Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum

Solutions for Session 1, Part B

See solutions for Problems: B2 | B3 | B4 | B5 | B6| B7 | B8 | B9 | B10 | B11

Problem B2

Here is the completed table:

 Number of Sheep in front of Eric Number of sheep shorn before Eric 4 2 5 2 6 2 7 3 8 3 9 3 10 4 11 4

 Problem B3 The table suggests that the number of sheep shorn before Eric goes up by one every three sheep, so there will be 17 sheep shorn before Eric if there are 50 in line ahead of him.

 Problem B4 You might consider making a table, or looking for a pattern and giving a description like the one above. The description is "algebraic" if it involved mathematical thinking tools. It would not be algebraic if you decided to build the above table and continue it until you could read off the answer.

Problem B5

Here is the completed table:

 Number of Sheep in front of Eric Number of sheep shorn before Eric 37 13 296 99 1000 334 7695 2565 37, 38, or 39 13 61, 62, or 63 21

Note that there is more than one answer to the last two questions (exactly three, actually). Also note that you need to think algebraically to answer the question for larger numbers.

 Problem B6 If Eric sneaks ahead of three sheep at a time, the table will appear different: The number shorn before Eric will grow by one for every four sheep shorn (instead of three). If he sneaks ahead of 10 sheep at a time, the number shorn before Eric grows by 1 for every 11 sheep shorn.

 Problem B7 You could build a formula, or algorithm. One algorithm is to round up the number S / (k + 1), where S is the number of sheep in front of Eric, and k is the number of sheep he sneaks past each time. The "1" in this algorithm accounts for the one sheep shorn at the front of the line.

 Problem B8 The rule becomes: Round up (S - 2) / (k + 1), because Eric sneaks past two sheep immediately. Another possible rule is to round down S / (k + 1).

Problem B9

In this situation, two sheep are shorn each time before Eric sneaks up two sheep. We get the following table:

 Number of sheep in front of Eric Sheep shorn before Eric 4 2 5 3 6 4 7 4 8 4 9 5 10 6 11 6 12 6 13 7

So, the rule is: If the number of sheep in front of Eric is not two more than a multiple of 4 (e.g., 5, 7, etc.) divide the number of sheep in front of Eric by 2 and round up. If the number of sheep in front of Eric is two more than a multiple of 4 (e.g., 6, 10, 14, etc.), divide by 2 and add 1.

 Problem B10 You may have used a graph, a table, or a description of the rule in words or algebraic symbols.

 Problem B11 Reasoning skills would be used to make a convincing argument that the number three will be involved in the construction of a general rule for Eric the Sheep's behavior, or to explain why the pattern will continue past the first dozen sheep.