SOLUTION: A, B and C together can complete a piece of work in 10 days. All the three started working at it together and after 4 days, A left then B and C together completed the work in 10 mo

Algebra ->  Rate-of-work-word-problems -> SOLUTION: A, B and C together can complete a piece of work in 10 days. All the three started working at it together and after 4 days, A left then B and C together completed the work in 10 mo      Log On


   



Question 962901: A, B and C together can complete a piece of work in 10 days. All the three started working at it together and after 4 days, A left then B and C together completed the work in 10 more days. A alone can complete the work in?
Answer by LinnW(1048) About Me  (Show Source):
You can put this solution on YOUR website!
Set a = the rate of A in (piece of work)/day
b = the rate of B in (piece of work)/day
c = the rate of C in (piece of work)/day
Below is the equation for all three working together.
( 10 days ) * ( a + b + c ) = 1(piece of work)
Next we have the equation were all work together for 4 days,
then only B and C continue for 10 more days.
( 4 days) * ( a + b + c ) + (10 days)( b + c) = 1(piece of work)
Since each left hand side is equal to 1 piece of work,
we have
( 10 days ) * ( a + b + c ) =
( 4 days) * ( a + b + c ) + (10 days)( b + c)
and
10( a + b + c ) = 4( a + b + c ) + 10( b + c)
10a + 10b + 10c = 4a + 4b + 4c + 10b + 10c
combining terms
10a + 10b + 10c = 4a + 14b + 14c
Notice that we can rewrite the above as
10a + 10(b + c) = 4a + 14(b + c)
if we subtract 10(b + c) from each side
10a = 4a + 4(b + c)
subtract 4a from each side
6a = 4(b + c)
divide each side by 4
6a/4 = (b + c)
Recall
( 10 days ) * ( a + b + c ) = 1(piece of work) or
10(a + b + c) = 1
so
10a + 10(b + c) = 1
Now substitute 6a/4 for (b + c) and we have
10a + 10(6a/4) = 1
10a + 15a = 1
25a = 1
This means that A at rate a will take 25 days to complete the work.