Hi Turkish,

This question is essentially a "weighted average" question, which means that it can be solved in a variety of ways. Here's an approach that uses the Weighted Average Formula:

We're mixing a certain amount of a 15% salt solution with a certain amount of a 25% salt solution and ending up with a total solution that is 21% salt.

X = # of ounces of 15% solution
Y = # of ounces of 25% solution

(.15X + .25Y)/(X+Y) = .21

.15X + .25Y = .21X + .21Y
.04Y = .06X
4Y = 6X
2Y = 3X

This ratio tells us that for every 2 ounces of 15% solution, we have 3 ounces of 25% solution. So, we can have ANY totals that maintain this ratio. For the sake of simplicity, I'm going to choose a TOTAL of 5 ounces (2 ounces of 15% and 3 ounces of 25%)

The original question is phrased in a way that implies that ALL the 5 ounces were 15% salt, but some of those ounces were removed/replaced with 25% salt. If we start with 5 total ounces and replace 3 of them with a 25% salt solution, then we get the 21% solution above. Thus 3/5 of the original solution was replaced.

Final Answer:

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Hi Celestial09,

In the first sentence of the question, we're told that a fraction of a solution was REPLACED, so the original number of ounces stays the same. We're NOT adding or subtracting ounces - we're replacing them.

In my example, the original mixture is 5 ounces total, so some fraction "out of 5" was replaced.... To get the 21% solution, 3 of those original ounces are replaced by the 25% solution, thus 3 of the 5 (or 3/5) of the original solution is replaced.

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Rich
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let x=fraction of solution replaced
.15-.15x+.25x=.21
x=3/5

I am sorry if it sounds stupid, but why is it 4:6 and not 6:4?

Check out this post. It will help you understand weighted average.

http://www.veritasprep.com/blog/2011/03 ... -averages/
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hi thanks for your valuable reply but couldnt understand your last line.
still can`t figure out how come 3/5th of "original" mixture.

Weighted average with one variable approach.

A is the original solution
B is the resultant solution
x = the amount (volume) replaced:

1) x is a volume of salt, both removed and replaced. The ingredient is the same. You don't need a second variable;

2) Volume removed = volume replaced;

3) However, the concentration (percentage) of salt is different in the volume removed and the volume replaced. Same volume, different concentrations of salt;

4) x, the volume removed and replaced is a part of the whole, A. Part x of whole A = fraction of A. Answers are all 0 < x < 1. Assume A's volume = 1

5) Volume of A and volume of B are equal.

Original, (salt concentration)(volume): (.15)(1)
Removed: (.15)(x)
Replaced: (.25)(x)
Resultant solution, (salt concen.)(volume): (.21)(1)

Original - removed + replaced = Resultant solution

(.15)(1) - .15(x) + .25(x) = .21(1)

Multiply all by 100 to make arithmetic easier.

(15)(1) - 15x + 25x = 21(1)
10x = 6
x = \(\frac{6}{10}=\frac{3}{5}\)

Answer B

--10x = 6 might seem odd. Fractional replacements are hard. Think in volumes first, then in fractions.
--Amount x = a volume. A has a volume of 1. x is the unknown volume removed and replaced. And it is not all of A.
--Fractions: all of A did not go. Only part of A's volume got removed and replaced. (Part) x of (whole) A got removed and replaced. \(\frac{part}{whole}\) = fraction.
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@karsihma , can you kindly explain the last statement " So for every 2 parts of 15% solution, we have 3 parts of 25% solution. So 3/5th of the original solution was replaced.
"

Originally there were 5 parts of only the 15 percent solution. At at the end there were 2 parts of the 15 percent solution and 3 parts of the 25 percent solution. So 3 parts out of
5 were replaced. 3/5 ths was replaced.

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Let the total solution be 100 ml. Suppose x ml was removed from Solution A.. So the remaining salt solution in A is (100-x) * 15%... Now in this solution 0.25x is added... So (100-x)*0.15 + 0.25x = 0.21
X = 60.. So the fraction of original solution replaced was 60/100 or 3/5.
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