \\ & = 240 \phantom{.} 000 \text{ cm}^3 \phantom{000000} [1 \text{ m}^3 = 1 \phantom{.} 000 \phantom{.} 000 \text{ cm}^3] \\ & = 240 \text{ litres} \phantom{00000000.} [1 \text{ litre} = 1000 \text{ cm}^3] \end{align*}

(i)

\begin{align*} 210 \text{ m/s} & = { 210 \text{ m} \over 1 \text{ second}} \\ & = { {210 \over 1000} \text{ km} \over {1 \over 3600} \text{ hour} } \\ & = 756 \text{ km/h} \end{align*}

(ii)

\begin{align*} \text{Time taken} & = {\text{Distance} \over \text{Speed}} \\ & = {3240 \text{ km} \over 756 \text{ km/h} } \\ & = 4{2 \over 7} \text{ h} \\ & = 4 \text{ hours } {2 \over 7} \times 60 \text{ minutes} \\ & \approx 4 \text{ hours } 17 \text{ minutes} \end{align*}

\begin{align*} \text{Circumference of circle} & = \pi d \\ \text{Circumference of wheel} & = \pi (58.4) \\ & = 58.4 \pi \text{ cm} \\ \\ \text{Distance covered by bicycle in one minute} & = 25 \div 60 \phantom{000000} [25 \text{ km/h} \implies 25 \text{ km in 1 hour}] \\ & = {5 \over 12} \text{ km} \\ & = 416{2 \over 3} \text{ m} \\ & = 41 \phantom{.} 666 {2 \over 3} \text{ cm} \\ \\ \text{Number of revolutions} & = { 41 \phantom{.} 666 {2 \over 3} \text{ cm} \over 58.4 (3.142) \text{ cm} } \\ & \approx 227.08 \\ & \approx 227 \end{align*}

\begin{align*} \text{Speed} & = { \text{Distance} \over \text{Time} } \\ \\ \text{Initial speed} & = { 57 \text{ km} \over 2{3 \over 4} \text{ h} } \\ & = 20{8 \over 11} \text{ km/h} \\ \\ \text{Required speed} & = { 117 - 57 \over 6.5 - 2{3 \over 4} } \\ & = 16 \text{ km/h} \\ \\ \text{Required reduction} & = 20{8 \over 11} - 16 \\ & = 4{8 \over 11} \text{ km/h} \end{align*}

(i)

\begin{align*} 298 \phantom{.} 000 \text{ km/h} & = { 298 \phantom{.} 000 \text{ km} \over 1 \text{ h} } \\ & = { 298 \phantom{.} 000 \times 1000 \text{ m} \over 3600 \text{ s} } \\ & = 82 \phantom{.} 777{7 \over 9} \text{ m/s} \\ & \approx 82 \phantom{.} 800 \text{ m/s (to 3 s.f.)} \\ & = 8.28 \times 10^4 \text{ m/s} \end{align*}

(ii)

\begin{align*} \text{Time taken} & = { \text{Distance} \over \text{Speed} } \\ & = { 372.5 \text{ m} \over 82 \phantom{.} 777 {7 \over 9} \text{ m/s}} \\ & = 0.0045 \text{ s} \\ \\ \text{Time difference} & = 0.0045 \times 2 \phantom{000000} [\text{Extra 372.5 m travelled there and back}] \\ & = 0.009 \text{ s} \\ & = 9.00 \times 10^{-3} \text{ s} \end{align*}

(a)

\begin{align*} 3(2x - 1) - 4(x - 7) & = 6x - 3 - 4x + 28 \\ & = 2x + 25 \end{align*}

(b)

\begin{align*} 14 - 3(5 - 4x) + 6x & = 14 - 15 + 12x + 6x \\ & = -1 + 18x \end{align*}

(c)

\begin{align*} 7(2y + 3) - 4(3 - y) & = 14y + 21 - 12 + 4y \\ & = 18y + 9 \end{align*}

(d)

\begin{align*} 9(5p - 6) + 4(7 - 13p) & = 45p - 54 + 28 - 52p \\ & = - 7p - 26 \end{align*}

(e)

\begin{align*} 5 - 3(q + r) - 6(3r - 2q) & = 5 - 3q - 3r - 18r + 12q \\ & = 5 + 9q - 21r \end{align*}

(f)

\begin{align*} (a + 2b)^2 - (a - 2b)^2 & = \underbrace{ (a)^2 + 2(a)(2b) + (2b)^2 }_{ (a + b)^2 = a^2 + 2ab + b^2 } - [ \underbrace{ (a)^2 - 2(a)(2b) + (2b)^2 }_{ (a - b)^2 = a^2 - 2ab + b^2 } ] \\ & = a^2 + 4ab + 4b^2 - (a^2 - 4ab + 4b^2) \\ & = a^2 + 4ab + 4b^2 - a^2 + 4ab - 4b^2 \\ & = 8ab \end{align*}

(a)

\begin{align*} 5x^2 - 20x^2y & = 5x^2 (1 - 4y) \end{align*}

(b)

\begin{align*} x^2 - 4xy + 4y^2 & = (x - 2y)(x - 2y) \\ & = (x - 2y)^2 \end{align*}

(c)

\begin{align*} (3x + 4y)^2 - 9z^2 & = (3x + 4y)^2 - (3z)^2 \\ & = (3x + 4y + 3z)(3x + 4y - 3z) \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \end{align*}

(d)

\begin{align*} 5p^2 + 11p + 2 & = (5p + 1)(p + 2) \end{align*}

(e)

\begin{align*} 6q^2 - 31q + 35 & = (3q - 5)(2q - 7) \end{align*}

(a)

\begin{align*} 2[3a - 2(3a - 1) + 4(a + 1)] & = 2(3a - 6a + 2 + 4a + 4) \\ & = 2(a + 6) \\ & = 2a + 12 \end{align*}

(b)

\begin{align*} 8(x - y) - [x - y - 3(y - z - x)] & = 8x - 8y - (x - y - 3y + 3z + 3x) \\ & = 8x - 8y - (4x - 4y + 3z) \\ & = 8x - 8y - 4x + 4y - 3z \\ & = 4x - 4y - 3z \end{align*}

(c)

\begin{align*} 2b(c - a) - [3c(a - b) - 3a(b + c)] & = 2bc - 2ab - (3ac - 3bc - 3ab - 3ac) \\ & = 2bc - 2ab - (-3bc - 3ab) \\ & = 2bc - 2ab + 3bc + 3ab \\ & = 5bc + ab \end{align*}

(d)

\begin{align*} 3(a - c) - \{ 5(2a - 3b) - [5a - 7(a - b)] \} & = 3a - 3c - [ 10a - 15b - (5a - 7a + 7b) ] \\ & = 3a - 3c - [ 10a - 15b - (-2a + 7b) ] \\ & = 3a - 3c - ( 10a - 15b + 2a - 7b ) \\ & = 3a - 3c - ( 12a - 22b ) \\ & = 3a - 3c - 12a + 22b \\ & = -9a + 22b - 3c \end{align*}

(a)

\begin{align*} {25 a^2 \over b^2 c} \times {bc^3 \over 100a^3} \div {15 \over c^2} & = {25 a^2 b c^3 \over 100 a^3 b^2 c} \times {c^2 \over 15} \\ & = {25 a^2 b c^5 \over 1500 a^3 b^2 c} \phantom{000000} [a^m \times a^n = a^{m + n} ] \\ & = { c^4 \over 60 a b} \end{align*}

(b)

\begin{align*} {8 a^5 b^2 c \over (-2ab)^2 } & = { 8a^5 b^2 c \over 4a^2 b^2 } \phantom{000000} [ (-2ab)^2 = (-2ab) \times (-2ab)] \\ & = { 2 a^3 c \over 1} \\ & = 2a^3 c \end{align*}

(c)

\begin{align*} {2a^2 b^3 \over 3b} \div { (2a)^2 \over 15ab^2 } & = { 2 a^2 b^2 \over 3} \times { 15ab^2 \over (2a)^2 } \\ & = { 2 a^2 b^2 \over 3} \times { 15 ab^2 \over 4a^2 } \phantom{000000} [ (2a)^2 = 2a \times 2a ] \\ & = { 30 a^3 b^4 \over 12 a^2 } \\ & = { 5 a b^4 \over 2} \end{align*}

(d)

\begin{align*} {a - 1 \over a - b} \div {1 - a \over a^2 - b^2 } & = { a - 1 \over a - b} \times {a^2 - b^2 \over 1 - a} \\ & = { (a - 1)(a^2 - b^2) \over (a - b)(1 - a)} \\ & = { - (1 - a)(a + b)(a - b) \over (a - b)(1 - a)} \\ & = -(a + b) \end{align*}

(e)

\begin{align*} {2x^2 + 11x + 15 \over x^2 - 9} & = { (2x + 5)(x + 3) \over x^2 - 3^2 } \\ & = { (2x + 5)(x + 3) \over (x + 3)(x - 3) } \phantom{000000} [ a^2 - b^2 = (a + b)(a - b)] \\ & = { 2x + 5 \over x - 3} \end{align*}

(a)

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

(b)

\begin{align*} ab - bc - ac + c^2 & = b(a - c) - c(a - c) \\ & = (a - c)(b - c) \end{align*}

(c)

\begin{align*} ax - kx - ah + kh & = ax - ah - kx + kh \\ & = a(x - h) - k (x - h) \\ & = (x - h)(a - k) \end{align*}

(d)

\begin{align*} 20 ac - 4ad - 15kc + 3kd & = 4a (5c - d) - 3k (5c - d) \\ & = (5c - d)(4a - 3k) \end{align*}

(e)

\begin{align*} 6a^2 + 3ab - 8ka - 4kb & = 3a(2a + b) - 4k(2a + b) \\ & = (2a + b)(3a - 4k) \end{align*}

(a)

\begin{align*} ax^2 + bx + c & = 0 \\ bx & = - ax^2 - c \\ b & = {-ax^2 - c \over x} \phantom{00000} [\text{Can stop here}] \\ b & = - {ax^2 + c \over x} \end{align*}

(b)

\begin{align*} {1 \over a} + {b \over 2} + {3 \over c} & = k \\ {2c \over 2ac} + { abc \over 2ac} + {6a \over 2ac} & = k \\ {2c + abc + 6a \over 2ac} & = {k \over 1} \\ 2c + abc + 6a & = 2ac k \phantom{000000} [\text{Cross-multiply}] \\ 2c + abc - 2ack & = - 6a \\ c(2 + ab - 2ak) & = - 6a \\ c & = {-6a \over 2 + ab - 2ak} \phantom{000000} [\text{Can stop here}] \\ c & = {6a \over -(2 + ab - 2ak)} \\ c & = {6a \over -2 -ab + 2ak} \end{align*}

(c)

\begin{align*} \sqrt{ 4x^2 - 5k } & = 2x + 3 \\ 4x^2 - 5k & = (2x + 3)^2 \\ 4x^2 - 5k & = \underbrace{ (2x)^2 + 2(2x)(3) + (3)^2 }_{ (a + b)^2 = a^2 + 2ab + b^2 } \\ 4x^2 - 5k & = 4x^2 + 12x + 9 \\ 4x^2 - 4x^2 - 12x & = 9 + 5k \\ -12x & = 9 + 5k \\ x & = {9 + 5k \over -12} \\ x & = -{9 + 5k \over 12} \end{align*}

(d)

\begin{align*} v^2 & = u^2 + 2as \\ - u^2 & = 2as - v^2 \\ u^2 & = v^2 - 2as \\ u & = \pm \sqrt{ v^2 - 2as} \end{align*}

(e)

\begin{align*} x & = \sqrt[3]{a \over b - a} \\ x^3 & = {a \over b - a} \\ {x^3 \over 1} & = {a \over b - a} \\ x^3 (b - a) & = a \phantom{000000} [\text{Cross-multiply}] \\ bx^3 - ax^3 & = a \\ -ax^3 - a & = - bx^3 \\ ax^3 + a & = bx^3 \\ a(x^3 + 1) & = bx^3 \\ a & = {bx^3 \over x^3 + 1} \end{align*}

(f)

\begin{align*} V & = {2\pi \over 3} (h^3 - d^3) \\ 3V & = 2\pi (h^3 - d^3) \\ {3V \over 2\pi} & = h^3 - d^3 \\ {3V \over 2\pi} + d^3 & = h^3 \\ \\ h & = \sqrt[3]{ {3V \over 2\pi} + d^3 } \phantom{000000} [\text{Can stop here}] \\ & = \sqrt[3]{ {3V \over 2\pi} + {2\pi d^3 \over 2\pi} } \\ & = \sqrt[3] { {3V + 2\pi d^3 \over 2\pi } } \end{align*}

(a)

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

(b)

\begin{align*} {2y + 3 \over 9y^2 - 1} - {5 \over 3y - 1} & = {2y + 3 \over (3y)^2 - (1)^2} - {5 \over 3y - 1} \\ & = {2y + 3 \over (3y + 1)(3y - 1)} - {5 \over 3y - 1} \phantom{000000} [ a^2 - b^2 = (a + b)(a - b)] \\ & = {2y + 3 \over (3y + 1)(3y - 1)} - {5(3y + 1) \over (3y + 1)(3y - 1)} \\ & = {2y + 3 - 5(3y + 1) \over (3y + 1)(3y - 1)} \\ & = {2y + 3 - 15y - 5 \over (3y + 1)(3y - 1)} \\ & = {-13y - 2 \over (3y + 1)(3y - 1)} \phantom{000000} [\text{Can stop here}] \\ & = - {13y + 2 \over (3y + 1)(3y - 1)} \end{align*}

(c)

\begin{align*} {3a \over a - 3} + {2 \over a + 4} & = {3a(a + 4) \over (a - 3)(a + 4)} + {2(a - 3) \over (a - 3)(a + 4)} \\ & = {3a(a + 4) + 2(a - 3) \over (a - 3)(a + 4)} \\ & = {3a^2 + 12a + 2a - 6 \over (a - 3)(a + 4)} \\ & = {3a^2 + 14a - 6 \over (a - 3)(a + 4)} \end{align*}

(d)

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

(e)

\begin{align*} {5x \over 2x - y} + {y \over 3x - y} & = {5x(3x - y) \over (2x - y)(3x - y)} + {y(2x - y) \over (2x - y)(3x - y)} \\ & = {5x(3x - y) + y(2x - y) \over (2x - y)(3x - y)} \\ & = {15x^2 - 5xy + 2xy - y^2 \over (2x -y)(3x - y)} \\ & = {15x^2 - 3xy - y^2 \over (2x - y)(3x - y)} \end{align*}

(a)

\begin{align*} x^2 - 4 & = x^2 - 2^2 \\ & = (x + 2)(x - 2) \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \\ \\ {3 \over x + 2} - {x - 5 \over (x + 2)(x - 2)} + {1 \over x - 2} & = {3(x - 2) \over (x + 2)(x - 2)} - {x - 5 \over (x + 2)(x - 2)} + {x + 2 \over (x + 2)(x - 2)} \\ & = {3(x - 2) - (x - 5) + x + 2 \over (x + 2)(x - 2)} \\ & = {3x - 6 - x + 5 + x + 2 \over (x + 2)(x - 2)} \\ & = {3x + 1 \over (x + 2)(x - 2)} \end{align*}

(b)

\begin{align*} x^2 - 1 & = -(1 - x^2) \\ & = -(1^2 - x^2) \\ & = -(1 + x)(1 - x) \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \\ \\ {1 \over 1 - x} + {2 \over 1 + x} + {2x \over -(1 + x)(1 - x)} & = {1 + x \over (1 + x)(1 - x)} + {2(1 - x) \over (1 + x)(1 - x)} - {2x \over (1 + x)(1 - x)} \\ & = {1 + x + 2(1 - x) - 2x \over (1 + x)(1 - x)} \\ & = {1 + x + 2 - 2x - 2x \over (1 + x)(1 - x)} \\ & = {3 - 3x \over (1 + x)(1 - x)} \\ & = {3(1 - x) \over (1 + x)(1 - x)} \\ & = {3 \over 1 + x} \end{align*}

(c)

\begin{align*} 2x^2 + x - 6 & = (2x - 3)(x + 2) \\ \\ {1 \over 2x - 3} - {2 \over x + 2} - {2x - x^2 \over (2x - 3)(x + 2)} & = {x + 2 \over (2x - 3)(x + 2)} - {2(2x - 3) \over (2x - 3)(x + 2)} - {2x - x^2 \over (2x - 3)(x + 2)} \\ & = {x + 2 - 2(2x - 3) - (2x - x^2) \over (2x - 3)(x + 2)} \\ & = {x + 2 - 4x + 6 - 2x + x^2 \over (2x - 3)(x + 2)} \\ & = {x^2 - 5x + 8 \over (2x - 3)(x + 2)} \end{align*}

(d)

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

(i)

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

(ii)

\begin{align*} b & = {2ac \over 3c - a} \\ & = {2 \left(2{1 \over 5}\right)(-3) \over 3(-3) - \left(2{1 \over 5}\right)} \\ & = {-{66 \over 5} \over -{56 \over 5}} \\ & = {33 \over 28} \\ & = 1 {5 \over 28} \end{align*}

(a)

\begin{align*} V & = {b^2 \over 2} (a + 3h) \\ 2V & = b^2 (a + 3h) \\ {2V \over b^2} & = a + 3h \\ -3h & = a - {2V \over b^2} \\ 3h & = {2V \over b^2} - a \\ h & = {1 \over 3} \left( {2V \over b^2} - a \right) \\ h & = {2V \over 3b^2} - {1 \over 3}a \end{align*}

(b)(i)

\begin{align*} V & = {b^2 \over 2} (a + 3h) \\ & = {(1.2)^2 \over 2} [ 0.2 + 3(1.5) ] \\ & = 3.384 \text{ m}^3 \end{align*}

(b)(ii)

\begin{align*} h & = {2V \over 3b^2} - {1 \over 3}a \\ & = {2(12) \over 3(0.4)^2} - {1 \over 3}(2.5) \\ & = 49 {1 \over 6} \text{ m} \end{align*}

(b)(iii)

\begin{align*} V & = {b^2 \over 2} (a + 3h) \\ 15 & = {b^2 \over 2} [0.25 + 3(2.4)] \\ 15 & = {b^2 \over 2} (7.45) \\ {15 \over 7.45} & = {b^2 \over 2} \\ 2(15) & = 7.45 b^2 \\ 30 & = 7.45b^2 \\ \\ b^2 & = {30 \over 7.45} \\ b & = \pm \sqrt{30 \over 7.45} \\ b & \approx \pm 2.01 \end{align*}

(i)

\begin{align*} {x^2 \over a^2} - {y^2 \over b^2} & = 1 \\ {b^2 x^2 \over a^2 b^2} - {a^2 y^2 \over a^2 b^2} & = 1 \\ {b^2 x^2 - a^2 y^2 \over a^2 b^2 } & = {1 \over 1} \\ b^2 x^2 - a^2 y^2 & = a^2 b^2 \phantom{000000} [\text{Cross-multiply}] \\ -a^2 y^2 & = a^2 b^2 - b^2 x^2 \\ a^2 y^2 & = b^2 x^2 - a^2 b^2 \\ y^2 & = {b^2 x^2 - a^2 b^2 \over a^2} \\ y & = \pm \sqrt{ b^2 x^2 - a^2 b^2 \over a^2 } \end{align*}

(ii)

\begin{align*} y & = \pm \sqrt{ b^2 x^2 - a^2 b^2 \over a^2 } \\ & = \pm \sqrt{ (2)^2 (-5)^2 - (3)^2 (2)^2 \over (3)^2 } \\ & = \pm \sqrt{ 64 \over 9 } \\ & = \pm 2 {2 \over 3} \end{align*}

(i)

\begin{align*} \text{Fraction of pool (A)} & = {1 \over x} \end{align*}

(ii)

\begin{align*} \text{Fraction of pool in two hours (B)} & = {1 \over x + 1.5} \times {2 \over 1} \\ & = {2 \over x + 1.5} \end{align*}

(iii)

\begin{align*} \text{Fraction of pool in one hour (A and B)} & = {1 \over x} + {1 \over x + 1.5} \end{align*}

(a)

\begin{align*} & {2 \over 12}, {5 \over 12}, {8 \over 12}, {11 \over 12}, \underline{ {14 \over 12} = {7 \over 6}, {17 \over 12} } \end{align*}

(b)

\begin{align*} 0, -1, -8, -27, -64, ... & = (0)^3, (-1)^3, (-2)^3, (-3)^3, (-4)^3, ... \\ & = (1 - 1)^3, (1 - 2)^3, (1 - 3)^3, (1 - 4)^3, (1 - 5)^3, ... \\ \\ T_n & = (1 - n)^3 \end{align*}

(a)

\begin{align*} \text{Common difference, } d & = {42 - 78 \over 4 } \\ & = -9 \\ \\ a & = 78 - 9 = 69 \\ b & = 69 - 9 = 60 \\ c & = 60 - 9 = 51 \end{align*}

(b)

\begin{align*} T_n & = T_1 + (n - 1)d \phantom{000000} [T_1: \text{ first term, } d: \text{ common difference}] \\ & = 78 + (n - 1)(-9) \\ & = 78 - 9(n - 1) \\ & = 78 - 9n + 9 \\ & = 87 - 9n \end{align*}

(c)

\begin{align*} T_n & < 0 \\ 87 - 9n & < 0 \\ -9n & < -87 \\ n & > {-87 \over -9} \\ n & > 9 {2 \over 3} \\ \\ \text{First } n & = 10 \\ \\ T_{10} & = 87 - 9(10) \\ & = -3 \end{align*}

(i)

\begin{align*} u_5 & = 7^2 - 5 + 3 = 47 \\ \\ u_6 & = 8^2 - 6 + 3 = 61 \end{align*}

(ii)

\begin{align*} u_n & = (n + 2)^2 - n + 3 \end{align*}

(iii)

\begin{align*} u_n & = (n + 2)^2 - n + 3 \\ & = \underbrace{ (n)^2 + 2(n)(2) + (2)^2 }_{ (a + b)^2 = a^2 + 2ab + b^2 } - n + 3 \\ & = n^2 + 4n + 4 - n + 3 \\ & = n^2 + 3n + 7 \\ \\ \text{Let } u_k & = 161 \\ k^2 + 3k + 7 & = 161 \\ k^2 + 3k + 7 - 161 & = 0 \\ k^2 + 3k - 154 & = 0 \\ (k - 11)(k + 14) & = 0 \end{align*} \begin{align*} k - 11 & = 0 && \text{ or } & k + 14 & = 0 \\ k & = 11 &&& k & = -14 \text{ (Reject, since } k > 0) \end{align*}

(iv)

\begin{align*} u_p & = p^2 + 3p + 7 \\ \\ u_{p + 1} & = (p + 1)^2 + 3(p + 1) + 7 \\ & = \underbrace{ (p)^2 + 2(p)(1) + (1)^2 }_{ (a + b)^2 = a^2 + 2ab + b^2 } + 3p + 3 + 7 \\ & = p^2 + 2p + 1 + 3p + 3 + 7 \\ & = p^2 + 5p + 11 \\ \\ u_{p + 1} - u_p & = p^2 + 5p + 11 - (p^2 + 3p + 7) \\ & = p^2 + 5p + 11 - p^2 - 3p - 7 \\ & = 2p + 4 \end{align*}

(v)

\begin{align*} \text{From (iv), } u_{p + 1} - u_p & = 2p + 4 \\ & = 2(p + 2) \phantom{000000} [\text{Multiple of 2, so always even}] \\ \\ \text{Since } u_{p + 1} - u_p \text{ is always even } & \text{and } 3 \text{ is odd,} \\ \text{two consecutive terms } & \text{cannot have a difference of } 3 \end{align*}

(a)

\begin{align*} \text{Total no. of children} & = x + 1 + y + 2 \\ & = x + y + 3 \\ \\ \text{Total age of children} & = q(x + y + 3) \\ & = qx + qy + 3q \\ \\ \text{Total age of boys} & = p(x + 1) \\ & = px + p \\ \\ \text{Total age of girls} & = qx + qy + 3q - (px + p) \\ & = qx + qy + 3q - px - p \\ \\ \text{Average age of girls} & = { qx + qy + 3q - px - p \over y + 2} \end{align*}

(b)

\begin{align*} \text{Sum of ages in 2 years' time} & = x - 3 - 3 \phantom{000000} [\text{Minus 3 for each person}] \\ & = x - 6 \\ \\ \text{Vani's age in 2 years' time} & = {x - 6 \over 3} \phantom{000000000} [\text{Vani: 1 unit, Siti: 2 units}] \\ \\ \text{Siti's age in 2 years' time} & = {x - 6 \over 3} \times 2 \\ & = {2(x - 6) \over 3} \\ \\ \text{Siti's present age} & = {2(x - 6) \over 3} - 2 \\ & = {2x - 12 \over 3} - {2 \over 1} \\ & = {2x - 12 \over 3} - {6 \over 3} \\ & = {2x - 18 \over 3} \end{align*}

(i)

\begin{align*} S_3 & = (1 + 2 + 3)^2 = 36 \\ \\ S_4 & = (1 + 2 + 3 + 4)^2 = 100 \\ \\ S_5 & = (1 + 2 + 3 + 4 + 5)^2 = 225 \end{align*}

(ii)

\begin{align*} S_1 & = \left[ {1 \over 2}(1)(2) \right]^2 = 1^2 \\ S_2 & = \left[ {1 \over 2}(2)(3) \right]^2 = 3^2 \\ S_3 & = \left[ {1 \over 2}(3)(4) \right]^2 = 6^2 \\ \\ S_n & = \left[ {1 \over 2}n(n + 1) \right]^2 \end{align*}

(iii)

\begin{align*} S_k & = \left[ {1 \over 2}k(k + 1) \right]^2 \\ 44 \phantom{.} 100 & = \left[ {1 \over 2}k(k + 1) \right]^2 \\ \sqrt{ 44 \phantom{.} 100 } & = {1 \over 2}k (k + 1) \\ 210 & = {1 \over 2}k(k + 1) \\ 210 & = {1 \over 2}k^2 + {1 \over 2}k \\ 420 & = k^2 + k \\ 0 & = k^2 + k - 420 \\ 0 & = (k - 20)(k + 21) \end{align*} \begin{align*} k - 20 & = 0 && \text{ or } & k + 21 & = 0 \\ k & = 20 &&& k & = -21 \text{ (Reject, since } k > 0) \end{align*}

(a)

\begin{align*} a & = 8 \phantom{0000000000} [\text{Cubes at the corners}] \\ \\ b & = 12 \times (4 - 2) = 24 \phantom{000000} [\text{Cubes along the 12 edges, not corners}] \\ \\ c & = 6 \times (4 - 2)^2 = 24 \phantom{000000} [\text{Cubes in the centre of the 6 faces}] \\ \\ d & = 2^3 = 8 \phantom{000000} [\text{'Inner' cube after removing outside layer}] \end{align*}

(b)

$$ \text{No. of cubes} = 8 \phantom{0000000000} [\text{Cubes at the corners}] $$

(c)

\begin{align*} \text{No. of cubes not painted} & = 3^3 = 27 \end{align*}

(d)(i)

\begin{align*} \text{Pattern for 2 faces painted} & = \underbrace{0}_{n = 2}, 12, 24, ... \\ \\ T_n & = a + (n - 1)d \phantom{000000} [a \text{: first term, } d \text{: difference}] \\ & = -12 + (n - 1)(12) \phantom{000000} [\text{Extend pattern back to } n = 1: 0 - 12 = -12] \\ & = -12 + 12(n - 1) \\ & = -12 + 12n - 12 \\ & = 12n - 24 \\ & = 12(n - 2) \end{align*}

(d)(ii)

\begin{align*} \text{Pattern for 1 face painted} & = \underbrace{0}_{n = 2}, 6, 24, ... \\ & = 6(0)^2, 6(1)^2, 6(2)^2, ... \\ & = 6(2 - 2)^2, 6(3 - 2)^2, 6(4 - 2)^2, ... \\ \\ T_n & = 6(n - 2)^2 \end{align*}

(d)(iii)

\begin{align*} \text{Pattern for no faces painted} & = \underbrace{0}_{n = 2}, 1, 8, ... \\ & = 0^3, 1^3, 2^3, ... \\ & = (2 - 2)^3, (3 - 2)^3, (4 - 2)^3, ... \\ \\ T_n & = (n - 2)^3 \end{align*}

(i)

\begin{align*} l & = 40 + 20 = 60 \\ \\ m & = 5^2 = 25 \\ \\ n & = (5 + 1)^2 = 36 \end{align*}

(ii)

\begin{align*} T & = S + P - 1 \end{align*}

(iii)

\begin{align*} T & = S + P - 1 \\ 364 & = 169 + P - 1 \\ 364 - 169 + 1 & = P \\ 196 & = P \end{align*}

(iv)

\begin{align*} S & = n^2 \text{ and } P = (n + 1)^2 \phantom{000000} [\text{Both columns contain only perfect squares}] \\ \\ \sqrt{112} & \approx 10.583 \\ \\ 112 & \text{ is not a perfect square, so it cannot appear in either column} \end{align*}

(v)

\begin{align*} T & = S + P - 1 \\ & = n^2 + (n + 1)^2 - 1 \\ & = n^2 + (n)^2 + 2(n)(1) + (1)^2 - 1 \\ & = n^2 + n^2 + 2n + 1 - 1 \\ & = 2n^2 + 2n \\ \\ \text{Let } 4442 & = 2n^2 + 2n \\ 0 & = 2n^2 + 2n - 4442 \\ 0 & = n^2 + n - 2221 \\ \\ n & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-1 \pm \sqrt{(1)^2 - 4(1)(-2221)} \over 2(1)} \\ & = {-1 \pm \sqrt{8885} \over 2} \\ & \approx 46.63 \text{ or } -47.63 \\ \\ \text{Since } n \text{ is not an} & \text{ integer, }4442 \text{ cannot appear in the } T \text{ column} \end{align*}

(a)

\begin{align*} p & = 4, q = 5 \\ \\ \text{As no. of sides increases by } & 1, \text{ no. of diagonals from } V \text{ increases by 1} \end{align*}

(b)

\begin{align*} \text{Differences in } d & : 2, \phantom{.} 3, \phantom{.} 4, \phantom{.} 5, \phantom{.} 6 \\ \\ r & = 9 + 5 = 14 \\ \\ s & = 14 + 6 = 20 \end{align*}

(c)

\begin{align*} v & = n - 3 \end{align*}

(d)(i)

\begin{align*} 0 & = {0 \times 3 \over 2} \\ 2 & = {1 \times 4 \over 2} \\ 5 & = {2 \times 5 \over 2} \\ 9 & = {3 \times 6 \over 2} \\ \\ d & = {vn \over 2} \end{align*}

(d)(ii)

\begin{align*} d & = {vn \over 2} \\ \\ \text{Since } & v = n - 3, \\ d & = { (n - 3)n \over 2} \\ d & = { n(n - 3) \over 2 } \end{align*}

(d)(iii)

\begin{align*} d & = { n(n - 3) \over 2 } \\ & = {30(30 - 3) \over 2} \\ & = 405 \end{align*}

(i)

\begin{align*} p & = {256 \over 2} = 128 \end{align*}

(ii)

\begin{align*} 1024 & = 2^{10} = 2^{11 - 1} \\ 512 & = 2^{9} = 2^{11 - 2} \\ 256 & = 2^8 = 2^{11 - 3} \\ \\ T_n & = 2^{11 - n} \end{align*}

(iii)

\begin{align*} T_k & = 2^{11 - k} \\ {1 \over 4} & = 2^{11 - k} \\ {1 \over 2^2} & = 2^{11 - k} \\ 2^{-2} & = 2^{11 - k} \phantom{000000} \left[ a^{-n} = {1 \over a^n} \right] \\ \\ \therefore -2 & = 11 - k \\ k & = 11 + 2 \\ k & = 13 \end{align*}

(i)

$$ x = 7 $$

(ii)

$$ x = - 2 $$

(iii)

$$ x = 7 $$

(a)

\begin{align*} 9x - 7 & \le 12 \\ 9x & \le 12 + 7 \\ 9x & \le 19 \\ x & \le {19 \over 9} \\ x & \le 2{1 \over 9} \end{align*}

(b)

\begin{align*} 7 - 2x & > 2 \\ -2x & > 2 - 7 \\ -2x & > -5 \\ x & < {-5 \over -2} \\ x & < 2.5 \end{align*}

(c)

\begin{align*} 3 + 5x & \ge 32 \\ 5x & \ge 32 - 3 \\ 5x & \ge 29 \\ x & \ge {29 \over 5} \\ x & \ge 5.8 \end{align*}

(d)

\begin{align*} 3x - 4 & \ge {1 \over 3}x - 2 \\ 3x - {1 \over 3}x & \ge -2 + 4 \\ 2{2 \over 3}x & \ge 2 \\ x & \ge 2 \div 2{2 \over 3} \\ x & \ge 0.75 \end{align*}

(e)

\begin{align*} 12 & < 3x - 1 &&& 3x - 1 & < 27 \\ -3x & < -1 - 12 &&& 3x & < 27 + 1 \\ -3x & < -13 &&& 3x & < 28 \\ x & > {-13 \over -3} &&& x & < {28 \over 3} \\ x & > 4{1 \over 3} &&& x & < 9{1 \over 3} \end{align*}
$$ 4{1 \over 3} < x < 9{1 \over 3} $$

(a)

\begin{align*} x + 2y & = 8 \phantom{0} \text{--- (1)} \\ \\ 3x + 2y & = 12 \phantom{0} \text{--- (2)} \\ \\ (1) & - (2), \\ x + 2y - (3x + 2y) & = 8 - 12 \\ x + 2y - 3x - 2y & = -4 \\ -2x & = -4 \\ x & = {-4 \over -2} \\ x & = 2 \\ \\ \text{Substitute } & x = 2 \text{ into (1),} \\ 2 + 2y & = 8 \\ 2y & = 8 - 2 \\ 2y & = 6 \\ y & = {6 \over 2} \\ y & = 3 \\ \\ \therefore x & = 2, y = 3 \end{align*}

(b)

\begin{align*} x + y & = {5 \over 6} \phantom{0} \text{--- (1)} \\ \\ x - y & = {1 \over 6} \phantom{0} \text{--- (2)} \\ \\ (1) & - (2), \\ x + y - (x - y) & = {5 \over 6} - {1 \over 6} \\ x + y - x + y & = {2 \over 3} \\ 2y & = {2 \over 3} \\ y & = {2 \over 3} \div 2 \\ y & = {1 \over 3} \\ \\ \text{Substitute } & y = {1 \over 3} \text{ into (1),} \\ x + {1 \over 3} & = {5 \over 6} \\ x & = {5 \over 6} - {1 \over 3} \\ x & = {1 \over 2} \\ \\ \therefore x & = {1 \over 2}, y = {1 \over 3} \end{align*}

(c)

\begin{align*} 3a - 2b & = 1 \phantom{0} \text{--- (1)} \\ \\ 5a + 3b & = -11 \\ 5a & = -3b - 11 \\ a & = {1 \over 5}(-3b - 11) \\ a & = -{3 \over 5}b - {11 \over 5} \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 3 \left(-{3 \over 5}b - {11 \over 5}\right) - 2b & = 1 \\ -{9 \over 5}b - {33 \over 5} - 2b & = 1 \\ -{9 \over 5}b - 2b & = 1 + {33 \over 5} \\ -3.8b & = 7.6 \\ b & = {7.6 \over -3.8} \\ b & = -2 \\ \\ \text{Substitute } & b = -2 \text{ into (2),} \\ a & = -{3 \over 5}(-2) - {11 \over 5} \\ a & = -1 \\ \\ \therefore a & = -1, b = -2 \end{align*}

(d)

\begin{align*} 3p - 4q - 24 & = 0 \phantom{0} \text{--- (1)} \\ \\ 5p - 6q - 38 & = 0 \\ 5p & = 6q + 38 \\ p & = {1 \over 5}(6q + 38) \\ p & = 1.2q + 7.6 \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 3(1.2q + 7.6) - 4q - 24 & = 0 \\ 3.6q + 22.8 - 4q - 24 & = 0 \\ 3.6q - 4q & = -22.8 + 24 \\ -0.4q & = 1.2 \\ q & = {1.2 \over -0.4} \\ q & = -3 \\ \\ \text{Substitute } & q = -3 \text{ into (2),} \\ p & = 1.2(-3) + 7.6 \\ p & = 4 \\ \\ \therefore p & = 4, q = -3 \end{align*}

(e)

\begin{align*} {1 \over 4}x + {3 \over 5}y & = -4 \phantom{0} \text{--- (1)} \\ \\ {1 \over 5}x + {1 \over 4}y & = -{9 \over 10} \\ 5 \left({1 \over 5}x + {1 \over 4}y\right) & = 5 \left(-{9 \over 10}\right) \\ x + 1.25y & = -4.5 \\ x & = -1.25y - 4.5 \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ {1 \over 4}(-1.25y - 4.5) + {3 \over 5}y & = -4 \\ -0.3125y - 1.125 + 0.6y & = -4 \\ -0.3125y + 0.6y & = -4 + 1.125 \\ 0.2875y & = -2.875 \\ y & = {-2.875 \over 0.2875} \\ y & = -10 \\ \\ \text{Substitute } & y = -10 \text{ into (2),} \\ x & = -1.25 (-10) - 4.5 \\ x & = 8 \\ \\ \therefore x & = 8, y = -10 \end{align*}

(a)

\begin{align*} 3x - 2 & = 0 && \text{ or } & 2x + 7 & = 0 \\ 3x & = 2 &&& 2x & = -7 \\ x & = {2 \over 3} &&& x & = -{7 \over 2} \end{align*}

(b)

\begin{align*} 2x^2 + 5x - 12 & = 0 \\ (2x - 3)(x + 4) & = 0 \end{align*} \begin{align*} 2x - 3 & = 0 && \text{ or } & x + 4 & = 0 \\ 2x & = 3 &&& x & = -4 \\ x & = {3 \over 2} \end{align*}

(c)

\begin{align*} 4x^2 - 3x & = 7 \\ 4x^2 - 3x - 7 & = 0 \\ (4x - 7)(x + 1) & = 0 \end{align*} \begin{align*} 4x - 7 & = 0 && \text{ or } & x + 1 & = 0 \\ 4x & = 7 &&& x & = -1 \\ x & = {7 \over 4} \end{align*}

(d)

\begin{align*} 7x^2 & + 15x - 3 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-15 \pm \sqrt{(15)^2 - 4(7)(-3)} \over 2(7)} \\ & = {-15 \pm \sqrt{309} \over 14} \\ & \approx 0.1842 \text{ or } -2.327 \\ & \approx 0.184 \text{ or } -2.33 \end{align*}

(e)

\begin{align*} (x - 5)(x + 3) & = 7 \\ x^2 + 3x - 5x - 15 & = 7 \\ x^2 - 2x - 15 & = 7 \\ x^2 - 2x - 15 - 7 & = 0 \\ x^2 - 2x - 22 & = 0 \end{align*}
\begin{align*} x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-2) \pm \sqrt{(-2)^2 - 4(1)(-22)} \over 2(1)} \\ & = {2 \pm \sqrt{92} \over 2} \\ & \approx 5.796 \text{ or } -3.796 \\ & \approx 5.80 \text{ or } -3.80 \end{align*}

(a)

\begin{align*} 3x(x + 4) + 28(x + 2) + 21 & = 0 \\ 3x^2 + 12x + 28x + 56 + 21 & = 0 \\ 3x^2 + 40x + 77 & = 0 \\ (3x + 7)(x + 11) & = 0 \end{align*} \begin{align*} 3x + 7 & = 0 && \text{ or } & x + 11 & = 0 \\ 3x & = -7 &&& x & = -11 \\ x & = -{7 \over 3} \end{align*}

(b)

\begin{align*} 2(4x^2 + 23x) & = 105 \\ 8x^2 + 46x & = 105 \\ 8x^2 + 46x - 105 & = 0 \\ (2x + 15)(4x - 7) & = 0 \end{align*} \begin{align*} 2x + 15 & = 0 && \text{ or } & 4x - 7 & = 0 \\ 2x & = - 15 &&& 4x & = 7 \\ x & = -{15 \over 2} &&& x & = {7 \over 4} \end{align*}

(c)

\begin{align*} (x - 1)^2 - 16 & = 0 \\ (x - 1)^2 & = 16 \\ x - 1 & = \pm \sqrt{16} \\ x - 1 & = \pm 4 \end{align*} \begin{align*} x - 1 & = 4 && \text{ or } & x - 1 & =- 4 \\ x & = 4 + 1 &&& x & = -4 + 1 \\ x & = 5 &&& x & = -3 \end{align*}

(d)

\begin{align*} {5 \over x - 1} & = {3 \over 1} + {4 \over x} \\ {5 \over x - 1} & = {3x \over x} + {4 \over x} \\ {5 \over x - 1} & = {3x + 4 \over x} \\ {5x \over x(x - 1)} & = { (3x + 4)(x - 1) \over x(x - 1)} \\ 5x & = (3x + 4)(x - 1) \\ 5x & = 3x^2 - 3x + 4x - 4 \\ 5x & = 3x^2 + x - 4 \\ 0 & = 3x^2 + x - 5x - 4 \\ 0 & = 3x^2 - 4x - 4 \\ 0 & = (3x + 2)(x - 2) \end{align*} \begin{align*} 3x + 2 & = 0 && \text{ or } & x - 2 & = 0 \\ 3x & = -2 &&& x & = 2 \\ x & = -{2 \over 3} \end{align*}

(e)

\begin{align*} {x \over 1 + x} + {x + 1 \over 1 - 3x} & = {1 \over 4} \\ {4x(1 - 3x) \over 4(1 + x)(1 - 3x)} + { 4(x + 1)(1 + x) \over 4(1 + x)(1 - 3x)} & = {(1 + x)(1 - 3x) \over 4(1 + x)(1 - 3x)} \\ 4x(1 - 3x) + 4(x + 1)^2 & = (1 + x)(1 - 3x) \\ 4x - 12x^2 + 4 \underbrace{ [ (x)^2 + 2(x)(1) + (1)^2 ] }_{ (a + b)^2 = a^2 + 2ab + b^2 } & = 1 - 3x + x - 3x^2 \\ 4x - 12x^2 + 4(x^2 + 2x + 1) & = 1 - 2x - 3x^2 \\ 4x - 12x^2 + 4x^2 + 8x + 4 & = 1 - 2x - 3x^2 \\ - 8x^2 + 12x + 4 & = 1 - 2x - 3x^2 \\ -8x^2 + 3x^2 + 12x + 2x + 4 - 1 & = 0 \\ -5x^2 + 14x + 3 & = 0 \\ 5x^2 - 14x - 3 & = 0 \\ (5x + 1)(x - 3) & = 0 \end{align*} \begin{align*} 5x + 1 & = 0 && \text{ or } & x - 3 & = 0 \\ 5x & = -1 &&& x & = 3 \\ x & = -{1 \over 5} \end{align*}

(f)

\begin{align*} {2 \over 3x} & = {x - 2 \over 4(x + 3)} \\ { 2[4(x + 3)] \over 12x(x + 3)} & = { 3x(x - 2) \over 12x(x + 3)} \\ 2[4(x + 3)] & = 3x(x - 2) \\ 8(x + 3) & = 3x^2 - 6x \\ 8x + 24 & = 3x^2 - 6x \\ 0 & = 3x^2 - 6x - 8x - 24 \\ 0 & = 3x^2 - 14x - 24 \\ 0 & = (3x + 4)(x - 6) \end{align*} \begin{align*} 3x + 4 & = 0 && \text{ or } & x - 6 & = 0 \\ 3x & = -4 &&& x & = 6 \\ x & = -{4 \over 3} \end{align*}

(g)

\begin{align*} {5 \over x + 2} - {5 \over x^2 -4} & = 0 \\ {5 \over x + 2} & = {5 \over x^2 - 2^2} \\ {5 \over x + 2} & = {5 \over (x + 2)(x - 2)} \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \\ {5(x - 2) \over (x + 2)(x - 2)} & = {5 \over (x + 2)(x - 2)} \\ 5(x - 2) & = 5 \\ 5x - 10 & = 5 \\ 5x & = 5 + 10 \\ 5x & = 15 \\ x & = {15 \over 5} \\ x & = 3 \phantom{000000} [x \neq \pm 2, \text{ so } x = 3 \text{ is valid}] \end{align*}

(i)

\begin{align*} 2x^2 + qx - 2 & = 0 \\ \\ \text{Let } & x = -2, \\ 2(-2)^2 + q(-2) - 2 & = 0 \\ 8 - 2q - 2 & = 0 \\ -2q & = -8 + 2 \\ -2q & = -6 \\ q & = {-6 \over -2} \\ q & = 3 \end{align*}

(ii)

\begin{align*} 2x^2 + qx - 2 & = 0 \\ 2x^2 + 3x - 2 & = 0 \\ (2x - 1)(x + 2) & = 0 \end{align*} \begin{align*} 2x - 1 & = 0 && \text{ or } & x + 2 & = 0 \\ 2x & = 1 &&& x & =- 2 \\ x & = {1 \over 2} \end{align*}
$$ \text{Other solution is } x = {1 \over 2} $$

(a)

\begin{align*} \text{Smallest possible length} & = 8.5 \text{ cm} \phantom{000000} [8.5 \approx 9] \\ \\ \text{Smallest possible breadth} & = 5.5 \text{ cm} \phantom{000000} [5.5 \approx 6] \\ \\ \text{Smallest perimeter} & = 2(8.5) + 2(5.5) \\ & = 28 \text{ cm} \end{align*}

(b)

\begin{align*} \text{Largest possible length} & = 9.5 \text{ cm} \phantom{000000} [\text{Upper bound; } 9.4999... \approx 9] \\ \\ \text{Largest possible breadth} & = 6.5 \text{ cm} \phantom{000000} [\text{Upper bound; } 6.4999... \approx 6] \\ \\ \text{Largest possible area} & = 9.5 \times 6.5 \\ & = 61.75 \text{ cm}^2 \end{align*}

(a)

\begin{align*} -5 & < 12 - 3a &&& 12 - 3a & < -1 \\ 3a & < 12 + 5 &&& -3a & < -1 -12 \\ 3a & < 17 &&& -3a & < -13 \\ a & < {17 \over 3} &&& a & > {-13 \over -3} \\ a & < 5{2 \over 3} &&& a & > 4{1 \over 3} \end{align*} \begin{align*} 4{1 \over 3} & < a < 5{2 \over 3} \\ \\ a & = 5 \end{align*}

(b)

\begin{align*} b - 5 & \le 7 &&& 3b - 2 & \ge 11 \\ b & \le 7 + 5 &&& 3b & \ge 11 + 2 \\ b & \le 12 &&& 3b & \ge 13 \\ & &&& b & \ge {13 \over 3} \\ & &&& b & \ge 4{1 \over 3} \end{align*} \begin{align*} 4{1 \over 3} & \le b \le 12 \\ \\ \text{Odd integers: } & 5, 7, 9, 11 \end{align*}

(c)

\begin{align*} {2 \over 7} & < 2x - 1 &&& {2x - 1 \over 1} & < {9 + 2x \over 12} \\ -2x & < -1 - {2 \over 7} &&& {12(2x - 1) \over 12} & < {9 + 2x \over 12} \\ -2x & < -{9 \over 7} &&& 12(2x - 1) & < 9 + 2x \\ x & > -{9 \over 7} \div -2 &&& 24x - 12 & < 9 + 2x \\ x & > {9 \over 14} &&& 24x - 2x & <9 + 12 \\ & &&& 22x & < 21 \\ & &&& x & < {21 \over 22} \end{align*}
$$ {9 \over 14} < x < {21 \over 22} $$

(a)

\begin{align*} -5 & \le 4x + 1 &&& 4x + 1 & \le 2x + 9 \\ -4x & \le 1 + 5 &&& 4x - 2x & \le 9 - 1 \\ -4x & \le 6 &&& 2x & \le 8 \\ x & \ge {6 \over -4} &&& x & \le 4 \\ x & \ge -1.5 \end{align*} $$ -1.5 \le x \le 4 $$
\begin{align*} -6 & \le 2y - 2 &&& 2y - 2 & \le 8 \\ -2y & \le -2 + 6 &&& 2y & \le 8 + 2 \\ -2y & \le 4 &&& 2y & \le 10 \\ y & \ge {4 \over -2} &&& y & \le 5 \\ y & \ge -2 \end{align*} $$ -2 \le y \le 5 $$
\begin{align*} \text{Greatest value of } x - y & = 4 - (-2) \\ & = 6 \end{align*}

(b)

\begin{align*} (x + y)(x - y) & = x^2 - y^2 \\ \\ \text{Smallest } x^2 & = 0 \phantom{000000} [x = 0 \text{ is in the range}] \\ \text{Largest } y^2 & = 5^2 = 25 \phantom{000000} [-2 \le y \le 5] \\ \\ \text{Smallest value of } x^2 - y^2 & = (0)^2 - (5)^2 \\ & = -25 \end{align*}

(i)

\begin{align*} \text{Let } & d = 40, \\ 40 & = 35 + 7t - 2t^2 \\ 2t^2 - 7t + 40 - 35 & = 0 \\ 2t^2 - 7t + 5 & = 0 \\ (2t - 5)(t - 1) & = 0 \end{align*} \begin{align*} 2t - 5 & = 0 && \text{ or } & t - 1 & =0 \\ 2t & = 5 &&& t & = 1 \text{ second} \\ t & = {5 \over 2} \\ t & = 2.5 \text{ seconds} \end{align*}

(ii)

\begin{align*} \text{Let } & d = 0, \\ 0 & = 35 + 7t - 2t^2 \\ 2t^2 - 7t - 35 & = 0 \\ \\ t & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-7) \pm \sqrt{ (-7)^2 - 4(2)(-35)} \over 2(2)} \\ & = {7 \pm \sqrt{ 329 } \over 4 } \\ & \approx 6.2845 \text{ or } -2.7845 \text{ (NA, since } t \ge 0) \\ & \approx 6.28 \text{ seconds} \end{align*}

(i)

\begin{align*} \text{Greatest value of } 2x - y & = 2(5) - (-1) \\ & = 11 \end{align*}

(ii)

\begin{align*} \text{Least possible value of } 2xy & = 2(5)(-1) \\ & = -10 \end{align*}

(iii)

\begin{align*} \text{Greatest possible value of } {y \over x} & = { 7 \over 1 } \\ & = 7 \end{align*}

\begin{align*} {x \over x - y} - {y \over x - y} & = {x - y \over x - y} \\ & = 1 \\ \\ \text{Since } {x \over x - y} - {y \over x - y} & = 1 > 0, \\ \therefore {x \over x - y } & \text{ has a greater value} \end{align*}

\begin{align*} \text{Let no. of correct answers} & = x \\ \\ \text{No. of incorrect answers} & = 21 - x \\ \\ \text{Total marks} & = (2)(x) - (1)(21 - x) \\ & = 2x - 21 + x \\ & = 3x - 21 \\ \\ \text{Total marks} & \le 33 \phantom{000000} [\text{Not more than 33}] \\ 3x - 21 & \le 33 \\ 3x & \le 33 + 21 \\ 3x & \le 54 \\ x & \le 18 \\ \\ \text{Max. no. of correct answers} & = 18 \end{align*}

(i)

\begin{align*} \text{Length} & = (x + 3.4) \text{ cm} \\ \\ \text{Area} & = (x)(x + 3.4) \\ & = ( x^2 + 3.4x ) \text{ cm}^2 \\ \\ 125 & = x^2 + 3.4x \\ 0 & = x^2 + 3.4x - 125 \end{align*}

(ii)

\begin{align*} 0 & = x^2 + 3.4x - 125 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-3.4 \pm \sqrt{(3.4)^2 - 4(1)(-125)} \over 2(1)} \\ & = {-3.4 \pm \sqrt{511.56} \over 2} \\ & \approx 9.6088 \text{ or } -13.009 \text{ (NA, since } x > 0) \\ \\ \text{Length} & = 9.6088 + 3.4 \\ & = 13.0088 \\ & \approx 13.01 \text{ cm (2 d.p.)} \end{align*}

\begin{align*} 9 \text{ minutes 10 seconds} & = (9 \times 60) + 10 \\ & = 550 \text{ seconds} \\ \\ 2.4 \text{ km} & = 2.4 \times 1000 \\ & = 2400 \text{ m} \\ \\ \text{Let time taken for } A & = x \text{ seconds} \\ \\ \text{Distance for } A & = \text{Speed} \times \text{Time} \\ & = 4 \times x \\ & = 4x \text{ m} \\ \\ \text{Time taken for } B & = (550 - x) \text{ seconds} \\ \\ \text{Distance for } B & = 5 \times (550 - x) \\ & = ( 2750 - 5x ) \text{ m} \\ \\ \text{Total distance} & = 4x + 2750 - 5x \\ 2400 & = 2750 - x \\ x & = 2750 - 2400 \\ x & = 350 \end{align*} \begin{align*} \text{Distance for } A & = 4(350) &&& \text{Distance for } B & = 2750 - 5(350) \\ & = 1400 \text{ m} &&& & = 1000 \text{ m} \\ & = 1.4 \text{ km} &&& & = 1 \text{ km} \end{align*}

\begin{align*} \text{Let first digit} & = x \\ \\ \text{Second digit} & = 11 - x \\ \\ \text{Value of original number} & = 10x + 11 - x \phantom{000000.} [\text{E.g. } 23 = 20 + 3] \\ & = 9x + 11 \\ \\ \text{Value of reversed number} & = 10(11 - x) + x \phantom{00000}[\text{E.g. } 32 = 30 + 2] \\ & = 110 - 10x + x \\ & = 110 - 9x \\ \\ \text{Original number} - \text{Reversed number} & = 27 \\ 9x + 11 - (110 - 9x) & = 27 \\ 9x + 11 - 110 + 9x & = 27 \\ 9x + 9x & = 27 - 11 + 110 \\ 18x & = 126 \\ x & = {126 \over 18} \\ x & = 7 \\ \\ \text{First digit} = x = 7, & \text{ Second digit} = 11 - 7 = 4 \\ \\ \text{Original number} & = 74 \end{align*}

\begin{align*} \text{Let no. of postcards Cheryl has} & = x \\ \\ \text{No. of postcards Bernard has} & = 4x \\ \\ \text{No. of postcards Albert has} & = 2.5(4x) \\ & = 10x \\ \\ x + 4x + 10x & = 180 \\ 15x & = 180 \\ x & = {180 \over 15} \\ x & = 12 \\ \\ C: 12, B: 4(12) = 48, & \phantom{.} A: 10(12) = 120 \end{align*}

\begin{align*} \text{Let Yihao's initial amount} & = 3x \\ \\ \text{Raju's initial amount} & = 5x \\ \\ { \text{Yihao's current amount} \over \text{Raju's current amount} } & = {3x + 30 \over 5x - 30} \\ {9 \over 11} & = {3x + 30 \over 5x - 30} \\ {9(5x - 30) \over 11(5x - 30)} & = {11(3x + 30) \over 11(5x - 30)} \\ 9(5x - 30) & = 11(3x + 30) \\ 45x - 270 & = 33x + 330 \\ 45x - 33x & = 330 + 270 \\ 12x & = 600 \\ x & = {600 \over 12} \\ x & = 50 \\ \\ \text{Yihao's current amount} & = 3(50) + 30 \\ & = \$ 180 \end{align*}

(i)

\begin{align*} \text{Distance} & = \text{Speed} \times \text{Time} \\ \\ \text{Distance between B and D} & = (x)(2) + (x - 3)(1.5) \\ & = 2x + 1.5x - 4.5 \\ & = ( 3.5x - 4.5 ) \text{ km} \end{align*}

(ii)

\begin{align*} \text{Distance between B and D} & = (x + 1)(3) \\ & = (3x + 3) \text{ km} \end{align*}

(iii)

\begin{align*} 3.5x - 4.5 & = 3x + 3 \\ 3.5x - 3x & = 3 + 4.5 \\ 0.5x & = 7.5 \\ x & = {7.5 \over 0.5} \\ x & = 15 \\ \\ \text{Distance} & = 3(15) + 3 \\ & = 48 \text{ km} \end{align*}

(iv)

\begin{align*} \text{Average speed} & = { \text{Total distance} \over \text{Total time} } \\ & = { 2(48) \over 2 + 1.5 + 0.5 + 3 } \phantom{000000} [\text{Include half an hour rest time}] \\ & = 13{5 \over 7} \text{ km/h} \end{align*}

(i)

\begin{align*} \text{Time taken} & = { \text{Distance} \over \text{Speed} } \\ & = {200 \over x} \text{ h} \end{align*}

(ii)

\begin{align*} \text{Time taken} & = { 200 \over x + 5 } \text{ h} \end{align*}

(iii)

\begin{align*} 1 \text{ hour 15 minutes} & = 1 + {15 \over 60} \text{ hour} \\ & = {5 \over 4} \text{ hour} \\ \\ {200 \over x} - {200 \over x + 5} & = {5 \over 4} \\ {200(x + 5) \over x(x + 5)} - {200x \over x(x + 5)} & = {5 \over 4} \\ {200(x + 5) - 200x \over x(x + 5)} & = {5 \over 4} \\ {200x + 1000 - 200x \over x(x + 5)} & = {5 \over 4} \\ {1000 \over x(x + 5)} & = {5 \over 4} \\ {4(1000) \over 4x(x + 5)} & = {5x(x + 5) \over 4x(x + 5)} \\ 4(1000) & = 5x(x + 5) \\ 4000 & = 5x^2 + 25x \\ 0 & = 5x^2 + 25x - 4000 \\ 0 & = x^2 + 5x - 800 \phantom{0} \text{ (Shown)} \end{align*}

(iv)

\begin{align*} 0 & = x^2 + 5x - 800 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-5 \pm \sqrt{(5)^2 - 4(1)(-800)} \over 2(1)} \\ & = {-5 \pm \sqrt{3225} \over 2} \\ & \approx 25.894 \text{ or } -30.894 \text{ (NA, since } x > 0) \\ \\ \text{Speed from Segamat to Singapore} & = x + 5 \\ & = 25.894 + 5 \\ & = 30.894 \\ & \approx 30.9 \text{ km/h} \end{align*}

\begin{align*} \text{Let numerator} & = x \\ \\ \text{Denominator} & = x + 2 \\ \\ \text{Original fraction} & = {x \over x + 2} \\ \\ \text{New fraction} & = {x + 3 \over x + 2 + 3} \\ & = {x + 3 \over x + 5} \\ \\ {x + 3 \over x + 5} - {x \over x + 2} & = {3 \over 20} \\ {20(x + 2)(x + 3) \over 20(x + 5)(x + 2)} - {20x(x + 5) \over 20(x + 5)(x + 2)} & = {3(x + 5)(x + 2) \over 20(x + 5)(x + 2)} \\ 20(x + 2)(x + 3) - 20x(x + 5) & = 3(x + 5)(x + 2) \\ 20(x^2 + 3x + 2x + 6) - 20x^2 - 100x & = 3(x^2 + 2x + 5x + 10) \\ 20(x^2 + 5x + 6) - 20x^2 - 100x & = 3(x^2 + 7x + 10) \\ 20x^2 + 100x + 120 - 20x^2 - 100x & = 3x^2 + 21x + 30 \\ 120 & = 3x^2 + 21x + 30 \\ 0 & = 3x^2 + 21x + 30 - 120 \\ 0 & = 3x^2 + 21x - 90 \\ 0 & = x^2 + 7x - 30 \\ 0 & = (x - 3)(x + 10) \end{align*} \begin{align*} x - 3 & = 0 && \text{ or } & x + 10 & = 0 \\ x & = 3 &&& x & = -10 \text{ (NA, since } x > 0) \end{align*}
\begin{align*} \text{Original fraction} & = {3 \over 3 + 2} \\ & = {3 \over 5} \end{align*}

(a)(i)

\begin{align*} \text{Cost price} & = \$ \left( {32 \over x} \right) \end{align*}

(a)(ii)

\begin{align*} 5 \text{ cents} & = \$ 0.05 \\ & = \$ \left(1 \over 20\right) \\ \\ \text{Selling price} & = {32 \over x} + {1 \over 20} \\ & = {20(32) \over 20x} + {x \over 20x} \\ & = \$ \left( {640 + x \over 20x} \right) \end{align*}

(b)

\begin{align*} \text{Revenue} & = (x - 20) \left( 640 + x \over 20x\right) \\ & = \left( x - 20 \over 1\right)\left(640 + x \over 20x\right) \\ & = { (x - 20)(640 + x) \over 20x } \\ & = { 640x + x^2 - 12 \phantom{.} 800 - 20x \over 20 x }\\ & = \$ \left( { x^2 + 620x - 12 \phantom{.} 800 \over 20x } \right) \\ \\ 35 & = { x^2 + 620x - 12 \phantom{.} 800 \over 20x } \\ {35 \over 1} & = { x^2 + 620x - 12 \phantom{.} 800 \over 20x } \\ {20x(35) \over 20x} & = { x^2 + 620x - 12 \phantom{.} 800 \over 20x } \\ 20x(35) & = x^2 + 620x - 12 \phantom{.} 800 \\ 700x & = x^2 + 620x - 12 \phantom{.} 800 \\ 0 & = x^2 + 620x - 700x - 12 \phantom{.} 800 \\ 0 & = x^2 - 80x - 12 \phantom{.} 800 \phantom{0} \text{ (Shown)} \end{align*}

(c)

\begin{align*} 0 & = x^2 - 80x - 12 \phantom{.} 800 \\ 0 & = (x - 160)(x + 80) \end{align*} \begin{align*} x - 160 & = 0 && \text{ or } & x + 80 & = 0 \\ x & = 160 &&& x & = -80 \text{ (NA, since } x > 0) \end{align*}
$$ \text{Fruitseller bought 160 apples} $$

(a)

\begin{align*} A: x \text{ minutes} & \rightarrow \text{1 tank} \\ 1 \text{ minute} & \rightarrow {1 \over x} \text{ tank} \\ \\ B: (x + 40) \text{ minutes} & \rightarrow \text{1 tank} \\ 1 \text{ minute} & \rightarrow {1 \over x + 40} \text{ tank} \\ \\ A \&B: 48 \text{ minutes} & \rightarrow \text{1 tank} \\ 1 \text{ minute} & \rightarrow {1 \over 48} \text{ tank} \\ \\ {1 \over x} + {1 \over x + 40} & = {1 \over 48} \\ {48(x + 40) \over 48x(x + 40)} + {48x \over 48x(x + 40)} & = {x(x + 40) \over 48x(x + 40)} \\ 48(x + 40) + 48x & = x(x + 40) \\ 48x + 1920 + 48x & = x^2 + 40x \\ 96x + 1920 & = x^2 + 40x \\ 0 & = x^2 + 40x - 96x - 1920 \\ 0 & = x^2 - 56x - 1920 \phantom{0} \text{ (Shown)} \end{align*}

(b)(i)

\begin{align*} 0 & = x^2 - 56x - 1920 \\ 0 & = (x - 80)(x + 24) \end{align*} \begin{align*} x - 80 & = 0 && \text{ or } & x + 24 & = 0 \\ x & = 80 &&& x & = -24 \end{align*}

(b)(ii)

\begin{align*} \text{Cannot accept } x = -24 \text{ since time taken cannot be negative} \end{align*}

(c)

\begin{align*} \text{Time taken by } B & = x + 40 \\ & = 80 + 40 \\ & = 120 \text{ minutes} \\ & = 2 \text{ hours} \end{align*}

(a)

\begin{align*} x + y & = 3 \\ y & = -x + 3 \end{align*}

\(x\)	\(0\)	\(1\)	\(2\)
\(y\)	\(3\)	\(2\)	\(1\)

\begin{align*} x + 3y & = 1 \\ 3y & = - x + 1 \\ y & = -{1 \over 3}x + {1 \over 3} \end{align*}

\(x\)	\(-2\)	\(1\)	\(4\)
\(y\)	\(1\)	\(0\)	\(-1\)

$$ x= 4, y = -1 $$

(b)

\begin{align*} 2x - y & = 4 \\ -y & = -2x + 4 \\ y & = 2x - 4 \end{align*}

\(x\)	\(0\)	\(1\)	\(2\)
\(y\)	\(-4\)	\(-2\)	\(0\)

\begin{align*} 5x + 3y & = -1 \\ 3y & = -5x - 1 \\ y & = -{5 \over 3}x - {1 \over 3} \end{align*}

\(x\)	\(-2\)	\(1\)	\(2.5\)
\(y\)	\(3\)	\(-2\)	\(-4.5\)

$$ x = 1, y = -2 $$

(c)

\begin{align*} x + 3y - 12 & = 0 \\ 3y & = -x + 12 \\ y & = -{1 \over 3}x + 4 \end{align*}

\(x\)	\(-3\)	\(0\)	\(3\)
\(y\)	\(5\)	\(4\)	\(3\)

\begin{align*} 2x - 2y & = 0 \\ -2y & = -2x \\ y & = x \end{align*}

\(x\)	\(-1\)	\(0\)	\(1\)
\(y\)	\(-1\)	\(0\)	\(1\)

$$ x = 3, y = 3 $$

(d)

\begin{align*} 2x + 5y + 2 & = 0 \\ 5y & = -2x - 2 \\ y & = -{2 \over 5}x - {2 \over 5} \end{align*}

\(x\)	\(-3.5\)	\(-1\)	\(1.5\)
\(y\)	\(1\)	\(0\)	\(-1\)

\begin{align*} x + 5y - 3 & = 0 \\ 5y & = -x + 3 \\ y & = -{1 \over 5}x + {3 \over 5} \end{align*}

\(x\)	\(-2\)	\(0\)	\(3\)
\(y\)	\(1\)	\(0.6\)	\(0\)

$$ x = -5, y = 1.6 $$

(i)

\begin{align*} y & = (x + 1)(2 - x) \\ \\ \text{Let } & x = 0, \\ y & = (0 + 1)(2 - 0) \\ y & = 2 \\ \\ \implies & B(0, 2) \\ \\ \text{Let } & y = 0, \\ 0 & = (x + 1)(2 - x) \end{align*} \begin{align*} x + 1 & = 0 && \text{ or } & 2 - x & = 0 \\ x & = -1 &&& -x & = -2 \\ & &&& x & = 2 \\ \\ \implies & A(-1, 0) &&& \implies & C(2, 0) \end{align*}

(ii)

\begin{align*} \text{Line of symmetry: } x & = {-1 + 2 \over 2} \\ x & = 0.5 \end{align*}

(i)

\begin{align*} y & = x^2 + hx - 5 \\ \\ \text{Let } & x = 0, \\ y & = (0)^2 + h(0) - 5 \\ y & = -5 \\ \\ \implies & A(0, -5) \end{align*}

(ii)

\begin{align*} y & = x^2 + hx - 5 \\ \\ \text{Using } & (1, -2), \\ -2 & = (1)^2 + h(1) - 5 \\ -2 & = 1 + h - 5 \\ -h & = 1 - 5 + 2 \\ -h & = -2 \\ h & = 2 \end{align*}

\begin{align*} \text{When } & y = 0, \end{align*} \begin{align*} x & = 1 && \text{ or } & x & = 4 \\ x - 1 & =0 &&& x - 4 & =0 \end{align*} \begin{align*} y & = (x - 1)(x - 4) \\ y & = x^2 - 4x - x + 4 \\ y & = x^2 - 5x + 4 \\ \\ \therefore p & = -5, q = 4 \end{align*}

(a)

\begin{align*} y & = x^3 + 4 \end{align*}

(b)

\begin{align*} y & = 4^x \end{align*}

(c)

\begin{align*} y & = {4 \over x} \end{align*}

\begin{align*} y & = x^{-2} \\ \\ n & = -2 \end{align*}

\begin{align*} y & = ka^x \\ \\ \text{Using } & (0, 3), \\ 3 & = k a^0 \\ 3 & = k(1) \\ 3 & = k \\ \\ y & = 3 a^x \\ \\ \text{Using } & (4, 48), \\ 48 & = 3 a^4 \\ {48 \over 3} & = a^4 \\ 16 & = a^4 \\ \pm \sqrt[4]{16} & = a \end{align*} \begin{align*} a & = 2 && \text{ or } & a & = -2 \text{ (Reject, since } a > 0) \end{align*} $$ \therefore a = 2, k = 3 $$

(a)

\begin{align*} y & = x^2 + 2 \\ \\ \text{Let } & x = 0,\\ y & = (0)^2 + 2 \\ y & = 2 \phantom{000000} [y \text{-intercept}] \end{align*}

(b)

\begin{align*} x + y & = 2 \\ y & = -x + 2 \phantom{000000} [y = mx + c] \\ \\ y \text{-inter} & \text{cept is } 2 \\ \\ \text{Let } & y = 0, \\ 0 & = -x + 2 \\ x & = 2 \phantom{00000000000} [x \text{-intercept}] \end{align*}

\begin{align*} y & = 9 - 4x^2 \\ \\ \text{Let } & x = 0, \\ y & = 9 - 4(0)^2 \\ y & = 9 \\ \\ \implies & P(0, 9) \\ \\ \text{Let } & y = 0, \\ 0 & = 9 - 4x^2 \\ 4x^2 & = 9 \\ x^2 & = {9 \over 4} \\ x & = \pm \sqrt{9 \over 4} \\ x & = \pm 1.5 \\ \\ \implies & Q(-1.5, 0), R(1.5, 0) \end{align*}

(a)

\begin{align*} y & = (x + 1)(x - 5) \phantom{000000} [\text{Minimum curve since coefficient of } x^2 \text{ is positive}] \\ \\ \text{Let } & x = 0, \\ y & = (0 + 1)(0 - 5) \\ y & = -5 \phantom{0000000000000000}[ y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = (x + 1)(x - 5) \end{align*} \begin{align*} x + 1 & = 0 && \text{ or } & x - 5 & = 0 \\ x & = -1 &&& x & = 5 \phantom{0000000} [x \text{-intercepts}] \end{align*}
\begin{align*} \text{Line of symmetry: } x & = {-1 + 5 \over 2} \\ x & = 2 \\ \\ \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = (2 + 1)(2 - 5) \\ y & = -9 \\ \\ \text{Turning point: } & (2, -9) \end{align*}

(b)

$$ x = 2 $$

(c)

$$ (2, -9) $$

(i)

\begin{align*} y & = -(x + 3)^2 + 1 \\ \\ \text{Let } & y = 0, \\ 0 & = -(x + 3)^2 + 1 \\ (x + 3)^2 & = 1 \\ x + 3 & = \pm \sqrt{1} \end{align*} \begin{align*} x + 3 & = 1 && \text{ or } & x + 3 & = - 1 \\ x & = 1 - 3 &&& x & = -1 - 3 \\ x & = -2 &&& x & = -4 \\ & (-2, 0) &&& & (-4, 0) \end{align*}
\begin{align*} y & = -(x + 3)^2 + 1 \\ \\ \text{Let } & x = 0, \\ y & = -(0 + 3)^2 + 1 \\ y & = -8 \\ \\ & (0, -8) \end{align*}

(ii)

\begin{align*} y & = -(x + 3)^2 + 1 \\ \\ \text{Max. } & \text{point: } (-3, 1) \end{align*}

(iii)

\begin{align*} y & = -(x + 3)^2 + 1 \\ \\ \text{Shape: } & \text{Maximum curve } (\cap) \\ \\ y \text{-intercept: } & y = -8 \\ \\ x \text{-intercepts: } & x = -2, -4 \\ \\ \text{Turning point: } & (-3, 1) \end{align*}

(iv)

$$ x = - 3 $$

(i)

\begin{align*} x^2 - 6x + 4 & = x^2 - 6x + \left(6 \over 2\right)^2 - \left(6 \over 2\right)^2 + 4 \phantom{000000} [\text{Complete the square}] \\ & = x^2 - 6x + 3^2 - 9 + 4 \\ & = (x - 3)(x - 3) - 5 \\ & = (x - 3)^2 - 5 \end{align*}

(ii)

\begin{align*} y & = x^2 - 6x + 4 \\ y & = (x - 3)^2 - 5 \\ \\ TP: & (3, -5) \end{align*}

(iii)

\begin{align*} y & = (x - 3)^2 - 5 \phantom{000000} [\text{Minimum curve } \cup] \\ \\ TP: & (3, -5) \\ \\ \text{Let } & x = 0, \\ y & = (0 - 3)^2 - 5 \\ y & = 4 \phantom{000000000000000.} [y \text{-intercept}] \end{align*}

(iv)

$$ x = 3 $$

(i)

\begin{align*} x^2 + x + 3 & = x^2 + x + \left(1 \over 2\right)^2 - \left(1 \over 2\right)^2 + 3 \phantom{000000} [\text{Complete the square}] \\ & = \left(x + {1 \over 2}\right)^2 - {1 \over 4} + 3 \\ & = \left(x + {1 \over 2}\right)^2 + {11 \over 4} \end{align*}

(ii)

\begin{align*} y & = x^2 + x + 3 \\ & = \left(x + {1 \over 2}\right)^2 + {11 \over 4} \\ & = (x + 0.5)^2 + 2.75 \\ \\ \text{TP: } & (-0.5, 2.75) \end{align*}

(iii)

\begin{align*} y & = (x + 0.5)^2 + 2.75 \phantom{000} [\text{Minimum curve } \cup] \\ \\ \text{TP: } & (-0.5, 2.75) \\ \\ \text{Let } & x = 0, \\ y & = (0 + 0.5)^2 + 2.75 \\ y & = 3 \phantom{00000000000000000} [y \text{-intercept}] \end{align*}

(iv)

$$ x = -0.5 $$

(a)

(b)

(c)(i)

$$ \text{See graph in (b)} $$

(c)(ii)

\begin{align*} \text{Area of trapezium} & = {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} \\ & = {1 \over 2} \times (2 + 1) \times 2 \\ & = 3 \text{ units}^2 \end{align*}

(a)

\(x\)	\(-1.5\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(2.5\)
\(y\)	\(-4.646\)	\(-4.5\)	\(-4\)	\(-3\)	\(-1\)	\(0.657\)

(b)

$$ x = 1.8 $$

(c)

\begin{align*} \text{For all real} & \text{ values of } x, \\ 2^x & > 0 \\ 2^x - 5 & > - 5 \\ \\ \therefore \text{Curve of } y = 2^x - 5 & \text{ always lies above } y = -5 \end{align*}

(a)

\(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(y\)	\(10\)	\(0\)	\(-6\)	\(-8\)	\(-6\)	\(0\)	\(10\)

(b)

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {2 - (-10) \over 4 - 1} \\ & = {12 \over 3} \\ & = 4 \end{align*}

(c)(i)

(c)(ii)

$$ x = -0.3 \text{ or } 3.3 $$

(c)(iii)

$$ -0.3 \le x \le 3.3 $$

(a)

\(x\)	\(0.5\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)
\(y\)	\(8\)	\(4\)	\(3.5\)	\(4.67\)	\(6.25\)	\(8\)	\(9.83\)	\(11.7\)

(b)

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {11 - 1.2 \over 7.4 - 0.6} \\ & \approx 1.4412 \\ & \approx 1.44 \end{align*}

(c)(i)

\begin{align*} {5 \over x} + 2x - 8 & = 0 \\ {5 \over x} + 2x - 8 + 5 & = 0 + 5 \\ \underbrace{ {5 \over x} + 2x - 3}_\text{Curve} & = 5 \\ \\ \text{Draw } & y = 5 \\ \\ \therefore x & = 0.8 \text{ or } 3.2 \end{align*}

(c)(ii)

(a)

(b)

Bottom half of the graph (more relevant for subsequent parts):

(c)

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = { 0 - (-3) \over 3 - (-1) } \\ & = 0.75 \end{align*}

(d)

\begin{align*} \underbrace{ x^2(x - 2) }_\text{Curve} & = -1 \\ \\ \text{Draw } & y = -1 \\ \\ \text{From graph, } x & = -0.6 \text{ or } 1 \text{ or } 1.6 \end{align*}

(e)

(f)(i)

$$ x = -1, 1, 2 $$

(f)(ii)

\begin{align*} y & = x^2(x - 2) \phantom{0} \text{--- (1)} \\ \\ y & = x - 2 \phantom{0} \text{--- (2)} \\ \\ \text{Sub } & \text{(1) into (2),} \\ x^2(x - 2) & = x - 2 \\ x^3 - 2x^2 & = x - 2 \\ x^3 - 2x^2 - x + 2& = 0 \\ \\ A = -2, B& = -1, C = 2 \end{align*}

(a)

(b)(i)

\begin{align*} x^2 + 3x & = 8 \\ x^2 + 3x - 3 & = 8 - 3 \\ \underbrace{x^2 + 3x - 3}_\text{Curve} & = 5 \\ \\ \text{Draw } & y = 5 \\ \\ \text{From graph, } x & = -4.7 \text{ or } 1.7 \end{align*}

(b)(ii)

(b)(iii)

(c)

\begin{align*} \text{Line from (b)(iii) } y & = x \text{ has gradient} = 1 \phantom{000000} [y = mx + c] \\ \\ \therefore \text{Tangent must be } & \text{parallel to } y = x \\ \\ & \therefore (-1, -5) \end{align*}

(a)

\begin{align*} \text{Area of square} & = x \times x \\ & = x^2 \\ \\ \text{Volume} & = x^2 \times h \\ 35 & = hx^2 \\ {35 \over x^2 } & = h \\ \\ \text{Height} & = {35 \over x^2} \text{ m} \end{align*}

(b)

\begin{align*} A & = x^2 + 4(x \times h) \\ & = x^2 + 4xh \\ & = x^2 + 4x \left(35 \over x^2\right) \phantom{000000} [\text{From (a)}] \\ & = x^2 + \left(4x \over 1\right)\left(35 \over x^2\right) \\ & = x^2 + {140x \over x^2} \\ & = x^2 + {140 \over x} \phantom{0} \text{ (Shown)} \end{align*}

(c)

(d)

Bottom half of the graph:

(e)(i)

$$ x \approx 3.07 \text{ m} \text{ or } 5.4 \text{ m} $$

(e)(ii)

$$ \text{Minimum value of } A = 51 \text{ m}^2 $$

(e)(iii)

\begin{align*} \text{From graph, } x & = 4.1 \\ \\ h & = {35 \over x^2} \\ & = {35 \over 4.1^2} \\ & \approx 2.0821 \\ & \approx 2.08 \text{ m} \end{align*}

First part:

\begin{align*} \text{Initial } (t = 2) \text{ speed of car} & = 0 \text{ m/s} \implies (2, 0) \\ t = 3, \text{ Speed of car} & = 3 \text{ m/s} \implies (3, 3) \\ t = 4, \text{ Speed of car} & = 6 \text{ m/s} \implies (4, 6) \end{align*}

(a)

$$ \text{From graph, } t = 6.4 \text{ seconds} $$

(b) On the speed-time graph, gradient of curve = acceleration of object

\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {14.1 - 9 \over 7.5 - 0} \\ & = 0.68 \\ \\ \text{Acceleration} & = 0.68 \text{ m/s}^2 \end{align*}

(a)

\begin{align*} \text{Base of box} & = (16 - 2x) \text{ cm} \\ \\ \text{Height of box} & = x \text{ cm} \\ \\ \text{Volume} & = (16 - 2x) \times (16 - 2x) \times x \\ & = x(16 - 2x)^2 \\ & = x[ (16)^2 - 2(16)(2x) + (2x)^2 ] \phantom{000000} [ (a - b)^2 = a^2 - 2ab + b^2 ] \\ & = x (256 - 64x + 4x^2) \\ & = 256x - 64x^2 + 4x^3 \\ V & = ( 4x^3 - 64x^2 + 256x) \text{ cm}^3 \phantom{0} \text{ (Shown)} \end{align*}

(b)

(c)

(d)(i)

$$ V = 304 \phantom{000000} [y \text{-coordinate of maximum point}] $$

(d)(ii)

\begin{align*} \text{Base of box} & = 16 - 2x \\ & = 16 - 2(2.65) \phantom{000000} [x \text{-coordinate of maximum point}] \\ & = 10.7 \\ \\ \text{Base area} & = 10.7 \times 10.7 \\ & = 114.49 \\ & \approx 114 \text{ cm}^2 \text{ (to 3 s.f.)} \end{align*}

(e)

\begin{align*} \text{Base of box} & = (16 - 2x) \text{ cm} \\ \\ \text{If } x > 8, \text{ base } & \text{of box will have negative value} \end{align*}

(a)

\begin{align*} \text{Radius of semicircle} & = {1 \over 2}x \text{ m} \\ \\ \text{Circumference of semicircle} & = {1 \over 2} \times 2 \pi r \\ & = \pi r \\ & = \pi \left({1 \over 2}x\right) \\ & = {1 \over 2} \pi x \text{ m} \\ \\ \text{Perimeter} & = {1 \over 2} \pi x + 2y + x \\ 12 & = {1 \over 2} \pi x + 2y + x \\ -2y & = {1 \over 2} \pi x + x - 12 \\ 2y & = 12 - {1 \over 2} \pi x - x \\ y & = {1 \over 2} \left(12 - {1 \over 2} \pi x - x \right) \\ y & = 6 - {1 \over 4} \pi x - {1 \over 2} x \end{align*}

(b)

\begin{align*} \text{Area of semicircle} & = {1 \over 2} \pi r^2 \\ & = {1 \over 2} \pi \left({1 \over 2}x\right)^2 \\ & = {1 \over 2} \pi \left({1 \over 4}x^2\right) \phantom{000000} \left[ {1 \over 2} x \times {1 \over 2}x = {1 \over 4} x^2 \right] \\ & = {1 \over 8} \pi x^2 \\ \\ \text{Area of rectangle} & = xy \\ & = x \left(6 - {1 \over 4} \pi x - {1 \over 2}x \right) \\ & = 6x - {1 \over 4} \pi x^2 - {1 \over 2}x^2 \\ \\ A & = {1 \over 8} \pi x^2 + 6x - {1 \over 4} \pi x^2 - {1 \over 2}x^2 \\ & = 6x - {1 \over 2}x^2 + {1 \over 8} \pi x^2 - {1 \over 4} \pi x^2 \\ & = \left( 6x - {1 \over 2}x^2 - {1 \over 8} \pi x^2 \right) \text{ m}^2 \phantom{0} \text{ (Shown)} \end{align*}

(c)

(d)

(e)(i)

$$ A = 10.1 \text{ m}^2 \phantom{000000} [y \text{-coordinate of maximum point}] $$

(e)(ii)

\begin{align*} \text{Arc length} & = \text{Circumference of semicircle} \\ & = {1 \over 2} \pi x \\ & = {1 \over 2} \pi (3.35) \phantom{000000} [x \text{-coordinate of maximum point}] \\ & \approx 5.2621 \\ & \approx 5.26 \text{ m} \end{align*}

(a)(i)

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

(a)(ii)

\begin{align*} \text{S} \$ 16 & \rightarrow \text{HK} \$ 100 \\ \text{S} \$ 1 & \rightarrow \text{HK} \$ {100 \over 16} = \text{HK} \$ 6.25 \\ \text{S} \$ 20 & \rightarrow \text{HK} \$ 6.25 \times 20 = \text{HK} \$ 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*} f & = {9 \over 5}c + 32 \\ c & = {5 \over 9}(f - 32) \\ \\ \text{Increase in } c & = {5 \over 9}(120 - 32) - {5 \over 9}(70 - 32) \\ & = {5 \over 9}(120 - 70) \\ & = 27.777... \\ & \approx 27.8 \\ \\ \text{Check from graph: } & 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) and (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} \\ & \approx 48.412 \\ & \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 the water rises, 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 \phantom{000000} [18 \text{ km is in the 15 to 20 km band}] \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} \\ & \approx 3.1429 \\ & \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} \phantom{000000} [1000 \text{ to } 1030 \text{ and } 1230 \text{ to } 1400] \\ & = 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{ hour 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{At 1130, Car A is } & 80 \text{ km and Car B is } 40 \text{ km from Pattaya} \\ \\ \text{Distance apart} & = 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)

$$ 12.5 \text{ cm} $$

(i)

\begin{align*} A & = \{ 2, 3, 5, 7 \} \\ \\ B & = \{3, 5, 7, 9 \} \\ \\ A \cup B & = \{ 2, 3, 5, 7, 9 \} \phantom{000000} [\text{Elements in } A \text{ or } B \text{ or both}] \end{align*}

(ii)

\begin{align*} A & = \{ 2, 3, 5, 7 \} \\ \\ B & = \{3, 5, 7, 9 \} \\ \\ B' & = \{ 2, 4, 6, 8 \} \\ \\ A \cap B' & = \{ 2 \} \end{align*}

(i)

\begin{align*} A & = \{ \text{a, e} \} \\ \\ A \cap B & = \{ \text{a} \} \end{align*}

(ii)

\begin{align*} A & = \{ \text{a, e} \} \\ \\ A \cup B & = \{ \text{a, c, e, f} \} \phantom{000000} [\text{Letters in } A \text{ or } B \text{ or both}] \end{align*}

(iii)

\begin{align*} A & = \{ \text{a, e} \} \\ \\ B' & = \{ \text{b, d, e} \} \\ \\ A \cap B' & = \{ \text{e} \} \end{align*}

(i)

\begin{align*} A & = \{ 2, 3, 5, 7 \} \\ \\ B & = \{ 2, 4, 6, 8, 10 \} \\ \\ C & = \{ 3, 6, 9 \} \\ \\ A \cap B & = \{ 2 \} \end{align*}

(ii)

\begin{align*} B' & = \{ 1, 3, 5, 7, 9 \} \\ \\ A \cup B' & = \{ 1, 2, 3, 5, 7, 9 \} \phantom{000000} [\text{Elements in } A \text{ or } B' \text{ or both}] \end{align*}

(iii)

\begin{align*} C' & = \{ 1, 2, 4, 5, 7, 8, 10 \} \\ \\ A \cap C' & = \{ 2, 5, 7 \} \end{align*}

(i)

\begin{align*} A & = \{ 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32 \} \end{align*}

(ii)

\begin{align*} B & = \{ 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33 \} \end{align*}

(iii)

\begin{align*} C & = \{ 11, 13, 17, 19, 23, 29, 31 \} \end{align*}

(iv)

\begin{align*} A \cap B & = \{ 15, 17, 19, 21, 23, 25, 27, 29, 31 \} \end{align*}

(v)

\begin{align*} B \cap C & = \{ 11, 13, 17, 19, 23, 29, 31 \} \end{align*}

(vi)

\begin{align*} C' & = \{ 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33 \} \\ \\ B \cap C' & = \{ 15, 21, 25, 27, 33 \} \end{align*}

(i)

\begin{align*} A \cap B & = A \end{align*}

(ii)

\begin{align*} A \cup B & = B \end{align*}

(i)

\begin{align*} C & = \{ 1, 4, 9, 16, ... \} \phantom{000000} [\text{Perfect squares, since } \sqrt{x} \text{ is a positive integer}] \\ \\ A \cap C & = \{ 64, 81, 100 \} \end{align*}

(ii)

\begin{align*} B & = \{ \sqrt{64}, \sqrt{81}, \sqrt{100} \} \\ & = \{ 8, 9, 10 \} \end{align*}

(i)

\begin{align*} A & = \{ 1, 2, 3, 4, 6, 12 \} \end{align*}

(ii)

\begin{align*} A' & = \{ 5, 7, 8, 9, 10, 11 \} \\ \\ B & = \{ 1, 9 \} \phantom{000000} [\sqrt{1} = 1, \sqrt{9} = 3] \\ \\ A' \cap B & = \{ 9 \} \end{align*}

(iii)

\begin{align*} A \cup B & = \{ 1, 2, 3, 4, 6, 9, 12 \} \end{align*}

(a)

(b)(i)

\begin{align*} A \cap B & = \{ 2, 4, 14 \} \end{align*}

(b)(ii)

\begin{align*} A' \cup B & = \{ 1, 2, 3, 4, 5, 7, 9, 11, 13, 14, 15, 17, 19 \} \end{align*}

(b)(iii)

\begin{align*} A' \cap B' & = \{ 3, 5, 9, 11, 13, 15, 17, 19 \} \end{align*}

(i)

\begin{align*} C & = \{ 1, 2, 3, 4, 6, 8, 12, 24 \} \end{align*}

(ii)

\begin{align*} A & = \{ 2, 3, 5, 7, 11, 13, 17, 19, 23 \} \\ \\ A \cap C & = \{ 2, 3 \} \end{align*}

(iii)

\begin{align*} B & = \{ 3, 6, 9, 12, 15, 18, 21, 24 \} \\ \\ B' & = \{ 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23 \} \\ \\ B' \cap C & = \{ 1, 2, 4, 8 \} \end{align*}

(iv)

\begin{align*} A \cap B' & = \{ 2, 5, 7, 11, 13, 17, 19, 23 \} \end{align*}

(i)

(ii)

Note: \( P \cap Q \ne \emptyset\) means that sets \(P\) and \(Q\) do intersect!

(i)

(ii)

(a)

$$ A \cap B' $$

(b)

$$ (A \cup B)' $$

(c)

$$ A \cap B' $$

(d)

$$ (A \cap B)' $$

(i) and (iii)

(ii)

\begin{align*} A \subset B \end{align*}

(i)

\begin{align*} 14 - x & > 3 \\ -x & > 3 - 14 \\ -x & > - 11 \\ x & < 11 \\ \\ A & = \{ -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \} \\ \\ 1 - 2x & < 10 \\ -2x & < 10 - 1 \\ -2x & < 9 \\ x & > {9 \over -2} \\ x & > -4.5 \\ \\ B & = \{ -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 \} \\ \\ A \cup B & = \{ -10, -9, ..., 14, 15 \} = \xi \\ \\ (A \cup B)' & = \{ \} = \emptyset \end{align*}

(ii)

\begin{align*} A \cap B & = \{ -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \} \end{align*}

(iii)

\begin{align*} B' & = \{ -10, -9, -8, -7, -6, -5 \} \\ \\ A \cap B' & = \{ -10, -9, -8, -7, -6, -5 \} \end{align*}

(a)(i)

$$ \text{No, since rational numbers such as } {3 \over 5} \text{ are not integers.} $$

(a)(ii)

\begin{align*} & \text{No. Since } B \subset A, \phantom{.} A \cup B = A. \\ & \text{Irrational numbers such as } \pi \text{ are real numbers but are not in } A \cup B. \end{align*}

(b)

(c)(i)

\begin{align*} A' & = \{ x: x \text{ is an irrational number} \} \\ \\ & \text{Element: } \pi \end{align*}

(c)(ii)

\begin{align*} B' & = \{ x: x \text{ is not an integer} \} \\ \\ A \cap B' & = \{ x: x \text{ is a rational number that is not an integer} \} \\ \\ & \text{Element: } {1 \over 3} \end{align*}

(a)

\begin{align*} \{ 2, 4 \} \subset A \phantom{000000} [ \{ 2, 4\} \text{ is a set}] \end{align*}

(b)

\begin{align*} A \cap (B \cup C) & = \{ 6 , 12, 20 \} \\ \\ 9 & \notin A \cap (B \cup C) \end{align*}

(c)

\begin{align*} A \cap B \cap C & = \emptyset \end{align*}

(d)

\begin{align*} A \cup B \cup C & = \xi \end{align*}

(a)

\begin{align*} \text{P(Student does not own cat or dog)} & = { 7 \over 7 + 4x + 3x - 6 + x^2 + 3x} \\ {7 \over 40} & = {7 \over x^2 + 10x + 1} \\ \\ 40 & = x^2 + 10x + 1 \\ 0 & = x^2 + 10x + 1 - 40 \\ 0 & = x^2 + 10x - 39 \\ 0 & = (x + 13)(x - 3) \end{align*} \begin{align*} x + 13 & = 0 && \text{ or } & x - 3 & = 0 \\ x & = -13 \text{ (Reject, since } 4x \ge 0) &&& x & = 3 \end{align*}

(b)

\begin{align*} n (C \cap D) & = 3x - 6 \\ & = 3(3) - 6 \\ & = 3 \end{align*}

(c)

\begin{align*} \text{No. of students who own a dog only} & = x^2 + 3x \\ & = (3)^2 + 3(3) \\ & = 18 \end{align*}

(a)

\begin{align*} \left( \begin{matrix} -1 & 1 \\ -2 & 5 \end{matrix} \right) + \left( \begin{matrix} 6 & 0 \\ 1 & 4 \end{matrix} \right) & = \left( \begin{matrix} 5 & 1 \\ -1 & 9 \end{matrix} \right) \end{align*}

(b)

\begin{align*} \underset{3 \times 1}{ \left( \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right) } - \underset{3 \times 1}{ \left( \begin{matrix} 3 \\ 2 \\ 1 \end{matrix} \right) } & + \underset{2 \times 1}{ \left( \begin{matrix} -1 \\ -2 \end{matrix} \right) } \\ \\ \text{Not possible since } & \text{matrices have different order} \end{align*}

(c)

\begin{align*} \left( \begin{matrix} 1 & 2 \end{matrix} \right) + \left( \begin{matrix} -1 & -3 \end{matrix} \right) - \left( \begin{matrix} -2 & 5 \end{matrix} \right) & = \left( \begin{matrix} 0 & -1 \end{matrix} \right) - \left( \begin{matrix} -2 & 5 \end{matrix} \right) \\ & = \left( \begin{matrix} 2 & - 6 \end{matrix} \right) \end{align*}

(d)

\begin{align*} (3) + {1 \over 2}(2) - (0) & = (3) + (1) - (0) \\ & = (4) \end{align*}

(a)

\begin{align*} \underset{2 \times 1}{ \left( \begin{matrix} 5 \\ -1 \end{matrix} \right) } \underset{1 \times 2}{ \left( \begin{matrix} 3 & 4 \end{matrix} \right) } & = \underset{2 \times 2}{ \left( \begin{matrix} 5 \times 3 & 5 \times 4 \\ (-1) \times 3 & (-1) \times 4 \end{matrix} \right) } \\ & = \left( \begin{matrix} 15 & 20 \\ -3 & -4 \end{matrix} \right) \end{align*}

(b)

\begin{align*} \underset{1 \times 2}{ \left( \begin{matrix} 3 & 2 \end{matrix} \right) } \underset{2 \times 1}{ \left( \begin{matrix} 4 \\ -1 \end{matrix} \right) } & = \underset{1 \times 1}{ \left( \begin{matrix} (3 \times 4) + [2 \times (-1)] \end{matrix} \right) } \\ & = \left( \begin{matrix} 10 \end{matrix} \right) \end{align*}

(c)

\begin{align*} \underset{3 \times 2}{ \left( \begin{matrix} 3 & 4 \\ 1 & 5 \\ 2 & 6 \end{matrix} \right) } & \underset{3 \times 1}{ \left( \begin{matrix} 3 \\ 1 \\ 5 \end{matrix} \right) } \\ \\ \text{Not possible since the number of columns of the first matrix } (2) & \text{ is not equal to the number of rows of the second matrix } (3) \end{align*}

(d)

\begin{align*} \underset{1 \times 3}{ \left( \begin{matrix} 2 & -1 & 4 \end{matrix} \right) } \underset{3 \times 1}{ \left( \begin{matrix} 0 \\ -3 \\ {1 \over 2} \end{matrix} \right) } & = \underset{1 \times 1}{ \left( \begin{matrix} (2 \times 0) + [(-1) \times (-3)] + (4 \times {1 \over 2}) \end{matrix} \right) } \\ & = \left( \begin{matrix} 5 \end{matrix} \right) \end{align*}

(e)

\begin{align*} \underset{2 \times 2}{ \left( \begin{matrix} 4 & 1 \\ 0 & 2 \end{matrix} \right) } \underset{2 \times 2}{ \left( \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right) } & = \underset{2 \times 2}{ \left( \begin{matrix} (4 \times 1) + (1 \times 3) & (4 \times 2) + (1 \times 4) \\ (0 \times 1) + (2 \times 3) & (0 \times 2) + (2 \times 4) \end{matrix} \right) } \\ & = \left( \begin{matrix} 7 & 12 \\ 6 & 8 \end{matrix} \right) \end{align*}

(i)

\begin{align*} 2 \left( \begin{matrix} 2 & -1 \\ 1 & 3 \end{matrix} \right) + \left( \begin{matrix} 5 & w \\ y & 4 \end{matrix} \right) & = \left( \begin{matrix} x & 6 \\ 4 & z \end{matrix} \right) \\ \left( \begin{matrix} 4 & -2 \\ 2 & 6 \end{matrix} \right) + \left( \begin{matrix} 5 & w \\ y & 4 \end{matrix} \right) & = \left( \begin{matrix} x & 6 \\ 4 & z \end{matrix} \right) \\ \left( \begin{matrix} 9 & - 2 + w \\ 2 + y & 10 \end{matrix} \right) & = \left( \begin{matrix} x & 6 \\ 4 & z \end{matrix} \right) \end{align*} \begin{align*} x & = 9 &&& -2 + w & = 6 &&& 2 + y & = 4 &&& z & = 10 \\ & &&& w & = 8 &&& y & = 2 &&& \end{align*}

(ii)

\begin{align*} 3 \left( \begin{matrix} 2 & -1 \\ 1 & 3 \end{matrix} \right) - 2 \left( \begin{matrix} 5 & w \\ y & 4 \end{matrix} \right) & = 4 \left( \begin{matrix} x & 6 \\ 4 & z \end{matrix} \right) \\ \left( \begin{matrix} 6 & - 3 \\ 3 & 9 \end{matrix} \right) - \left( \begin{matrix} 10 & 2w \\ 2y & 8 \end{matrix} \right) & = \left( \begin{matrix} 4x & 24 \\ 16 & 4z \end{matrix} \right) \\ \left( \begin{matrix} -4 & -3 - 2w \\ 3 - 2y & 1 \end{matrix} \right) & = \left( \begin{matrix} 4x & 24 \\ 16 & 4z \end{matrix} \right) \end{align*} \begin{align*} -4 & = 4x &&& -3 - 2w & = 24 &&& 3 - 2y & = 16 &&& 1 & = 4z \\ -1 & = x &&& -2w & = 27 &&& -2y & = 13 &&& {1 \over 4} & = z \\ & &&& w & = -{27 \over 2} &&& y & = -{13 \over 2} \end{align*}

(a)

\begin{align*} \underset{2 \times 2}{ \left( \begin{matrix} 3 & -1 \\ 0 & 2 \end{matrix} \right) } \underset{2 \times 1}{ \left( \begin{matrix} 2 \\ 3 \end{matrix} \right) } & = \underset{2 \times 1}{ \left( \begin{matrix} (3 \times 2) + (-1 \times 3) \\ (0 \times 2) + (2 \times 3) \end{matrix} \right) } \\ & = \left( \begin{matrix} 3 \\ 6 \end{matrix} \right) \\ \\ \left( \begin{matrix} x \\ y \end{matrix} \right) & = \left( \begin{matrix} 3 \\ 6 \end{matrix} \right) + \left( \begin{matrix} 5 \\ 2 \end{matrix} \right) \\ & = \left( \begin{matrix} 8 \\ 8 \end{matrix} \right) \\ \\ x & = 8, y = 8 \end{align*}

(b)

\begin{align*} 2 \left( \begin{matrix} x \\ y \end{matrix} \right) - 4 \left( \begin{matrix} 2 \\ 5 \end{matrix} \right) & = \left( \begin{matrix} 4x \\ 3y \end{matrix} \right) \\ \left( \begin{matrix} 2x \\ 2y \end{matrix} \right) - \left( \begin{matrix} 8 \\ 20 \end{matrix} \right) & = \left( \begin{matrix} 4x \\ 3y \end{matrix} \right) \\ \left( \begin{matrix} 2x - 8 \\ 2y - 20 \end{matrix} \right) & = \left( \begin{matrix} 4x \\ 3y \end{matrix} \right) \end{align*} \begin{align*} 2x - 8 & = 4x &&& 2y - 20 & = 3y \\ 2x - 4x & = 8 &&& 2y - 3y & = 20 \\ -2x & = 8 &&& -y & = 20 \\ x & = -4 &&& y & = -20 \end{align*}

(c)

\begin{align*} x \left( \begin{matrix} 2 \\ 4 \end{matrix} \right) + y \left( \begin{matrix} 5 \\ 7 \end{matrix} \right) & = \left( \begin{matrix} -5 \\ -1 \end{matrix} \right) \\ \left( \begin{matrix} 2x \\ 4x \end{matrix} \right) + \left( \begin{matrix} 5y \\ 7y \end{matrix} \right) & = \left( \begin{matrix} -5 \\ -1 \end{matrix} \right) \\ \left( \begin{matrix} 2x + 5y \\ 4x + 7y \end{matrix} \right) & = \left( \begin{matrix} -5 \\ -1 \end{matrix} \right) \end{align*} \begin{align*} 2x + 5y & = -5 &&& 4x + 7y & = -1 \phantom{0} \text{--- (1)} \\ 2x & = -5 - 5y \\ x & = -2.5 - 2.5y \phantom{0} \text{--- (2)} \end{align*} \begin{align*} \text{Substitute } & \text{(2) into (1),} \\ 4(-2.5 - 2.5y) + 7y & = -1 \\ -10 - 10y + 7y & = -1 \\ -10y + 7y & = -1 + 10 \\ -3y & = 9 \\ y & = -3 \\ \\ \text{Substitute } & y = -3 \text{ into (2),} \\ x & = -2.5 - 2.5(-3) \\ x & = 5 \\ \\ \therefore x & = 5, y = -3 \end{align*}

(a)

\begin{align*} \underset{2 \times 2}{ \left( \begin{matrix} 0 & 2 \\ -1 & 4 \end{matrix} \right) } \underset{2 \times 1}{ \left( \begin{matrix} x \\ y \end{matrix} \right) } & = \left( \begin{matrix} 2 \\ 3 \end{matrix} \right) \\ \underset{2 \times 1}{ \left( \begin{matrix} (0 \times x) + (2 \times y) \\ (-1 \times x) + (4 \times y) \end{matrix} \right) } & = \left( \begin{matrix} 2 \\ 3 \end{matrix} \right) \\ \left( \begin{matrix} 2y \\ -x + 4y \end{matrix} \right) & = \left( \begin{matrix} 2 \\ 3 \end{matrix} \right) \end{align*} \begin{align*} 2y & = 2 &&& -x + 4y & = 3 \\ y & = 1 &&& -x + 4(1) & = 3 \\ & &&& -x + 4 & = 3 \\ & &&& - x & = 3 - 4 \\ & &&& - x & = -1 \\ & &&& x & = 1 \end{align*}

(b)

\begin{align*} \underset{2 \times 2}{ \left( \begin{matrix} 2 & 1 \\ 4 & x \end{matrix} \right) } \underset{2 \times 1}{ \left( \begin{matrix} y \\ 2 \end{matrix} \right) } & = \left( \begin{matrix} 6 \\ 10 \end{matrix} \right) \\ \underset{2 \times 1}{ \left( \begin{matrix} (2 \times y) + (1 \times 2) \\ (4 \times y) + (x \times 2) \end{matrix} \right) } & = \left( \begin{matrix} 6 \\ 10 \end{matrix} \right) \\ \left( \begin{matrix} 2y + 2 \\ 4y + 2x \end{matrix} \right) & = \left( \begin{matrix} 6 \\ 10 \end{matrix} \right) \end{align*} \begin{align*} 2y + 2 & = 6 &&& 4y + 2x & = 10 \\ 2y & = 4 &&& 2x & = 10 - 4y \\ y & = 2 &&& 2x & = 10 - 4(2) \\ & &&& 2x & = 2 \\ & &&& x & = 1 \end{align*}

(c)

\begin{align*} \underset{2 \times 2}{ \left( \begin{matrix} 3 & 1 \\ x & 4 \end{matrix} \right) } \underset{2 \times 1}{ \left( \begin{matrix} y \\ 1 \end{matrix} \right) } & = \left( \begin{matrix} -2 \\ 1 \end{matrix} \right) \\ \underset{2 \times 1}{ \left( \begin{matrix} (3 \times y) + (1 \times 1) \\ (x \times y) + (4 \times 1) \end{matrix} \right) } & = \left( \begin{matrix} -2 \\ 1 \end{matrix} \right) \\ \left( \begin{matrix} 3y + 1 \\ xy + 4 \end{matrix} \right) & = \left( \begin{matrix} -2 \\ 1 \end{matrix} \right) \end{align*} \begin{align*} 3y + 1 & = -2 &&& xy + 4 & = 1 \\ 3y & = -3 &&& xy& = -3 \\ y & = -1 &&& x(-1) & = -3 \\ & &&& -x & = -3 \\ & &&& x & = 3 \end{align*}

\begin{align*} \underset{2 \times 3}{ \left( \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 4 \end{matrix} \right) } \underset{3 \times 1}{ \left( \begin{matrix} p \\ 0 \\ 2 \end{matrix} \right) } & = \left( \begin{matrix} 2 \\ 2q \end{matrix} \right) \\ \underset{2 \times 1}{ \left( \begin{matrix} (1 \times p) + (2 \times 0) + (3 \times 2) \\ (0 \times p) + (1 \times 0) + (4 \times 2) \end{matrix} \right) } & = \left( \begin{matrix} 2 \\ 2q \end{matrix} \right) \\ \left( \begin{matrix} p + 6 \\ 8 \end{matrix} \right) & = \left( \begin{matrix} 2 \\ 2q \end{matrix} \right) \end{align*} \begin{align*} p + 6 & = 2 &&& 2q & = 8 \\ p & = -4 &&& q & = 4 \end{align*}

\begin{align*} \textbf{AB} & = \textbf{A} + \textbf{B} \\ \underset{2 \times 2}{ \left( \begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix} \right) } \underset{2 \times 2}{ \left( \begin{matrix} a & b \\ 0 & c \end{matrix} \right) } & = \left( \begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix} \right) + \left( \begin{matrix} a & b \\ 0 & c \end{matrix} \right) \\ \underset{2 \times 2}{ \left( \begin{matrix} (2 \times a) + (0 \times 0) & (2 \times b) + (0 \times c) \\ (0 \times a) + (3 \times 0) & (0 \times b) + (3 \times c) \end{matrix} \right) } & = \left( \begin{matrix} 2 + a & 0 \\ 0 & 3 + c \end{matrix} \right) \\ \left( \begin{matrix} 2a & 2b \\ 0 & 3c \end{matrix} \right) & = \left( \begin{matrix} 2 + a & 0 \\ 0 & 3 + c \end{matrix} \right) \end{align*} \begin{align*} 2a & = 2 + a &&& 2b & = 0 &&& 3c & = 3 + c \\ a & = 2 &&& b & = 0 &&& 2c & = 3 \\ & &&& & &&& c & = {3 \over 2} \end{align*}

(i)

\begin{align*} 2 \left( \begin{matrix} -2 & 1 \\ -1 & 4 \end{matrix} \right) + \left( \begin{matrix} 3 & 1 \\ -2 & - 5 \end{matrix} \right) - \left( \begin{matrix} 3 & -1 \\ 4 & - 2 \end{matrix} \right) & = \left( \begin{matrix} -4 & 2 \\ -2 & 8 \end{matrix} \right) + \left( \begin{matrix} 3 & 1 \\ -2 & - 5 \end{matrix} \right) - \left( \begin{matrix} 3 & -1 \\ 4 & - 2 \end{matrix} \right) \\ & = \left( \begin{matrix} -1 & 3 \\ -4 & 3 \end{matrix} \right) - \left( \begin{matrix} 3 & -1 \\ 4 & - 2 \end{matrix} \right) \\ & = \left( \begin{matrix} -4 & 4 \\ -8 & 5 \end{matrix} \right) \end{align*}

(ii)

\begin{align*} \left( \begin{matrix} 3 & 1 \\ -2 & - 5 \end{matrix} \right) + \underset{2 \times 2}{ \left( \begin{matrix} -2 & 1 \\ -1 & 4 \end{matrix} \right) } \underset{2 \times 2}{ \left( \begin{matrix} 3 & - 1 \\ 4 & - 2 \end{matrix} \right) } & = \left( \begin{matrix} 3 & 1 \\ -2 & - 5 \end{matrix} \right) + \underset{2 \times 2}{ \left( \begin{matrix} (-2 \times 3) + (1 \times 4) & (-2 \times -1) + (1 \times -2) \\ (-1 \times 3) + (4 \times 4) & (-1 \times -1) + (4 \times -2) \end{matrix} \right) } \\ & = \left( \begin{matrix} 3 & 1 \\ -2 & - 5 \end{matrix} \right) + \left( \begin{matrix} -2 & 0 \\ 13 & -7 \end{matrix} \right) \\ & = \left( \begin{matrix} 1 & 1 \\ 11 & -12 \end{matrix} \right) \end{align*}

(iii)

\begin{align*} \left( \begin{matrix} -2 & 1 \\ -1 & 4 \end{matrix} \right) + \underset{2 \times 2}{ \left( \begin{matrix} 3 & 1 \\ -2 & - 5 \end{matrix} \right) } \underset{2 \times 2}{ \left( \begin{matrix} 3 & - 1 \\ 4 & - 2 \end{matrix} \right) } & = \left( \begin{matrix} -2 & 1 \\ -1 & 4 \end{matrix} \right) + \underset{2 \times 2}{ \left( \begin{matrix} (3 \times 3) + (1 \times 4) & (3 \times -1) + (1 \times -2) \\ (-2 \times 3) + (-5 \times 4) & (-2 \times -1) + (-5 \times -2) \end{matrix} \right) } \\ & = \left( \begin{matrix} -2 & 1 \\ -1 & 4 \end{matrix} \right) + \left( \begin{matrix} 13 & -5 \\ -26 & 12 \end{matrix} \right) \\ & = \left( \begin{matrix} 11 & -4 \\ -27 & 16 \end{matrix} \right) \end{align*}

(a)

\begin{align*} \underset{2 \times 2}{ \left( \begin{matrix} 0 & 5 \\ -2 & 4 \end{matrix} \right) } \underset{2 \times 1}{ \left( \begin{matrix} x \\ y \end{matrix} \right) } & = \left( \begin{matrix} -15 \\ 28 \end{matrix} \right) \\ \underset{2 \times 1}{ \left( \begin{matrix} (0 \times x) + (5 \times y) \\ (-2 \times x) + (4 \times y) \end{matrix} \right) } & = \left( \begin{matrix} -15 \\ 28 \end{matrix} \right) \\ \left( \begin{matrix} 5y \\ -2x + 4y \end{matrix} \right) & = \left( \begin{matrix} -15 \\ 28 \end{matrix} \right) \end{align*} \begin{align*} 5y & = -15 &&& -2x + 4y & = 28 \\ y & = -3 &&& -2x & = 28 - 4y \\ & &&& -2x & = 28 - 4(-3) \\ & &&& -2x & = 40 \\ & &&& x & = - 20 \end{align*}

(b)(i)

\begin{align*} \textbf{AB} & = \underset{2 \times 2}{ \left( \begin{matrix} 3 & 2 \\ 1 & 4 \end{matrix} \right) } \underset{2 \times 1}{ \left( \begin{matrix} 1 \\ 3 \end{matrix} \right) } \\ & = \underset{2 \times 1}{ \left( \begin{matrix} (3 \times 1) + (2 \times 3) \\ (1 \times 1) + (4 \times 3) \end{matrix} \right) } \\ & = \left( \begin{matrix} 9 \\ 13 \end{matrix} \right) \end{align*}

(b)(ii)

\begin{align*} \textbf{AC} & = \underset{2 \times 2}{ \left( \begin{matrix} 3 & 2 \\ 1 & 4 \end{matrix} \right) } \underset{1 \times 2}{ \left( \begin{matrix} 2 & 1 \end{matrix} \right) } \\ \\ & \text{Not possible since } \textbf{A} \text{ has 2 columns but } \textbf{C} \text{ has only 1 row} \end{align*}

(b)(iii)

\begin{align*} \textbf{BC} & = \underset{2 \times 1}{ \left( \begin{matrix} 1 \\ 3 \end{matrix} \right) } \underset{1 \times 2}{ \left( \begin{matrix} 2 & 1 \end{matrix} \right) } \\ & = \underset{2 \times 2}{ \left( \begin{matrix} 1 \times 2 & 1 \times 1 \\ 3 \times 2 & 3 \times 1 \end{matrix} \right) } \\ & = \left( \begin{matrix} 2 & 1 \\ 6 & 3 \end{matrix} \right) \end{align*}

(b)(iv)

\begin{align*} \textbf{CB} & = \underset{1 \times 2}{ \left( \begin{matrix} 2 & 1 \end{matrix} \right) } \underset{2 \times 1}{ \left( \begin{matrix} 1 \\ 3 \end{matrix} \right) } \\ & = \underset{1 \times 1}{ \left( \begin{matrix} (2 \times 1) + (1 \times 3) \end{matrix} \right) } \\ & = \left( \begin{matrix} 5 \end{matrix} \right) \end{align*}

(a)

\begin{align*} \underset{2 \times 2}{ \left( \begin{matrix} 4 & 2 \\ -1 & p \end{matrix} \right) } \underset{2 \times 2}{ \left( \begin{matrix} 2 & 3 \\ 0 & 1 \end{matrix} \right) } & = \left( \begin{matrix} q & 14 \\ -2 & 7 \end{matrix} \right) \\ \underset{2 \times 2}{ \left( \begin{matrix} (4 \times 2) + (2 \times 0) & (4 \times 3) + (2 \times 1) \\ (-1 \times 2) + (p \times 0) & (-1 \times 3) + (p \times 1) \end{matrix} \right) } & = \left( \begin{matrix} q & 14 \\ -2 & 7 \end{matrix} \right) \\ \left( \begin{matrix} 8 & 14 \\ -2 & -3 + p \end{matrix} \right) & = \left( \begin{matrix} q & 14 \\ -2 & 7 \end{matrix} \right) \end{align*} \begin{align*} q & = 8 &&& -3 + p & = 7 \\ & &&& p & = 10 \end{align*}

(b)(i)

\begin{align*} \textbf{AB} & = \underset{2 \times 2}{ \left( \begin{matrix} 3 & 1 \\ 0 & 2 \end{matrix} \right) } \underset{2 \times 2}{ \left( \begin{matrix} 2 & 6 \\ 0 & k \end{matrix} \right) } \\ & = \underset{2 \times 2}{ \left( \begin{matrix} (3 \times 2) + (1 \times 0) & (3 \times 6) + (1 \times k) \\ (0 \times 2) + (2 \times 0) & (0 \times 6) + (2 \times k) \end{matrix} \right) } \\ & = \left( \begin{matrix} 6 & 18 + k \\ 0 & 2k \end{matrix} \right) \\ \\ \textbf{BA} & = \underset{2 \times 2}{ \left( \begin{matrix} 2 & 6 \\ 0 & k \end{matrix} \right) } \underset{2 \times 2}{ \left( \begin{matrix} 3 & 1 \\ 0 & 2 \end{matrix} \right) } \\ & = \underset{2 \times 2}{ \left( \begin{matrix} (2 \times 3) + (6 \times 0) & (2 \times 1) + (6 \times 2) \\ (0 \times 3) + (k \times 0) & (0 \times 1) + (k \times 2) \end{matrix} \right) } \\ & = \left( \begin{matrix} 6 & 14 \\ 0 & 2k \end{matrix} \right) \end{align*}

(b)(ii)

\begin{align*} \textbf{AB} & = \textbf{BA} \\ \left( \begin{matrix} 6 & 18 + k \\ 0 & 2k \end{matrix} \right) & = \left( \begin{matrix} 6 & 14 \\ 0 & 2k \end{matrix} \right) \\ \\ 18 + k & = 14 \\ k & = -4 \end{align*}

(i)

\begin{align*} \textbf{R} & = \textbf{AM} \\ & = \underset{2 \times 3}{ \left( \begin{matrix} 39 & 5 & 6 \\ 29 & 8 & 13 \end{matrix} \right) } \underset{3 \times 1}{ \left( \begin{matrix} 2 \\ 0 \\ -1 \end{matrix} \right) } \\ & = \underset{2 \times 1}{ \left( \begin{matrix} (39 \times 2) + (5 \times 0) + (6 \times -1) \\ (29 \times 2) + (8 \times 0) + (13 \times -1) \end{matrix} \right) } \\ & = \left( \begin{matrix} 72 \\ 45 \end{matrix} \right) \end{align*}

(ii)

\begin{align*} \textbf{R} & = \left( \begin{matrix} 72 \\ 45 \end{matrix} \right) \begin{matrix} \text{Li Ting's marks} \\ \text{Bernard's marks} \end{matrix} \\ \\ \text{The marks obtained} & \text{ by Li Ting and Bernard } \textbf{respectively}. \end{align*}

\begin{align*} \textbf{AB} & = \textbf{BA} \\ \underset{2 \times 2}{ \left( \begin{matrix} -4 & p \\ -1 & 2 \end{matrix} \right) } \underset{2 \times 2}{ \left( \begin{matrix} q & 0 \\ 2 & 3 \end{matrix} \right) } & = \underset{2 \times 2}{ \left( \begin{matrix} q & 0 \\ 2 & 3 \end{matrix} \right) } \underset{2 \times 2}{ \left( \begin{matrix} -4 & p \\ -1 & 2 \end{matrix} \right) } \\ \underset{2 \times 2}{ \left( \begin{matrix} (-4 \times q) + (p \times 2) & (-4 \times 0) + (p \times 3) \\ (-1 \times q) + (2 \times 2) & (-1 \times 0) + (2 \times 3) \end{matrix} \right) } & = \underset{2 \times 2}{ \left( \begin{matrix} (q \times -4) + (0 \times -1) & (q \times p) + (0 \times 2) \\ (2 \times -4) + (3 \times -1) & (2 \times p) + (3 \times 2) \end{matrix} \right) } \\ \left( \begin{matrix} -4q + 2p & 3p \\ -q + 4 & 6 \end{matrix} \right) & = \left( \begin{matrix} -4q & pq \\ -11 & 2p + 6 \end{matrix} \right) \end{align*} \begin{align*} -4q + 2p & = -4q &&& -q + 4 & = -11 \\ 2p & = -4q + 4q &&& -q & = -15 \\ 2p & = 0 &&& q & = 15 \\ p & = 0 \end{align*} \begin{align*} \text{Check: } 3p & = pq \implies 0 = 0, \phantom{000} 6 = 2p + 6 \implies 6 = 6 \end{align*}

(i)

\begin{align*} 12 \textbf{A} & = 12 \left( \begin{matrix} 120 \\ 95 \\ 102 \end{matrix} \right) \\ & = \left( \begin{matrix} 1440 \\ 1140 \\ 1224 \end{matrix} \right) \end{align*}

(ii)

\begin{align*} \textbf{B} & = 12 \textbf{A} + \textbf{W} \\ & = \left( \begin{matrix} 1440 \\ 1140 \\ 1224 \end{matrix} \right) + \left( \begin{matrix} 60 \\ 75 \\ 62 \end{matrix} \right) \\ & = \left( \begin{matrix} 1500 \\ 1215 \\ 1286 \end{matrix} \right) \\ \\ \textbf{B} \text{ represents the total cost of a 1-year subscription} & \text{ plus the World Cup package for each cable provider respectively} \end{align*}

(i)

\begin{align*} \textbf{P} & = \left( \begin{matrix} 3 \\ 1 \\ 0 \end{matrix} \right) \end{align*}

(ii)

\begin{align*} \textbf{RP} & = \underset{3 \times 3} { \left( \begin{matrix} 12 & 5 & 3 \\ 3 & 8 & 7 \\ 9 & 4 & 4 \end{matrix} \right) } \underset{3 \times 1} { \left( \begin{matrix} 3 \\ 1 \\ 0 \end{matrix} \right) } \\ & = \underset{3 \times 1} { \left( \begin{matrix} (12 \times 3) + (5 \times 1) + (3 \times 0) \\ (3 \times 3) + (8 \times 1) + (7 \times 0) \\ (9 \times 3) + (4 \times 1) + (4 \times 0) \end{matrix} \right) } \\ & = \left( \begin{matrix} 41 \\ 17 \\ 31 \end{matrix} \right) \end{align*}

(iii)

\begin{align*} \textbf{RP}\text{ represents the total points of Wanderers, United and Saints respectively.} \end{align*}

(i)

\begin{align*} \textbf{K} & = 5 \textbf{A} \\ & = 5 \left( \begin{matrix} 13 & 14 \\ 10 & 12 \end{matrix} \right) \\ & = \left( \begin{matrix} 65 & 70 \\ 50 & 60 \end{matrix} \right) \end{align*}

(ii)

\begin{align*} \textbf{L} & = \underset{2 \times 2} { \left( \begin{matrix} 65 & 70 \\ 50 & 60 \end{matrix} \right) } \underset{2 \times 1} { \left( \begin{matrix} 2.50 \\ 1.80 \end{matrix} \right) } \\ & = \underset{2 \times 1} { \left( \begin{matrix} (65 \times 2.50) + (70 \times 1.80) \\ (50 \times 2.50) + (60 \times 1.80) \end{matrix} \right) } \\ & = \left( \begin{matrix} 288.50 \\ 233 \end{matrix} \right) \end{align*}

(iii)

\begin{align*} \text{They represent the total fees (\$) collected from boys and girls respectively in a week.} \end{align*}

(i)

\begin{align*} \textbf{PQ} & = \underset{2 \times 3} { \left( \begin{matrix} 75 & 84 & 135 \\ 88 & 95 & 140 \end{matrix} \right) } \underset{3 \times 1} { \left( \begin{matrix} 0.70 \\ 0.85 \\ 1 \end{matrix} \right) } \\ & = \underset{2 \times 1} { \left( \begin{matrix} (75 \times 0.7) + (84 \times 0.85) + (135 \times 1) \\ (88 \times 0.7) + (95 \times 0.85) + (140 \times 1) \end{matrix} \right) } \\ & = \left( \begin{matrix} 258.90 \\ 282.35 \end{matrix} \right) \\ \\ \text{Total cost of } & \text{beverages sold for Cafe A and Cafe B respectively} \end{align*}

(ii)

\begin{align*} \textbf{R} - \textbf{Q} & = \left( \begin{matrix} 3.20 \\ 4 \\ 4.50 \end{matrix} \right) - \left( \begin{matrix} 0.70 \\ 0.85 \\ 1 \end{matrix} \right) \\ & = \left( \begin{matrix} 2.50 \\ 3.15 \\ 3.50 \end{matrix} \right) \\ \\ \text{Profit per cup of latte} & \text{, mocha and iced lemon tea respectively.} \end{align*}

(iii)

\begin{align*} \textbf{P} (\textbf{R} - \textbf{Q}) & = \underset{2 \times 3} { \left( \begin{matrix} 75 & 84 & 135 \\ 88 & 95 & 140 \end{matrix} \right) } \underset{3 \times 1} { \left( \begin{matrix} 2.50 \\ 3.15 \\ 3.50 \end{matrix} \right) } \\ & = \underset{2 \times 1} { \left( \begin{matrix} (75 \times 2.5) + (84 \times 3.15) + (135 \times 3.5) \\ (88 \times 2.5) + (95 \times 3.15) + (140 \times 3.5) \end{matrix} \right) } \\ & = \left( \begin{matrix} 924.60 \\ 1 \phantom{.} 009.25 \end{matrix} \right) \\ \\ \text{Profit earned by } & \text{Cafe A and Cafe B on the particular day respectively.} \end{align*}

(i)

\begin{align*} \underset{3 \times 4} { \left( \begin{matrix} 50 & 60 & 70 & 40 \\ 30 & 40 & 50 & 30 \\ 40 & 30 & 60 & 50 \end{matrix} \right) } \underset{4 \times 1} { \left( \begin{matrix} 3.2 \\ 3.1 \\ 3 \\ 3.3 \end{matrix} \right) } & = \underset{3 \times 1} { \left( \begin{matrix} (50 \times 3.2) + (60 \times 3.1) + (70 \times 3) + (40 \times 3.3) \\ (30 \times 3.2) + (40 \times 3.1) + (50 \times 3) + (30 \times 3.3) \\ (40 \times 3.2) + (30 \times 3.1) + (60 \times 3) + (50 \times 3.3) \end{matrix} \right) } \\ & = \left( \begin{matrix} 688 \\ 469 \\ 566 \end{matrix} \right) \end{align*}

(ii)

\begin{align*} & \phantom{=0} \underset{1 \times 3} { \left( \begin{matrix} 25 & 12 & 15 \end{matrix} \right) } \underset{3 \times 4} { \left( \begin{matrix} 50 & 60 & 70 & 40 \\ 30 & 40 & 50 & 30 \\ 40 & 30 & 60 & 50 \end{matrix} \right) } \\ & = \underset{1 \times 4} { \left( \begin{matrix} (25 \times 50) + (12 \times 30) + (15 \times 40) & (25 \times 60) + (12 \times 40) + (15 \times 30) & (25 \times 70) + (12 \times 50) + (15 \times 60) & (25 \times 40) + (12 \times 30) + (15 \times 50) \end{matrix} \right) } \\ & = \left( \begin{matrix} 2210 & 2430 & 3250 & 2110 \end{matrix} \right) \end{align*}

(iii)

\begin{align*} \text{Total no. of tubs} & = 2210 + 2430 + 3250 + 2110 \\ & = 10 \phantom{.} 000 \end{align*}

(i)

\begin{align*} \textbf{X} & = \underset{3 \times 3} { \left( \begin{matrix} 65 & 45 & 46 \\ 60 & 56 & 58 \\ 78 & 56 & 54 \end{matrix} \right) } \underset{3 \times 1} { \left( \begin{matrix} 10 \\ 20 \\ 50 \end{matrix} \right) } \\ & = \underset{3 \times 1} { \left( \begin{matrix} (65 \times 10) + (45 \times 20) + (46 \times 50) \\ (60 \times 10) + (56 \times 20) + (58 \times 50) \\ (78 \times 10) + (56 \times 20) + (54 \times 50) \end{matrix} \right) } \\ & = \left( \begin{matrix} 3850 \\ 4620 \\ 4600 \end{matrix} \right) \\ \\ \text{Total value (in cents) of } & \text{coins collected on Friday, Saturday and Sunday respectively.} \end{align*}

(ii)

\begin{align*} \textbf{Y} & = \underset{3 \times 4} { \left( \begin{matrix} 32 & 26 & 18 & 16 \\ 45 & 34 & 20 & 10 \\ 38 & 22 & 25 & 24 \end{matrix} \right) } \underset{4 \times 1} { \left( \begin{matrix} 1 \\ 2 \\ 5 \\ 10 \end{matrix} \right) } \\ & = \underset{3 \times 1} { \left( \begin{matrix} (32 \times 1) + (26 \times 2) + (18 \times 5) + (16 \times 10) \\ (45 \times 1) + (34 \times 2) + (20 \times 5) + (10 \times 10) \\ (38 \times 1) + (22 \times 2) + (25 \times 5) + (24 \times 10) \end{matrix} \right) } \\ & = \left( \begin{matrix} 334 \\ 313 \\ 447 \end{matrix} \right) \\ \\ \text{Total value (in dollars) of } & \text{notes collected on Friday, Saturday and Sunday respectively.} \end{align*}

(iii)

\begin{align*} \textbf{S} & = {1 \over 100} \underset{1 \times 3} { \left( \begin{matrix} 1 & 1 & 1 \end{matrix} \right) } \underset{3 \times 1} { \left( \begin{matrix} 3850 \\ 4620 \\ 4600 \end{matrix} \right) } \\ & = {1 \over 100} \underset{1 \times 1} { \left( \begin{matrix} 3850 + 4620 + 4600 \end{matrix} \right) } \\ & = {1 \over 100} \left( \begin{matrix} 13 \phantom{.} 070 \end{matrix} \right) \\ & = \left( \begin{matrix} 130.70 \end{matrix} \right) \\ \\ \text{The total value of } & \text{10 cents, 20 cents and 50 cents coins collected in dollars.} \end{align*}

(iv)

\begin{align*} \textbf{T} & = \underset{1 \times 3} { \left( \begin{matrix} 1 & 1 & 1 \end{matrix} \right) } \underset{3 \times 1} { \left( \begin{matrix} 334 \\ 313 \\ 447 \end{matrix} \right) } + \left( \begin{matrix} 130.70 \end{matrix} \right) \\ & = \underset{1 \times 1} { \left( \begin{matrix} 334 + 313 + 447 \end{matrix} \right) } + \left( \begin{matrix} 130.70 \end{matrix} \right) \\ & = \left( \begin{matrix} 1094 \end{matrix} \right) + \left( \begin{matrix} 130.70 \end{matrix} \right) \\ & = \left( \begin{matrix} 1224.70 \end{matrix} \right) \\ \\ \text{The total } & \text{amount collected from the charity event} \end{align*}

(i)

\begin{align*} \underset{3 \times 3} { \left( \begin{matrix} 12 & 8 & 11 \\ 18 & 11 & 7 \\ 8 & 9 & 15 \end{matrix} \right) } \underset{3 \times 1} { \left( \begin{matrix} 48 \\ 32 \\ 26 \end{matrix} \right) } & = \underset{3 \times 1} { \left( \begin{matrix} (12 \times 48) + (8 \times 32) + (11 \times 26) \\ (18 \times 48) + (11 \times 32) + (7 \times 26) \\ (8 \times 48) + (9 \times 32) + (15 \times 26) \end{matrix} \right) } \begin{matrix} \text{Robert} \\ \text{Star} \\ \text{Mido} \end{matrix} \\ & = \left( \begin{matrix} 1118 \\ 1398 \\ 1062 \end{matrix} \right) \end{align*}

(ii)

\begin{align*} \underset{1 \times 3} { \left( \begin{matrix} 9.8 & 10.4 & 9.9 \end{matrix} \right) } \underset{3 \times 1} { \left( \begin{matrix} 1118 \\ 1398 \\ 1062 \end{matrix} \right) } & = \underset{1 \times 1} { \left( \begin{matrix} (9.8 \times 1118) + (10.4 \times 1398) + (9.9 \times 1062) \end{matrix} \right) } \\ & = \left( \begin{matrix} 36 \phantom{.} 009.40 \end{matrix} \right) \end{align*}

(i)

\begin{align*} \text{HK\$} 100 & = \text{S\$} 16 \\ \text{HK\$} 1 & = \text{S\$} {16 \over 100} \\ \\ \text{RM100} & = \text{S\$}40 \\ \text{RM1} & = \text{S\$} {40 \over 100} \\ \\ \underset{3 \times 2} { \left( \begin{matrix} 125 & 320 \\ 200 & 160 \\ 90 & 450 \end{matrix} \right) } \underset{2 \times 1} { \left( \begin{matrix} {16 \over 100} \\ {40 \over 100} \end{matrix} \right) } & = \underset{3 \times 1} { \left( \begin{matrix} (125 \times {16 \over 100}) + (320 \times {40 \over 100}) \\ (200 \times {16 \over 100}) + (160 \times {40 \over 100}) \\ (90 \times {16 \over 100}) + (450 \times {40 \over 100}) \end{matrix} \right) } \\ & = \left( \begin{matrix} 148 \\ 96 \\ 194.40 \end{matrix} \right) \end{align*}

(ii)

\begin{align*} \underset{1 \times 3} { \left( \begin{matrix} 1 & 1 & 1 \end{matrix} \right) } \underset{3 \times 1} { \left( \begin{matrix} 148 \\ 96 \\ 194.40 \end{matrix} \right) } & = \underset{1 \times 1} { \left( \begin{matrix} 148 + 96 + 194.40 \end{matrix} \right) } \\ & = \left( \begin{matrix} 438.40 \end{matrix} \right) \end{align*}

(a)

\begin{align*} \textbf{Q} & = \left( \begin{matrix} 4 & 3 & 2 \\ 4 & 4 & 0 \end{matrix} \right) \end{align*}

(b)

\begin{align*} \textbf{R} & = \underset{2 \times 3} { \left( \begin{matrix} 4 & 3 & 2 \\ 4 & 4 & 0 \end{matrix} \right) } \underset{3 \times 2} { \left( \begin{matrix} 42 & -3 \\ 48 & 5 \\ 56 & -4 \end{matrix} \right) } \\ & = \underset{2 \times 2} { \left( \begin{matrix} (4 \times 42) + (3 \times 48) + (2 \times 56) & (4 \times -3) + (3 \times 5) + (2 \times -4) \\ (4 \times 42) + (4 \times 48) + (0 \times 56) & (4 \times -3) + (4 \times 5) + (0 \times -4) \end{matrix} \right) } \\ & = \left( \begin{matrix} 424 & -5 \\ 360 & 8 \end{matrix} \right) \begin{matrix} \text{Siti} \\ \text{Vani} \end{matrix} \end{align*}

(c)

\begin{align*} \textbf{R} & = \left( \begin{matrix} 424 & -5 \\ 360 & 8 \end{matrix} \right) \begin{matrix} \text{Siti} \\ \text{Vani} \end{matrix} \\ \\ \text{Siti should } & \text{buy from Bakery B as it is \$ 5 cheaper} \end{align*}

(d)

\begin{align*} \textbf{RS} & = \underset{2 \times 2} { \left( \begin{matrix} 424 & -5 \\ 360 & 8 \end{matrix} \right) } \underset{2 \times 1} { \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) } \\ \\ \textbf{S} & = \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) \end{align*}