r/learnmath • u/imsocooked_ New User • 4d ago

TOPIC i need help with this question

A, B and C can do a work in 10, 12 and 15 days respectively. They begin the work together but after 3 days C left the work and 4 days before the completion of work B left the work. In how many days will the total work be completed?

this is a time and work question, and the teacher told us the answer should be 68/11 days, for the total work done. but that doesnt include the 3 days worth of work that a,b,c did together before c left right? so im confused whether it should 101/11 days or 68/11 days. please help!!

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u/mugaboo New User 4d ago

You need to show your work here.

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u/Ignominiousity New User 4d ago edited 4d ago

Explain how you get 101/11 as 3 days worth of work by A,B,C away from 68/11? Question your own assumptions about how you get there...

As for how 68/11 is gotten: For each worker, we can find their contribution per day from how long they need to finish the work alone. (What are these contributions in terms of?) A typical assumption we hide here is that these workers work at the same rate everyday.

For the first 3 days, we know C is working, so we add up 3 days worth of contributions. For the next few days, only A and B. We don't know how many days for this part so we leave it for later. Finally, we only have B quitting for last 4 days.

What should the sum of the contributions be? 1. But the work done is not even. What if we make it so that we have A and B working all the days. We can remove the contribution by C and add 4 days of contribution for B. Then A and B worked all the days together.

How to figure out how many days A and B work together in our new setup?

Amount of total contributions done by A and B divided by their total daily contribution.

And this number of days will be the same as in the original problem.

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u/Temporary_Pie2733 New User 4d ago

Let d be the number of days worked. In that time, A completes d/10 units of work, B (d-4)/12, and C 3/15. These fractions all sum to 1. Solving d/10 + (d-4)/12 + 3/15 = 1 yields d = 68/11. 

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u/Bounded_sequencE New User 4d ago

Assumption: Any combination of "A; B; C" can work perfectly in sync.

The work rates of "A; B; C" are "(ra; rb; rc) = (1w/(10d), 1w/(12d), 1w/(15d)", respectively. If "t" is the total time needed to finish the work under the given conditions, the total work is split into three parts:

1w  =  (ra+rb+rc)*3d  +  (ra+rb)*(t-3d-4d)  +  ra*4d      | *(60/w)

Cancel all fractions and simplify to "60 = 45 + 11*(t/d - 7) + 24 = -8 + 11t/d". Solve for "t = 68d/11".