Question 1207792: 1. Two pumps of different sizes
2. How much water should be added to 64 ounces of a 10% salt solution to make a 2% salt solution?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39620)   (Show Source):
You can put this solution on YOUR website! #2, very very routine exercise
Water with nothing dissolved in it for practical purposes, 0% salt.
If adding v ounces, (but you likely should mean, "fluid ounces") then
#1
Larger Pump,  , its pump-out rate, unit of TANK per HOUR
Smaller Pump rate, 
BOTH pumps together, rate, 
Description give that the two pump working together do the job in 5 hours.
Answer by ikleyn(52810)  (Show Source):
You can put this solution on YOUR website! .
2. How much water should be added to 64 ounces of a 10% salt solution to make a 2% salt solution?
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Let V be the final volume of the mixture.
The mass of the dissolved salt is the same (it is not changed).
It leads to equation
0.1*64 = 0.02*V
which gives V =  =  = 5*64 = 320 ounces.
It means that 320-64 = 256 ounces of water should be added. ANSWER
Solved.
To see many other similar (and different) problems of the same type
on dilution acid solutions by adding water, look into the lesson
- Special type mixture problems on DILUTION adding water
in this site.
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I just asked you once, and I repeat it again:
- please do not pack more than one problem per post.
It is the rule and the policy of this forum:
One and only one problem per post.
This rule is printed at web-page
https://www.algebra.com/tutors/students/ask.mpl?action=ask_question&topic=Equations&return_url=http://www.algebra.com/algebra/homework/equations/
from which you post your problems to this forum.
So, it is assumed that you read it before you post, are familiar with it and follow this rule.
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I saw, that you just re-posted first problem in a separate post.
I welcome this and approve this your action.
At the same time, I delete this first problem so that other tutors do not worry in vain.
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