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Solutions for Session 4, Part C
See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6| C7
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Problem C1 | |
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Here, we add six red chips to four black chips. After doing this, four zero pairs can be removed, leaving two red chips. So +4 + (-6) = -2.
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Problem C2 | |
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Here, we add six red chips to four red chips. After doing this, we have 10 red chips. So -4 + (-6) = -10.
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Problem C3 | |
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Start with five black chips. We want to subtract (take away) six red chips. Since there are no red chips to take away, we must add a number of zero pairs (one black and one red chip each). Add six of these pairs so that there are six red chips to take away. Eleven black chips remain, so +5 - (- 6) = +11.
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Problem C4 | |
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Start with five red chips. We want to subtract (take away) six red chips. Since there are not enough red chips to take away, we must add one zero pair (one black and one red chip each). Now there are six red chips to take away. Removing them leaves one black chip, so -5 - (-6) = +1.
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Problem C5 | |
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Start with an empty circle. We want to take away three groups (the first -3) of three red chips (the second -3). Since there are no red chips to take away, we must add three zero pairs for each group, or nine zero pairs in all. Taking away three red chips from the group leaves three black chips in each group. Since there are three groups, a total of nine black chips remain. So -3 • (-3) = +9.
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Problem C6 | |
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Start with an empty circle. We want to add three groups (+3) of three red chips (-3). We can do this directly. After doing this, there are nine red chips in the circle, so +3 • (-3) = -9.
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Problem C7 | |
a. | Using the partitive method, there are two black chips in each of four groups: (+8) / (+4) = +2.
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b. | Using the partitive method, there are two red chips in each of the two groups: (-4) / (+2) = -2.
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c. | Using the quotative method, there are four sets of two red chips: (-8) / (-2) = +4.
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d. | This division problem cannot be modeled using the colored-chip model. |
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