Question

One job can be finished by 30 workers for 60 days. The work was started by 20 workers, and after 10 days came 5 workers. For how long will the whole work be done?

Answer 1

`color(brown)("64 +10 = 74 days. ")`

This is not an efficient way of solving it but demonstrates what is actually going on. My other solution is more efficient.

Explanation:

Assumption: all workers have the same rate of work per day.

Let me introduce you to a concept that may be new to you.

Effort verses time = total work done.

Let the amount of work effort be `w`
Let the generic time in days be `t`
Set time for 10 days as `t_10`
Set unknown time as `t_x`

Let the total amount of effort for the job be `W`
Let the unknown element of time be

The actual value of `W` does not matter in this case

`color(blue)("Initial condition")`

`wxxt=W`

Given that the count of workers is 30 and that the time they worked was 60 days giving:

`30wxx60=W" ".................................Equation(1)`
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`color(blue)("Determine the work rate for 1 worker")`

From `Eqn(1)` we may determine the amount of effort (`w`) contributed by 1 worker for 1 day

`30wxx60=W`

`30w=W/60`

`w=W/(30xx60) = W/1800`
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`color(blue)("Combining the work force")`

Known that total time spent working as a team was: `t_10+t_x`

20 workers worked for 10 days on their own `->20wxxt_10`

They we joined by 5 workers so the effort they contributed was `5wxxt_x`

But the initial 20 workers worked alongside the additional 5 for the same time of `t_x`. So now we have:

`20w(t_10+t_x)+(5wxxt_x)=W" "......................Equation(2)`
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We know that `t_10=10" days"`
We know that `w=W/1800`

Substitute these values into `Eqn(2)`

`[20(W/1800)(10+t_x)]+[5(W/1800)t_x]=W`

`color(white)("d")W/90(10+t_x)color(white)("ddddd")+color(white)("ddd")(W/360)t_xcolor(white)("d")=W`

`color(white)("dd")W/9 +W/90 t_xcolor(white)("ddd.d")+color(white)("dddd")W/360 t_xcolor(white)("ddd")=W`

`color(white)("d")`
`color(white)("d")`

`W/9+W/72 t_x=W`

`W/72t_x=W-W/9`

`t_x=(72(9cancel(W)-cancel(W)))/(9cancel(W))color(brown)(larr" This is why the value of "W)` `color(brown)(color(white)("dddddddddddddddddddddd")"does not matter")`

`t_x=(72xx8)/9 = 64 larr" Days"`
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`color(brown)("Get the equations correct and the "W" should always cancel")`

Answer 2

By special request: By 'proportional' method

64 + 10 =74 days

Explanation:

`color(blue)("Important fact about being 'proportional'")`

You have probably come across this before

`"directly proportional "->y=kx`
`"inversely proportional "->y=k/x`

Where `k` is some constant value.

Consider the inversely proportions `y=k/x`
From this we have `k=yx`

Set `y="count of workes " and x=" count of days"`

Then for the units we have:
`yxx x = "count of workes "xx" count of days"= "workerdays"`

workerdays is a sort of index number that just sets the relationship between the variables `x and y`

`color(brown)("This is what I am going to use")`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Answering the question")`

Let the unknown count of days be `x`

Total amount of effort `=30xx60 = 1800color(white)("dd")"worker days"`

Worker days would be an equivalent measure of effort. A sort of index value.

Note that the initial count of workers (20) worked for 10 days + the time (`x`) that the 5 worked. Giving:

`(20xx10)+(20xx x)+(5xx x)=1800 " workerdays"`

`200+20x+5x=1800`

`200+25x=1800`

`25x=1800-200=1600`

`x=1600/25 = 64`

So total time is `10+x=10+64=74`
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