Arizona AIMS Mathematics High School Sample Test Annotated Solutions Manual

This question is testing your ability to calculate permutations.

They are asking for the number of permutations of n things taken n at a time.

Specifically, the are asking for number of ways you can arrange 6 students in EVERY order.

Mathematician use factorial notation to represent the number of possible permutations for situations like this.

There are 6 seats and 6 students. Six seats taken at six students at a time.

₆P₆ = 6!

6! = (6)(5)(4)(3)(2)(1)

As you can see by multiplying the first 3 factors, the only possible answer is Choice A.

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3 days. 3 times. 2 gyms…

possible outcomes?

3 x 3 x 2 = 18

18 different outcomes are possible

choice C is the answer.

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Algorithm I finds the median value of this data set of 25 values.

Algorithm II finds the median value of this data set of 25 values.

Algorithm III finds the maximum value.

Algorithm IV finds the mininum value.

Only I and II are equivalent. Your correct answer is Choice A.

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Sample Space?

Basically, the test item is a simple question that is worded in such a way that makes the exam writer seem to be smart. (What a windbag!)

If two people go see the movies together, then the people will hold two tickets.

One ticket for Movie E and one ticket for Movie E. (EE).

One ticket for Movie F and one ticket for Movie F. (FF).

One ticket for Movie G and one ticket for Movie G. (GG).

Thus our sample space is simply {EE, FF, GG}.

Choice C is the correct answer.

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choice A makes no sense. if figure ABCD is a rectangle, thus the upper base angles and the lower base angles are exactly 90^o and the figure CANNOT be a trapezoid.

choice B is not valid becauase John is not necessarily a freshman.

choice C is not valid. all rectangles are in fact parallelograms. however, if the figure is a parallelogram it is not necessarily a rectangle.

choice D is the only choice remaining.

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This question is a time waster. Looking at the “algorithms”, we see that two outcomes I and II show us parallel lines.

The third outcome shows perpendicular lines.

It is safe to say that you cannot prove perpendicular lines with I or II.

Thus, choices B, C, and D are all eliminated.

Only choice A is vaild.

Note that both algorithms yield a proof of parallel lines.

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