Let assume we have 2 machines M1 and M2
M1 can finish a task in say x hours (lets say x = 2)
M2 can finish a task in say y hours (lets say y = 3)

One approach is to assume TOTAL WORK to be done as some multiple of X and Y hours. Lets say total work = 12

M1 works on a task for 1 hour and then M2 joins M1 to finsh the work in Z hours, Calculate Z.

Now we know from above that M1 can do 12 units of WORK in 2 hours; so in 1 hour he completes 12/2 = 6 units of work.
similarly M2 can do 12 units of WORK in 3 hours; so in 1 hour he completes 12/3 = 4 units of work.

Now we know M1 worked alone for an hour. in that he must have finished 6 units
work remaining = 12 - 6 = 6 units
M1 + M2 together do 6 + 4 = 10 units of work in 1 hour (=60 mins)
to do 6 units of work, they will need: 60 * 6 / 10 = 36 mins

hence Z = 36 mins or 3/5 hours

With an example it will be easier...
It takes computer A 36 hours to process data. If Computer A start working at 7pm on monday and after half the data are processed computerB 3 times as fast as A joins in the task, at what time the process will be finish on tuesday?
I am looking for the algebraic way to solve this please.
Thx

claudayst wrote:What is the formula or rather the rationale for computing a combined rate but just for a portion of the work, for example two machines working together for a certain amount of time and then one of the machine stops while the other continues. The way around as well: one machine start working and then another one joins. Many thx in advance!

I think these examples should serve as an answer to your question:

Example 1--
A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days B had to leave and A alone completed the remaining work. The whole work was completed in :
Explanation:

(A + B)'s 1 day's work = ( 1/15 + 1/10 ) = 1/6

Work done by A and B in 2 days = ( 1/6 x 2 ) = 1/3

Remaining work = ( 1 - 1/3 ) = 2/3

Now, 1/15 work is done by A in 1 day.

Therefore 2/3 work will be done by A in ( 15 x 2/3 ) = 10 days.

Hence, the total time taken = (10 + 2) = 12 days.


Example 2--

X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?

Explanation:

Work done by X in 4 days = (1/20 x 4 ) = 1/5

Remaining work = (1 - 1/5) = 4/5

(X + Y)'s 1 day's work = (1/20 + 1/12) = 8/60 = 2/15

Now, 2/15 work is done by X and Y in 1 day.

So, 4/5 work will be done by X and Y in 15/2 x 4/5 = 6 days.
Hence, total time taken = (6 + 4) days = 10 days.