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

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?

1.5

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

• 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? Commented Aug 27, 2019 at 6:29

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?

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?

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 $$q=\frac{0.15}{0.10}=1.5.$$

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