Question 684987: A can do a piece of work in 24 days, B in 32 days and C in 64 days. all begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of work. how many days did the work last ?
Answer by Jolliano(16)   (Show Source):
You can put this solution on YOUR website! A = 24 days
B = 32 days
C = 64 days.
Fraction of work done in one day
by A = 1/24
by B = 1/32
by C = 1/64.
During the six days before A leaves,the fraction of the work that has been done is
6/24 = 1/4 by A
6/32 = 3/16 by B
6/64 = 3/32 by C.
Total fraction of work done in First 6 days =
1/4 + 3/16 + 3/32
=(8 + 6 + 3)/32
= 17/32.
Fraction of work left = 1 -17/32 = 15/32.
Since B leaves 6 days before the completion of the work,it means that C works alone for 6 days. The fraction of work C does alone in those 6 days is
6/64 = 3/32.
Therefore fraction of work done before B's departure =
15/32 - 3/32 = 12/32
=3/8.
Let X be the number of days in which B and C did 3/8 of the work.
Therefore,
x/32 is done by B
and x/64 is done by C.
This shows that
x/32 + x/64 = 3/8
Multiply through by 64
2x + x = 3 * 8
3x = 24
x = 24/3 = 8 days.
Total number of days
= 6 + 8 + 6 = 20 days.
|
|
|
