SOLUTION: A,B, and C could finish a job alone in 5 hours. The three started together at 8AM but B rested at 9AM and resumed working at 9:30 AM. At what time they will finish the work?

The other tutor with his equation

t/5 + (t-.5)/5 + t/5 = 1

is wrong, first because the first term 
should be 1/5, not t/5, because from 8 till 9 
is 1 hour, not t hours.  His second term is 
wrong because B was not working during that 
half hour from 9 till 9:30, so he couldn't use 
1/5.  The last term is wrong because it is not 
known that they worked as long after 9:30 as 
they worked before 9. 

It is not known how long it would takes B working
alone to do the job.  This is another variable.  
If B takes x hours to do the job, then he does 
1/xths of the joh in one hour.  So if B had been 
working that half hour instead of resting, they 
would have completed (1/2)(1/5) or 1/(10)ths of 
the job during that half hour.  However, since B 
didn't work during that half hour we must subtract 
the portion of the job he would have done, which 
is (1/2)(1/x) = 1/(2x) so from 9:00 to 9:30 A and C
only did [1/10 - 1/(2x)] = (2x-10)/(2x) = 2(x-5)/(2x) 
= (x-5)/x of the job.
was done.

Now we don't know how long it will take after 9:30
to finish the job, for that depends on how fast B
works.   So we have to use another unknown
for that, y hours.  Let y = the time after 9:30 that 
it takes all three to finish the job.

1/5 + (x-5)/x + y/5 = 1

That simplifies to

y = (25-x)/x 

There is not enough information to give just 
one possible answer. There are many possible
answers.  Here's why.

Suppose B is such a slow worker that it would 
take him a long time to do the whole job by 
himself (maybe it would even take him as long 
as 10 years!) So A and B would do virtually all 
the work.  B's contribution to the work effort
of the A,B,C team is practically nothing.  So 
without B, the pair or workers A and C could 
virtually do the job in a tiny bit over 5 hours 
and finish at a tiny bit after 1:00PM.  
It wouldn't matter about B resting, since he
works so slow in this case.   

Now look at another extreme case.  Suppose B 
is the fast worker and it takes A and C ten 
years to do the job.   So it's practically the 
same as if B works 1/5th of the job in the
hour from 8 till 9, rests for a half hour 
and does the remaining 4/5th of a job in 4 
hours. He'd lose the half hour he rested, 
and finish at a little after 1:30PM.

Let's take an intermediate case, where A,B, and 
C all three work at the same rate. Then it would 
take each 3 times as long to do the job alone or 
3×5 or 15 hours each to do it alone.

So from 8:00AM till 9:00AM, they do 3 fifteenths
or 3/15ths of the job.  Then in the 1/2 hour from 
9:00 till 9:30, they do 
(1/2)(2/15) = 1/15 of the job.  So that's a total 
of 4/15ths of the job at 9:30AM. They still have 
11/15ths of the job left to do, and since they can 
do 3/15ths in one hour it will take them 
(11/15)÷(3/15) or 11/3 or 3 2/3 hours to finish. 
That's 3 hours and 20 minutes after 9:30AM so 
they'll finish at 12:50AM.

So there are many possibiloities.  Check the problem 
again.  I'll bet something was left out. But don't
go with the other tutor's solution.  
 
Edwin