Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum

Solutions for Session 5, Part C

See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6| C7| C8

 Problem C1 The arrangement of the coins should look like this: Here is the corresponding line plot: Note that there are nine dots, one for each stack, which are placed in the line plot according to the number of coins in each stack.

 Problem C2 The arrangement of the coins should look like this: Here is the corresponding line plot:

 Problem C3 The arrangement of the coins should look like this: Here is the corresponding line plot:

 Problem C4 There are multiple solutions to each of these problems, so answers will vary. Here are some possible solutions: a. Here is a different line plot with a mean equal to 5: b. Here is a line plot with a mean equal to 5 that has exactly 2 stacks of 5 coins: c. Here is a line plot with a mean equal to 5 but a median not equal to 5: d. Here is a line plot with a mean equal to 5 that has no 5-coin stacks: e. Here is a line plot with a mean equal to 5 that has two 5-coin stacks, 4 stacks with more than 5 coins, and 3 stacks with fewer than 5 coins: f. Here is a line plot with a mean equal to 5 that has two 5-coin stacks, 5 stacks with more than 5 coins, and 2 stacks with fewer than 5 coins: g. Here is a line plot with a mean equal to 5 that has two 5-coin stacks, two 10-coin stacks, and 5 stacks with fewer than 5 coins:

 Problem C5 You would need to remove 3 coins from one of the stacks of 5, resulting in a stack of 2 coins. This would reset the mean, since adding 3 coins to a stack changed the mean.

 Problem C6 Yes, you could remove 1 coin from one stack and 2 coins from another. Alternately, you could remove 1 coin from each of three stacks. You could even add 1 coin to a stack and remove 4 from another. The only requirement is that you subtract a total of 3 coins from the total number of coins (to counterbalance the 3 coins you added previously).

Problem C7

Answers will vary. Here are some possible solutions:

 • You could add 4 coins to the stack of 3 to create a stack of 7 coins. • You could add 2 coins to each of 2 stacks of 5 to create 2 stacks of 7 coins each. • You could add 1 coin apiece to 4 of the stacks of 5 to create 4 stacks of 6 coins each.

The only requirement is that you must add 4 coins to the total number of coins to return the sum to 45, which will return the mean to 45 / 9 = 5.

 Problem C8 As with Problem C4, answers will vary. See Problem C4 for some sample solutions.