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Session 4 Part A Part B Part C Part D Homework
 
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Notes for Session 4, Part B


Note 3

We will now look at ways to compare ratios in the context of mixing colors. In these problems, it is important to understand that we'll be thinking about these mixtures without using the common algorithms for comparing fractions. That is, we need to assume we don't have the skills for finding a common denominator to compare -- for example, 4/7 and 3/8. This forces us to compare ratios by relying on a more basic understanding of what a ratio is.

<< back to Part B: The Mixture Blues


 

Note 4

Groups: Work in pairs on Problems B1-B13. Discuss solutions to Problem B5 in particular.

Consider the following solution:

In mixture A, there is one more blue beaker than clear, and in mxture B, there is one more blue beaker than clear. Therefore, the mixtures are the same color.

Try to find the fault in this explanation. This is the same reasoning, by the way, that students use when they claim that 2/3 is the same as 3/4, since the numerator in each fraction is one less than the denominator. They are, in fact, using additive strategies.

A lovely way of solving mixture problems like this is by comparing the effect these containers have on the entire mixture. For example, if mixture A had 2 clear containers and 3 blue, and mixture B had 99 clear containers and 100 blue, the extra blue in A would have a greater effect on the entire mixture than the extra blue in B which is spread over 199 total containers. Therefore, A would be bluer (3/5 compared to 100/199).

Some people may use a canceling technique to compare these ratios. They will form blue/clear container pairs and cross them out. For example, in Problem B6, they will cancel all the containers in A after forming 2 blue-clear pairs, and do the same in B with 1 clear left over, claiming that B is less blue. This technique will not work in Problem B5, however, since after using the "canceling" technique, each mixture has 1 blue container left, suggesting they're the same shade of blue. Once again, though, the effect of the extra blue in A is greater than the effect in B because there are fewer containers in A.

Another approach to these problems is to equalize the number of containers in each situation by creating multiples of mixtures. For example, in Problem B7, if you double mixture B, you get the same number of containers as in mixture A, making them easy to compare.

<< back to Part B: The Mixture Blues


 

Note 5

Problem B10 is an interesting one and may require spending some time thinking about how to prove this assertion. In fact, A U B can never be bluer than both A and B. You can use some algebra to prove this. Find a relationship between a, b, c, and d such that: (a + c) / (b + d) is greater than a/b and (a + c) / (b + d) is greater than c/d. These two inequalities become: c / d is greater than a / c and a / b is greater than c / d, a contradiction. Therefore, A U B can never be bluer than both A and B.

<< back to Part B: The Mixture Blues

 

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