Question 1064251: Machine A can do a certain job in 8 hours. machine B can do the same job in 10 hours. machine C can do the same job in 12 hours. all three machines start the job at 9.00 am, machine B breaks down at 11.00am and the other two machines finish the job. approximately at what time would the job be completed?
(please explain the method)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! in one hour
A does 1/8, B does 1/10, C does 1/12
x is the number of total hours worked.
B worked 2 hours
2/8+2/10+2/12=30/120+24/120+20/120=74/120 of the job done by all three.
The other two account for 46/120; that comes from the fact that if 74/120 was done by all 3, the remaining 46/120 (74+46=120) were done by the two remaining machines.
x/8+x/12=46/120
15x/120+10x/120=46/120
15x+10x=46
25x=46
x=1.84 hours or 12:50.24 ANSWER
check
A worked 3.84 hours, B 2 hours and C 3.84 hours
3.84/8+2/10+3.84/12=
common denominator=120
57.6+24+38.4 is the numerator, which adds to 120.
|
|
|
