Diluting a Cleaning Solution
A custodian has 12 liters of a 30% bleach solution and wants to dilute it with water (0% bleach) to obtain a mixture that is no more than 10% bleach but at least 6% bleach, so it is strong enough to sanitize yet safe for the floor. He plans to add $x$ liters of water. He also has a second container with 5 liters of 45% solution that he could blend in, but he is unsure whether adding it would push the concentration out of the acceptable range.
1. Write and solve the inequality for the amount of water x that yields a concentration between 6% and 10%.
Read the worked answer
Bleach amount is fixed at $0.30(12) = 3.6$ L. Concentration after adding $x$ L water: $\frac{3.6}{12 + x}$. Require $0.06 \le \frac{3.6}{12+x} \le 0.10$. Right side: $\frac{3.6}{12+x} \le 0.10 \Rightarrow 3.6 \le 0.10(12+x) \Rightarrow 3.6 \le 1.2 + 0.10x \Rightarrow x \ge 24$. Left side: $\frac{3.6}{12+x} \ge 0.06 \Rightarrow 3.6 \ge 0.06(12+x) \Rightarrow 3.6 \ge 0.72 + 0.06x \Rightarrow x \le 48$. So $24 \le x \le 48$ liters of water.
2. If he adds the entire 5 L of 45% solution to the original 12 L (no water), what is the resulting concentration, and is it acceptable?
Read the worked answer
Bleach: $3.6 + 0.45(5) = 3.6 + 2.25 = 5.85$ L. Total volume: $17$ L. Concentration: $5.85/17 \approx 0.344 = 34.4%$, far above 10%, so it is not acceptable. He would need substantial dilution, not blending, to bring it into range.
3. Suppose he first adds 30 L of water, then adds the 5 L of 45% solution. Is the final concentration within range?
Read the worked answer
After water: bleach $3.6$ L in $42$ L. Adding 5 L of 45%: bleach $3.6 + 2.25 = 5.85$ L, volume $47$ L. Concentration $5.85/47 \approx 0.1245 = 12.4%$, which exceeds 10%. So the blend pushes it out of range; he would need more water (about $x$ solving $5.85/(42+5+x) \le 0.10 \Rightarrow x \ge 11.5$ L more).