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MTEL General Curriculum Mathematics Practice
Question 1 
There are six gumballs in a bag — two red and four green. Six children take turns picking a gumball out of the bag without looking. They do not return any gumballs to the bag. What is the probability that the first two children to pick from the bag pick the red gumballs?
\( \large \dfrac{1}{3}\) Hint: This is the probability that the first child picks a red gumball, but not that the first two children pick red gumballs.  
\( \large \dfrac{1}{8}\) Hint: Are you adding things that you should be multiplying?  
\( \large \dfrac{1}{9}\) Hint: This would be the probability if the gumballs were returned to the bag.  
\( \large \dfrac{1}{15}\) Hint: The probability that the first child picks red is 2/6 = 1/3. Then there are 5 gumballs in the bag, one red, so the probability that the second child picks red is 1/5. Thus 1/5 of the time, after the first child picks red, the second does too, so the probability is 1/5 x 1/3 = 1/15. 
Question 2 
Which of the following is not possible?
An equiangular triangle that is not equilateral.Hint: The AAA property of triangles states that all triangles with corresponding angles congruent are similar. Thus all triangles with three equal angles are similar, and are equilateral.  
An equiangular quadrilateral that is not equilateral.Hint: A rectangle is equiangular (all angles the same measure), but if it's not a square, it's not equilateral (all sides the same length).  
An equilateral quadrilateral that is not equiangular.Hint: This rhombus has equal sides, but it doesn't have equal angles:  
An equiangular hexagon that is not equilateral.Hint: This hexagon has equal angles, but it doesn't have equal sides: 
Question 3 
What is the greatest common factor of 540 and 216?
\( \large{{2}^{2}}\cdot {{3}^{3}}\) Hint: One way to solve this is to factor both numbers: \(540=2^2 \cdot 3^3 \cdot 5\) and \(216=2^3 \cdot 3^3\). Then take the smaller power for each prime that is a factor of both numbers.  
\( \large2\cdot 3\) Hint: This is a common factor of both numbers, but it's not the greatest common factor.  
\( \large{{2}^{3}}\cdot {{3}^{3}}\) Hint: \(2^3 = 8\) is not a factor of 540.  
\( \large{{2}^{2}}\cdot {{3}^{2}}\) Hint: This is a common factor of both numbers, but it's not the greatest common factor. 
Question 4 
The Americans with Disabilties Act (ADA) regulations state that the maximum slope for a wheelchair ramp in new construction is 1:12, although slopes between 1:16 and 1:20 are preferred. The maximum rise for any run is 30 inches. The graph below shows the rise and runs of four different wheelchair ramps. Which ramp is in compliance with the ADA regulations for new construction?
AHint: Rise is more than 30 inches.  
BHint: Run is almost 24 feet, so rise can be almost 2 feet.  
CHint: Run is 12 feet, so rise can be at most 1 foot.  
DHint: Slope is 1:10  too steep. 
Question 5 
Kendra is trying to decide which fraction is greater, \( \dfrac{4}{7}\) or \( \dfrac{5}{8}\). Which of the following answers shows the best reasoning?
\( \dfrac{4}{7}\) is \( \dfrac{3}{7}\)away from 1, and \( \dfrac{5}{8}\) is \( \dfrac{3}{8}\)away from 1. Since eighth‘s are smaller than seventh‘s, \( \dfrac{5}{8}\) is closer to 1, and is the greater of the two fractions.  
\( 74=3\) and \( 85=3\), so the fractions are equal.Hint: Not how to compare fractions. By this logic, 1/2 and 3/4 are equal, but 1/2 and 2/4 are not.  
\( 4\times 8=32\) and \( 7\times 5=35\). Since \( 32<35\) , \( \dfrac{5}{8}<\dfrac{4}{7}\)Hint: Starts out as something that works, but the conclusion is wrong. 4/7 = 32/56 and 5/8 = 35/56. The cross multiplication gives the numerators, and 35/56 is bigger.  
\( 4<5\) and \( 7<8\), so \( \dfrac{4}{7}<\dfrac{5}{8}\)Hint: Conclusion is correct, logic is wrong. With this reasoning, 1/2 would be less than 2/100,000. 
Question 6 
The histogram below shows the frequency of a class's scores on a 4 question quiz.
What was the mean score on the quiz?
\( \large 2.75\) Hint: There were 20 students who took the quiz. Total points earned: \(2 \times 1+6 \times 2+ 7\times 3+5 \times 4=55\), and 55/20 = 2.75.  
\( \large 2\) Hint: How many students are there total? Did you count them all?  
\( \large 3\) Hint: How many students are there total? Did you count them all? Be sure you're finding the mean, not the median or the mode.  
\( \large 2.5\) Hint: How many students are there total? Did you count them all? Don't just take the mean of 1, 2, 3, 4  you have to weight them properly. 
Question 7 
Which of the numbers below is a fraction equivalent to \( 0.\bar{6}\)?
\( \large \dfrac{4}{6}\) Hint: \( 0.\bar{6}=\dfrac{2}{3}=\dfrac{4}{6}\)  
\( \large \dfrac{3}{5}\) Hint: This is equal to 0.6, without the repeating decimal. Answer is equivalent to choice c, which is another way to tell that it's wrong.  
\( \large \dfrac{6}{10}\) Hint: This is equal to 0.6, without the repeating decimal. Answer is equivalent to choice b, which is another way to tell that it's wrong.  
\( \large \dfrac{1}{6}\) Hint: This is less than a half, and \( 0.\bar{6}\) is greater than a half. 
Question 8 
Use the graph below to answer the question that follows:
The graph above best matches which of the following scenarios:
George left home at 10:00 and drove to work on a crooked path. He was stopped in traffic at 10:30 and 10:45. He drove 30 miles total.Hint: Just because he ended up 30 miles from home doesn't mean he drove 30 miles total.  
George drove to work. On the way to work there is a little hill and a big hill. He slowed down for them. He made it to work at 11:15.Hint: The graph is not a picture of the roads.  
George left home at 10:15. He drove 10 miles, then realized he‘d forgotten something at home. He turned back and got what he‘d forgotten. Then he drove in a straight line, at many different speeds, until he got to work around 11:15.Hint: A straight line on a distance versus time graph means constant speed.  
George left home at 10:15. He drove 10 miles, then realized he‘d forgotten something at home. He turned back and got what he‘d forgotten. Then he drove at a constant speed until he got to work around 11:15. 
Question 9 
Each individual cube that makes up the rectangular solid depicted below has 6 inch sides. What is the surface area of the solid in square feet?
\( \large 11\text{ f}{{\text{t}}^{2}}\) Hint: Check your units and make sure you're using feet and inches consistently.  
\( \large 16.5\text{ f}{{\text{t}}^{2}}\) Hint: Each square has surface area \(\dfrac{1}{2} \times \dfrac {1}{2}=\dfrac {1}{4}\) sq feet. There are 9 squares on the top and bottom, and 12 on each of 4 sides, for a total of 66 squares. 66 squares \(\times \dfrac {1}{4}\) sq feet/square =16.5 sq feet.  
\( \large 66\text{ f}{{\text{t}}^{2}}\) Hint: The area of each square is not 1.  
\( \large 2376\text{ f}{{\text{t}}^{2}}\) Hint: Read the question more carefully  the answer is supposed to be in sq feet, not sq inches.

Question 10 
The student used a method that worked for this problem and can be generalized to any subtraction problem.Hint: Note that this algorithm is taught as the "standard" algorithm in much of Europe (it's where the term "borrowing" came from  you borrow on top and "pay back" on the bottom).  
The student used a method that worked for this problem and that will work for any subtraction problem that only requires one regrouping; it will not work if more regrouping is required.Hint: Try some more examples.  
The student used a method that worked for this problem and will work for all threedigit subtraction problems, but will not work for larger problems.Hint: Try some more examples.  
The student used a method that does not work. The student made two mistakes that cancelled each other out and was lucky to get the right answer for this problem.Hint: Remember, there are many ways to do subtraction; there is no one "right" algorithm. 
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