Story Problems

DSC_0731Back in the good old days, an algebraic story problem could very well be John has 4 apples, Jane has two less than Tim, but Tim has twice as many apples as John. How many apples has Jane? A slightly more advanced story problem could be to ask for the final price of a £10 hat after a 10% discount and 15% VAT added.

Apples are a timeless feature without doubt, and the VAT hasn’t disappeared either, but I wonder which story problems modern textbooks have to offer.

How’s this?

John, Jane and Tim each begin downloading Boom! The Final Massacre at 16:00. By 16:20, John has finished downloading 25%.  Tim’s Internet connection is twice as fast as Johns, but Jane’s is one third slower as Tim’s. When will Jane’s download complete?

Is Three Times Seven The Optimum?

wheelsYou will remember that dice are made of six square sides, carrying one to six pips on each sides. The pips (or points, or eyes) are arranged such that pips on opposing sides add up to seven: one opposes six, two opposes five, and three opposes four. Each pair adds up to seven.

Why is that, I wonder?

What stops me from manufacturing regular dice with sides adding up to six (one, opposing five, and two, opposing four) and nine (three, opposing six)? Or four (one, opposing three), six (two, opposing four), and eleven (five, opposing six)?

I can only think of one reason: maybe, designing dice such that the sum of pips on opposing sides equals seven, is not at all about the sum being seven. Maybe, the argument is that all pairs of opposing sides add up to the same number results in fairly balanced dice (the weight across all axis levels, so that the centre of gravity is in the geometrical centre of the body)?

I tried a few alternative designs, and have not yet found a design that produces equal sums of pips on all pairs of opposing sides, other than seven.

Can you?

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