# A gardener has 26 lb. of a mixture of fertilizer and weed killer. If 1 lb. of the mixture is replaced by weed killer, the result is 5% weed killer. What % of the original mixture was the weed killer?

Aug 6, 2015

The original mixture was 1.15% weed killer.

#### Explanation:

So, you know that you have mixture containing weed killer, let's say $x$, and fetilizer, let's say $y$.

Moreover, you know that this mixture has a mass of 26 pountds. Now, you remove one pound from the original mixture and add one pound of weed killer.

Your mixture is now 5% weed killer. The key to this problem is to use this percent concentration by mass and the fact that the mass of the mixture remains unchanged to determine how many pounds of weed killer you have in this new mixture.

m_"weed killer"/m_"mixture" * 100 = 5%

${m}_{\text{weed killer" = (5 * m_"mixture")/100 = (5 * 26)/100 = "1.3 lbs}}$

Now, to get the mass of the weed killer in the original mixture, simply subtract the added 1 lb.

${m}_{\text{weed killer original" = 1.3 - 1 = "0.3 lbs}}$

SInce the mass of the mixture is unchanged, the original percent concentration of the mixture was

m_"weed killer original"/m_"mixture" * 100 = (0.3 color(red)cancelcolor(black)("lbs."))/(26color(red)cancelcolor(black)("lbs.")) * 100 = color(green)("1.15%")