A is more efficient than B and they together can complete a work in 24 days. Had A done 50% of the work and then B, the remaining work, then the work would have been done in 50 days. B alone will complete 40% of the same work in ?
Solution:
Let A alone take x days
And B alone take y days
x/2+ y/2= 50
x+y = 100
x = 100-y
So 1/{100-y} + 1/y = 1/24
24y + 24{100-y} = y{100-y}
24y + 2400 -24y = 100y - y^2
y^2 -100y + 2400 = 0
{y-60} {y-40} = 0
y = 40 OR 60
Since A is more efficient , B takes 60 days A takes 40 days
So 40% of the work will take B 40/100 X 60 = 24 days
Alternative method:
ReplyDeleteLet the number of days required by A and B to complete the work be a and b, respectively.
A will do 50% of the wok in a/2 days and B in b/2 days.
We are given, a/2 + b/2 = 50
=> a + b = 100 or a = 100 - b
In one day, A will do is 1/a th of the work and B will do 1/b th of the work.
Working together, in one day they will do 1/a + 1/b of the work which is 1/24 of the total work as they take 24 days to finish the work together.
So, 1/a + 1/b = 1/24
=> (a + b)/ab = 1/24
Replace a + b as 100 and a as 100 - b in the above equation,
100/b(100 - b) = 1/24
=> 100/b(100 - b) = 100/2400
=> b(100 - b) = 2400
Factorizing 2400 such that the two factors have the above relation,
b(100 - b) = 40 x 60
=> b = 40 or 60
as a + b = 100,
If b is 40, a is 60 an if b is 60 a is 40.
But we are given that A is more efficient than B. That means A takes lesser time to complete the work.
=> a < b
=> a = 40 and b = 60
So B takes 60 days to finish the complete work
To complete 40% of the work, b would take 40% of 60 days,
= 24 days.
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