Math Problem of the Day

By Deane Barker on September 19, 2005

SAT Math Problem of the Day: This post may prove me to be a hopeless dork, but nonetheless —

I’ve been concerned lately that my math skills are declining. You don’t do deliberate math everyday, just incidental math — math related to some other problem. While this is fine, I realized the other day that I had suddenly forgotten the difference between a mean and a mode, and that distressed me.

So I bookmarked the link above. Everyday, it has a new SAT-level math problem, and they’re trickier than you think. I missed today’s question, for instance, over something I would have remembered intuitively 10 years ago, but now I’m rusty.

I challenge everyone to bookmark the link above, and do the problem faithfully every day for 30 days. You’ll be a better person for it.



  1. While I recalled mean and mode, I fell for the -3 is smaller than -12 trick. Its early. No coffee yet. (Its not like I work on financial applications for a bank or anything.)

    And while this probably belongs in the Sudoku post, I did notice that Barnes & Noble has a Sudoku display set up toward the rear of the store.

  2. I fell for the -3 is smaller than -12 trick.

    That’s exactly what got me, and I feel like a total tool because of it. Thus, the need to do these problems every day.

  3. In our defense, they asked for the smaller number, not the lesser. Three negatives are less negatives than twelve negatives.

    I got nailed on that too, but I like your reasoning.

  4. Its a trick question and impossible to calculate with the information provided. What state(s) do they work in? What about taxes, 401K, etc?

  5. dz – You also forgot to inquire of how much of their salary is off the books. An does using the helicopter count as income?

  6. I got today’s, but not easily. I ended up reducing both equations as far as I could, then doing trial and error on the possible answers (cheating, perhaps)?

    The method they offer in the solution seems much easier, but it’s not something I remember from school.

  7. All you need to do is solve one equation for variable and then plug that into the other equation so that it is expressed in terms of all one variable and then solve for that variable. I started out by mutiplying everything by 10 to get rid of the decimal places.

  8. All you need to do is solve one equation for variable

    How can you do that when you have two variables? I can’t solve for one without knowing the value of the other, right?

  9. You can express the value of y in terms of x, so solve for y and plug it back into the other equation.

    Their solution seemed simpler.

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