Question 1193648: A scientist mixes water (containing no salt) with a solution that contains 55% salt. She wants to obtain 220 ounces of a mixture that is 30% salt. How many ounces of water and how many ounces of the 55% salt solution should she use?
Water= Ounces
Salt Solution= Ounces
Found 3 solutions by math_tutor2020, ikleyn, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
x = amount of the 55% salt solution, in ounces
It isn't clear if your teacher is referring to dry ounces or fluid ounces. Though the math will lead to the same result either way.
If we had x ounces of the 55% salt solution (i.e. 55% salt, the rest water), then we have 0.55x ounces of pure salt.
We aim to have 220 ounces of total mixture, after adding in the water to dilute things. We want to dilute to a 30% salt concentration.
So,
(amount of pure salt)/(total mix) = salt concentration
(0.55x)/(220) = 0.30
Multiply both sides by 220, then divide both sides by 0.55 as the steps show below
(0.55x)/(220) = 0.30
0.55x = 220*0.30
0.55x = 66
x = 66/0.55
x = 120
We need 120 ounces of the 55% salt solution
This leaves 220-x = 220-120 = 100 ounces of pure water
Answers:
Water = 100 ounces
Salt solution = 120 ounces
Answer by ikleyn(52786)  (Show Source):
You can put this solution on YOUR website! .
For those who composes Math problems without having necessary knowledge on the subject,
it is useful to know that 55% salt (NaCl) solution does not exist in the nature, as well as 30% salt solution.
It is because somewhere at 26% - 28% concentration, the mixture becomes SATURATED and does not dissolve more salt.
Usually, these facts are known to 7-th grade students from Science.
You can read about it from this Wikipedia article
https://en.wikipedia.org/wiki/Saline_water#:~:text=The%20saturation%20level%20is%20only,of%2028.1%25%20w%2Fw.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Ignore the fact that a 30% or 55% salt solution is not physically possible. This is a math problem for giving the student practice in solving mixture problems.
The formal algebraic solution provided by the other tutor uses a good traditional method.
Here is a quick informal way to solve any 2-part mixture problem like this.
Think of this as starting with a 55% solution and adding a 0% solution to obtain a 30% solution. The percentage of the mixture starts at 55% and moves towards 0%, stopping when it reaches 30%.
30% is 5/11 of the way from 55% to 0%. (55 to 0 is a change of 55; 55 to 30 is a change of 25; 25/55 = 5/11.)
That means 5/11 of the mixture is the 0% solution (water) that she is adding.
So the amount of water to be added to make 220 ounces of the mixture is 5/11 of 220 ounces, which is 100 ounces. So then the number of ounces of 55% solution she needs is 220-100 = 120.
ANSWER: 120 ounces of the 55% solution; 100 ounces of water.
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