(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} = 8.9925 \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 \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 \\ \\ 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 \\ \\ l & = 3 \\ \\ b & = 2^2 = 4 \\ \\ h & = 2^3 = 8 \\ \\ \text{Dimensions: } & 3 \text{ cm by 4 cm by 8 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} & = 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} \\ & = 11.775 \\ & \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} \\ & = 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 & = 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}] \\ & = 3 \% \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 & = 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 \\ & = \$ 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 = 1000 \phantom{.} 000 \text{ cm}^3 \\ \\ 4 \phantom{.} 200 \phantom{.} 000 \text{ m}^3 & = 4 \phantom{.} 200 \phantom{.} 000 \times 1000 \phantom{.} 000 \\ & = 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 \\ & = 304.761 \\ & \approx 305 \% \end{align*}

(a)

\begin{align*} 2.74 \times 10^8 & = 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} \\ & = 1.4087 \\ & \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 \% \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*}