# How many liters of a solution containing 45% acid must be mixed with a solution containing 60% acid to obtain 40 liters of a 48% acid solution?

##### 1 Answer

#### Answer:

#### Explanation:

In order to solve this problem, you need to set up two equations with two unknowns, the *volume of the solutions* and the *volume of the solutes*.

Use the percent concentration by volume and the volume of the target solution to figure out how much acid it must contain

#40 color(red)(cancel(color(black)("L solution"))) * "48 L acid"/(100color(red)(cancel(color(black)("L")))) = "19.2 L acid"#

Now, if

#x + y= 40#

#overbrace(45/100 * x)^(color(red)("acid from 1"^"st" "solution")) + overbrace(60/100 * y)^(color(green)("acid from 2"^"nd" "solution")) = overbrace(19.2)^(color(blue)("acid in target solution"))#

This means that you have

#x = 40 - y#

Which gets you

#45/100 * (40-y) + 60/100 * y = 19.2#

#1800 - 45y + 60y = 1920#

#15y = 1920 - 1800 implies y = 120/15 = 8#

Therefore,

#x = 40 - 8 = 32#

So, to get