Sec 4 E Maths: Ratio, Proportion, Speed and Map Scales, practice questions & worked solutions

Method 1

$$\begin{aligned}P&=kQ^3\\k&=\dfrac P{Q^3}\\P_{new}&=k(Q_{new})^3\\8P&=k(Q_{new})^3\\8P&=\dfrac P{Q^3}(Q_{new})^3\\(Q_{new})^3&=8Q^3\\Q_{new}&=\sqrt[3]{8Q^3}\\Q_{new}&=2Q\end{aligned}$$

Percentage Increase

$$\begin{aligned}&=\dfrac{2Q-Q}Q\times100\%\\&=100\%\end{aligned}$$

Method 2

$$\begin{aligned}P&=kQ^3\\Q^3&=\dfrac Pk\\Q_{new}{}^3&=\dfrac{P_{new}}k\\&=\dfrac{8P}k\end{aligned}$$

Percentage Increase

$$\begin{aligned}&=\dfrac{\sqrt[3]{\dfrac{8P}k}-\sqrt[3]{\dfrac Pk}}{\sqrt[3]{\dfrac Pk}}\times100\%\\&=\dfrac{\sqrt[3]8-1}1\times100\%\\&=100\%\end{aligned}$$

Answer: $100\%$

$$\begin{array}{cc}S:M&S:L\\3:8&5:2\\15:40&15:6\end{array}$$

$$S:M:L=15:40:6$$

$$\begin{aligned}\text{Total}&=\left(\frac{306}{40-6}\right)(61)\\&=549\end{aligned}$$

Answer: $549$

(a)

$$\begin{aligned}1\text{ cm}&:10\text{ km}\\1\text{ cm}^2&:100\text{ km}^2\\\text{Area (map)}&=\frac{225}{100}\\&=2.25\text{ cm}^2\end{aligned}$$

(b)

$$\begin{aligned}36\text{ cm}^2&:225\text{ km}^2\\6\text{ cm}&:15\text{ km}\\1\text{ cm}&:2.5\text{ km}\\1\text{ cm}&:250000\text{ cm}\end{aligned}$$

Map scale is $1:250000$

Answer: (a) $2.25$ cm$^2$ (b) $1:250000$

(a)

$$\frac{12.5\text{ m}}{1\text{ s}}=\frac{0.0125\text{ km}}{\frac1{3600}\text{ h}}=45\text{ km/h}$$

(b)

Let $d$ be distance each travel to school.

$$\begin{aligned}\frac d5&=\frac d{45}+\frac{20}{60}\\\frac{9d}{45}-\frac d{45}&=\frac13\\\frac{8d}{45}&=\frac13\\24d&=45\\d&=1.875\text{ km}\end{aligned}$$

Answer: (a) $45$ km/h (b) $1.875$ km

Remaining house to be painted $=1-\dfrac56=\dfrac16$.
Sam can paint $\dfrac16$ house in 3 days, he will paint the entire house in 18 days.
Rate at which Sam & Wilson take to paint the house $=\dfrac{1\text{ house}}{6\text{ days}}$.
Rate at which Wilson take to paint he house $=\dfrac{1\text{ house}}{6\text{ days}}-\dfrac{1\text{ house}}{18\text{ days}}=\dfrac{1\text{ house}}{9\text{ days}}$.
Time taken for Wilson to paint the house on his own $=1\div\dfrac19=9$ days.

Answer: $9$ days

Let $x$ be the total distance travelled.

$$\frac{x}{\frac{0.35x}{20}+\frac{0.65x}{30}}=x\div\left[\frac{1.05x+1.3x}{60}\right]$$

$$=x\div\frac{2.35x}{60}$$

$$=25.5\text{ km/h (to 3 sf)}$$

Answer: $25.5\text{ km/h}$

$$\begin{aligned}P&=\frac k{\sqrt Q}\\\frac k{\sqrt9}+\frac k{\sqrt{16}}&=21\\\frac k3+\frac k4&=21\\\frac{7k}{12}&=21\\7k&=252\\k&=36\\\frac{36}{\sqrt Q}&=100\\Q&=\left(\frac{36}{100}\right)^2=0.1296\qquad\left[\text{Accept }\frac{81}{625}\right]\end{aligned}$$

Answer: $Q=0.1296$

(a)

$$\begin{aligned}&I=\frac k{d^2}\\[10pt]&8=\frac k{3^2}\end{aligned}$$

$k=72$

$$\begin{aligned}&I=\frac{72}{d^2}\end{aligned}$$

(b)

$$\begin{aligned}&I=\frac{72}{(0.25d)^2}\\[10pt]&I=\frac{72}{0.0625d^2}\end{aligned}$$

$$\begin{aligned}&I=\frac{1152}{d^2}\end{aligned}$$

$$\begin{aligned}&\frac{1152}{72}=16\ \textit{times}\\[10pt]&\%\text{ change}=\frac{1152-72}{72}\times100\%=1500\%\end{aligned}$$

Answer: (a) $I=\dfrac{72}{d^2}$ (b) $1500\%$

Let Ching's and Lex's original savings be $$3u$ and $$5u$ respectively.

$$\begin{aligned}\frac{3u-60}{5u-60}&=\frac47\\7(3u-60)&=4(5u-60)\\21u-420&=20u-240\\21u-20u&=420-240\\u&=180\end{aligned}$$

$\therefore$ Total amount of money now

$$\begin{aligned}&=180(3+5)-2(60)\\&=\$1320\end{aligned}$$

Answer: $1320

Let the amount of money shared by $ $x$.

$$\text{Amy's original share}:\ \frac29x$$

$$\text{Amy's final share}:\ \frac34\left(\frac29x\right)=\frac16x$$

$$\text{Amy's transferred share}:\ \frac14\left(\frac29x\right)=\frac1{18}x$$

$$\text{Ben's original share}:\ \frac39x=\frac13x$$

$$\text{Ben's final share}:\ \frac23\left(\frac13x+\frac1{18}x\right)=\frac7{27}x$$

$$\text{Ben's transferred share}:\ \frac13\left(\frac13x+\frac1{18}x\right)=\frac7{54}x$$

$$\text{Cal's original share}:\ \frac49x$$

$$\text{Cal's final share}:\ \frac49x+\frac7{54}x=\frac{31}{54}x$$

Final ratio

$$=\frac16x:\frac7{27}x:\frac{31}{54}x$$

$$=9:14:31$$

Answer: $9:14:31$

(a)(i)

$1000$

(a)(ii)

$$\begin{aligned}&0.573\text{mg}\times2=1.146\text{mg}=1146\,\mu\text{g}\end{aligned}$$

Yes, he met the daily recommended intake.

(b)(i)

$2.55\text{km}$

(b)(ii)

Area scale $1\text{cm}^2:0.09\text{km}^2$
Area on map $=0.36\div0.09=4\text{ cm}^2$

(b)(iii)

$0.5625\text{ cm}^2:0.36\text{ km}^2$
$1\text{ cm}^2:0.64\text{ km}^2$
$1\text{ cm}:0.8\text{ km}$
$n=80000$

Answer: (a)(i) $p=1000$   (ii) yes: one cup has $2\times0.573=1.146$ mg $=1146$ micrograms, more than 900 micrograms   (b)(i) $2.55$ km   (ii) $4$ cm$^2$   (iii) $n=80\,000$

(a)

$\dfrac1x$.

(b)

$\dfrac1{x-25}$.

(c)

$$\begin{aligned}\frac1x+\frac1{x-25}&=\frac16\\\frac{x+x-25}{x(x-25)}&=\frac16\\6(2x-25)&=x(x-25)\\12x-150&=x^2-25x\\x^2-25x-12x+150&=0\\x^2-37x+150&=0\quad\text{(shown)}\end{aligned}$$

(d)

$$\begin{aligned}x^2-37x+150&=0\\x&=\frac{-(-37)\pm\sqrt{(-37)^2-4(1)(150)}}{2(1)}\\x&=32.365\text{ or }x=4.634\\x&=32.4\text{ or }x=4.63\ (3\text{ s.f.})\end{aligned}$$

(e)

Reject $x=4.63$ as the time taken, $x-25$, cannot be negative.
Time Tap Y takes to fill up the tank completely $=32.365-25=7.365$
$=7$ hours 22 minutes (nearest minute).

Answer: (a) $\dfrac1x$ (b) $\dfrac1{x-25}$ (d) $x=32.4$ or $4.63$ (e) $7$ hours $22$ minutes