A and B can do a piece of work in 12 days B and C in 15 days and C and a in 20 days how much time will A alone take to finish the work
Answers
Step-by-step explanation:
A and B can do a piece of work in 12 days, B and C in 15 days, and C and A in 20 days. How much time will A alone take to finish the job?
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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
Given:
A and B can do a piece of work in = 12 days
Work done by A and B in 1 day = 1/12
B and C can do a piece of work in = 15 days
Work done by B and C in 1 day = 1/15
A and C can do a piece of work in = 20 days
Work done by A and C in 1 day = 1/20
Solution:
On adding A, B and C we get,
2(A+B+C)’s one day work
= 1/12 + 1/15 + 1/20
= (5+4+3)/60 (by taking LCM for 12, 15 and 20 which is 60)
= 12/60
= 1/5
A+B+C one day work = 1/(5×2) = 1/10
Knowing that:
Work A can do in 1 day
= (A+B+C)’s 1 day work – (B+C)’s 1 day work
= 1/10 – 1/15
= (3-2)/30 (by taking LCM for 10 and 15 which is 30)
= 1/30
Answer:
∴ A alone can finish the work in = 1/(1/30) = 30days.