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Race To 100 Algebra, Level 3 Problem What is this problem about? This is about number patterns and also about solving simple algebraic problems. Of course, it can be done going laboriously from jump to jump. If your children can only do it this way thats fine, however, you should point out to them that there are other ways to do it. Its worth noting that in problems like this, it is always worth asking the class to estimate what will happen before they try to work it out. It helps them to get a feel for the problem and it also helps their intuition. The extra twist in Extension 1, is that both Fred and Freda dont land exactly on the number 100. Extension 2 is hard unless done by a constructive approach like making a table or by guess and check. It can be done by solving two equations but this is beyond this Level. The extension problems here could be development band work. Achievement Objectives Algebra (Level 3) - describe in words, rules for continuing number and spatial sequence patterns - state the general rule for a set of practical problems Mathematical Processes Resources Number lines (1-100) (or use metre rulers) Blackline master of the problem (English) Blackline master of the problem (Maaori) Specific learning outcomes The children will be able to: - describe in words number patterns - work systematically to solve a problem involving number patterns Teaching sequence
Extension to the problem In Extension 1, on what number does the overtaking take place? Solution
So from the table you can see that Fred gets to the 100th square and had to wait two jumps for Freda to catch up. The other thing that you and the children can see from the table is that this is a tedious way to do this problem. If you note that Freda is jumping on the squares numbered 2# + 30, then she gets to the 100th square when 2# + 30 = 100. This is when # = 35 (check this with the table). On the other hand, Fred is using the pattern 3# + 1. So he gets to 100 when 3# + 1 = 100. In other words when 3# = 99 or when # = 33. As we saw in the table, Fred gets to the 100th square in 33 jumps, two ahead of Freda. Extensions: Using the equation 2# + 51 = 100 for Freda, we can use guess and check to see that # must be bigger than 24 (2 x 24 + 51 = 99) and less than 25 (2 x 25 + 51 = 101). So Freda will need 25 steps to get to the 100th square. For Fred we get the equation 4# + 1= 100. In this case using guess and check or a table or some other means we can see that # must be more than 24 (4 x 24 + 1 = 97) and less than 25 (4 x 25 + 1 = 101). So what happened? Well they both landed on the 101st square on their 25th jump. As it happens I think that a photo finish would have shown Freda just ahead. But what do you think? (Does it help that if we put 2# + 51 = 4# + 1. Then we get 2# = 50, so # = 25. They land together at the end of the 25th jump but that is the first time that they are together. Freda is ahead up to that point.) |
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