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A factory has a batch of hand sanitiser with 50% alcohol content. How much pure alcohol should they include in a 600 mL bottle to make a 68% mixture?

I thought: simple enough, the factory has $$ 0.5 = \frac{300}{600}. $$ They need to figure out how much more alcohol to mix; therefore, $$ 0.68 = \frac{300+x}{600} \Rightarrow x = 108\text{ mL}. $$ But the answer is $$ 0.68 = \frac{300+0.5x}{600} \Rightarrow x = 216\text{ mL}. $$ I can't figure out what I have done wrong.

ryang
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scribe
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3 Answers3

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To keep the volume constant when you add $x$ ml of pure alcohol, you must remove $x$ ml of the sanitizer. Since half of the sanitizer is alcohol, you are removing $0.5x$ ml of alcohol before adding $x$ ml of alcohol. The net alcohol content is $300-0.5x+x=300+0.5x$ ml.

jcaron
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Daniel Mathias
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A factory has a batch of hand sanitiser with 50% alcohol content. How much pure alcohol should they include in a 600 mL bottle to make a 68% mixture?

Let $x$ mL be the volume of pure alcohol to be mixed with the given hand sanitiser to obtain the required mixture.

Since the given hand sanitiser is $50\%$ ABV, it contributes $\frac12(600-x)$ mL of alcohol.

The required mixture is 68% ABV and of volume 600 mL: $$\frac{x+\frac12(600-x)}{600}=0.68\\x=216.$$

$$0.68 = \frac{300+x}{600} \quad\Rightarrow\quad x = 108\text{ mL}.$$

  1. Is $x$ a pure number or a physical quantity with unit mL ?

  2. The symbol ⇒ means 'implies', not 'therefore'; your presentation just says that the right side is the answer if the left side is true, but doesn't actually assert that the right side is the answer.

    Better (albeit not the correct answer): $$0.68 = \frac{300+x}{600}; \text{ therefore, }x = 108$$ or just $$0.68 = \frac{300+x}{600}\\x = 108.$$

ryang
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  • I just meant that "if $f(x) = a$ holds, then $x=b$. In this case we might even get away with an if and only if. – scribe Oct 22 '24 at 22:50
  • @scribe Observe that neither "I'm going to the pictures iff it is Monday" nor "if it's Monday I'm going to the pictures" actually asserts that I'm going to the pictures. The information given in the above exercise is a given, so doesn't need to be presented as a hypothetical (as if there are other hypothetical cases to consider). "A implies B" is a logically weaker assertion than "we know that A is true; therefore B must also be true (since A implies B)". Using ⇒ to mean therefore seems to be the later equivalent of using = to mean output (1 + 2 = 3 + 4 = 7). – ryang Oct 23 '24 at 03:24
  • Doesn't an addition of $x$ amount of alcohol increase the total volume? Shouldn't it be $\frac{300+x}{600+x}=.68\implies .32x=108\implies x= 337.5$? – John Douma Dec 17 '24 at 03:18
  • @JohnDouma You've misread the question as "How much pure alcohol should they add to a 600 mL bottle to make a 68% mixture?" – ryang Dec 17 '24 at 08:01
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It can be done without Algebra, using inverse proportion , as the ratio of pure alcohol$\,:\,$sanitizer needed to get the target
will be in inverse proportion to their distance from target, so $(68-50):(100-68) = 18:32 = 9:16$

Thus Pure alcohol needed $= \frac{9}{(9+16)} \times 600 = 216 ml$

$\underline{\texttt{Addressing OP's Query}}$

Suppose you have stocks of $60\%$ alcohol and $30\%$ alcohol and want to make a $200$ ml cocktail with $45\%$ alcohol

Alcohol: $\small{30\% ---------- 45\% - - - - - - - - - - 60\%}$

Since $45$ is the mid-point of $30$ and $60$ it is obvious that $100$ ml of each will be needed

But what if we want the cocktail to have $50\%$ alcohol ?

Alcohol: $\small{30\% -------------50\% - - - - - - - - 60\%}$

Obviously now we shall need more of the $60\%$ alcohol and the new ratio won't be $15:15$ but $20:10 = \; 2:1$, so in inverse proportion to their respective distances from the target, directly giving the answer of $\frac23\cdot200 ml$ of $60\%$ alcohol

This method is not only faster, but less error prone than using Algebra, where students are more prone to making mistakes.

true blue anil
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