# How many liters of a 30% alcohol solution must be mixed with 20 liters of a 80% solution to get a 70% solution?

Aug 2, 2017

$5$ $\text{L}$

#### Explanation:

I'll assume the given percentages are those by volume (i.e. $\text{v/v}$)..

We're asked to find the volume, in $\text{L}$, of a 30% $\text{v/v}$ alcohol solution must be added to $20$ $\text{L}$ of a 80% $\text{v/v}$ to obtain a solution that is 70% $\text{v/v}$.

To do this, we can set up sort of an algebraic equation that represents this situation.

Here, we have

• $x$ liters of $0.3$ fractional volume solution

• $20$ liters of $0.8$ fractional volume solution

These two added together must be a solution with fractional volume $0.7$ (the volume will be $20$ $\text{L}$ $+ x$), so we can write an equation

ul(overbrace(x)^"volume of 30%"(0.3) + overbrace((20))^"volume of 80%"(0.8) = overbrace((x+20))^"total volume"(0.7)

Now, we simply solve for $x$:

$0.3 x + 16 = 0.7 x + 14$

$0.4 x = 2$

color(red)(ul(x = 5

So the required volume of the 30% by volume solution is

color(red)(ulbar(|stackrel(" ")(" "5color(white)(l)"L"" ")|)