Hummingbirds and Bees

In May 2023, a Facebook post presented the following dilemma:

“A dear friend gives you a Mason jar that is 2 cups sugar to 1 quart water—prepared for her bees. The ratio for hummingbirds is 1 part sugar to 4 parts water. How much water do I add to the bee feed to make it good for hummingbirds?”

It's a really cool question, and I love that the person posting cared enough about the hummingbirds to get it right!

NOTE: This first answer uses the assumption that one part plus two parts equals three parts. This is actually not true for sugar dissolved in water (see the Addendum below), but is a necessary assumption in order to get a mathematical solution (without additional information). Also, we'll suppose that we don't know how much bee solution is in that Mason jar gift—we only know the ratio of sugar to water.

Note first that the bee ratio is:

$$ \begin{align} &\frac{2 \text{ cups sugar}}{1 \text{ qt water}}\cr\cr &\quad = \frac{2 \text{ cups sugar}}{4 \text{ cups water}}\cr\cr &\quad = \frac{1 \text{ part sugar}}{2 \text{ parts water}} \end{align} $$

Now, do this real (or thought) experiment: Go to the kitchen, take any size measuring cup (call this ‘a part’), and combine:

This gives you a (bee) mixture with a $\ \displaystyle\frac{1 \text{ part sugar}}{2 \text{ parts water}}\ $ ratio.

KEY OBSERVATION:
** $\displaystyle\frac13$ of this mixture is sugar and $\displaystyle\frac23$ is water **

If you want a hummingbird mixture instead, then you have to add $\,2\,$ more parts of water. That is, you have to add an additional two-thirds of your existing solution (by volume) of water, to achieve the hummingbird mixture.

So, here's what you need to do. Pour your bee mixture into a volume measure. Call this initial amount of bee solution $\,V\,$ (for ‘Volume’). Compute $\,\frac 23V\,.$ Then, add this much additional water to the bee mixture, and you'll have your hummingbird mixture.

Note: If it's a (full) one quart Mason jar, then you need to add $$\frac 23(1 \text{ qt}) = \frac23(4 \text{ cups}) = 2\frac23 \text{ cups}$$ of additional water.

For the hard-to-convince, here's some additional verification that this is the correct solution:

Again, let $\,V\,$ be the initial volume of bee solution. We know that $\,\frac 13V\,$ of this solution is sugar and $\,\frac 23V\,$ is water. Now, add $\,\frac 23V\,$ additional water. What is the ratio of sugar to water in the final solution?

$$ \begin{align} &\text{final solution ratio of sugar to water} \cr\cr &\qquad = \frac{\frac 13V}{\frac 23V + \frac 23V}\cr\cr &\qquad = \frac{\frac 13V}{\frac 43V}\cr\cr &\qquad = \frac 14 \end{align} $$

Perfect!

By the way: I did go to my kitchen, combined $\,\frac 14\,$ cup sugar and $\,\frac 24\,$ cups water, and ended up with (approximately—see below) $\,\frac 34\,$ cups solution. I would have had to add $\,\frac 24\,$ more cups water to make it work for the hummingbirds ...

An Addendum: Two Problem Interpretations

Here's an addendum, written up after talking with my genius husband Ray (and after having now used $\,\frac 12\,$ cup of sugar in science experiments).

I (Dr. Burns) approached this as a math problem, and I made the assumption that when you combine one part plus two parts, you get three parts. This is true sometimes, and not true other times. With sugar dissolved in water, it is not true. Indeed, one part sugar (by volume) added to two parts water (by volume) gives less than three parts, by volume. (From our own experimentation, it gives approximately $\,2\frac 23\,$ parts, by volume.) This has to do with the air between the sugar grains, the crystalline structure of sugar, and how the sugar and water molecules interact when dissolved. Borrowing an enlightening sentence from this physics discussion: sugar has a lot of empty space in between the sugar grains, and water can fill up that space (think about adding water to a full cup of sugar—you can add quite a bit before it starts to overflow).

Here's an interpretation of the Facebook problem that has an exact solution, without any assumptions. Suppose it is known that the (large) Mason jar contains precisely one quart of water and two cups of sugar . In this case, it is indeed correct to add one additional quart of water (which will likely need to be done in a larger container). Note that if the original volume of the two cups of sugar and one quart of water is six cups (which it will not be), then this answer is consistent with that given above, since $$ \frac23(6 \text{ cups}) = 4 \text{ cups} $$ of additional water needs to be added.

However, if all we know is that the mixture is a ‘one part sugar to two parts water’ solution, then we cannot calculate the correct amount of additional water to add without additional information. We also need to know about volume changes when sugar is mixed with water. Now it's in the realm of a science question!

A Refined Solution: Taking Into Account Pre-Mix versus After-Mix Volume Change

Here's an answer that takes into account the results of our at-home (imprecise) experimentation comparing the volumes of the unmixed sugar and water with the volume of the mixed solution.

We found, by experimentation, that:

$$ \begin{gather} \overbrace{(1 \text{ part sugar}) + (2 \text{ parts water})}^{\text{before mixing}}\cr\cr \text{ is approximately equal to }\cr\cr 2\frac23 \text{ parts mixed solution} \end{gather} $$

The answer presented next could certainly be replicated with more accurate data.

Intuition: That Mason jar has some volume, $\,V\,$, of bee solution in it ($\,1\,$ part sugar to $\,2\,$ parts water). But, before the sugar and water were mixed, the sum of their individual volumes was slightly more than the volume of the final solution that you're looking at. Thus, you'll actually need to add a little bit more than the $\,\frac 23V\,$ additional water from the answer above.

So, how much more? The answer that follows gives a great illustration of an important theme emphasized throughout this website: expressions have lots of different names, and different names are good for different things!

Let $\,P\,$ denote ‘a part’ (by volume). We're going to rename the ratio of the sum of the unmixed volumes to the mixed volume:

$$ \begin{align} &\frac{\text{unmixed}}{\text{mixed}}\cr\cr &\qquad = \frac{\overbrace{P}^{\text{sugar}} + \overbrace{2P}^{\text{water}}}{(2+\frac 23)P}\cr &\qquad\quad\ \text{(the mixed number $\,2\frac 23\,$ is $\,2 + \frac 23\,$)}\cr\cr &\qquad = \frac{3P}{\frac83 P}\cr &\qquad\quad\ \text{(rename $\,2\frac23\,$ as an improper fraction)}\cr\cr &\qquad = \frac{9P}{8P}\cr &\qquad\quad\ \text{(since $\,3\div\frac83 = 3\cdot\frac38 = \frac 98)$}\cr\cr &\qquad = \frac{\frac98P}{P}\cr &\qquad\quad\ \text{(multiply both top and bottom by $\,\frac 18\,$)}\cr\cr &\qquad = \frac{P + \frac 18P}{P}\cr &\qquad\quad\ \text{(rename $\,\frac98\,$ as $\,1 + \frac 18\,$)} \end{align} $$

This last name for the ratio, $\,\frac{P + \frac 18P}{P}\,$, clearly shows how to get from the final (mixed) volume to the original (bigger) volume: increase the final volume by one-eighth of itself!

This is exactly the information needed:

Does this agree with our intuition? We expected the amount of additional water to be a little bit more than the ‘two-thirds’ of the prior answer. Indeed, $\,\frac 34\,$ is a little more than $\,\frac 23\,$: $$\frac 34 - \frac 23 \ =\ \frac9{12} - \frac8{12} \ =\ \frac1{12} $$ It's always good to check your intuition!