SOLUTION: A does 3/10 of a work in 6 days .He then calls in B and they finish the work in 7 days more .How long would B take to do the work by himself?
A,B,C working together ,can do
Algebra ->
Customizable Word Problem Solvers
-> Misc
-> SOLUTION: A does 3/10 of a work in 6 days .He then calls in B and they finish the work in 7 days more .How long would B take to do the work by himself?
A,B,C working together ,can do
Log On
Question 609447: A does 3/10 of a work in 6 days .He then calls in B and they finish the work in 7 days more .How long would B take to do the work by himself?
A,B,C working together ,can do a piece of work in 18 days .A can do it alone in 72 days .After working for 6 days ,A fell ill.How long will B and C together take to finish the remaining work? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A does 3/10 of a work in 6 days .He then calls in B and they finish the work in 7 days more .
How long would B take to do the work by himself?
:
Find how long (a), it takes for A to do the job by himself a = 6
3a = 6(10)
3a = 60
a = 60/3
a = 20 days, A by himself
:
Let b = B's time by himself
Let the completed job = 1
A works a total of 13 days so we have: + = 1
multiply by 20b
13b + 20(7) = 20b
140 = 20b - 13b
140 = 7b
b = 140/7
b = 20 days, B by himself
;
:
A,B,C working together ,can do a piece of work in 18 days. A can do it alone in 72 days.
:
Find how long (bc) for B&C to do the job (we can treat B&C as a single person) + = 1
Multiply by 72bc
18bc + 18(72) = 72bc
1296 = 72bc - 18bc
1296 = 54bc
bc = 1296/54
bc = 24 days B*C only working
:
After working for 6 days, A fell ill.
How long will B and C together take to finish the remaining work?
:
Let t = time worked by B*C to finish the job + = 1
multiply by 72, results
6 + 3(t+6) = 72
6 + 3t + 18 = 72
3t = 72 -24
3t = 48
t = 48/3
t = 16 days for B&C to finish the job, when A leaves after 6 days
