SOLUTION: If three pipes are all opened, they can fill an empty swimming pool in 3 hours. The largest pipe alone takes 1/3 the time that the smallest pipe takes and half the time the other p

Question 807710: If three pipes are all opened, they can fill an empty swimming pool in 3 hours. The largest pipe alone takes 1/3 the time that the smallest pipe takes and half the time the other pipe takes. How long would it take each pipe to fill the pool by itself?
Can you be a little specific because i'm really confused with this problem, thanks!
Found 3 solutions by josmiceli, amalm06, ikleyn:
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Add the rates of filling to get the rate of all 3
pipes filling together
Let = the rate of filling of the largest pipe
Let = the rate of filling of the middle size pipe
Let = the rate of filling of the smallest pipe
---------------------------------------------
If three pipes are all opened, the total rate is
( 1 pool filled ) / ( 3 hrs ) = , so
(1)
(2)
( The largest pipe's rate is 3 times the smallest pipe )
(3)
( The largest pipe's rate is 2 times the medium size pipe )
--------------------------------------------------
(2)
and
(3)
----------------------
Substitute these results into (1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
and, since
(2)
(2)
(2)
and
(3)
(3)
(3)
----------------------
The largest pipe can fill ( 2 pools ) / ( 33 hrs ), so
it can fill ( 1 pool ) / ( 16.5 hrs )
----------------------------
The medium size pipe can fill ( 1 pool ) / ( 11 hrs )
----------------------------
The smallest pipe can fiill ( 2 pools ) / ( 11 hrs )
It can fill ( 1 pool ) / ( 5.5 hrs )
-----------------------------------------
Check the answer:
(1)
(1)
Multiply both sides by
(1)
(1)
(1)
OK

Answer by amalm06(224)   (Show Source): You can put this solution on YOUR website!
[(1/L1)+(1/L2)+(1/L3)](3)=1
L1=(1/3)L3 ---> L3=3L1
L1=(1/2)L2 ---> L2=2L1
(1/L1)+(1/2L1)+(1/3L1)(3)=1
11/2L1 = 1 so L1 = 11/2
L1=5.5 hrs (smallest)
L2=11 hrs (medium)
L3=16.5 (largest)
Answer by ikleyn(52780)   (Show Source): You can put this solution on YOUR website!
.

Let x be the rate of the largest pipe. Then, according to the condition, the rate of the smallest pipe is , while the rate of the medium pipe is of the tank volume per hour. The equation is + + = , or + + = , = ====> Multiply both sides by 6 ====> 6x + 3x + 2x = 2 ====> 11x = 2 ====> x = . Thus the largest pipe will fill the tank in = 5.5 hours. Then the smallest pipe will fill it in 5.5*3 = 16.5 hours, and the medium pipe will fill it in 5.5*2 = 11 hours.

Answer. 5.5 hours for the largest pipe;   16.5 hours for the smallest pipe;   and 11 hours for the medium pipe.

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It is a typical joint work problem.

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