Prompt: Is d greater than or equal to 0?
Statement #1: d is the median of d, 1/d, -d
OK, that's interesting. We are going to consider four cases:
(a) positive numbers greater than 1
(b) positive fractions between 0 and 1
(c) negative numbers less than -1
(d) negative fractions between -1 and 0
(e) zero itself
Case (a): let d = 3
d = 3, 1/d = 1/3, and -d = -3. Of these, 1/d = 1/3 is the median, so this case is not a possibility.
Case (b): let d = 1/2
d = 1/2, 1/d = 2, and -d = -1/2. Of these, d = 1/2 is the median, so this case is a possibility.
Case (c): let d = -5
d = -5, 1/d = -1/5, and -d = 5. Of these, 1/d = -1/5 is the median, so this case is not a possibility.
Case (d): let d = -1/4
d = -1/4, 1/d = -4, and -d = 1/4. Of these, d = -1/4 is the median, so this case is a possibility.
Case (e): let d = 0
d = 0, -d = 0, and 1/d is undefined. It's unclear how to talk about median when one of the three numbers is undefined, but insofar as we just consider the list {0, 0}, the median is 0, which is equal to d.
So, given this statement, d could be a positive fraction between 0 and 1, or a negative fraction between -1 and 0, or zero itself. Since d could be positive or negative, we don't have enough information to answer the question. Statement #1, by itself, is insufficient.
Statement #1: d^3 is the median of d, d^2, d^3
Same four cases (those cases are useful in a wide variety problems!)
Case (a): let d = 3
d = 3, d^2 = 9, d^3 = 27. Of these, d^2 = 9 is the median, so this case is not a possibility.
Case (b): let d = 1/2
d = 1/2, d^2 = 1/4, d^3 = 1/8. Of these, d^2 = 1/4 is the median, so this case is not a possibility.
Case (c): let d = -5
d = -5, d^2 = +25, d^3 = -125. Of these, d = -5 is the median, so this case is not a possibility.
Case (d): let d = -1/4.
d = -1/4, d^2 = +1/16, d^3 = -1/64. Of these, d^3 = -1/64 is the median, so only this case is a possibility.
Case (e): let d = 0
d = 0, d^2 = 0, d^3 = 0. Since they're all equal, and of them could be the median. So, here it is true, among other things, that d^3 is the median.
Case (d) allows for a negative possibility. Case (e) allows for possibility greater than or equal to zero, namely, zero itself. Therefore, we still cannot give a definitive answer to the question. Statement #2, by itself, is insufficient.
Combined statements #! & #2:
Even taken together, both statements allow case (d), negative fractions, and both cases allow (e), zero. Therefore, even taken together, we cannot answer the question definitively. Both statements together are insufficient.
Answer = E.
Does that make sense? Please let me know if I can answer any questions about what I've said here.
Mike
