Answer these questions:

1. A bat and a ball cost $1.10 in total. The bat costs $1.00 more than the ball. How much does the ball cost? _____ cents
2. If it takes 5 machines 5 minutes to make 5 widgets, how long would it take 100 machines to make 100 widgets? _____ minutes
3. In a lake, there is a patch of lily pads. Every day, the patch doubles in size. If it takes 48 days for the patch to cover the entire lake, how long would it take for the patch to cover half of the lake? _____ days

The correct answers are 5 cents, 5 minutes, and 47 days, but each question also invites a quick, intuitive response that’s wrong. In order to succeed, you have to suppress your “gut” response and reflect on your own cognition deeply enough to see the error. Psychologist Shane Frederick devised the three-question test in 2005 to illustrate these two modes of thought, unreflective and reflective, which he called System 1 and System 2.

Scores on the CRT correlate with various measures of intelligence, patience, and deliberation, but cognitive ability alone isn’t strongly correlated with CRT scores: If you’re not disposed to answer impulsively then the problems aren’t hard, and if you do answer impulsively then cognitive ability won’t help you. A sample of students at MIT averaged 2.18 correct answers, Princeton 1.63, Carnegie Mellon 1.51, Harvard 1.43; see the link below for more.

(Shane Frederick, “Cognitive Reflection and Decision Making,” Journal of Economic Perspectives 19:4 [2005], 25-42.) (Thanks, Drake.)

So many more men seem to say that they may soon try to stay at home so as to see or hear the same one man try to meet the team on the moon as he has at the other ten tests.

This ungainly but grammatical 41-word sentence was constructed by Anton Pavlis of Guelph, Ontario, in 1983. It’s an alphametic: If each letter is replaced with a digit (EOMSYHNART = 0123456789), then you get a valid equation:

   SO     31
 MANY   2764
 MORE   2180
  MEN    206
 SEEM   3002
   TO     91
  SAY	 374
 THAT   9579
 THEY   9504
  MAY    274
 SOON   3116
  TRY    984
   TO     91
 STAY   3974
   AT     79
 HOME   5120
   SO     31
   AS     73
   TO     91
  SEE    300
   OR     18
 HEAR   5078
  THE    950
 SAME   3720
  ONE    160
  MAN    276
  TRY    984
   TO     91
 MEET   2009
  THE    950
 TEAM   9072
   ON     16
  THE    950
 MOON   2116
   AS     73
   HE     50
  HAS    573
   AT     79
  THE    950
OTHER  19508
+ TEN    906

TESTS  90393

Apparently this appeared in the Journal of Recreational Mathematics in 1972; I found the reference in the April 1983 issue of Crux Mathematicorum, which confirmed (by computer) that the solution is unique.

French mathematician Joseph Bertrand offered this observation in his Calcul des probabilités (1889). Inscribe an equilateral triangle in a circle, and then choose a chord of the circle at random. What is the probability that this chord is longer than a side of the triangle? There seem to be three different answers:

1. Choose two random points on the circle and join them, then rotate the triangle until one of its vertices coincides with one of these points. Now the chord is longer than a side of the triangle when its farther end falls on the arc between the other two vertices of the triangle. That arc is one third of the total circumference of the circle, so by this argument the probability is 1/3.

2. Choose a radius of the circle, choose a point on that radius, and draw a chord through that point that’s perpendicular to the radius. Now imagine rotating the triangle so that one of its sides also intersects the radius perpendicularly. Our chord will be longer than a side of the triangle if the point we chose is closer to the circle’s center than the point where the triangle’s side intersects the radius. The triangle’s side bisects the radius, so by this argument the probability is 1/2.

3. Choose a point anywhere in the circle and draw the chord for which this is the midpoint. This chord will be longer than a side of the triangle if the point we chose falls within a concentric circle whose radius is half the radius of the larger circle. That smaller circle has one-fourth the area of the larger circle, so by this argument the probability is 1/4.

Further methods yield still further solutions. After more than a century, the implications of Bertrand’s conundrum are still being discussed.

What is the shape of the moon’s path around the sun? The moon orbits the earth, and the earth orbits the sun, so many of us imagine it looks something like the image on the left, a looping motion in which the moon periodically slides “backward” during its progress around the larger body.

But it’s not! The shape is closer to a 13-gon with rounded corners; there are no loops. Helmer Aslaksen, a mathematician at the National University of Singapore, writes, “I like to visualize this as follows. Imagine you’re driving on a circular race track. You overtake a car on the right, and immediately slow down and go into the left lane. When the other car passes you, you speed up and overtake on the right again. You will then be making circles around the other car, but when seen from above, both of you are driving forward all the time and your path will be convex.”

More at his page.

(Thanks, Drake.)

From the Royal Society of Chemistry’s Chemistry World blog: In 1955, when impish graduate student A.T. Wilson published a paper with his humorless but brilliant supervisor, Melvin Calvin, Wilson made a wager with a department secretary that he could sneak a picture of a man fishing into one of the paper’s diagrams. He won the wager — can you find the fisherman?