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

My attempt; In $10$ days, $A$ can do $1$ work

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

Also, In $1$ day, $B$ can do $\frac{1}{20}$ work And, In $1$ day, $C$ can do $\frac{1}{30}$ work. Now, Whay should I do after this?

Iaamuser user
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2 Answers2

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You're definitely on track to the correct solution. The first day, $\frac1{10} + \frac{1}{20} = \frac{3}{20}$ work is done. The next day, $\frac1{10} + \frac{1}{30} = \frac{4}{30}$ work is done. After two days, therefore, $\frac{3}{20} + \frac{4}{30} = \frac{17}{60}$ work is done in total. Can you see how to finish from here?

Arthur
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  • could you give a hint to the next step. – Iaamuser user Feb 13 '16 at 09:43
  • @Iaamuseruser How much work is done the third day? The fourth day? Carry on until you reach a total of $\frac{60}{60}$ work done. – Arthur Feb 13 '16 at 09:58
  • but is it not a bit boring and lengthy.? – Iaamuser user Feb 13 '16 at 10:01
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    @Iaamuseruser What is wrong with that? It won't be that lengthy. OK, the (slightly) shorter version: You know that after two days, one day with help from $B$ and one day with help from $C$, $\frac{17}{60}$ work has been done. After two more days (identical to the first two days), $\frac{34}{60}$ work has been done. Two more days, and $\frac{51}{60}$ has been done, and you know that you're close. – Arthur Feb 13 '16 at 10:06
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In first 2 days work done by them = 2/10 + 1/20 + 1/30 = (2*6+1*3+2)/60 = 17/60 ( on day1 - 1/10 + 1/20 = 9/60 and on day 2 = 1/10 + 1/30 = 8/60) this cycle will be repeated until the work will get completed. So if the total work is 60 units each day work will be 9, 8,9,8 9...and so on' where 9+8+9+8+9+8+9 = 60 so work will be completed in 7 days.

Kouts Km
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