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I understand the rules of adding fractions perfectly well. I know how to find common denominators, and understand why adding fractions without common denominators doesn't make sense.

But, today someone asked me about adding $\frac{5}{6}$ and $\frac{21}{28}$. They were wondering why dividing the two factions and taking the average ($\frac{0.8333 + 0.75}{2} = 0.7917$) was giving them a different value than adding them together, and then dividing ($\frac{5}{6} + \frac{21}{28} = \frac{26}{34} = 0.7647$).

My initial reaction was the same as yours: I was taken aback by someone adding fractions in this way, and I gave a quick refresher of why this doesn't make sense, and how to properly add fractions.

I thought I had helped them fix their problem, until they gave more more context: The $\frac{5}{6}$ was five people in a group of six who were observed washing their hands after a certain activity. The $\frac{21}{28}$ was twenty-one out of twenty-eight people who were observed washing their hands after a certain activity. The goal was to find the total number of people who had washed their hands, as a fraction of the total number of observed people. So, $\frac{26}{34}$ is actually correct in this case.

But, I'm still having trouble reconciling this with what I know about fractions. At least, what I think I know. Is there a term for these sorts of fractions? This isn't like having five slices of a six-slice pizza combined with twenty-one slices of a twenty-eight-slice pizza. This is like having one oven holding six pizzas, five of which have mushrooms, and another (huge) oven holding twenty-eight pizzas, twenty-one of which have mushrooms. What fraction of the pizzas have mushrooms? Is it $\frac{5}{6} + \frac{21}{28}$ ? I don't think so.

Is there some terminology I'm forgetting here? Why am I getting so tripped up by this?

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  • $\begingroup$ In statistics, there is a reason researchers strive to have samples of the same size, and this is one of them. $\endgroup$ – scott Jul 5 '16 at 1:21
  • $\begingroup$ The question you give isn't solved by adding. 5/6 refers to 5/6 of all people, not just the 6. And 21/28 (which should be 3/4) refers to 21/28 of all people. Not just the 28. So the answer isn't to be found by adding the fractions. It's by adding the fractions of total. Ie answer is (5/6x6 +21/28)/(28+6) $\ne $ 5/6 + 21/28$. It does make a nice simplification that it reduces to $(5+21)/(6+28) $ but that is a different concept. $\endgroup$ – fleablood Jul 5 '16 at 4:05
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This is called the mediant of the two fractions - simply add the denominator and numerator separately.

The other approach, of adding the two fractions and dividing by two, is called the arithmetic mean.

The reason that the mediant is appropriate and the arithmetic mean is not is because you want to find the total number of pizzas with mushrooms out of both sets combined. Each of the fractions you mentioned gives you the proportion of pizzas with mushrooms just with respect to that individual sunset of pizzas. So why would you add the two together, give them equal weighting towards the end result, and then divide by two?

If the two subsets of pizzas don't have the same size, you want to give the larger one more "weight" towards the end result, and the mediant does that properly. For a more generalized look at this topic, look at weighted averages.

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    $\begingroup$ Sorry, but the phrase "sunset of pizzas" produces some pretty funny mental imagery. :) $\endgroup$ – scott Jul 5 '16 at 1:20
  • $\begingroup$ I attempted to edit that, but of course it's not a big enough change for a suggested edit. $\endgroup$ – WhiteHotLoveTiger Jul 7 '16 at 17:01
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There is such a thing as a weighted average. A weighted average of the two fractions, with weights proportional to the sizes of the groups, gives the right answer.

The sizes of the groups are $6$ and $28$, so the weights are $\dfrac 6 {6+28} \approx 0.1764706$ and $\dfrac{28}{6+28} \approx 0.8235294$.

The two fractions are $5/6\approx 0.83333$ and $21/28 = 3/4 = 0.75$. The weighted average is \begin{align} (\text{first weight}\times\text{first value}) & {} + (\text{second weight}\times\text{second value}) \\[10pt] = \left( \frac 6 {34} \times \frac 5 6 \right) & {} + \left( \frac{28}{34}\times\frac {21}{28} \right). \end{align} The $6$s cancel and the $28$s cancel and you get $$ \frac 5 {34} + \frac{21}{34} = \frac{26}{34}. $$

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