Answer Key 9.10
[latexpage]
- \(\begin{array}{l}
\\ \\
\text{Father}=\text{Bill}-2\text{ h}\\ \\
\therefore \dfrac{1}{B-2\text{ h}}+\dfrac{1}{B}=\dfrac{1}{2\text{ h } 24\text{ min}}
\end{array}\) - \(\begin{array}{l}
\\ \\
\text{Smaller}=\text{Larger}+4\text{ h}\\ \\
\therefore \dfrac{1}{L+4\text{ h}}+\dfrac{1}{L}=\dfrac{1}{3\text{ h }45\text{ min}}
\end{array}\) - \(\begin{array}{l}
\\ \\
\text{Jack}=\text{Bob}-1\text{ h} \\ \\
\therefore \dfrac{1}{B-1\text{ h}}+\dfrac{1}{B}=\dfrac{1}{1.2\text{ h}}
\end{array}\) - \(\begin{array}{l}
\\ \\ \\ \\ \\ \\
\dfrac{1}{Y}+\dfrac{1}{B}=\dfrac{1}{T}\\ \\
Y=6\text{ d}\\
B=4\text{ d}\\ \\
\therefore \dfrac{1}{6}+\dfrac{1}{4}=\dfrac{1}{T}
\end{array}\) - \(\begin{array}{l}
\\ \\
\text{John}=\text{Carlos}+8\text{ h}\\ \\
\dfrac{1}{C+8\text{ h}}+\dfrac{1}{C}=\dfrac{1}{3\text{ h}}
\end{array}\) - \(\begin{array}{l}
\\ \\ \\ \\ \\ \\ \\
M=3\text{ d} \\
N=4\text{ d} \\
E=5\text{ d} \\ \\
\dfrac{1}{M}+\dfrac{1}{N}+\dfrac{1}{E}=\dfrac{1}{T} \\ \\
\dfrac{1}{3\text{ d}}+\dfrac{1}{4\text{ d}}+\dfrac{1}{5\text{ d}}=\dfrac{1}{T}
\end{array}\) - \(\begin{array}{l}
\\ \\ \\ \\ \\
\text{Raj}=4 \text{ d} \\ \\
\text{Rubi}=\dfrac{1}{2}\text{ Raj or }2\text{ d} \\ \\
\therefore \dfrac{1}{4 \text{ d}}+\dfrac{1}{2\text{ d}}=\dfrac{1}{T}
\end{array}\) - \(\dfrac{1}{20\text{ min}}+\dfrac{1}{30\text{ min}}=\dfrac{1}{T}\)
- \(\begin{array}{rrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\dfrac{1}{24\text{ d}}&+&\dfrac{1}{I}&=&\dfrac{1}{6\text{ d}} \\ \\
&&\dfrac{1}{I}&=&\dfrac{1}{6\text{ d}}-\dfrac{1}{24\text{ d}} \\ \\
&&\dfrac{1}{I}&=&\dfrac{4}{24\text{ d}}-\dfrac{1}{24\text{ d}} \\ \\
&&\dfrac{1}{I}&=&\dfrac{1}{8\text{ d}} \\ \\
&&\therefore I&=&8\text{ days}
\end{array}\) - \(\begin{array}{rrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\dfrac{1}{C}&+&\dfrac{1}{A}&=&\dfrac{1}{3.75\text{ d}} \\ \\
\dfrac{1}{5\text{ d}}&+&\dfrac{1}{A}&=&\dfrac{1}{3.75\text{ d}} \\ \\
&&\dfrac{1}{A}&=&\dfrac{1}{3.75\text{ d}}-\dfrac{1}{5\text{ d}} \\ \\
&&\dfrac{1}{A}&=&\dfrac{4}{15\text{ d}}-\dfrac{3}{15\text{ d}} \\ \\
&&\dfrac{1}{A}&=&\dfrac{1}{15\text{ d}} \\ \\
&&\therefore A&=&15\text{ days}
\end{array}\) - \(\begin{array}{rrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\dfrac{1}{S}&+&\dfrac{1}{F}&=&\dfrac{1}{\text{job}} \\ \\
\dfrac{1}{3\text{ d}}&+&\dfrac{1}{6\text{ d}}&=&\dfrac{1}{\text{job}} \\ \\
\dfrac{2}{6\text{ d}}&+&\dfrac{1}{6\text{ d}}&=&\dfrac{1}{\text{job}} \\ \\
&&\dfrac{3}{6\text{ d}}&=&\dfrac{1}{\text{job}} \\ \\
&&\dfrac{1}{2\text{ d}}&=&\dfrac{1}{\text{job}} \\ \\
&&\therefore \text{job}&=&2 \text{ days}
\end{array}\) - \(\begin{array}{rrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\dfrac{1}{T}&+&\dfrac{1}{J}&=&\dfrac{1}{\text{job}} \\ \\
\dfrac{1}{10\text{ h}}&+&\dfrac{1}{8\text{ h}}&=&\dfrac{1}{\text{job}} \\ \\
\dfrac{4}{40\text{ h}}&+&\dfrac{5}{40\text{ h}}&=&\dfrac{1}{\text{job}} \\ \\
&&\dfrac{9}{40\text{ h}}&=&\dfrac{1}{\text{job}} \\ \\
&&\therefore \text{job}&=&\dfrac{40\text{ h}}{9} \\ \\
&&\text{job}&=&4\dfrac{4}{9}\text{ h}= 4.\bar{4}\text{ h}
\end{array}\) - \(\begin{array}{rrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&&\text{fast}&=&2\times \text{slow} \\ \\
\dfrac{1}{F}&+&\dfrac{1}{S}&=&\dfrac{1}{6\text{ h}} \\ \\
\dfrac{1}{2S}&+&\dfrac{1}{S}&=&\dfrac{1}{6\text{ h}} \\ \\
\dfrac{1}{2S}&+&\dfrac{2}{2S}&=&\dfrac{1}{6\text{ h}} \\ \\
&&\dfrac{3}{2S}&=&\dfrac{1}{6\text{ h}} \\ \\
&&\dfrac{2S}{3}&=&6\text{ h} \\ \\
&&S&=&\dfrac{6(3)}{2} \\ \\
&&S&=&9\text{ h}
\end{array}\) - \(\begin{array}{rrcrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&&\text{slower}&=&3\times \text{faster} \\ \\
\dfrac{1}{F}&+&\dfrac{1}{S}&=&\dfrac{1}{3\text{ h}} \\ \\
\dfrac{1}{F}&+&\dfrac{1}{3F}&=&\dfrac{1}{3\text{ h}} \\ \\
\dfrac{3}{3F}&+&\dfrac{1}{3F}&=&\dfrac{1}{3\text{ h}} \\ \\
&&\dfrac{4}{3F}&=&\dfrac{1}{3\text{ h}} \\ \\
&&\therefore \dfrac{4}{3}&=&\dfrac{F}{3\text{ h}} \\ \\
&&F&=&4\text{ h}
\end{array}\) - \(\begin{array}{rrcrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&&\text{full}&=&8\text{ h} \\ \\
&&\text{empty}&=&2\times\text{full or }16\text{ h} \\ \\
\dfrac{1}{F}&-&\dfrac{1}{E}&=&\dfrac{1}{T} \\ \\
\dfrac{1}{8\text{ h}}&-&\dfrac{1}{16\text{ h}}&=&\dfrac{1}{T} \\ \\
\dfrac{2}{16\text{ h}}&-&\dfrac{1}{16\text{ h}}&=&\dfrac{1}{T} \\ \\
&&\therefore \dfrac{1}{16\text{ h}}&=&\dfrac{1}{T} \\ \\
&&T&=&16\text{ h}}
\end{array}\) - \(\begin{array}{rrcrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\dfrac{1}{E}&-&\dfrac{1}{F}&=&\dfrac{1}{T} \\ \\
\dfrac{1}{3}&-&\dfrac{1}{5}&=&\dfrac{1}{T} \\ \\
\dfrac{5}{15}&-&\dfrac{3}{15}&=&\dfrac{1}{T} \\ \\
&&\dfrac{2}{15\text{ min}}&=&\dfrac{1}{T} \\ \\
&&\therefore T&=&\dfrac{15\text{ min}}{2}=7.5\text{ min}
\end{array}\) - \(\begin{array}{rrcrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\dfrac{1}{\text{full}}&-&\dfrac{1}{\text{empty}}&=&\dfrac{1}{2 T} \\ \\
\dfrac{1}{10\text{ h}}&-&\dfrac{1}{15\text{ h}}&=&\dfrac{1}{2 T} \\ \\
\dfrac{3}{30\text{ h}}&-&\dfrac{2}{30\text{ h}}&=&\dfrac{1}{2 T} \\ \\
&&\dfrac{1}{30\text{ h}}&=&\dfrac{1}{2 T} \\ \\
&&2T&=&30\text{ h} \\ \\
&&T&=&\dfrac{30\text{ h}}{2} \\ \\
&&T&=&15\text{ h}
\end{array}\) - \(\begin{array}{rrcrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\dfrac{1}{\text{full}}&-&\dfrac{1}{\text{empty}}&=&\dfrac{3}{4T} \\ \\
\dfrac{1}{6\text{ min}}&-&\dfrac{1}{8\text{ min}}&=&\dfrac{3}{4T} \\ \\
\dfrac{4}{24\text{ min}}&-&\dfrac{3}{24\text{ min}}&=&\dfrac{3}{4T} \\ \\
&&\therefore \dfrac{1}{24\text{ min}}&=&\dfrac{3}{4T} \\ \\
&&T&=&\dfrac{3}{4}(24\text{ min}) \\ \\
&&T&=&18\text{ min}
\end{array}\) - \(\begin{array}{rrcrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\dfrac{1}{H}&+&\dfrac{1}{C}&=&\dfrac{1}{T} \\ \\
\dfrac{1}{H}&+&\dfrac{1}{3.5\text{ min}}&=&\dfrac{1}{2.1\text{ min}} \\ \\
&&\dfrac{1}{H}&=&\dfrac{1}{2.1\text{ min}}-\dfrac{1}{3.5\text{ min}} \\ \\
&&\dfrac{1}{H}&=&\dfrac{50}{105\text{ min}}-\dfrac{30}{105\text{ min}} \\ \\
&&\dfrac{1}{H}&=&\dfrac{20}{105\text{ min}} \\ \\
&&\dfrac{1}{H}&=&\dfrac{4}{21\text{ min}} \\ \\
&&H&=&\dfrac{21}{4}\text{ min} \\ \\
&&H&=&5.25 \text{ min}
\end{array}\) - \(\begin{array}{rrcrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\dfrac{1}{A}&+&\dfrac{1}{B}&=&\dfrac{1}{T} \\ \\
\dfrac{1}{4.5\text{ h}}&+&\dfrac{1}{B}&=&\dfrac{1}{2\text{ h}} \\ \\
&&\dfrac{1}{B}&=&\dfrac{1}{2\text{ h}}-\dfrac{1}{4.5\text{ h}} \\ \\
&&\dfrac{1}{B}&=&\dfrac{9}{18\text{ h}}-\dfrac{4}{18\text{ h}} \\ \\
&&\dfrac{1}{B}&=&\dfrac{5}{18\text{ h}} \\ \\
&&B&=&\dfrac{18\text{ h}}{5} \\ \\
&&B&=&3.6\text{ h}
\end{array}\)
