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.

To begin work on this, get all the information in the bullet points into equations.

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:

And now we solve for y:

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: ruchita patel says:

y = a^3 Rushil Patel says:

Y = 1/(a^5) Mike McClenathan says:

I have tried to reword this to make it more clear…I’m worried the question confused people when it first posted. Jack Cornish says:

y = 1/a is the answer ? Rushil Patel says:

AWw well then. I change my answer is Y = 1/a Jack Cornish says:

You can have the book if I won, I already have one. I just wanted to see if I got it correct. Rushil Patel says:

Sweet thanks man. Mike McClenathan says:

Nobody’s got it yet Rushil Patel says:

Ahhaah. y = a Mike McClenathan says:

🙂 NOW you’re doing it. i’ll email you about the book. Guest says:

wait no no. it is y = 1/(a^3) Guest says:

. Peter says:

Here’s an explanation. 🙂

p/q is constant, so (p/q)=k
xy is constant, so xy=j

Since j=k, we know that (p/q)=xy.

Rearranging this, we get p=qxy.

Since p=a^3, we have a^3=qxy.
Since q=a, we have a^3=axy.
Since x=a, we have a^3=(a)(a)y.

Which is a^3=(a^2)y

We divide both sides by a^2 to get a=y. Mike McClenathan says:

Nice! I also would have accepted y = a^(-3), since you could just as easily have interpreted this as q/p = k. Guest says:

y = 1/a^2