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$A$, $B$ and $C$ can do a piece of work in $30$, $40$ and $50$ days respectively. If $A$ and $B$ work in alternate days started by $A$ and they get the assistance of $C$ all the days, find in how many days the whole work will be finished?

My Attempt:

In $30$ days, $A$ does $1$ work. In $1$ day, $A$ does $\frac {1}{30}$ work.

In $40$ days, $B$ does $1$ work. In $1$ day, $B$ does $\frac {1}{40}$ work.

In $50$ days, $C$ does $1$ work. In $1$ day, $C$ does $\frac {1}{50}$ work.

In $1$ day, $(A+C)$ do $\frac {4}{75}$ work.

In $1$ day, $(B+C)$ do $\frac {9}{200}$ work.

I could not solve from here. Please help.

pi-π
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3 Answers3

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Suppose there are $\;x\;$ days $\;A+C\;$ work together, and $\;y\;$ days $\;B+C\;$ work together, then

$$\frac{4x}{75}+\frac{9y}{200}=1\stackrel{\cdot600}\implies 32x+27y=600$$

There are, of course, infinite solutions, but we'd like $\;x,y\;$ to be positive integers, and this restricts seriously the possibilities: $\;x=12\;,\;\;y=8\;$ fulfill the conditions.

Further explanation: since $\;32x\;$ is always even, observe that $\;y\;$ must be even too.

I'm now paying attention to the data that the working days are alternating, and thus Pieter21's solution is the way to go, though there is no "exact* number of full work days.

DonAntonio
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  • Could you please explain.how did you get the values of $x$ and $y$? – pi-π Oct 12 '16 at 17:28
  • @user354073 In this particular case, by try and error...and some estimation. As we need that $;32x=600-27y;$, check the very first even values of $;y;$ .... Observe that you won't have to go very high as the values wanted are positive , and $;27\cdot22;$ is close already to $;600;$ , so at most you'll need to check $;\frac{22}2=11;$ multiples of $;27;$ . – DonAntonio Oct 12 '16 at 17:31
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Divide the work in $600$ pieces (the least common multiple of the dividers).

  • $A$ does $\frac{600}{30} = 20$ pieces a day every other day.

  • $B$ does $\frac{600}{40} = 15$ pieces a day every other day.

  • $C$ does $\frac{600}{50} = 12$ pieces each day.

So alternating, $32 = 20+12 = A + C$, and $27 = 15+12 = B + C$ pieces of work get done each day.

We need to get $600$ pieces of work done, and because of the alternation, we get $59 = 32+27$ pieces done in two days.

After $\lfloor{\frac{600}{59}}\rfloor = 10$ alternations ($20$ days), $10$ pieces are left ($600 - 59 \times 10$).

The final $10$ pieces get done in $1$ day, as it is less than $32$.

So in total: $21$ days.

Pieter21
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  • I did not understand? Could you please clarify a bit more? – pi-π Oct 12 '16 at 16:03
  • I edited the answer. If still not clear, please try to explain which part is not clear. – Pieter21 Oct 12 '16 at 17:05
  • How did you get: Adoes 20 a day every other day,B does 15 a day every other day, and C does 12 each day.? – pi-π Oct 12 '16 at 17:13
  • @Pieter21 You've been a member for over two years and have posted some 139 answers. I think it is about time to learn how to type with MathJax, lest your answers will go unappreciated because it is hard to read them. – DonAntonio Oct 12 '16 at 17:21
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A takes-> 30 days, B takes-> 40 days, C takes-> 50 days Total unit of work they have to finish is(LCM of 30,40,50) i.e 600 units It means A does -> 20 unit of work in 1 day, B does -> 15 unit in 1 day and C does 12 units in 1 day.

On 1st day A+C does 32 units out of 600, On 2nd day B+C does next 27 units of remaining work.

It means they do 59 unit of work in 2 days. In 20 days 590 units will be done by them. Remaining 10 units are done by A+C in \frac{10}{32}$$ days.

Total days they took to finish the work 600/59 =10 \frac{5}{16}$$ = 20 \frac{5}{16}$$ days.

Akshay
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