Fermi Questions

Fermi Questions will once again be a Division C event for the 2013 competition year.

What is a Fermi Question?
A fermi question is one where you estimate the answer. Questions like "How many licks does it take to get to the center of a tootsie roll pop", where you have little or no data to use and have to make assumptions. Fermi questions are named after Enrico Fermi, a physicist who was very good at these types of questions.

In Science Olympiad, answers to fermi questions are given in powers of ten. For example, if you estimate that it would take 400 licks to get to the center of a tootsie roll pop, then you put it in scientific notation (4 * 10^2) and use the exponent on the ten as your answer, yielding 2. If the you got 600 licks (6 * 10^2), then your answer would be 3, as you round up. Generally, fives can round either way, depending on if you think your answer is high or low.

Points are usually given as follows: 5 points for the correct power of ten, 3 points for one away from the correct power of ten, and 1 point for two away from the correct power of ten. For example, if the correct answer to the number of licks to the center of a tootsie roll pop is 2, and you answer 2, you get five points. If you answer 3 or 1, you get 3 points, and if you answer 4 or 0, you get one point.

Sample Question
Here is a fermi question, worked out, with explanations.

How many pieces of paper could a package of pencil lead cover?

determine the facts

The first step in solving a fermi question is to determine what facts you need. In this one, you need to know:


 * How much paper a piece of lead can cover.
 * How large a piece of paper is.
 * How many pieces of lead are in a package.

Now, the size of a piece of paper is 8.5 x 11 inches, which most people should know. The number of pieces of lead in a package could be known, but if not, would have to be assumed. However, the area a piece of lead can cover is not known, and must be assumed.

making the assumptions

The key to fermi questions is having good assumptions. Obviously, the best assumptions are the ones you don't have to assume, like the size of a piece of paper. However, more often than not, you will have to assume.

The area a piece of lead can cover is based on the width and length of the line it can draw. Using common knowledge, we know that a piece of lead is either .5 mm or .7 mm. However, we choose to use .5 mm because 5 is an easier number to deal with than 7.

Now that we know the width of the line, we need to know the length of the line.

In order to do this, you have to use common sense. If you say it can draw 1 meter, then that's too small. If you say it can draw 100 km, then that's too large. It may be 10 meters, but that seems a little short. It could be 1 km, but that seems a little long. 100 m still seems a little short, but close than 1000 m (1 km). Therefore, we will use 300 m, keeping in mind it could still be small.

So we take $$5\cdot 10-4$$ times $$1 \cdot 3\cdot 102$$, and get $$15\cdot 10-2$$, or $$2\cdot 10-1$$ (going up because the 300 m could have been small) now, this is, in plain English, 2 tenths of a square meter, or 2000 square centimeters, per piece of lead.

Now, we must assume the number of pieces of lead in a package. The number of pieces of lead in a package vary greatly, so it's not very important to be accurate. 10 pieces seems low, and 1000 seems too high. 100 still seems a little high, so we choose 70 as the number of pieces of lead in a package.

$$2\cdot 103\text{ cm}^2$$ times $$60= 1.4\cdot 105$$, which we will change to $$2\cdot 105$$ in order to simplify the math. This can be done because first, the number comes from assumptions that may not be right, second, because the final answer will only be an exponent, and third, because we will round down later to compensate.

The last step is the divide this by the number of square centimeters in a piece of paper. 8.5 and 11 are awkward numbers, so we simply change them to ten. This will remain somewhat accurate, because when we rounded down on one, we rounded up on the other. Note that this will always result in a larger answer.

Now we need to convert 1 square foot into square centimeters we can do this quite simply if we remember that foot long rulers can fit 30 cm on them. 30 * 30 is 900, so we say there are 1000 cm2 in a piece of paper. This value is high, but if we remember that we're dividing, we'll realize this will cancel out the other rounding up (1.5 to 2) we did earlier.

So, $$\frac{2\cdot 10^5}{1\cdot 10^2}$$ is $$2\cdot 10^3$$ sheets of paper. Therefore, our answer is 3.

Magnitude notation
The above question was worked out sequentially with numbers in to allow for explanation. This works well for shorter problems, but with longer problems with longer numbers, this can take a while, and can become a problem with many questions and limited time and space.

When doing such problems, it is better to note numbers in a format that is easier to work with.

This format, which is called magnitude notation*, works as follows.


 * $$1\cdot 10^3$$ would be E3
 * $$4\cdot 10^3$$ would be +E3
 * $$7\cdot 10^3$$ would be -E4

Basically, you put the number in exponential notation (like your calculator does) and round it so there is only one digit.

Then, if the one digit is 0, 1 or 2, leave the exponent part (E3), but do not put a plus or minus. If the digit is an 8 or 9, at one to the exponent and do not put a plus or minus. If the digit is a 3 or 4, then you leave the exponent and put a plus. If the digit is a 6 or 7 add one to the exponent and put a minus. If 5 is rounded up, put add one to the exponent and put a minus. If five is rounded down, then put a plus.

With magnitude notation, when two numbers with plus signs are multiplied, the pluses are removed and one is added to the resulting exponent. for example, +E5 times +E7 equals E13.

When two numbers, one with a +, and one with a -, are multiplied, the signs are canceled without changing the exponent.

When two numbers, both with minus signs, are multiplied, one is subtracted from the exponent and the signs removed.

With division, same signs cancel. Opposing signs are removed, adding one to the exponent of the value with the + sign.

In addition and subtraction, if the exponents are equal and if the signs are two pluses, add one to the exponent and remove the signs. If there are no signs, put a plus (exponents still equal). If both signs are minus, remove the signs (exponents still equal).

If one exponent is one larger, remove a minus sign if there is one on the larger one, add a plus sign is there is no sign on the larger one, and increase the exponent on the larger one while removing the sign if there is a plus sign on the larger one.

If the exponents are 2 or more apart, simply ignore the smaller number.


 * Magnitude notation was created by Justin Chuang of Virgil I. Grissom High School, specifically for Science Olympiad

Although magnitude notation appears more complicated, it is faster and neater to work with, and thus easier to use with limited time and space.

Quick Equations
In Fermi questions, there isn't much need for accurate complex equations. therefore, use the following formulas.

Volume of a sphere: $$\frac{4}{3}\pi r^3$$ Area of a circle: $$\pi r^2$$

For other figures, make up formulas on the spot. For example, use the triangle formula for a parabola, slightly increasing the base and height of the triangle to compensate for the curve, and for an ellipsoid, use a cylinder, decreasing the length to compensate for the curve.

So how do I do well?
You do well in Fermi Questions by being as accurate as possible with the most questions. This can be done certain ways:

Partner pairing
In Fermi questions, the partner pairing is important. You generally want two people with different interests and personalities. The ideal pair would be the estimator, and the number cruncher.

Estimator
For the estimator, you want a visual/kinetic (but more visual) person with a good memory. They need to be able estimate the dimensions of, say, a football stadium based on their memory of one. They need to be able to estimate the weights of objects they may or may not have held before, but know the visual size of.

The estimator needs to be able to, on the spot, assume values and get them close to actuality. He (or she) should also know random facts, like the frequency of a cordless phone, or the number of Crayola colors.

Number Cruncher
The number cruncher is someone quick to perform the calculations, although if magnitude notation is used, it is not as essential. The number cruncher should know physical values and conversion factors, like how many pounds are in a kilogram, the speed of light, speed of sound, etc.

If one member needs to be left out, make it the number cruncher. The estimator can memorize the values, but the number cruncher cannot easily learn how to estimate.

Use facts
If an estimator cannot be found, then this is your only option. You must find as many facts as you can, like the volume of a stadium.

Facts that turn up

 * physics facts, like the speed of sound, or the wavelength of a certain color light.
 * human body facts.
 * state facts from the state you're in
 * facts about the U.S.A.
 * facts about the world.

Sometimes fermi questions simply ask for an obscure fact, such as the mass of a swallow and the mass of a coconut. As it is impossible to know everything, just guess. You're only using powers of ten as answers, and therefore, you can be off by a lot and still get the answer right.

Things to remember

 * You can be off by as much as 250% and still get points. Being off by 50% will still give you full points for that problem.


 * Guess away. There are no penalty for being wrong. Just don't guess blindly. Don't say 3 elephants weigh as much as the titanic.


 * When you round something down, round something else up (or down if you're dividing).


 * These are exponents. E3 plus +E7 is +E7 (this is magnitude notation)


 * The competition is timed. Don't dwell on a single problem unless it's your last.


 * Keep your calculations neat and simple. Often times you're limited in space.


 * Use common sense.