# unitary method

• Aug 22nd 2006, 08:09 PM
joshirabi
unitary method
please do this problem using chain rule :)
A, B and C can finish a piece of work in 30, 40, and 60 days respectively. 10 days after they started to work together B leaves. 4 days after B left, A leaves and C completes the remaining work. Find how many days C had worked altogether.
• Aug 22nd 2006, 08:52 PM
CaptainBlack
Quote:

Originally Posted by joshirabi
please do this problem using chain rule :)
A, B and C can finish a piece of work in 30, 40, and 60 days respectively. 10 days after they started to work together B leaves. 4 days after B left, A leaves and C completes the remaining work. Find how many days C had worked altogether.

I don't know what you think the chain rule is but here is a solution.

A does 1/30 of the work per day, B does 1/40 and C 1/60.

Total work done by B is 10/40=1/4. Total work done by A is 14/30=7/15.

So total work done by A and B is 1/4+7/15=43/60, leaving 17/60 of the
work to be done by C. However C does 1/60 of the work per day, hence
works for a total of 17 days.

RonL
• Aug 22nd 2006, 09:00 PM
joshirabi
thanks
chain rule is
making fraction of ratio
take two product of fractions
equlize the product
suppose the value by x
making two product equlize
and find the value of x
• Aug 22nd 2006, 09:02 PM
CaptainBlack
Quote:

Originally Posted by joshirabi
chain rule is
making fraction of ratio
take two product of fractions
equlize the product
suppose the value by x
making two product equlize
and find the value of x

And there was I thinking it was something to do with taking derivatives :)

RonL