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:
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:
Comments (15)
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