Steve Kass

I’m a mathematics teacher, and I have some arithmetic problems for you to solve.

1. Mitt drives to the beach and back.

Q. Mitt Romney drives 50 miles from his home one of his homes to the beach and back one summer, with or without a dog on the roof. He averages 50 miles an hour on the way out, and he averages 10 miles an hour on the way back. What was Mitt’s average speed on the road?

A. It was 16-2/3 mph¹, but Mitt’s campaign insists he’s no slowpoke and that he averaged 30 mph. (They also say the trip explains his tan.)

2. “Going Dutch” with Mitt.

Q. You and Mitt Romney have lunch together at Chick-fil-A. The bill is $15, and he only has a $10 bill, so he pays 2/3 of the bill. Later, you have dinner together at the country club, and the bill is $300. He puts in $100, and this time you pay 2/3 of the bill. Are you even-Steven?

A. Of course not, but Mitt says you are, because he paid 1/3 once and 2/3 once.

3. Mitt Sees Red.

Q. In the four squares below, about what percentage of the area overall is red?

A. About 10%. Certainly way less than half. Mitt disagrees. He says the percentage of red is over 75%, and he surprises you by providing actual details instead of saying he has a secret calculation that solves the problem and comes out the way he wants it to. He says there are three 100% red squares and one 5% red square, for an average of 305%/4 or about 76%. Ann tells you to stop giving him a hard time.

Earlier today, the Romney campaign website posted a letter from PriceWaterhouseCoopers “PWC” LLP summarizing the Romneys’ tax returns for the years 1990-2009. The carefully-worded letter² states that

The average of the annual “effective federal personal income tax rates” as computed based on the returns as prepared during the period is 20.20%.

PWC computed each year’s “effective federal personal income tax rate” as instructed by the Romneys.

As you requested, we compute each annual “effective federal personal income tax rate” as total taxes owed divided by adjusted gross income as shown on the federal income tax returns as prepared.

First of all, the Romneys’ “adjusted gross income” (AGI) may be less than what most of us think of as simply “what the Romneys made.” For example, the expenses of carrying on a trade or business (such as dressage) can be deducted from total income. The Romneys’ tax rate on their total income is lower than that on their adjusted gross income. Maybe a little, maybe a lot.

More importantly, however, is how Mitt and Ann instructed PWC to compute a number they could call an “average.”

Despite appearances, no one has said that the Romneys paid 20.20% of their adjusted gross income in federal income taxes during the period 1990 – 2009. They probably paid less.

The 20.20% figure is not the Romney’s federal income tax rate for income they earned during the period 1990 – 2009. It’s the average of 20 separate annual rates. Mathematically, it’s a weighted average of their annual rates.

If you think the average of 20 annual rates is the same as the 20-year average annual rate, you’re wrong. By that flawed logic, Mitt averaged 30 mph, you and he are even-Steven for splitting meals, and the colored squares in the picture above are 75% red. Go away.

Further blurring the issue, the campaign today also posted a letter from Brad Malt, the “trustee of the Romney’s [sic] blind trust,” about Mitt’s taxes. Malt tries to say that PWC calculated something that it didn’t calculate.

Regarding the PWC letter covering the Romneys’  tax filings over 20 years, from 1990 – 2009:

Over the entire 20-year period, the average annual effective federal tax rate was 20.20%.

Malt omitted two short and crucial words: of the. PWC did not say that Romneys’ average annual effective federal tax rate was 20.20%. They said that if you average the 20 separate effective rates on their 20 federal tax returns, you get 20.20%. It’s a silly thing to calculate, but it’s what the Romneys told them (and presumably paid them) to do.

The Romneys’ overall tax rate might have been lower than 20.20%. It might have been higher, too, but given the outrage over the relatively low rates they pay compared to working Americans, if it were higher, he’d have said so. And Mitt won’t release his tax returns. We can only assume the worst.

If you value the truth, vote for Barack Obama on November 6, 2012.

¹ Let’s figure it out. Mitt drove a total of 100 miles. It took him 1 hour (50 miles at 50 mph) to drive the first half of the trip and 5 hours (50 miles at 10 mph) to drive the second half. He drove a total of 100 miles in 6 hours, so his average speed was 16-2/3 miles per hour.

² The letter is signed personally, in blue, by PriceWaterhouseCoopers LLP, because why wouldn’t they sign it? Corporations are people.

Last night, as a few of us not backing up Ray Davies in Philadelphia gathered for dinner in Hoboken, I spotted this holiday mispostrophe (ssp. dyspostrophe).

It probably wasn’t intentional on Trader Joe’s’s part, but the mispostrophe distracted me from the numbers on the box — especially 24 and 50. But only briefly; the pressing question quickly loomed.

If 24 Milk Chocolates weigh 50 grams altogether, aren’t they too small?

The appropriate comparison was obvious: M&M’s®. Little did I suspect it would be something of a challenge to find out the true weight of one regular M&M.

Disregarding outliers like “I think it is about 15g; 15 grams is perhaps the answer,” answers on the web (to the question of an M&M’s weight) generally fell into two camps. There was a handful of a-bit-less-than-a-gram answers, like “There are about 500 Plain M&M’s per pound,” and there was also a handful of around-2-grams answers, like “After an experiment, of weighing M&M’s, here were the results. 1) 2.208 g 2) 1.882 g 3) 1.904 g 4) 2.438 g.”

After considerable “research,” but no direct measurement, I’m swayed, not by any attestations of milligram precision, but by the preponderance of evidence [and 1] that one regular M&M weighs a bit less than a gram. Which conclusion is consistent with my personal experiences as a candy sorter (when I can find an uncluttered flat surface, which isn’t very often).

From the web’s many M&M Q&A (or should I say Q&“A”?) a few examples:

• Q: What is the weight of one M-and-M candy? [link]
  A: I think it is about 15g; 15 grams is perhaps the answer
   
• Q: How much does an M and M weigh? [link]
  A: When we counted the number of M&M’s in a 12.6oz bag, we got 404, which means there are 32.06 M&M’s/oz, which means that each M&M weighs 1.13 grams. [SK: If you divide backwardsly, perhaps. Otherwise, each M&M weighs (on average) about 0.88 grams.]
   
• Q: How many m&m’s do you reckon are in 7oz? I’m ordering custom m&ms, and they come in 7oz bags. I need about 1000 m&ms, total. how many bags should I order? [link]
  A1: [Best Answer] 10 bags, maybe around 75 or 100 in each bag. [SK: Better safe than sorry.]
  A2: 2 or 3.
   
• Q: How much does a single plain m&m weigh? [link]
  A: After an experiment, of weighing M&M’s, here were the results. 1) 2.208 g 2) 1.882 g 3) 1.904 g 4) 2.438 g.

As for the pressing question, I’ll cautiously answer it “No” and hope Toby and Theo agree. Two or three M&M’s-worth of chocolate every day for most of a month — for those endless days, those sacred days, believe me — is not so bad. Despite anyone’s opinion that one serving of M&Ms comprises 208 grams (and 1023 calories).

This, this, and this. Specifically, foolish nonsense from someone named Todd (Henderson).

The toddtipping point? Right after Nobel Prize-winning economist Paul Krugman suggests indirectly that the Drs. Henderson earn about $450,000 a year, which could subject them to $10,000/year in additional taxes under Obama’s proposal to postpone the upcoming expiration of the Bush tax cuts, but only on the first $250,000 of income.

Todd (who would prefer a bigger tax cut for 2011 than Obama’s legislation provides and who threatens to fire his $20/month legal Mexican gardener if he doesn’t get his way) scrumbles¹. Within a single paragraph, Todd refudiates refutes Krugman’s estimate of his salary (“not even close to our income on the high side”) yet sees no contradiction in describing the injury he and his wife would sustain from $10,000/year in additional taxes, which he just implied he won’t have to pay (because his salary is “not even close” to high enough to result in that increase).

Professor Henderson careens further out of control a paragraph later, when he inflates the fictive $10,000 figure by 20%, to $12,000.

All this from someone with degrees in both engineering and law, whose ability to explain (when it suits him, apparently) was recently rewarded with tenure as a professor. In light of the facts of his education, Todd’s behavior doesn’t pass the smell test. I’m calling it toddfoolery. Either something tragic has happened to Todd’s mind since he received his degrees and tenure, or he’s a disingenuous liar. At least those are the only explanations I can imagine.

Update (21 Sep 2010): Yesterday, Todd removed his tomfoolerific posts, along with readers’ comments to those posts,from Truth on the Market, where they had appeared. Todd explains.

Update (21 Sep 2010): Today, Todd “hung up his blogging hat”.

[Note: The links at the beginning of this post are no longer valid.]

Update (12 Nov 2011): An alternate spelling of Toddfoolery (Todfoolery) is now available here: Pity the 1%, and Their Tod(d)foolery.

¹ The verb scrumble will be coined in a future installment of “Word of the Day.”

Google Maps Street View has a nice shot of the house across the street from where I grew up. The current owners moved in almost four years ago, and the outdoor wall lamps (installed by the owners previous to them, who flipped the house never having lived in it) have been lit day and night since.

How much have the current owners paid to brighten their Mulberry Drive neighborhood?

[Note: There should be a Google Maps Street View image here, but unfortunately Google Maps Street View can’t correctly produce embedding code. Pretend there’s an image of a house with some lights on the front or click the link below.]
View Larger Map

The pinpoint light sources shine brightly through the Phoenix sun, suggesting that each fixture houses a 60-watt (or brighter) incandescent lamp. Three 60-watt bulbs running continuously consume a kilowatt-hour of electricity every 5½ hours. That’s 4.32 kWh per day or just over 130 kWh a month. The electric utility for the area, Salt River Project, offers several rate plans, so it’s impossible to say exactly how much the lamps cost to run. Because the lights run day and night, SRP’s Basic Plan, which has no peak/off-peak pricing, would be the best value. During the seven summer months (May–October), SRP charges about 11¢/kWh. In “winter” the cost drops to about 8¢. Under any plan, the annualized cost will exceed 10¢/kWh, or $13/month. The cost of electricity hasn’t changed much over the past several years, so come December, when the owners celebrate four years in their house, they’ll have spent over $600.00 for the outdoor lighting.

On the brighter side, hydroelectric and nuclear power produce much of Arizona’s electricity, so the carbon footprint from running these lights isn’t as big as it might be. The big truck, on the other hand…

Another interesting question to answer is this: How much ice could these lights have melted? One kilowatt-hour equals about 860,000 calories (that’s 860 food “Calories”), or enough to melt a little more than 22 pounds of ice. The energy that runs these lights is all converted to heat, and if that 180 kW heat were used to melt ice instead of heat the Phoenix air, it would melt a lot of ice over four years — 68 or 69 tons, in fact, which would produce enough water to fill a good-sized in-ground swimming pool.

Facts
  The reflecting pool on the National Mall covers an area of about 8 acres. [reference]
  An American football field including the endzones covers about 1.3 acres. [reference]
  The seating area of Michigan Stadium covers about 6 or 7 acres. [reference]
  The capacity of Michigan Stadium is about 100,000 people. [reference]

Observations
  At the most crowded locations, the density of people on the Capitol lawn today was no more than in a packed stadium. [references: lawn, stadium]
  Most of the people on the mall today were within an area of two or three reflecting pools in size (and white). [reference]

Estimate
  (2.5 reflecting pools) × (8 acres per reflecting pool) ÷ (6.5 acres per stadium) × (100,000 people per stadium) × (average 0.5 density) = 150,000 people. More than 87,000, perhaps. Hundreds of thousands, as in more than 200,000? Not too likely. A million? No.

While my part of the world was embroiled in a heat wave last week, brutal midwinter cold was gripping Vostok Station high atop the mountains and ice of central Antarctica. At times, it was more than two hundred degrees colder in Vostok than it was here in the U.S. Northeast.

A few days earlier, by coincidence, I’d installed the last of my three SodaStream carbonators and ordered two refills. On the 7^th, in the middle of the heat wave, the local distributor exchanged my empties, leaving me with four pounds of freshly compressed carbon dioxide and two questions.

1. Is four pounds of CO₂ a lot compared to the amount already inside my condo, not counting the basement?
2. When the temperature drops below -109.3 °F, the sublimation point of carbon dioxide, does it snow dry ice?

I’m happy to share with you the answers.

1. Yes, quite a lot. The volume of my condo, not counting the basement, is about 7,000 cubic feet, or about 200,000 liters, and while I have far too much junk, my place is still mostly full of air. And like air most places, the air in my condo is mostly nitrogen, oxygen, and argon, and about 400 parts per million of it (by volume) is carbon dioxide. My condo contains a fifth of a million liters of air, I figured, so I’ve got a fifth of 400 liters of CO₂, or about 80 liters.

At room temperature, CO₂ is effectively an ideal gas, so I don’t need Google to convert 80 liters to moles, which is going to help with the calculation. A mole of an ideal gas at standard temperature and pressure¹ occupies 22.4 liters, so I’m living with about 4 moles of CO₂. (And maybe a few voles, too, but that’s not relevant here.) A mole of any chemical compound means 600 sextillion molecules of the stuff. Fortunately, we don’t have to do any sextillionary arithmetic. What makes a mole a mole is that it’s the number of Daltons (neutron or proton+electron masses, more or less) in a gram, so the mass of a mole of a chemical compound is simply its molecular weight in grams. The molecular weight of CO₂ is² 12+2×16, or 44, so a mole of CO₂ has a mass of about 44 grams or an ounce and a half.

Four moles of CO₂ in my condo — that’s about six ounces, or less than a tenth of what’s squeezed into my two new Sodastream carbonators.

2. No. The temperature -109.3 °F is the sublimation point of CO₂ at one atmosphere. That’s the temperature at which dry ice in your house or cooler chest (which is pure CO₂ solid under an atmosphere of air pressure) sublimates; it’s not the temperature at which CO₂ in plain air would precipitate into “snow.” Carbon dioxide won’t precipitate out of the air unless it’s cooled to the sublimation point for its partial pressure in air. That’s (see above) about 400 millionths of an atmosphere at sea level, and roughly two-thirds of that at Vostok Station, elevation 11,000 feet. Smart guy David R. Cook at Argonne National Laboratory has generously used some tax dollars to spread the word. He proposes that the temperature would have to be much colder than –110 — in fact about 220 degrees below zero on the Fahrenheit scale (and no, that’s not “twice as cold” as -110) for CO₂ to precipitate at Vostok Station.

Typical partial pressures of CO₂ on earth are unfortunately off the scale of the phase diagrams I could find, so I can’t provide more details or confirm what David says. You can, however, see the trend in the phase diagram below. The boundary between the periwinkle and salmon regions is where CO₂ sublimates or “snows.” The lower the pressure, the lower the temperature.

Public domain image from Wikimedia Commons.

The phase diagram also answers a question I didn’t think to ask.

• Are full SodaStream carbonators filled with liquid or gas? Liquid.

They’re about a liter big, which (using the calculations above) translates to an internal pressure of several hundred atmospheres, which is well within the green region. How cool is that? Liquid CO₂ right here in my house.

Call it chemistry or call it physics, this is all great fun.

Related links: How Much CO₂ is in a Bottle of Soda?, The Influence of CO₂ on the Chemistry of Soda, Soda’s Contribution to Global Warming

¹ Standard temperature and pressure is 20 °C and one (sea level) atmosphere of pressure. That’s close enough to the conditions in my condo. Around room temperature, gas expands at a rate of about 1% per 4 °C rise in temperature. It also expands as you climb above sea level, by about 1% for each 300 feet. The second derivatives are small, and the contributions from the partial derivative of volume with respect to temperature and altitude add, so yadda yadda yadda, we’re within a few percent. It was worth a quick thinking through, though.

² There are a some simplifications here, but again, they don’t affect the overall calculation by more than a percent or two. The value 44 assumes that every molecule of CO₂ is made up of the predominant isotopes of carbon and oxygen, carbon 12 and oxygen 16, and that single atoms of these isotopes weigh exactly 12 and 16 Daltons, respectively. The “official” atomic weights as periodically reviewed by IUPAC would be more accurate, and they give a molecular weight of 44.01. The discrepancy reflects the distribution of isotopes on earth as well as other details that probably involve higher physics and chemistry well beyond my understanding. In addition, it would seem to me that this measurement ought to be called the molecular mass, but when measuring any property of such tiny things, you’re going to run up against quantum mechanics and other complications, so I suppose whoever understands all of this can call it what they want.