$A$, $B$ and $C$ can do a ...

Question

$A$, $B$ 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.

Answer 1

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.

Answer 2

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.

Answer 3

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.