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How many liters of a $25\%$ percent saline solution must be added to $3$ liters of a $10\%$ percent saline solution to obtain a $15\%$ percent saline solution?

Answer:

1.5

But I don't know how to solve it. Help me, please.

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    $\begingroup$ You always have to first calm your mind and write down what you know, in mathematical terms. First, you have solution $A$ which has 3 litres of $10~\%$ saline solution. That means that it's a container that has $0.3$ litres of saline and the rest is water (or something else?)... So why don't you clreate some variables and assign their values? $\endgroup$
    – Matti P.
    Commented Aug 27, 2019 at 6:29

4 Answers 4

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Let $x$ represent the number of liters of $25\%$ saline solution that is added to the three liters of $10\%$ saline solution. Then the total volume of the $15\%$ saline solution will be $3 + x$.

The volume of saline in the $3$ liters of $10\%$ saline solution is $(0.1)(3~\text{L})$.

The volume of saline in the $x$ liters of $25\%$ saline solution is $(0.25)(x~\text{L})$.

The volume of saline in the $3 + x$ liters of $15\%$ saline solution that is obtained is $(0.15)[(3 + x)~\text{L}]$.

Since combining the $10\%$ saline solution with the $25\%$ saline solution yields the $15\%$ saline solution, the volume of saline in the $15\%$ solution must be the sum of the volumes of the saline in the $10\%$ solution and the $25\%$ solution, which yields the equation $$(0.1)(3~\text{L}) + (0.25)(x~\text{L}) = (0.15)[(3 + x)~\text{L}]$$ Can you take it from here?

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Let $l$ the liters that you need.

Then, $\underbrace{\frac{1}{4} \cdot l}_\text{l liters of a 25%} + \underbrace{\frac{1}{10}\cdot 3}_\text{3 liters of a 10%} = \underbrace{\frac{15}{100} \cdot (l+3)}_\text{(l+3) liters of a 15%}$

Can you take it from here?

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We want to find how much $q$ (in $\ell$) of solution $A$ with $25\%$ salt we must add with $3\ell$ of solution $B$ with $10\%$ salt to give a solution $A+B$ (this has nothing to do with the usual addition; it's just notation!) that has $15\%$ salt.

The invariant here is the saline content.

Thus, per litre, $A$ has $0.25q$ of salt, $B$ has $0.1×3$ of salt, and $A+B$ has $0.15(q+3)$ salt. Since the saline content is conserved, we must have that $$0.25q+0.3=0.15(q+3)$$ is true. This gives us $(0.25-0.15)q=0.45-0.3=0.15,$ or $$q=\frac{0.15}{0.10}=1.5.$$

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If 1 liter of 25% is added to 1 liter of 10% you get 2 liters of 17.5%...extra 2.5%/liter is enough to convert 2nd 10% liter to 15%. One 25% was enough to convert 2 liters, you only need .5 to convert the third liter. Total of 1.5 liters.

It’s the same as “how many tests do you have to score 25% in to bring up your average to 15% if you’ve already taken 3 tests at 10%”

Averages

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