# How many pounds of water must be added to 40 pounds of a 10% boric acid to reduce it to a 4% solution?

##### 1 Answer

#### Answer:

#### Explanation:

Think about it like this. Your current solution is **10 parts** boric acid and **90 parts** water.

You need to figure out how much water you must *add* in order to reduce this ratio to **4 parts** boric acid and **96 parts** water.

The important thing to keep in mind here is that the amount of boric acid **remains unchanged** because you're only adding water to the initial solution - you're essentially *diluting* this solution.

So, the initial solution contains

#40color(red)(cancel(color(black)("lbs. solution"))) * "10 lbs. boric acid"/(100color(red)(cancel(color(black)("lbs. solution")))) = "4 lbs. boric acid"#

Let's say that

#(4color(red)(cancel(color(black)("lbs."))))/((40 + x)color(red)(cancel(color(black)("lbs.")))) xx 100 = 4%#

Solve this equation for

#4 * (40 + x) = 400#

#160 + 4x = 400#

#4x = 240 implies x = 240/4 = "60 lbs."#

Therefore, if you add