This question was featured here on the blog awhile back, but it’s a good one, so I’ve decided to repost it. Plus, it makes math fun, and that’s never a bad thing.

PAT R. FROM WICHITA, KANSAS, ASKS: We are planning a neighborhood party for about 500 people. How many average-sized watermelons would we need to serve melon wedges?

Wow! 500 people sounds like quite a party. Mind if I come? I’m just kidding (sort of).

Let’s do some math here. We know that the average 20-pound watermelon yields about 53 6-ounce wedges (3/4 of an inch thick). That’s about two-and-a-half wedges per pound. Using that formula, you’ll need roughly 200 pounds of watermelon. In other words, 20 10-pound watermelons or 13-and-a-third 15-pound watermelons. (I wouldn’t recommend cutting the watermelon in the store to get that extra third of a watermelon. They might frown on that.)

Of course, that’s assuming everyone eats only one slice. Some will eat two or three (or ten) slices. Then again, others might not eat any, but that’s their loss, right? To be safe, go ahead and pick up a few extra watermelons, just in case. After all, like I told David a little while ago, there’s no such thing as “too much watermelon.”

Hope this helps. Let me know how the party turns out, and if you want me to stop by, you know how to reach me.


Chef Harry

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