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I have a statement that says:

In doing a job, Emilio and Sebastián take 2 months, Emilio and Rolando take 3 months, Sebastián and Rolando take 6 months. So, how long will the three in doing this work, if they work together?

I tried to make a system of equations, but I came up with an erroneous result.

I tried to solve with reasons, but I did not get anywhere. It is also assumed, that I should solve this in less than 2 minutes, but I have been trying for 45 minutes and I can not find the form, how could I then solve it?

I also know the equation to state the time it will take, which is: $\frac{1}{t} = \frac{1}{s} + \frac{1}{r} + \frac {1}{e}$, where $ t = $ total time among those, $ s = $ Sebastian's time, $ e = $ Emilio's time, $ r = $ Rolando's time.

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  • $\begingroup$ "Emilio and Sebastián take 2 months" means $\frac1e+\frac1s=\frac12$. Also, i would suggest solving for $\frac1e,\frac1s,\frac1r$ as the unknowns instead of $e,s,r$. It's a lot easier. $\endgroup$ – Arthur Apr 28 '18 at 5:12
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    $\begingroup$ Could one of them be very lazy ? $\endgroup$ – Claude Leibovici Apr 28 '18 at 5:14
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    $\begingroup$ As mentioned by Arthur, you could take the reciprocals of the time taken, add them up and divide them by two. $\endgroup$ – Helix Apr 28 '18 at 5:15
  • $\begingroup$ Helix, why he did that ? $\endgroup$ – Eduardo S. May 1 '18 at 1:21
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method 1:

let rates of working of Emilio , Sebastian and Rolando be E,S and R respectively

$E+S=\dfrac{1}{2} $

$S+R=\dfrac{1}{6}$

$R+E=\dfrac{1}{3}$

adding all of them

$2(E+S+R)=1$

$E+S+R=\dfrac{1}{2}$

so, if they (all three) work togeather the work will be finished in $2$ months

alternative method:

i am taking work in terms of "days " not in "months"

assume total units of work = $LCM (2\times 30,6\times30,3\times30)=6 \times 30$ units

since $E$ and $S$ together finish work in $60 $ days,work done by them in a day=$180/60=3 unit/day$

$S$ and $R$ together finish work in $180 $ days,work done by them in a day=$180/180=1 unit/day$

$R$ and $E$ together finish work in $90$ days,work done by them in a day=$180/90=2 unit/day$

If they all work togeather work done by them in a day =$3\dfrac{units} {day}$

so, all of them will together finish work of $180$ units in =$ \dfrac{180}{3} $days=$2$ months

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  • $\begingroup$ My problem to understand the first method, is just after: adding all of them, How did you manage to put them together? Where did that $ 2 $ come from? Could you explain the reasoning or method you used? $\endgroup$ – Eduardo S. Apr 28 '18 at 15:56
  • $\begingroup$ The same with the second method, my problem of not understanding it is right in the step: If they all work togeather work done by them in a day, how did you get your work together? $\endgroup$ – Eduardo S. Apr 28 '18 at 15:58
  • $\begingroup$ Help me @veeresh $\endgroup$ – Eduardo S. May 1 '18 at 1:20
  • $\begingroup$ His answer already tells you the answer $\endgroup$ – QuIcKmAtHs May 2 '18 at 0:58
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You can write, $$\frac{1}{e}+\frac{1}{s}=\frac{1}{2}$$ $$\frac{1}{e}+\frac{1}{r}=\frac{1}{3}$$ $$\frac{1}{s}+\frac{1}{r}=\frac{1}{6}$$

From these series of simultaneous equations, you can easily get your desired answer.

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    $\begingroup$ Unfortunately, it's not such an easy thing. The important thing to notice is that Rolando does no work, in which case $1/r$ would have to equal zero. I assume this is what the asker meant when they mention an erroneous result. Veeresh Pandey had another answer that focused on rates rather than completion times that avoids this issue, but unfortunately he's deleted it. $\endgroup$ – user296602 Apr 28 '18 at 5:27
  • $\begingroup$ thanks a lot @ T . Bongers. $\endgroup$ – Faraday Pathak Apr 28 '18 at 5:43
  • $\begingroup$ @T.Bongers would it then seem more understandable if i replace 1/r with r and each corresponding with the reciprocal as veeresh has done? $\endgroup$ – QuIcKmAtHs Apr 28 '18 at 9:18
  • $\begingroup$ Ah yes, parallel impedances. $\endgroup$ – Mateen Ulhaq Apr 28 '18 at 11:43
  • $\begingroup$ @T.Bongers I would appreciate, if you tell me how you got the conclusion that rolando does not work. $\endgroup$ – Eduardo S. Apr 28 '18 at 15:49

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