Reverse percentages

Question

My mother recently started doing a distance learning course, and is struggling with her mathematical questions. I'm trying to explain to my mother how to answer the following question. Despite my trying, I simply can't phrase the question in such a way that someone who hasn't done mathematics in a couple of decades and only has a grip of basic arithmetic could understand.

What I'm really looking for is a good (preferably visual) explanation of how reverse percentages work. Can anyone help me?

Question

A person consumes wine with alcoholic content 13%.  14.9g of alcohol is the maximum amount of wine allowed to be consumed each day. Assume 1 ml alcohol has a mass of 0.789g.

How many ml of wine could they drink each day if their consumption was upper limit?

My answer:

1. Figure out how much 14.9g (weight) of alcohol is in space (volume):

   $$14.9 / 0.789 = 18.88466413181242~\text{ml}$$

2. We know that the maximum space the alcohol can take up out of the whole of amount of wine is 18.88466413181242ml

   So 13% of that wine must not exceed 18.8...ml, but since we're finding the maximum we'll say that 13% = the 18.8...ml

   18.88466413181242        - Maximum amount of ml of alcohol to be drunk
   13% = 18.88466413181242 ml - Assume the 13% of the wine is the maximum 18.8 ml

3. Since 13% of the wine is 18.8ml we'll now calculate what 100% of the wine is:

   18.88466413181242ml / 13 = 1.452666471677878
                 - Divide by 13 and × by 100 to get the 100% of the volume in ml

   1.452666471677878 × 100 = 145ml (rounded)

Answer 1

The upshot is that $x\%$ of a number is $x\%$--that is $\frac{x}{100}$--times the number. Try some concrete examples with her, like "$4$ is $50\%$ of what number?" or "$3$ is $10\%$ of what number?"

Answer 2

The main concern would be the reason why she can't do the problem. The question that you gave is somewhat complicated, so it can be hard to understand why she doesn't know how to do the problem. Is she afraid of percentages? Is she scared off by the lack of integers? Does she not understand what the question is asking for? Does she know how to approach a simpler problem?

At the heart of it, the problem is one about ratios. I'd start her off with ratio problems that are solved by cross-multiplying and dividing. For example

"If 4 teaspoons of baking soda are needed to bake 50 cookies, how many teaspoons of baking soda are need to bake 75 (or 125) cookies".

When she is comfortable with this, move to non-integer values.

"If 1ml of alcohol has a mass of 0.789 g, how much volume does 14.9 g of alcohol take up?"

Then show her another similar version:

"If 100ml of this spirit has 13ml of alcohol, how much volume of it will have X amount of alcohol" (where X is the answer to the previous question).

Then, combine both of the previous:

"If 100ml of this spirit has 13ml of alcohol, and 1ml of alcohol has a mass of 0.789 g, how much volume does 14.9 g of alcohol take up?"

Answer 3

First off, I'd avoid using all those irrelevant decimal places which add nothing to the argument, and will confuse and worry people who aren't confident with numbers.

Then I'd look at using very simple examples to illustrate the structure of the argument (so if the thing comes up in an exam there is a simple and obvious and memorable model to follow).

For example if we start with $100$ units, add $50\%$ and then subtract $50\%$ we go from $100$ to $150$ to $75$. To get back from $150$ to $100$ we need to subtract one third. If we start with $100$ and reduce by $50\%$ and increase by $50%$ we get $50$ then $75$ - and we have to increase by $100\%$ (double) to get back to $100$.

[NB it is no coincidence that we get to $75$ both ways]

So are we adding a percentage, taking away a percentage or returning to a base figure? Are we increasing or decreasing? Use the simple model to work out which calculation is required.

And I would start, as you did, at the end - what we definitely know is that we can consume up to $14.9$g of alcohol, but the percentages are in terms of volumes. The first thing we need is to get everything in terms of volumes.

This is nothing to do with percentages - instead it is a conversion factor. $0.789$g represents $1$ml, so we need to divide our $14.9$g into units of $0.789$g giving (as you have, but rounded to three significant figures) $18.9$ml.

This $18.9$ml is $13\%$ of the volume, so we are working from a reduced figure back up to a base figure (we identify the kind of calculation we need) - so we divide $18.9$ by $0.13$ to obtain $145$ml.

There is more to do to embed the ideas, and maybe $50\%$ is not the best example, but it is memorable and simple, and I can test very quickly "would my calculation work in that case"?

But note also how the language we can use in explaining the case can be simple and accurate - for example in the conversion calculation to express the task as to divide [ ] into units of [ ] explicitly suggests a division sum, with everything set up to do the right thing.

When I am teaching this kind of thing (very occasionally) - I make sure that the person I am teaching goes away knowing the kind of questions to ask. "What are the most convenient units?" "How do I get those g into ml?" "Which way does the percentage go?"

Actually, I normally sit with someone and ask them the questions, and when I find the right question to help them, I make a mental note. Most often, they do the whole thing themselves, with only the questions to guide them. I then point out that they have done all the work themselves, and review the questions which helped them get there. The notes above illustrate the way I might go about shaping questions in this case.