Question

A and B can do a peice of work in 12 days A and C can do that work in 15 days B and C can do that work in 10 days so in how many days does A alone will do that work ​

Answer 1

Answer:

A and B do 1/12th of the work in a day.

B and C do 1/15th of the work in a day.

C and A do 1/20th of the work in a day.

So 2A + 2B + 2C do (1/12)+(1/15)+(1/20). LCM of 12, 15 and 20 is 300.

(1/12)+(1/15)+(1/20) = (25/300)+(20/300)+(15/300) = 60/300 or 1/5.

So A + B + C do 1/(2*5) = 1/10th.

To get A’s output subtract B and C’ output from the combined output of a, B and C, or (1/10)-(1/15) = (3/30)-(2/30) = 1/30th.

So A can do the work in 30 days, working alone.

Similarly B can do (1/10)-(1/20) = (2–1)/20 = 1/20th of the work in day. So he takes 20 days to do the work, working alone.

And C can do (1/10)-(1/12) = (6/60)-(5/60) = 1/60th of the work in a day. So C takes 60 days to do the work, working alone.

A takes 30 days, B takes 20 days and C takes 60 days, each working alone, to complete the work.