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.

pandqare directly proportional to each other; their proportionality constant iskxandyare inversely proportional to each other; their proportionality constant isjk=jais a constant greater than 1According to the conditions above, what is one possible value of

yin terms ofawhenp=a^{3},q=a, andx=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* = *a*^{3}, *q* = *a*, and *x* = *a*. So here we go:

*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:

y = a^3

Y = 1/(a^5)

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

y = 1/a is the answer ?

AWw well then. I change my answer is Y = 1/a

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

Sweet thanks man.

Nobody’s got it yet

Ahhaah. y = a

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

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

.

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.

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

y = 1/a^2