This is something I learned in my class last night, which I thought was very helpful for an average problem. The questions read: * *

*The average score by the class on a test was 90. If there were 8 males who averaged 87, and x number of females who averaged 92, how many females took the test? *

This type of question can be answered really quickly using a simple method we can call *The Seesaw*, here goes…

Males: 90-87 = 3 * 8 = 24 Average: 90 Females: 92-90 = 2, 24/2 = 12

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———————————————–>^ (this is the only way I can get the “fulcrum” to stay here in the post)

This poor attempt at a drawing is a seesaw with a fulcrum at 90 because that is the average. We need to find out how far from the average the males are. 3 points is the difference between the 87 and 90, which we then need to multiply by 8 (the # of males in the class). Since we have 24 on the left side of the fulcrum, we need 24 on the right side to “balance” out the seesaw. We see that the difference between the female scores and the average is 2, so there must be 12 females in the class to give the females a difference of 24 and balance out the seesaw.

I thought this to be a very quick and easy method to use on average problems. Anyone else have a trick they use for a certain problem that they would like to share? Think this is a silly way to do this and the algebra is easier? Have a good Thursday, I might be back later with some more stuff.

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Filed under: Test Prep | Tagged: Test Prep |

Rocky Balboa, on July 16, 2009 at 7:58 am said:for a polygon with n sides, sum of interior angles = (n-2)*180.

Eg. for a hexagon, sum of all interior angles = (6-2)*180 = 720 degrees.