Math Inferences Drill

    • June 25, 2017 at 8:35 pm #3144


      I’m having trouble with lesson 12 in the LSAT trainer. The Situation: Five of eight students will be selected to take part in a math competition. The students are from classroom A, B, and C. The rules of the game have given us some information about the number of students from each class, and that’s been added into the chart already. Your job is to fill in the possibilities for the blank spaces. The first line has been filled in for you.

      I’m confused with these inferences and I don’t understand how to get the possibilities. Can you please explain this exercise in more detail.


    • June 26, 2017 at 4:20 pm #3147
      Mike Kim

      Hi Jon — here are a few different ways to think about it / a few suggestions — hope they help and let me know if you have any follow-up —

      1) The math behind it

      if we think of A, B, and C as variables representing the number of students from each class, the equation behind that drill is

      A + B + C = 5.

      From there, if, for example, we are told A = 1 and C = 2, then we know

      1 + B + 2 = 5,

      And we can infer that B must then equal 2.

      If we are told an inequality, something like A = 1 and B is less than 3, then we know:

      A must be 1, and B can either be 0, 1, or 2.

      If A is 1 and B is 0, C = 4
      If A is 1 and B is 1, C = 3
      If A is 1 and B is 2, C = 2

      So, in this case, C can be 2, 3, or 4.

      2) a visual representation —

      One way to get really good at these inferences really fast is to practice making them using coins — use quarters for A, nickels for B etc — and, knowing that you have to end up at 5 coins total, just walk through the various possibilities using the coins — do this until you feel comfortable and I think it makes it much easier for your brain to literally “see” these inferences.

      3) in thinking about the range of options, try to figure out the max and the min —

      So, for example, if we are told A < 2 and B =1 or 2, and we are asked to solve for C -- We can start by trying to get as big a C as possible -- (we can make A = 0 and B = 1, smallest possible values, which would make C =4), and then as small a C as possible (we can make A = 1 and B = 2, biggest possible values, making C = 2). This tells us that the range for C is from 2 to 4. The second exercise on the page has some more stipulations, but it requires the same type of thinking/inferences --HTH and again if you have any follow-up or need anything else let me know -- Best, Mike

    • July 22, 2017 at 4:21 pm #3170

      Simple enough! Thanks Mike!

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