How do you figure out pipe filling pool problems?

Question

When two pipes fill a pool together, they can finish in 4 hours. If one of the pipes fills half the pool then the other takes over and finishes filling the pool, it will take them 9 hours. How long will it take each pipe to fill the pool if it were working alone?

How many hours for one pipe to fill the pool? How many hours for the other pipe to fill the pool?

Answer 1

Pipe A fills 1/x of the pool per hour, and Pipe B fills 1/y of the pool per hour. In other words, A takes x hours to fill the pool and B takes y hours to fill the pool.

 

(For example, if A took 3 hours to fill the pool, it would fill 1/3 of the pool per hour. It's usually a good idea to take whatever information you're given in a problem like this and turn it into rates, like 1/x and 1/y. Then, you can multiply the rates by hours worked to find out what portion of the work has been done after that many hours.)

 

Together, they fill 1/x + 1/y of the pool per hour. So, 4 hours times that rate should equal 1 full pool, since together they take 4 hours to fill the pool. That gives you your first equation:

 

4(1/x + 1/y) = 1

 

You're also told that if one pipe fills half the pool and the other takes over to fill the rest, it will take 9 hours. We don't know how long Pipe A works, so call that number of hours t. Pipe B must work for 9 - t hours.

 

So, Pipe A, working for t hours, fills half the pool:

 

t(1/x) = 1/2

 

And Pipe B, working for 9 - t hours, fills half the pool:

 

(9-t)(1/y) = 1/2

 

 

 

You now have a system of three equations (because this is an exceptionally complicated pipe-filling-pool problem):

 

4(1/x + 1/y) = 1

t(1/x) = 1/2

(9-t)(1/y) = 1/2

 

Solve for t in that second equation, then plug it into the third equation:

 

t(1/x) = 1/2

t = x(1/2)

t = x/2

 

(9 - x/2)(1/y) = 1/2

 

 

 

Now you've got a system of two equations:

 

4(1/x + 1/y) = 1

(9 - x/2)(1/y) = 1/2

 

Solve for y in the second equation, then plug that into the first equation:

 

(9 - x/2)(1/y) = 1/2

(9 - x/2) = y(1/2)

2(9 - x/2) = y

18 - x = y    (this is a really nice-looking form of that equation; we'll use it again later)

 

4(1/x + 1/y) = 1

4(1/x + 1/(18-x)) = 1

4/x + 4/(18-x) = 1

(4(18-x))/(x(18-x)) + (4x)/(x(18-x)) = 1

(4(18-x) + 4x)/(x(18-x)) = 1

4(18-x) + 4x = x(18-x)

72 - 4x + 4x = 18x - x²

72 = 18x - x²

x² - 18x + 72 = 0

(x - 6)(x - 12) = 0

x - 6 = 0    OR    x - 12 = 0

x = 6    OR    x = 12

 

So, Pipe A takes either x = 6 or x = 12 hours to fill the pool alone.

 

 

 

Plug x = 6 and x = 12 into our equation from earlier to get:

 

18 - x = y

18 - 6 = y

12 = y

 

OR

 

18 - x = y

18 - 12 = y

6 = y

 

When Pipe A takes 6 hours to fill the pool, Pipe B takes 12. When Pipe A takes 12 hours to fill the pool, Pipe B takes 6. These cases are equivalent, since the original problem makes no distinction between the pipes. So:

 

One pipe takes 6 hours to fill the pool and the other takes 12.