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PISA Problem Problem

December 7, 2013 am31 10:30 am
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Ever count ceiling tiles during a boring test? I have. And I run into a problem (besides that people notice me staring goofy at the ceiling). The tiles never fit. How do I count the broken tiles?  Does each broken tile count as a full tile, since a full tile had to be consumed to make it? Or should I just be counting the area? My geek game, my geek rules – I count both ways.

One of the released PISA math questions sets up a similar problem – and then does not give even partial credit for a partial geek answer.

Before I share the question, I had every intention of fully ignoring PISA. The results are used to do things like justify attacks on teachers local unions in the US, or to set up charter schools whose kids would never learn to agonize over going with the Geek answer or the normal answer. The test is used to harm us, to create panic over a single data point (is there still a country between Sweden and Russia?)

But I was curious about what the questions looked like. Most of them seemed fine. But here’s the ceiling tile question:

PATIO

Nick wants to pave the rectangular patio of his new house. The patio has length 5.25 metres and width 3.00 metres. He needs 81 bricks per square metre.

Calculate how many bricks Nick needs for the whole patio.

Hmm. The 5 x3.25 patio, there are 5 x 3 = 15 nice squares that could be filled. 15 x 81 = 1215. But now we have a strip a quarter of a meter wide (hate when they have fractional remainders, 1/4, for a metric problem. feels like mixed units), and 3 meters long. We should think about the shape of the bricks.

(by the way, the geek analysis is about to cost me credit).

Since there are 81 bricks to the square meter, the bricks could be little squares. One ninth (yeeks!) of a meter on a side. So two rows of them would be about .22 m, and we would have to get another row of partial bricks to fill our .25 strip (and lead to some wastage. or is it waste?) So it would take 3 x 9 = 27 bricks to complete each quarter square meter, or 27 x 3 = 81 of them to eat up the full remaining strip, so 1215 + 81 = 1296. My first answer. And I would not have gotten full credit. Because I forgot something.

What if the bricks were a different shape?  We are tiling square meters, and the problem implies we can do that well. With 81 things – doesn’t that mean we are stuck with rectangles?  1 x 4 rectangles could work. That’s 1/18 of a meter by 4/18 of a meter. With one row long-ways and one row side-ways we could create a strip 5/18 m. wide. 5/18 ≈ .278, and now we are getting somewhere. Now, 54 long-ways bricks (stacked long side against long side) would create a strip 3 m x .22 m, and 14 side ways bricks would put a layer on top that was 3.11 m x 0.56 m, so there still is waste, but much less. 54 + 14 + 1215 = 1283.

And I still don’t have full credit. But now I have an approach to refine my answer.

Imagine bricks 3 m long and 1/81 of a meter wide. 81 of them cover a square meter. 20 of them would be just under a quarter of a square meter, so let’s go with 21 once, twice, three times. That’s a strip 3 m long and 21/81 ≈ .259 m wide, with minimal possible waste. Can’t get a better answer than this. That’s 3 x 21 = 63 + 1215 = 1278 bricks. Perfect answer.

So if the bricks are squares, we need 1296, but as the rectangles get further and further from square shape, we need fewer and fewer, with the minimum possible 1278.

And PISA would:

SCORING PATIO: QUESTION 13

Full Credit

Code 2: 1275, 1276 or 1275.75 (unit not required).

Partial Credit

Code 1: 15.75 (units not required)

OR

  • 1215 bricks for 5m X 3m

(This score is used for students who are able to calculate the number of bricks for an integer number of square metres, but not for fractions of square metres.)

OR

  • Error in calculating the area, but multiplied by 81 correctly

OR

  • Rounded off the area and then multiplied by 81 correctly

No Credit

Code 0: Other responses

Code 9: Missing

PISA would give me partial credit, while a Finnish child who was 2 or 3 bricks short of a full load would get full credit.

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One Comment leave one →
  1. David-S permalink
    December 8, 2013 am31 10:00 am 10:00 am

    No flooring installer would calculate that way or lay tiles that way, and if they did, the owner would not pay. Training for this task would include examining the access and traffic flows and the symmetry of appearance of the floor. Then a center point or center line would be established and the number of tiles calculated to achieve the desired aesthetic. Perhaps the desired result was a weave, or diagonal, or herringbone. Probably just easier to order about 10-15% extra, or a little more if the edges are uneven. Don’t forget to consider the number of pieces in a box of tiles. You can’t generally buy tiles “loose”, and you need some for future fixes.

    I guess I get no credit on that question, but I do get paid by Nick, who loves the finished job and the input he had.

    P.S. I’m laughing while I read the problem as I remember the Mars satellite that crashed into Mars instead of having a soft landing. A mixture of metric and non metric values mishmashed together! A $300 million oops, plus the lost time which is incalculable.

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