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A factory has a batch of hand sanitizer with $50\%$ alcohol content. How much pure ( $100\%$ concentration) alcohol should they include in a $600\text{mL}$ bottle to make a $\newline$ $68\%$ mixture?

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Weighted averages: word problems

Full solution

Q. A factory has a batch of hand sanitizer with

50\%

alcohol content. How much pure (

100\%

concentration) alcohol should they include in a

600\text{mL}

bottle to make a

\newline

68\%

mixture?

Calculate Alcohol Amount: Calculate the amount of alcohol in the initial $600\,\text{mL}$ of $50\%$ alcohol solution. $\newline$ Calculation: $600\,\text{mL} \times 50\% = 600 \times 0.50 = 300\,\text{mL}$ of alcohol.
Determine Pure Alcohol Added: Let $x$ be the amount of pure alcohol ( $100\%$ concentration) added to the mixture. $\newline$ Reasoning: Adding $x$ mL of $100\%$ alcohol will increase the total alcohol content.
Calculate Total Volume: Calculate the total volume of the mixture after adding $x$ mL of pure alcohol. $\newline$ Calculation: Total volume = $600$ mL + $x$ mL.
Set Up Final Concentration Equation: Set up the equation for the final concentration of alcohol in the mixture to be $68\%$ . Calculation: $(300 \, \text{mL} + x \, \text{mL}) / (600 \, \text{mL} + x \, \text{mL}) = 68\% = 0.68$ .
Solve for x: Solve the equation for x. $\newline$ Calculation: $300 + x = 0.68 \times (600 + x)$ $\newline$ $300 + x = 408 + 0.68x$ $\newline$ $x - 0.68x = 408 - 300$ $\newline$ $0.32x = 108$ $\newline$ $x = \frac{108}{0.32}$ $\newline$ $x = 337.5 \text{ mL}$

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