I'm in Chicago without a computer so this will be short and sweet since I'm typing with my thumbs.
First correct and non-anonymous comment wins access to the Math Guide.
Chris bought a pumpkin last year that had a diameter of 10 inches and yielded the PERFECT amount of toasted seeds. His pumpkin this year has a diameter of 13 inches. Pumpkin skins generally have a thickness of one tenth of the radius of the pumpkin. Assume pumpkins are spherical and that each pumpkin has roughly equal amounts of seeds by volume. What will be the ratio of the amount of seeds Chris gets this year to the amount of seeds he got last year?
UPDATE: Nice work, Edward. Solution below the cut.
So the first thing you need to do is make sure you know how to find the volume of a sphere*:
Next, establish the inner volumes (the volumes that don't include the skins) of each pumpkin. Let's take them one at a time.
The pumpkin last year had a diameter of 10 inches, which means it had a radius of 5 inches. The skin, according to the problem, is therefore 5/10 of an inch, so the radius of the inner pumpkin (where the seeds are) is 4.5 inches. So the inner volume is:
Which works out to about 382 cubic inches. (With all the approximations and assumptions in this problem -- like pumpkins as spheres -- it's silly to be any more precise.)
This year's pumpkin has a diameter of 13 inches, so it's got a radius of 6.5 inches and, by the same calculations as above, a skin that's 0.65 inches thick. That means its inner radius is 5.85. Throw that into the formula to find the inner volume of this year's pumpkin:
That works out to about 839 cubic inches.
So the new pumpkin's volume is a lot bigger than last year's volume.
To find the ratio of this year's seed volume to last year's simply divide:
839/382 = 2.2 (again, roughly).
* Please note that, like many of the Weekend Challenge question you'll find on this site, this isn't something you need to be able to do for the SAT, and if you ever are asked to work with the volume of a sphere, you'll be provided the formula.