Math, asked by pala7338, 1 year ago

A,b,c can do a piece of work in 150 days,75 days,75/2 days respectively if all three work alternatively starting with a then b followed by c then how many days work completed

Answers

Answered by sohailraja71
2

A takes 150 days to complete the work

so A's one day work = 1/150

B takes 75 days to complete the work

so B's one day work = 1/75 = 2 / 150

C takes 75/2 days to complete the work

so C's one day work = 2/75 = 4 / 150

on first day , A completes 1/150 work

on second day B adds 2/150 work

and on third day C adds 4/150 work

so after 3 days together their work is = 7/150

so 3 days -------- 7 / 150 work

30 days -------- 70 / 150 work

60 days -------- 140 / 150 work

63 days -------- 147 / 150 work

on 64 th day A adds 1/150 work

completed work becomes , 148/150

on 65 th day B adds 2/150 work

completed work becomes , 150/150

so in 65 days whole work complete.

Answered by amitnrw
2

Answer:

65 Days

Step-by-step explanation:

A,b,c can do a piece of work in 150 days,75 days,75/2 days respectively

A's 1 day work = 1/150

B's 1 day work = 1/75

C's 1 Day work = 2/75

A + B + C ' together 1 day work = 1/150 + 1/75 + 2/75

=(1/150)(1 + 2 + 4)

= 7/150

But as they do not work together but alternately

so 3 days works = 7/150

7/150  work is done  in  3 days

1 work is done in  = 3 / (7/150)

= 450/7

= 64.29 days

This time is on average basis of a b & c work

but for 63 days each a , b & c worked for 63/3 = 21 days but after that each person did not work equally

so work done in 21 * 3 = 63 days =  (7/150) * 21 = 147/150

Work left after that = 1 - 147/150 = 3/150

in 64th day a worked =  1/150

so after 64th day work remained = 3/150 - 1/150 = 2/150 = 1/75

B's 1 day work = 1/75

So b complete work on 65th Day

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