 # K-12 estimation

Question: How can estimation be used to solve problems in K-12?

Estimation is often used to find the sum, product, etc. of two numbers.

A common activity is to fill a jar with jellybeans, then have kids estimate how many jellybeans are in the jar. Students are not allowed to open the jar, but they can pick it up, spin it around, or shake it.

What is the optimal strategy for counting the jelly beans in a jar?

Here's one possible strategy:

If the jar has a translucent bottom, you can count the beans on the bottom layer. Then, using the "height" of each bean, laying on its side, you can calculate the number of layers in the jar, then multiply. But maybe this doesn't give a good estimation, because perhaps the beans do not sit flat on the bottom layer. Then perhaps a jellybean can be approximated as two little cubes glued together. You could then count 1/2 jellybeans on the bottom row, according to how much of the bean is visible. If a bean is laying mostly on its side, it counts as 1, but if a bean is being held upright, then count it as 1/2. This would likely give a better estimate overall.

This strategy generalizes to any prism. For example, a cube of jellybeans, or even a triangular prism. Now that we have a strategy for prisms, we can make the problem harder. What if you're given a sphere of jellybeans? What if you're not a jar (cylinder) of jellybeans without a translucent bottom? How then might you estimate the number of beans?

The sphere and cylinder both have a circumference of 2 * pi * r. By counting the number of beans around the sphere or cylinder, we can obtain a circumference in terms of beans. Again, it's likely that counting beans in half units will yield a better estimation. Then use the circumference to find the radius of the sphere or cylinder. If it's a sphere, you can now compute an estimate for the volume in terms of beans, using the volume formula for spheres. If it's a cylinder, you must obtain the height, by counting along the center vertical line, before applying the formula.

I have not tried any of these methods. I wonder how well they fare. Can you think of a better way to estimate, using methods known to K-12 students, or otherwise?

Strangely, a quick Google search on "estimate to solve equations" didn't yield relevant results. Most pages referred to graphing to estimate solutions to equations. Also, "round to solve equations" yielded equally irrelevant results. In this case, most pages referred to rounding after finding the solution to an equation. But rounding can certainly be used to estimate solutions to certain equations. At very least, solutions to linear equations and inequalities can be estimated by rounding. However, it's important to note that estimates will typically sway further from the exact solution the more steps that are involved. The estimation to a one-step equation will likely be more accurate than a two-step, which in turn will likely be more accurate than a multi-step. However, this is not always the case, as it depends both on the precision of rounding, and the particular numbers involved.

Geometry:

Given some known length, students can estimate some unknown length. Given an angle, students can use increments of 45 degrees to estimate an unknown angle. For example, if some angle is 105 deg, then it's somewhere between 90 and 135. In fact, it's much closer to 90 deg, and so a reasonable estimate might be 112 deg, for example.

Fractions:

When adding or subtracting fractions with unlike denominators, students can round fractions to values that are easy to recognize. For example 21/60 + 14/27 is very nearly 1/3 + 1/2 = 5/6. If students know how to multiply by estimating, they should be able to multiply and divide fractions, like denominators or not, by rounding.

Square roots and logarithms:

Square roots and logarithms can be estimated by finding a lower and upper bound. For example $$\sqrt{103}$$ is somewhere between 10 and 11, because $$10^2 = 100,$$ and $$11^2 = 121.$$ I would guess it's about 10.1, because 103 is much closer to 100 than to 121. Logarithms can be estimated similarly. Here's a good video demonstrating that.