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?

2 Answers
May 30, 2018

#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)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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")#

May 31, 2018

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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