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A can do a piece of work in $10$ days, B in $20$ days and C in $30$ days. If A is assisted by B and C turn by turn in alternate days, in how many days the work might have been completed?

My Attempt:

In $1$ day, A can do $\frac {1}{10}$ work.

In $1$ day, B can do $\frac {1}{20}$ work.

In $1$ day, C can do $\frac {1}{30}$ work.

Now, how should I complete?

I know the question already has an answer, but I could not understand that. Can anyone give me a clear idea?

Thomas
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xxyy aazz
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    Fine, you have made a start. So now continue. On the first day what proportion of the total work is completed? What about the second day? – almagest Jun 08 '16 at 11:55

1 Answers1

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On the first day, $\frac{1}{10} + \frac{1}{20} = \frac{3}{20} = \frac{9}{60}$ work gets done. On the second day, $\frac{1}{10} + \frac{1}{30} = \frac{4}{30} = \frac{8}{60}$ more work gets done, making $\frac{17}{60}$ in total. Total work continues to increase alternatingly by $\frac{9}{60}$ and $\frac{8}{60}$, and on the third and following days we have $\frac{26}{60},~ \frac{34}{60},~ \frac{43}{60},~ \frac{51}{60},$ and $ \frac{60}{60}$ work done in total, respectively. The work is therefore completed after $7$ days.

Anon
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