SOLUTION: A large pump can fill a water tank in 6 hours, and a smaller pump can fill it in 10 hours. In order to fill an empty tank, the large pump is operated for 3 hours, and then the smal

Algebra ->  Rate-of-work-word-problems -> SOLUTION: A large pump can fill a water tank in 6 hours, and a smaller pump can fill it in 10 hours. In order to fill an empty tank, the large pump is operated for 3 hours, and then the smal      Log On


   

Question 901827: A large pump can fill a water tank in 6 hours, and a smaller pump can fill it in 10 hours. In order to fill an empty tank, the large pump is operated for 3 hours, and then the small pump is also turned on. In total, how many hours will be required to fill the tank in this way?
Found 2 solutions by lwsshak3, richwmiller:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
A large pump can fill a water tank in 6 hours, and a smaller pump can fill it in 10 hours. In order to fill an empty tank, the large pump is operated for 3 hours, and then the small pump is also turned on. In total, how many hours will be required to fill the tank in this way?
***
let x=time when both pumps are working
let(3+x)=total time required to fill tank
1/6=work rate of large pump
1/10=work rate of small pump
1/6+1/10=5/3+3/30=8/30=4/15=work rate with both pumps turned on
..
3%2F6%2Bx%2F10%2B4x%2F15=1
lcd=30
15+3x+8x=30
11x=15
x=15/11=1.36 hrs
3+x=4.36 hrs
how many hours will be required to fill the tank in this way? 4.36

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
x/10+x/6=1
x=15/4=x = 3.7500 for the two pumps to do the job together from the start
The large pump did half the job in the first 3 hours
3/6+x/10+x/6=1
x=15/8 = 1.8750 hours
So they need less than 2 hours to finish the job together
So it should take less than 3 hours for them both to finish the job.