Question

Ans
A is 40% more efficient than B and C is 20% less efficient than B. Working together, they can finish a work is 5 days. In
how many days will A alone complete 70% of that work?
8
10
7
9​

Answer 1

Answer:

8 days

Step-by-step explanation:

Given : 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.