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If $A$ can complete a job in $6$ days. $B$ can complete it in $10$ days.If they work on alternate days, in how many days the job would be completed ? Does it depend on who ($A$ or $B$) starts the job?

The solution suggested in my module is to find the work completed by both of them in $2$ days and if this is not an integer then the work will be completed faster if the faster worker starts the job. But I am not sure how/why this works.

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Let's suppose that the job entails 30 (the least common multiple of 6 and 10) equal units of work. Since $A$ can complete the job in 6 days, $A$ does 5 units of work per day. Since $B$ can complete the job in 10 days, $B$ does 3 units of work per day. So, if they alternate, in every 2-day period, they do 8 units of work, which is $\frac{8}{30}=\frac{4}{15}$ of the whole job.

To get at the "how/why" of starting with the faster worker, let's make two tables: $$\begin{matrix} \text{If A goes first...} & & \text{If B goes first...} \\ \begin{matrix} \text{day} & \text{units today} & \text{units total} \\ 1 & 5 & 5 \\ 2 & 3 & 8 \\ 3 & 5 & 13 \\ 4 & 3 & 16 \\ 5 & 5 & 21 \\ 6 & 3 & 24 \\ 7 & 5 & 29 \\ 8 & 3 & 32 \\ \end{matrix} & & \begin{matrix} \text{day} & \text{units today} & \text{units total} \\ 1 & 3 & 3\\ 2 & 5 & 8\\ 3 & 3 & 11\\ 4 & 5 & 16\\ 5 & 3 & 19\\ 6 & 5 & 24\\ 7 & 3 & 27\\ 8 & 5 & 32 \end{matrix}\end{matrix}$$

In the first case, going into the final day, there is 1 unit of work left to be done at a rate of 3 units per day, so it will take $\frac{1}{3}$ of that last day to finish. In the second case, going into the final day, there are 3 units of work left to be done at a rate of 5 units per day, so it will take $\frac{3}{5}$ of that last day to finish, which is longer. As you can see looking at the 7-day totals, if the job entailed 28 or 29 units of work (with the same 3 per day and 5 per day rates), the job could be finished in 7 days in the first case, but not in the second case.

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If a is the part of the total amount of work that $A$ gets done per day (i.e. $1/6=0.1666\dots$) and b is the part of the work that $B$ gets done per day (i.e. $1/10=0.1$), then $a+b$ is the amount of work that gets done every two days.

No matter who starts, the same amount of work will be done at the end of every two days, and the total amount of work done will look the same for every two-day cycle as well so we need only look at the two first days to see how the accumulated amount of work will progress.

In the interval between every two days, the accumulated amount of work done depends on who starts. If $f(t)$ is a function describing the work done when $A$ works first - then $B$, and $g(t)$ is a function describing the work done when $B$ starts first - then $A$ ($t$ being the number of days), we can for instance write them like this:

$$ \begin{align*} f (t) &= \begin{cases} a \cdot t, & t \in [0,1] \\ a + b, \cdot t & t \in (1,2] \end{cases} \\ g (t) & = \begin{cases} b \cdot t, & t \in [0,1] \\ b + a \cdot t, & t \in (1,2] \end{cases} \end{align*} $$

($t \in [0,1]$ meaning "for all times between $0$ and $1$", and likewise $t \in (1,2]$ "for all times between $1$ and $2$".)

By inspection, we see that as expected $g(0) = f(0)$, and $g(2) = f(2)$. However, in the interval between $0$ and $2$, the function beginning with the fastest worker (i.e. $f$, since $A$ is fastest) will always be greater than the other. So for any given $t$ that isn't a multiple of $2$, more work will have been done if the fastest worker works the first day.

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I don't really understand the function notations.May be a different approach of explaining might be helpful to me. –  Quixotic Jan 3 '11 at 13:01
    
I've marked up the equations so that they should look prettier now as well as added a couple of comments. –  josteinaj Jan 3 '11 at 23:34
    
The functions each consists of two "sub-equations". I would draw it, but I don't think math.stackexchange supports graphs in any way. Anyway, the essence is that the accumulated amount of work done is always greatest at any given time if the fastest worker goes first. However, every two whole days, or at any other time if they work equally fast, the accumulated amount of work done is equal. f and g only descibe one period of the two-day cycles, but if you change it to work for the entire timeline (add 2 ( a + b ) \cdot n or something) you could get the time of completion by solving f=1 and g=1 –  josteinaj Jan 3 '11 at 23:48
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A's 1 day work will be $\frac{1}{6}$ and B's 1 day work will be $\frac{1}{10}$. If they work together their combined work of 1 day will be $\frac{1}{6}+\frac{1}{10}=\frac{3}{10}$.

And they will together work in $\frac{10}{3}$ days.

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