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## MTEL General Curriculum Mathematics Practice

 Question 1

#### The picture below represents a board with pegs on it, where the closest distance between two pegs is 1 cm.  What is the area of the pentagon shown? A $$\large 8\text{ c}{{\text{m}}^{2}}$$Hint: Don't just count the dots inside, that doesn't give the area. Try adding segments so that the slanted lines become the diagonals of rectangles. B $$\large 11\text{ c}{{\text{m}}^{2}}$$Hint: Try adding segments so that the slanted lines become the diagonals of rectangles. C $$\large 11.5\text{ c}{{\text{m}}^{2}}$$Hint: An easy way to do this problem is to use Pick's Theorem (of course, it's better if you understand why Pick's theorem works): area = # pegs inside + half # pegs on the border - 1. In this case 8+9/2-1=11.5. A more appropriate strategy for elementary classrooms is to add segments; here's one way. There are 20 1x1 squares enclosed, and the total area of the triangles that need to be subtracted is 8.5 D $$\large 12.5\text{ c}{{\text{m}}^{2}}$$Hint: Try adding segments so that the slanted lines become the diagonals of rectangles.
Question 1 Explanation:
Topics: Calculate measurements and derive and use formulas for calculating the areas of geometric shapes and figures (Objective 0023).
 Question 2

#### Four children randomly line up, single file.  What is the probability that they are in height order, with the shortest child in front?   All of the children are different heights.

 A $$\large \dfrac{1}{4}$$Hint: Try a simpler question with 3 children -- call them big, medium, and small -- and list all the ways they could line up. Then see how to extend your logic to the problem with 4 children. B $$\large \dfrac{1}{256}$$Hint: Try a simpler question with 3 children -- call them big, medium, and small -- and list all the ways they could line up. Then see how to extend your logic to the problem with 4 children. C $$\large \dfrac{1}{16}$$Hint: Try a simpler question with 3 children -- call them big, medium, and small -- and list all the ways they could line up. Then see how to extend your logic to the problem with 4 children. D $$\large \dfrac{1}{24}$$Hint: The number of ways for the children to line up is $$4!=4 \times 3 \times 2 \times 1 =24$$ -- there are 4 choices for who is first in line, then 3 for who is second, etc. Only one of these lines has the children in the order specified.
Question 2 Explanation:
Topic: Apply knowledge of combinations and permutations to the computation of probabilities (Objective 0026).
 Question 3

#### A family has four children.  What is the probability that two children are girls and two are boys?  Assume the the probability of having a boy (or a girl) is 50%.

 A $$\large \dfrac{1}{2}$$Hint: How many different configurations are there from oldest to youngest, e.g. BGGG? How many of them have 2 boys and 2 girls? B $$\large \dfrac{1}{4}$$Hint: How many different configurations are there from oldest to youngest, e.g. BGGG? How many of them have 2 boys and 2 girls? C $$\large \dfrac{1}{5}$$Hint: Some configurations are more probable than others -- i.e. it's more likely to have two boys and two girls than all boys. Be sure you are weighting properly. D $$\large \dfrac{3}{8}$$Hint: There are two possibilities for each child, so there are $$2 \times 2 \times 2 \times 2 =16$$ different configurations, e.g. from oldest to youngest BBBG, BGGB, GBBB, etc. Of these configurations, there are 6 with two boys and two girls (this is the combination $$_{4}C_{2}$$ or "4 choose 2"): BBGG, BGBG, BGGB, GGBB, GBGB, and GBBG. Thus the probability is 6/16=3/8.
Question 3 Explanation:
Topic: Apply knowledge of combinations and permutations to the computation of probabilities (Objective 0026).
 Question 4

#### The histogram below shows the number of pairs of footware owned by a group of college students. #### The median number of pairs of footware owned is between 50 and 60 pairs.

Hint:
The same number of data points are less than the median as are greater than the median -- but on this histogram, clearly more than half the students own less than 50 pairs of shoes, so the median is less than 50.

#### The mode of the number of pairs of footware owned is 20.

Hint:
The mode is the most common number of pairs of footwear owned. We can't tell it from this histogram because each bar represents 10 different numbers-- perhaps 8 students each own each number from 10 to 19, but 40 students own exactly 6 pairs of shoes.... or perhaps not....

#### The mean number of pairs of footware owned is less than the median number of pairs of footware owned.

Hint:
This is a right skewed distribution, and so the mean is bigger than the median -- the few large values on the right pull up the mean, but have little effect on the median.

#### The median number of pairs of footware owned is between 10 and 20.

Hint:
There are approximately 230 students represented in this survey, and the 41st through 120th lowest values are between 10 and 20 -- thus the middle value is in that range.
Question 4 Explanation:
Topics: Analyze and interpret various graphic and data representations, and use measures of central tendency (e.g., mean, median, mode) and spread to describe and interpret real-world data (Objective 0025).
 Question 5

#### This student divides fractions by first finding a common denominator, then dividing the numerators.

$$\large \dfrac{2}{3} \div \dfrac{3}{4} \longrightarrow \dfrac{8}{12} \div \dfrac{9}{12} \longrightarrow 8 \div 9 = \dfrac {8}{9}$$ $$\large \dfrac{2}{5} \div \dfrac{7}{20} \longrightarrow \dfrac{8}{20} \div \dfrac{7}{20} \longrightarrow 8 \div 7 = \dfrac {8}{7}$$ $$\large \dfrac{7}{6} \div \dfrac{3}{4} \longrightarrow \dfrac{14}{12} \div \dfrac{9}{12} \longrightarrow 14 \div 9 = \dfrac {14}{9}$$

#### It is not valid. Common denominators are for adding and subtracting fractions, not for dividing them.

Hint:
Don't be so rigid! Usually there's more than one way to do something in math.

#### It got the right answer in these three cases, but it isn‘t valid for all rational numbers.

Hint:
Did you try some other examples? What makes you say it's not valid?

#### It is valid if the rational numbers in the division problem are in lowest terms and the divisor is not zero.

Hint:
Lowest terms doesn't affect this problem at all.

#### It is valid for all rational numbers, as long as the divisor is not zero.

Hint:
When we have common denominators, the problem is in the form a/b divided by c/b, and the answer is a/c, as the student's algorithm predicts.
Question 5 Explanation:
Topic: Analyze Non-Standard Computational Algorithms (Objective 0019).
There are 5 questions to complete.

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