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Pedestrian Mathematics

April17

In a drizzling rain, a 63-year-old professor, George Nobl, stood on a stretch of Times Square sidewalk the other day beside an easel blithely daring passers-by to solve a math problem.

Fred can paint a room in three hours,” the problem, printed in block letters on paper pinned to the easel, said. ”Mary can paint the same room in two hours. How long will it take them to paint the room together?

Solve the problem, and win a Snickers bar,” the message added.

In an area known for bullhorn-wielding sidewalk evangelists, three-card monte games and even the occasional chess face-off, some thought Professor Nobl was a con artist. Some figured he was promoting a candy bar. Many walked away, but quite a few lingered to study the problem.

It turned out Professor Nobl had nothing up his sleeve except a quirky effort to promote the fun of math. Every Wednesday at noon since last summer, he has stood on West 42nd Street between Fifth and Sixth Avenues challenging passers-by in this eccentric fashion.

If people want to answer those trivial questions in ‘Who Wants to Be a Millionaire,’ then why not give them a math problem?” he asked.

After a few minutes, several people in the crowd whispered answers into his ears by turns, with most of them guessing that the answer was two and a half hours.

If it takes one person two hours to paint the room, how come it would take longer for them to do the task together?” Professor Nobl asked tartly.

Tom Mansley, a nearby office worker on his way to lunch, was one of the few who got the correct answer. ”You don’t need a lot of mathematical background for this problem, but it’s tricky,” he said after calculating on a notepad for five minutes. ”I would have stood here all day in the rain to solve it. I like to meet challenges.

David Williams, a social worker who graduated from public schools in the Bronx and went on to college, was less lucky. ”When I see a word problem, I don’t know which rule to use,” Mr. Williams said. ”I remember a lot of rules, but I don’t know which one is the right rule.” Full article here.
So, what was the correct answer that the professor was looking for? (see Post Support)

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Post Support

NCEA Level 2 Algebra Problem. Using the information given, the shaded area = 9, that is:
y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?]

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