Thursday, February 22, 2007

Me vs. My Page-A-Day Mensa Puzzle Calendar - 02/22/07

QUESTION: "In the school lunchroom, one table and half a table ate a pizza and a half in a minute and a half. At the same rate, how many pizzas could two tables of hungry children eat in three minutes?"

ACTUAL (B.S.) ANSWER: "four pizzas (Each table eats a pizza in a minute and a half, so the cafeteria manager will need four pizzas.)"

CONSENSUS: Hogwash! Each table and a half eats a pizza in a minute and a half, not each table. Each table actually eats a pizza in a minute, which means that in three minutes they would eat three pizzas...and in three minutes two tables would eat six pizzas. The answer is six, and my Mensa calendar contains a typo. Who do I report this to? (Also, why did a character of "cafeteria manager" need to enter into this scenario? Feels extraneous.)

PuzzleManiac said...

No. Not six nor four.

If the "eating rate" is linear, then each table eat a pizza in 1 1/2 minutes (same as each 1 1/2 tables eat 1 1/2 pizza... prorated). So the solution would be 4 pizzas (Mensa's solution).

However the "eating rate" is NOT linear... People tend to eat slower when they are getting fuller. In addition, the "stomatch" is limited. The total "stomach" of the whole table may be less than 2 pizzas... which means they will never finish two pizzas in one meal.

Let's extend the problem... "How many pizzas could two tables of hungry kids eat in 1 hour???"

If the "eating rate" is linear and unlimited, then they could eat 80 (4 x 20) pizzas... assume each table has 10 kids... two tables have 20 kids... each kid need to eat 4 pizzas in one hour!!! That's not right!!! Each kid has limited "stomach", none of the kid can eat 4 pizzas in one meal.

This problem is simply ridiculous!!! Whoever came up with the question needs to re-examine their I.Q.!!!

Johnny said...

Huh. Good thing I never went ahead with that lawsuit.