 First of all, if you’re taking the May SAT tomorrow then I wish you the best of luck. You might want to sit this one out, at least until after your test. The prize this week for the first correct answer: domesticated animals will be able to understand what you say to them for 24 hours. Use this opportunity to tell dogs how awesome they are, cats what ingrates they are, and goldfish how sad and lonely their lives are as if they didn’t already know. I have no idea what to say to your ferret. Right, onwards:

If the difference between the areas of two circles is 10π and the sum of their diameters is 4√5, the radius of the larger circle is how much bigger than the radius of the smaller circle?

UPDATE: Commenter John Gutsch nailed it. Solution below the cut.

At first glance this looks like a circle question, but as you’ll see in a moment, it’s really only a circle question in a cosmetic sense. Really, it’s a solve for expressions question. Let’s begin.

Say the radius of the larger circle is m, and the radius of the smaller circle is n. Write an equation to represent the first thing the question tells you: the difference between the areas of the circles is 10π:

πm2 – πn2 = 10π
Oh snap! Looks like there’s a π in each term, so let’s divide that right out of there:
m2 – n2 = 10
Oh double-snap! Looks like we’ve got a difference of two squares! Factor it:
(n)(– n) = 10

Now what? Well, remember what else the question told us, that “the sum of their diameters is 4√5.” In other words:

2m + 2n = 4√5
2(m + n) = 4√5
m + n = 2√5

Wow. Substitute 2√5 for m + n, and you get:

(n)(– n) = 10
2√5(– n) = 10
– n = 10/(2√5)
m – n = √5

Or, as commenter John Gutsch put it, about 2.236. Great job! John Gutsch says:

2.236 Mike McClenathan says:

Great work, John! I’ll post the solution in a few minutes. JD says:

Hi,

How difficult would you rate this problem from an SAT standpoint?

Thanks Mike McClenathan says:

I’d say this is harder than I’d expect a problem on the SAT to be. If something like it did appear, it’d come at the very end of a multiple choice section, so a stumped student could at least attempt backsolving from the answer choices instead of solving algebraically.