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  1. For the question #81 solution, there is an even simpler approach than shown in the video. It isn’t too different, but simpler. If z is 2 multiples of x lower than z then x=z-10. If y is one multiple lower than z then y=z-5 so x + y = (z-10) + (z-5) = 2z – 15…it’s that simple.

    1. Question 69? There is no factor tree for 69. I assume you are talking about 67.

      There are 2 sets of numbers Kate.
      Immediately to the right of the tree at each step I quickly check divisibility. For example, determining the sum of the digits of a number in the factor tree is one way to assess quickly whether it is divisible by 3. We are looking for factors equal to 3 and 7 so I try to exhaust all the factors of 3 by doing this first. e.g. you will see a sum of digits of 12 and I write 12/3 = 4 which indicates the sum of the digits 1 + 0 + 2 + 9 = 12 which is divisible by 3 so the value 1,029 in the tree is also divisible by 3.

      Farther to the right I am essentially doing the long division with the factor 3 or 7, but showing it a little differently. 9,261 for example is divisible by 3 and I start with as many 1,000s x 3 that I can w/o going over. That is 3 x 3,000 = 9,000. That leaves 261 left over. Zero 100s x 3 go into that, but for the 10s…80 x 3=240 goes into 261 w/o going over. That leaves 261-240=21 left over. For the 1s…7 x 3 goes into 21 exactly. The resulting value for the next level down on the factor tree with 3 is 3,000 + 80 + 7 = 3,087. You basically do the same when you do long division. I lay it out in a way that is more intuitive and easy to check for me. You can choose to do it this way or not…whichever is easiest for you.

  4. Good question. I think someone else was curious about this too on a different exam.

    The top assumes 20 is the average. That is too low of course, but it helps make the calculation easier for the exam. Instead of multiplying each value by the count to get the sum we multiply the differences from 20 by the count to get a sum of differences. The addition and resulting division are much easier to calculate without a calculator in this case which is a good example where the tip comes in handy. You will almost certainly see a problem like this on the exam.

    The TestPrepSHSAT.com course has a lesson that goes through this in more detail.

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