Something you don’t often see in SAT prep materials is a pictograph, but when you actually look at a real SAT, pictographs are all over the place!

It’s not a conspiracy by a cabal of prep writers and SAT writers; pictograph questions are usually just really easy, and so prep writers don’t pay them much mind. This morning I decided to see if I could write a pictograph question that’s worthy of being called a Weekend Challenge. I guess you guys can judge whether or not I succeeded.

The prize this week: Same as the last few weeks. First correct response in the comments gets access to the PWN the SAT Math Guide Beta Program. One note: if you comment anonymously, I have no way of contacting you, or verifying who you are if you try to claim it was you later, so you can’t win. If you want to win, don’t post anonymously.

Sam is the kind of guy who keeps track of meaningless things in unnecessarily complicated charts. The chart above depicts the average number of times Sam finds himself laughing out loud during 3 distinct parts of the day. Sam has been keeping this chart for 4 days, and calculates that he will have to be made to LOL 10 more times than usual on the 5th night when his gassy uncle comes to family dinner in order to necessitate a change of one added face on his chart. On average, how many times a day has Sam been been made to LOL in school?

Put your answers in the comments! I’ll post the solution Monday.

UPDATE: Nice work, Kira! I hope you enjoy the Math Guide Beta.

Everyone else: The solution is below the cut.
So this might be a pictograph question, but it’s also a weighted average question, so the solution will involve the average table. If you’re a longtime reader, you should be nodding your head like yeah. (‘sup Miley?)

Basically, Sam averages 4n LOLs per family dinner, and it’ll take him 10 more LOLs than usual at dinner on the night in question to get his average up to 5n LOLs. Luckily, his flatulent uncle is on the way! Here’s the setup:

4 nights
4n LOLs per night
16n LOLs
+ 1 night
4n + 10 LOLs
+ 4n + 10 LOLs
= 5 nights
5n LOLs per night
= 20n + 10 LOLs
= 25n LOLs


There are two ways to fill in the bottom right field in the table: 1) add the right column and get 20n + 10 LOLs, or 2) multiply the bottom row to get 25n LOLs.

What that tells us is:

20n + 10 = 25n
10 = 5n
2 = n
If n = 2, then we know how many times Sam has been laughing out loud in school, on average. The pictograph tells us there are 6n LOLs per day, which means Sam’s friends must be hilarious. He averages 12 LOLs a day in school.

Comments (7)

Leave a Reply