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A & B can complete a job in 8 days and B & C can do it in 12 days. A & B commence the work and do it for 4 days, then A leaves. B continues for 2 days and leave. C starts working and finishes the job. How many days did C take to finish it?

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    $\begingroup$ Please read tag descrptions before you add them - your question clearly doesn't fall under (math-software). $\endgroup$ – Wojowu Apr 10 '16 at 13:41
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    $\begingroup$ Welcome to MSE ! On this site, you are expected to show the effort you have made to solve the problem; pl. show what you have tried and where you are stuck. $\endgroup$ – true blue anil Apr 10 '16 at 13:42
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Let $a$,$b$ and $c$ be the fraction of the job that $A$,$B$ and $C$ resp. can do in one day.

Then the first fact gives us that $a + b = \frac{1}{8}$, or $8a + 8b = 1$.

The second gives that $b + c = \frac{1}{12}$, or $12b + 12c = 1$.

So we have $4(a+b) + 2b$ fractions of the work before $C$ takes over.

Adding the first equations and dividing by 2 gives that $4a + 10b + 6c = 1$. And $C$ starts at $4a + 6b$, which leaves $4b + 6c$ of a job. This is $\frac{1}{3} + 2c$. But I cannot deduce how long $C$ would work over a third of a job by himself (which is what he has to do after 2 days of solo work). So I think it's an underdetermined problem as stated.

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