Combined Work problem , A, B & C

[sukh]

There are 3 workers A, B, & C ?

If A & B work together, the job will finish in 4 hours
If A & C work together, the job will finish in 6 hours
If B & C work together, the job will finish in 5 hours.
Find out if all 3 work together, in how many hours they will finish the job?
and how long would each take to do the job ?

kindly reply with any shortcut formulae etc

[Frankenstein]

Hi,
1/T = (1/t1) + (1/t2) + ..... + (1/tn)
Let t1,t2,t3 be the time taken for A,B,C to finish the work respectively
1/4 = (1/t1) + (1/t2)
1/6 = (1/t1) + (1/t3)
1/5 = (1/t2) + (1/t3)
Adding all 3 we get 2((1/t1) + (1/t2) + (1/t3)) = (30+20+24)/120
So, (1/t1) + (1/t2) + (1/t3) = 37/120
So, time taken for all to do finish the work together is (120/37) hrs
No, 1/t1 = (1/t1) + (1/t2) + (1/t3) - ((1/t2) + (1/t3)) = 37/120 - 1/5 = 13/120
So, A alone takes 120/13 hours to finish the work.
Similarly for B and C.

[sukh]

Thanx a lot .

[Troy B]

Frankenstein - would you mind expanding on 'adding all 3 we get 2((1/t1) + (1/t2) + (1/t3)) = (30+20+24)/120'

Thanks.

[Frankenstein]

Frankenstein - would you mind expanding on 'adding all 3 we get 2((1/t1) + (1/t2) + (1/t3)) = (30+20+24)/120'

Thanks.

Okay...
1/4 = (1/t1) + (1/t2) --- eqn(1)
1/6 = (1/t1) + (1/t3) --- eqn(2)
1/5 = (1/t2) + (1/t3) --- eqn(3)
Adding the above 3 equations we get 2((1/t1) + (1/t2) + (1/t3)) = 1/4 + 1/6 + 1/5 = (30+20+24)/120.