A can work 40% more than B. And B can work 20% less than C. If A take 9 days less than C to complete a work. Then how many days taken by B to complete that work. ?
Answers
Step-by-step explanation:
A is 40% more efficient than B , C is 20% less efficient than C
A,B,C together can finish work in 5 days
To find : How many days A alone will take to complete 70% of that work.
Consider complete work = 100%
Let A takes 'y' days to complete 70% work
Let B takes 'x' days to complete 100% work
Therefore B 's 1 day work = (100/x) % ---- (i)
Similarly A completes 140% work in 'x' days ----( given A's efficiency is 40% more than B )
Therefore A 's 1 day work = (140/x) % ----(ii)
Similarly C completes 80% work in 'x' days -- (given C's efficiency is 20% less than B)
Therefore C's 1 day work = (80/x)% -----(iii)
A,B and C combined work in 1 day = [(100/x) + (140/x) + (80/x)] %
Therefore,
A,B and C combined work in 5 days = 5* [(100/x) + (140/x) + (80/x)] % ---- (iv)
Given that A,B and C together completes the 100% work in 5 days ---(v)
Equating equation (iv) & (v) -- (Equating 5 days of work)
5*[(100/x) + (140/x) + (80/x)] % = 100 %
(100/x) + (140/x) + (80/x) = (100/5)
(100+140+80)/x = 20
(320)/x=20
x= 320/20
x = 16 days
Therefore B alone takes 16 days to complete 100 % work
Solving equation (ii) substitute x in (ii) ,
A 's 1 day work = (140/16) %
A's y days work = y*(140/16)% --(vi)
But A completes 70% work in y days
70 % = y *(140/16) %
y = (70*16)/140
y = 8 days
Therefore A takes 8 days to complete the 70 % work.