\begin{align*} \text{Let Rick's initial amount} & = 2x \\ \\ \text{Jane's initial amount} & = 3x \\ \\ {2x + 40 \over 3x + 40} & = {5 \over 7} \\ 7(2x + 40) & = 5(3x + 40) \phantom{000000} [\text{Cross-multiply}] \\ 14x + 280 & = 15x + 200 \\ 14x - 15x & = 200 - 280 \\ -x & = -80 \\ x & = 80 \\ \\ \text{Total amount (now)} & = 2x + 40 + 3x + 40 \\ & = 5x + 80 \\ & = 5(80) + 80 \\ & = \$ 480 \end{align*}

\begin{align*} \text{Felicia} & : \text{Grace} : \text{Helen} \\ 3 & : \phantom{00} 8 \phantom{00} : \\ & \phantom{000.} 6 \phantom{00} : 3 \\ 9 & : \phantom{00} 24 \phantom{0} : 12 \\ \\ \text{Percentage Helen owns} & = {12 \over 9 + 24 + 12} \times 100 \\ & = 26{2 \over 3} \% \end{align*}

(a)

\begin{align*} \text{Amount Queenie and Renee receives} & = 270 \times {7 \over 2 + 7} \\ & = \$ 210 \\ \\ \text{Amount Queenie gets} & = {210 - 60 \over 2} \\ & = \$ 75 \end{align*}

(b)

\begin{align*} \text{Amount Renee gets (initially)} & = 210 - 75 \\ & = \$ 135 \\ \\ \text{Amount Paul gets} & = 270 - 135 - 75 \\ & = \$ 60 \\ \\ \text{Paul} & : \text{Queenie} : \text{Renee} \\ 60 & : 75 + 15 \phantom{.} : 135 - 15 \\ 60 & : \phantom{00} 90 \phantom{000} : 120 \\ 6 & : \phantom{000} 9 \phantom{000} : 12 \\ 2 & : \phantom{000} 3 \phantom{000} : 4 \end{align*}

(a)

\begin{align*} 1.2 \text{ kg} & = 1200 \text{ g} \phantom{000000} [1 \text{ kg} = 1000 \text{ g} ] \\ \\ \text{Tomatoes} & : \text{Potatoes} : \text{Onions} \\ 400 & : \phantom{00} 1200 \phantom{0.} \\ 1 & : \phantom{000} 3 \\ 250 & : \phantom{0000000.} : 125 \\ 2 & : \phantom{0000000.} : 1 \\ 2 & : \phantom{000} 6 \phantom{00.0} : 1 \end{align*}

(b)

\begin{align*} \text{Mass of tomatoes} & = 1.8 \times {2 \over 2 + 6 + 1} \\ & = 0.4 \text{ kg} \end{align*}

(a)(i)

\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*}

(a)(ii)

\begin{align*} 1 \text{ cm} & : 0.2 \text{ km} \\ 17 \text{ cm} & : 3.4 \text{ km} \\ \\ \text{Actual distance} & = 3.4 \text{ km} \end{align*}

(b)

\begin{align*} 1 \text{ cm} & : 200 \text{ m} \\ 1^2 \text{ cm}^2 & : 200^2 \text{ m}^2 \\ 1 \text{ cm}^2 & : 40 \phantom{.} 000 \text{ m}^2 \\ 0.75 \text{ cm}^2 & : 30 \phantom{.} 000 \text{ m}^2 \\ \\ \text{Area on map} & = 0.75 \text{ cm}^2 \end{align*}

(a)(i)

\begin{align*} { \text{Population of Canada} \over \text{Population of Singapore} } & = { 38 \phantom{.} 090 \phantom{.} 225 \over 5 \phantom{.} 896 \phantom{.} 686} \\ & = 6.45959 \\ \\ \text{Ratio} & \approx 6.46 : 1 \end{align*}

(a)(ii)

\begin{align*} { \text{Area of Canada} \over \text{Area of Singapore} } & = { 9.985 \times 10^6 \over 716.1 } \\ & = 13 \phantom{.} 943 \\ \\ \text{Ratio} & \approx 13 \phantom{.} 900 : 1 \end{align*}

(b)

\begin{align*} \text{Let population} & = y \\ \\ \text{Let area} & = x \\ \\ y & = kx \\ {y \over x} & = k \\ \\ \text{When } y = 38 \phantom{.} 090 \phantom{.} 225 & \text{ and } x = 9.985 \times 10^6, \\ k & = { 38 \phantom{.} 090 \phantom{.} 225 \over 9.985 \times 10^6} \\ & = 3.8147 \\ \\ \text{When } y = 5 \phantom{.} 896 \phantom{.} 686 & \text{ and } x = 716.1, \\ k & = { 5 \phantom{.} 896 \phantom{.} 686 \over 716.1 } \\ & = 8234.4 \\ \\ \text{Since the value of } k & \text{ is different, population is not proportional to area} \end{align*}

(a)

\begin{align*} \text{Length (on book)} & = 4 + 1.5 + 0.5 + 3 + 1 \\ & = 10 \text{ cm} \\ \\ \text{Actual length} & = 10 \times 25 \phantom{.} 000 \\ & = 250 \phantom{.} 000 \text{ cm} \\ & = 2500 \text{ m} \phantom{00000000} [1 \text{ m} = 100 \text{ cm}] \\ & = 2.5 \text{ km} \phantom{00000000.} [1 \text{ km} = 1000 \text{ m}] \end{align*}

(b)

\begin{align*} \text{Time taken} & = { \text{Distance} \over \text{Speed} } \\ & = { 2.5 \text{ km} \times 2 \over 7 \text{ km/h} } \\ & = {5 \over 7} \text{ h} \\ & = 42.857 \text{ minutes} \\ & \approx 43 \text{ minutes} \end{align*}

(a)

\begin{align*} p & = k x^3 \\ \\ \text{When } x = R & \text{ and } p = 48, \\ 48 & = k R^3 \\ {48 \over R^3} & = k \\ \\ p & = \left(48 \over R^3\right) x^3 \\ \\ \text{When } & x = {R \over 2}, \\ p & = \left(48 \over R^3\right) \left(R \over 2\right)^3 \\ p & = \left(48 \over R^3\right) \left(R^3 \over 8\right) \\ p & = {48 R^3 \over 8 R^3} \\ p & = \$ 6 \end{align*}

(b)(i)

\begin{align*} p & = \left(48 \over R^3\right) x^3 \\ & = \left(48 \over 2^3\right) x^3 \\ & = 6x^3 \end{align*}

(b)(ii)

\begin{align*} p & = 6x^3 \\ \\ \text{When } & r = 3, \\ p & = 6(3)^3 \\ p & = \$ 162 \\ \\ \text{When } & r = 4, \\ p & = 6(4)^3 \\ p & = \$ 384 \\ \\ \text{Total price} & = 162 + 384 \\ & = \$ 546 \end{align*}

(a)

\begin{align*} I & = {k \over d^2} \\ \\ \text{Let initial value of } d & = x \\ \\ \text{When } d = x & \text{ and } I = 60, \\ 60 & = {k \over x^2} \\ 60x^2 & = k \\ \\ I & = {60x^2 \over d^2} \\ \\ \text{When } & d = 2x, \\ I & = {60x^2 \over (2x)^2} \\ & = {60x^2 \over 4x^2} \\ & = 15 \text{ units} \end{align*}

(b)(i)

\begin{align*} I & = {k \over d^2} \\ \\ \text{When } d = 20 & \text{ and } I = 90, \\ 90 & = {k \over 20^2} \\ 90 & = {k \over 400} \\ 400(90) & = k \\ 36 \phantom{.} 000 & = k \\ \\ I & = {36 \phantom{.} 000 \over d^2} \end{align*}

(b)(ii)

\begin{align*} I & = {36 \phantom{.} 000 \over d^2} \\ \\ \text{When } & d = 30, \\ I & = {36 \phantom{.} 000 \over 30^2} \\ & = 40 \text{ units} \end{align*}

\begin{align*} \text{Let frequency} & = y \text{ and wavelength} = x \\ \\ y & = {k \over x} \\ \\ \text{When } x = 6.25 \times 10^{-7} & \text{ and } y = 4.8 \times 10^{14}, \\ 4.8 \times 10^{14} & = {k \over 6.25 \times 10^{-7}} \\ (6.25 \times 10^{-7})(4.8 \times 10^{14}) & = k \\ 300 \phantom{.} 000 \phantom{.} 000 & = k \\ \\ y & = {300 \phantom{.} 000 \phantom{.} 000 \over x} \\ \\ \text{When } x & = 5.00 \times 10^{-10}, \\ y & = {300 \phantom{.} 000 \phantom{.} 000 \over 5.00 \times 10^{-10}} \\ y & = 6 \times 10^{17} \text{ Hz} \end{align*}

\begin{align*} V & = k (t + 273) \\ \\ \text{When } t = 0 & \text{ and } V = 560, \\ 560 & = k(0 + 273) \\ {560 \over 273} & = k \\ {80 \over 39} & = k \\ \\ V & = {80 \over 39} (t + 273) \\ \\ \text{When } & t = 39, \\ V & = {80 \over 39} (39 + 273) \\ & = 640 \text{ cm}^3 \end{align*}

(a)(i)

\begin{align*} & 6 \text{ workers take } 80 \text{ days to build } 1 \text{ house} \\ & 1 \text{ worker take } 80 \times 6 = 480 \text{ days to build 1 house} \phantom{000000} [\text{Less workers, more time taken}] \\ & 15 \text{ workers take } {480 \over 15} = 32 \text{ days to build 1 house} \end{align*}

(a)(ii)

\begin{align*} & 1 \text{ worker take } 480 \text{ days to build 1 house} \\ & 1 \times 24 = 24 \text{ workers take } {480 \over 24} = 20 \text{ days to build 1 house} \end{align*}

(b)

\begin{align*} & \text{Assume each worker works at the same rate} \end{align*}

(a)

\begin{align*} \text{Area on map} & = 5 \times 3 \\ & = 15 \text{ cm}^2 \\ \\ 15 \text{ cm}^2 & : 300 \text{ m}^2 \\ 1 \text{ cm}^2 & : 20 \text{ m}^2 \\ \sqrt{1} \text{ cm} & : \sqrt{20} \text{ m} \\ 1 \text{ cm} & : \sqrt{20} \text{ m} \\ \\ n & = \sqrt{20} \\ n & \approx 4.47 \end{align*}

(b)

\begin{align*} & 5 \text{ workers take } 4 \text{ hours to build area of } 60 \text{ m}^2 \\ & 1 \text{ worker take } 4 \times 5 = 20 \text{ hours to build area of } 60 \text{ m}^2 \\ & 1 \text{ worker take } 20 \times 5 = 100 \text{ hours to build area of } 300 \text{ m}^2 \\ & 8 \text{ workers take } {100 \over 8} = 12.5 \text{ hours to build area of } 300 \text{ m}^2 \end{align*}

(c)

\begin{align*} & \text{Assume each worker works at the same rate} \end{align*}

(a)

\begin{align*} T & = k \sqrt{L} \\ \\ \text{When } L = 0.9 & \text{ and } T = 1.9, \\ 1.9 & = k \sqrt{0.9} \\ {1.9 \over \sqrt{0.9}} & = k \\ \\ T & = \left(1.9 \over \sqrt{0.9}\right) \sqrt{L} \\ \\ \text{When } & T = 2.4, \\ 2.4 & = \left(1.9 \over \sqrt{0.9}\right) \sqrt{L} \\ 2.4 \div \left(1.9 \over \sqrt{0.9}\right) & = \sqrt{L} \\ \left[ 2.4 \div \left(1.9 \over \sqrt{0.9}\right) \right]^2 & = L \\ 1.436 & = L \\ \\ L & \approx 1.44 \text{ m} \end{align*}

(b)

\begin{align*} T & = \left(1.9 \over \sqrt{0.9}\right) \sqrt{L} \\ \\ \text{Let period of } A & = x \\ \\ x & = \left(1.9 \over \sqrt{0.9}\right) \sqrt{L} \\ x^2 & = \left(1.9 \over \sqrt{0.9}\right)^2 L \\ x^2 & = {361 \over 90} L \\ {90 \over 361}x^2 & = L \\ \\ \text{Length of } A & = {90 \over 361}x^2 \\ \\ \text{Period of } B & = {5 \over 3}x \\ \\ {5 \over 3}x & = \left(1.9 \over \sqrt{0.9}\right) \sqrt{L} \\ \left({5 \over 3}x\right)^2 & = \left(1.9 \over \sqrt{0.9}\right)^2 L \\ {25 \over 9}x^2 & = {361 \over 90} L \\ {25 \over 9}x^2 \div {361 \over 90} & = L \\ {250 \over 361}x^2 & = L \\ \\ \text{Length of } B & = {250 \over 361}x^2 \\ \\ { \text{Length of } A \over \text{Length of } B } & = { {90 \over 361}x^2 \over {250 \over 361}x^2 } \\ & = { {90 \over 361} \over {250 \over 361} } \\ & = { 9 \over 25} \\ \\ \therefore \text{Ratio} & = 9: 25 \end{align*}

(c)

\begin{align*} T & = k \sqrt{L} \\ \\ \text{Let initial length} & = y \\ \\ \text{Initial period, } T & = k \sqrt{y} \\ \\ \text{New length} & = y \times {100 + 19 \over 100} \\ & = 1.19 y \\ \\ \text{New period, } T & = k \sqrt{1.19 y} \\ & = k \sqrt{1.19} \sqrt{y} \\ \\ \text{Percentage change} & = { \text{Final} - \text{Initial} \over \text{Initial} } \times 100 \\ & = { k \sqrt{1.19} \sqrt{y} - k \sqrt{y} \over k \sqrt{y} } \times 100 \\ & = {k \sqrt{y} \left[ \sqrt{1.19} - 1 \right] \over k \sqrt{y} } \times 100 \\ & = (\sqrt{1.19} - 1) \times 100 \\ & = 9.0871 \\ & \approx 9.09 \% \end{align*}

(a)(i)

\begin{align*} \text{Length of upper part} & = { 7.5 \over 5 } \times 3 \\ & = 4.5 \text{ km} \end{align*}

(a)(ii)

\begin{align*} \text{Length of whole river} & = { 7.5 \over 5 } \times (5 + 4 + 3) \\ & = 18 \text{ km} \end{align*}

(b)(i)

\begin{align*} 6 \text{ cm} & : 18 \text{ km} \\ 1 \text{ cm} & : 3 \text{ km} \\ 1 \text{ cm} & : 3000 \text{ m} \phantom{000000} [1 \text{ km} = 1000 \text{ m} ] \\ 1 \text{ cm} & : 300 \phantom{.} 000 \text{ cm} \phantom{00.} [1 \text{ m} = 100 \text{ cm}] \\ 1 & : 300 \phantom{.} 000 \end{align*}

(b)(ii)

\begin{align*} 1 \text{ cm} & : 3 \text{ km} \\ 1^2 \text{ cm}^2 & : 3^2 \text{ km}^2 \\ 1 \text{ cm}^2 & : 9 \text{ km}^2 \\ 3 \text{ cm}^2 & : 27 \text{ km}^2 \\ \\ \text{Actual area} & = 27 \text{ km}^2 \end{align*}

(a)(i)

\begin{align*} y & = kx^2 \\ \\ \text{When } x = 5 & \text{ and } y = 10, \\ 10 & = k(5)^2 \\ 10 & = k(25) \\ {10 \over 25} & = k \\ 0.4 & = k \\ \\ y & = 0.4 x^2 \\ \\ \text{When } & x = 8, \\ y & = 0.4 (8)^2 \\ y & = 25.6 \end{align*}

(a)(ii)

\begin{align*} y & = 0.4 x^2 \\ \\ \text{When } & y = 14.4, \\ 14.4 & = 0.4 x^2 \\ {14.4 \over 0.4} & = x^2 \\ 36 & = x^2 \\ \pm \sqrt{36} & = x \\ \pm 6 & = x \\ \\ \text{Positive value of } x & = 6 \end{align*}

(b)

\begin{align*} y & = 0.4 x^2 \\ \\ \text{Let initial value of } x & = a \\ \\ \text{Initial value of } y & = 0.4 a^2 \\ \\ \text{New value of } x & = 3a \\ \\ \text{New value of } y & = 0.4 (3a)^2 \\ & = 0.4 (9a^2) \\ & = 3.6 a^2 \\ \\ c & = {3.6 a^2 \over 0.4 a^2} \\ c & = 9 \end{align*}