The rate of work for each contributor is the same
Let Alvin be by A
Let Simon be S
Let Theodore be T
Let the total amount of work done be w
Let time in hours be t
Let the rate of work be r
Which is a constant and attributed to each of them
Then #r(t_A+t_S+t_T)=w#
#color(blue)("~~~~~~~~~~~~~~~ Check units work ~~~~~~~~~~~~~~~~")#
the units for r is #("Amount if work")/("hour")#
The units for #t# is #("hour")#
so units#-> rt = ("Amount if work")/("hour") xx("hour") ="Amount if work "# which are the units for #w# so this works
#color(blue)("~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#
So
Let the start time for Alvin be #t_0 = 3:30#
Let the time start for Simon be #t_0 + 15/60 "minutes"#
Let the time start for Theodore be #t_0 +30/60 "minutes"#
Let the finish time be #t_f#
Consider each persons contribution
#A->r(t_f - 3 30/60) ->r(t_f -3 1/2)#
#S->r(t_f -(3 30/60 +15/60))->r(t_f -3 3/4)-#
#T->r(t_f-(3 30/60 +30/60))->r(t_f-4 ) #
so #r(t_f - 3 1/2) + r(t_f -3 3/4)+r(t_f-4) =w#
Then #r[(t_f - 3 1/2) + (t_f -3 3/4)+(t_f-4)] =w#
We know that the job can be completed in 4 hours for any one so the rate per hour (r) = 1/4
#color(green)("As each contributes a portion of the job they are working to")#
#color(green)("complete 1 job. Thus "color(red)(" w=1")#
So now we have:
#1/4[(t_f - 3 1/2) + (t_f -3 3/4)+(t_f-4)] =1#
Multiply both sides by 4
#(t_f - 3 1/2) + (t_f -3 3/4)+(t_f-4 ) =4#
Collecting like terms
#3t_f -11 1/4=4#
#color(blue)(3t_f)color(brown)(=4+11 1/4)color(blue)( =15 1/4color(brown)( =61/4 )#
divide both sides by 3
#t_f= 61/12# Remember this is for a 12 hour clock
So as we are counting from 12 o-clock the next time on the clock face is 1 o-clock. Thus we have to # color(blue)("start from zero")# to get 1 o-clock after 1 hour
#[0+ 5 +(1/12xx60)] "hours oclock"#
#t_f=# 5 minutes past 5
