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Solutions for Session 4, Part B
See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6| B7 | B8 | B9 | B10 | B11
B12 | B13
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Problem B1 | |
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A will be bluer, because B's mix is the same as A's, but with two additional clear beakers thrown in.
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Problem B2 | |
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B will be bluer. Its mix is the same as A's, but with an extra blue beaker thrown in.
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Problem B3 | |
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B will be bluer, for the same reason as in Problem B2. Or, consider what would happen if you tried to make equal doses of A and B. (Common denominators!)
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Problem B4 | |
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A will be bluer. One explanation is that B is half blue, while A is more than half blue.
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Problem B5 | |
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A will be bluer. A and B each have one more blue than clear beaker, but in A, each beaker makes more of a difference. In other words, A has 1/5 more blue than clear, and B has 1/7 more blue than clear, and since 1/5 is larger than 1/7, A is more blue. Another explanation is that to get B from A, you would need to add 1 blue and 1 clear beaker, which (as a mix) is lighter than A. Therefore, B will end up being lighter than A.
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Problem B6 | |
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A will be bluer. Its mix is the same as B's, but with one less beaker of clear water.
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Problem B7 | |
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They are identical in color. One explanation is that each mix has twice as much blue as clear. A second explanation is that mixture A can be made by doubling the ingredients of mixture B.
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Problem B8 | |
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The bluest is A. A is already bluer than B, so adding B's mix into A won't make it a deeper blue (think of what would happen with paint). This means that A U B cannot be bluer than A, so A is the bluest of the three. Another explanation is that A has one more blue than clear, and so does A U B, but in A U B the extra blue does not contribute as much to the mixture (since there are more beakers of liquid in it).
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Problem B9 | |
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If A U B is bluer than A, then B must be bluer than A. This is true because A U B is formed by adding B to A; if we add something to A and it becomes a deeper blue, whatever we added (B) must be bluer than what we started with (A). The reverse is also true: If A U B is bluer than B, then A is bluer than B.
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Problem B10 | |
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No such magical mixing device exists. Take the bluer of A and B; A U B can never be bluer than it. You could also solve this problem using algebra and common denominators.
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Problem B11 | |
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Yes, but only if A and B are equally blue. If A is bluer than B, adding B dilutes A, and the result (A U B) will not be as blue as A. If A and B are equally blue, A U B will be identical to both.
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Problem B12 | |
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Use the total number of blue beakers, divided by the total number of beakers used. If the BQ of A is M / N and the BQ of B is P / Q, then the BQ of A U B is (M + P) / (N + Q).
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Problem B13 | |
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There are several possible BQs for mixture B. Mixture B could contain only one beaker of blue (BQ = 1/1). A U B would then have BQ 2/4 = 1/2. If we wanted A U B to have BQ 3/6 (an equivalent fraction to 1/2), then mixture B would have to have BQ 2/3. Mixture B can have BQ 1/1, 2/3, 3/5, 4/7, 5/9, or any fraction of the form N / (2N - 1). Mixture B's only requirement is that it must have a BQ larger than 1/2 (see Problem B10 for the reasoning).
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