The ratio of 6th graders to 7th graders in a MATHCOUNTS club is 5:4. The ratio of 8th graders to 7th graders in the same club is
2 answers:
To solve this problem, KROW SDAWKCAB. (That's WORK BACKWARDS spelled backwards.)
We are told that there's a 3:2 ratio of 8th graders to 7th graders in the club. Let x be the number of 7th graders in the class. We set a proportion. The proportion puts the ratio on one side, how many students on the other. Tops are the 8th graders, bottoms are the 7th graders.
You can solve it by cross multipling or by noticing what happens between the fractions. On the top how did you go from 3 to 18? You multiplied by 6. So the same thing must happen on the bottoms, and 2 * 6 = 12.
By cross multiplying, 3x = 36 and when we divide both sides by 3, we find x = 12.
Now that we know we have 12 7th graders, we can find the number of 6th graders. The process is similar with 7th graders on bottom, 6th graders on top, and x for what we do not know.
We went from 4 to 12 by multiplying 3 on the bottom, and we repeat it on the top. So 5 * 3 = 15.
Thus, there are 15 6th graders in the club.
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I set this up as an inequality, 
. If you take the cubed root of 800, you get the lower bound of the side length, which is 9.2. Then I just worked my way up until I hit the first number that put me over a volume of 800. That number is 9.29, because 9.28 cubed is 799.1 (not high enough) and 9.29 cubed is 801.8. Therefore, the bounds of the sides exist within a conjunction: 
. That's the best I could come up with to help on that one. Wasn't sure if there was another method you were taught at school. I just used common sense more than any rule.