Someone requested recently at my Q&A page that I post a challenge question about direct and inverse variation. I quickly agreed, but it’s taken me longer than I wanted it to because I’ve been having a hard time hitting the right level of difficulty for a challenge question. I’m still not sure I’ve nailed it, but regardless, the wait is over. First correct response, as usual, gets a Math Guide shipped to their home (in the US) for free. Full contest rules, as always, apply.
- p and q are directly proportional to each other; their proportionality constant is k
- x and y are inversely proportional to each other; their proportionality constant is j
- k = j
- a is a constant greater than 1
According to the conditions above, what is one possible value of y in terms of a when p = a3, q = a, and x = a?
Put your answers in the comments. If you’re not registered with this site and your comment doesn’t appear immediately, don’t panic. I get comments in the order they’re submitted, but not everything shows up right away because I’m trying to prevent spam.
UPDATE: Rushil won the book, and Peter gave a nice explanation in the comments. My solution is below the cut.
If p and q are directly proportional, with proportionality constant k, there are actually two equations you could write:
This is because all that really needs to be true is that p and q always make the same fraction. I’m going to deal with the first one first, and then I’ll give a quick treatment to the second one. I would have accepted the answer that results from either one.
If x and y are inversely proportional, with proportionality constant j, there’s only one equation you can write:
Of course, if k and j are equal, then we can set up one equation:
And now we’re ready to substitute in our values. Remember, p = a3, q = a, and x = a. So here we go:
Had you used the second version of the p and q equation, you’d instead get y = a–3 here instead. I would have accepted that as an answer as well. Here’s how that’d work: