 Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum            Solutions for Session 4, Part B

See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6| B7 | B8 | B9 | B10 | B11
B12 | B13   Problem B1 A will be bluer, because B's mix is the same as A's, but with two additional clear beakers thrown in.   Problem B2 B will be bluer. Its mix is the same as A's, but with an extra blue beaker thrown in.   Problem B3 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!)   Problem B4 A will be bluer. One explanation is that B is half blue, while A is more than half blue.   Problem B5 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.   Problem B6 A will be bluer. Its mix is the same as B's, but with one less beaker of clear water.   Problem B7 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.   Problem B8 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).   Problem B9 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.   Problem B10 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.   Problem B11 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.   Problem B12 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).   Problem B13 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).     Session 4: Index | Notes | Solutions | Video