SOLUTION: A large pump can fill a tank in 28 minutes. Both a large pump and a small pump can fill a tank in 20 minutes. How long would it take the small pump to fill the tank by itself?

Algebra ->  Rate-of-work-word-problems -> SOLUTION: A large pump can fill a tank in 28 minutes. Both a large pump and a small pump can fill a tank in 20 minutes. How long would it take the small pump to fill the tank by itself?       Log On


   

Question 1040135: A large pump can fill a tank in 28 minutes. Both a large pump and a small pump can fill a tank in 20 minutes. How long would it take the small pump to fill the tank by itself?
Found 2 solutions by addingup, ikleyn:
Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
(28*x)/(x+28) = 20
28*x = 20(x+28)
28x = 20x+560
Subtract 20x from both sides:
8x = 560
x = 70 minutes

Answer by ikleyn(52795) About Me  (Show Source):
You can put this solution on YOUR website!
.
A large pump can fill a tank in 28 minutes. Both a large pump and a small pump can fill a tank in 20 minutes.
How long would it take the small pump to fill the tank by itself?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Two pumps working together fill  1%2F20  of the tank volume per minute.

One pump (large) fills  1%2F28  of the tank volume per minute.

Hence, the second pump fills  1%2F20+-+1%2F28 = 7%2F140+-+5%2F140 = %287-5%29%2F140 = 2%2F140 = 1%2F70  of the tank volume per minute.

It means that it will take 70 minutes for the second pump to fill the tank working alone.


      Lesson to learn from this solution:  use rates of work.
      You can add and distract them.  It does make sense.


For many other joint-of-work problems see the lessons
    - Using Fractions to solve word problems on joint work
    - Solving more complicated word problems on joint work
    - Selected joint-work word problems from the archive
in this site.