

Hi, dear supporters: James can finish a job in 2 hours while Hannah can finish the same job in 3 hours. If they start working together at $9a\phantom{\rule{0.12em}{0ex}}.m\phantom{\rule{0.12em}{0ex}}.,$ at what time will they finish the job? $O\phantom{\rule{0.12em}{0ex}}10:00$ am $o\phantom{\rule{0.12em}{0ex}}10:05$ am $o\phantom{\rule{0.12em}{0ex}}10:12$ am $o\phantom{\rule{0.12em}{0ex}}10:15$ am do you a formula for this type of problems? $i\phantom{\rule{0.12em}{0ex}}$ was given this answer but do not understand how it was obtained. I know what formula to use when the two persons are working together and the problem asks how long it will take for both of them working together to finish. I know that formula. but here $i\phantom{\rule{0.12em}{0ex}}$ am confused becasue the problem wants me to find at what time they will finish. Thanks for any help. James can do $\frac{1}{2}$ the work in an hour while Hannah can to $\frac{1}{3}$. Together they can do $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ of the work in an hour. So it will take $\frac{6}{5}$ hours $=1$ hour $12$ minutes They will finish at $10:12$ am Für alle, die mir helfen möchten (automatisch von OnlineMathe generiert): "Ich möchte die Lösung in Zusammenarbeit mit anderen erstellen." 

$i\phantom{\rule{0.12em}{0ex}}$ can apply this formula 1/Tab = 1/Ta $+$ 1/Tb but this is to find how long they will take to do the job working together. 

Hello. >> I know what formula to use when the two persons are working together and the problem asks how long it will take for both of them working together to finish ... becasue the problem wants me to find at what time they will finish. << I don't see a contradiction/problem. You have at start time. To evaluate the finish time you calculate first the duration time to finish the the job. Then $\text{finish time=start time + duration time}$ The term of the given solution seems the right one for me. Let's denote $\left[\frac{j\phantom{\rule{0.12em}{0ex}}}{h\phantom{\rule{0.12em}{0ex}}}\right]$ the unit "finished job per hour". So $0.7\left[\frac{j\phantom{\rule{0.12em}{0ex}}}{h\phantom{\rule{0.12em}{0ex}}}\right]$ means that you finish 70% of the job in one hour. Then the equation is $(\frac{1}{2}\left[\frac{j\phantom{\rule{0.12em}{0ex}}}{h\phantom{\rule{0.12em}{0ex}}}\right]+\frac{1}{3}\left[\frac{j\phantom{\rule{0.12em}{0ex}}}{h\phantom{\rule{0.12em}{0ex}}}\right])\cdot x\phantom{\rule{0.12em}{0ex}}\phantom{\rule{0.167em}{0ex}}\left[h\phantom{\rule{0.12em}{0ex}}\right]=1\left[j\phantom{\rule{0.12em}{0ex}}\right]$ LHS: You add the two fractions to get the proportion of the finished job in one hour. This sum you multiply by the number of hours $x\phantom{\rule{0.12em}{0ex}}\phantom{\rule{0.167em}{0ex}}\left[h\phantom{\rule{0.12em}{0ex}}\right]$ to finish the job. The unit $h\phantom{\rule{0.12em}{0ex}}$ is cancelling. RHS: $1\left[j\phantom{\rule{0.12em}{0ex}}\right]$ is the job which has to be finished. greetings, pivot 

$\frac{1}{2}+\frac{1}{3}=\frac{1}{x\phantom{\rule{0.12em}{0ex}}}\cdot 6x\phantom{\rule{0.12em}{0ex}}$ (common denominator) $3x\phantom{\rule{0.12em}{0ex}}+2x\phantom{\rule{0.12em}{0ex}}=6$ $5x\phantom{\rule{0.12em}{0ex}}=6$ $x\phantom{\rule{0.12em}{0ex}}=$ time needed if both do the job together $x\phantom{\rule{0.12em}{0ex}}=\frac{6}{5}$ hours $=1$ hour $12$ minutes $9+1h\phantom{\rule{0.12em}{0ex}}+20min=10:12,12$ past 9. 

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