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MTEL General Curriculum Mathematics Practice
Question 1 |
Which of the lists below is in order from least to greatest value?
\( \large -0.044,\quad -0.04,\quad 0.04,\quad 0.044\) Hint: These are easier to compare if you add trailing zeroes (this is finding a common denominator) -- all in thousandths, -0.044, -0.040,0 .040, 0.044. The middle two numbers, -0.040 and 0.040 can be modeled as owing 4 cents and having 4 cents. The outer two numbers are owing or having a bit more. | |
\( \large -0.04,\quad -0.044,\quad 0.044,\quad 0.04\) Hint: 0.04=0.040, which is less than 0.044. | |
\( \large -0.04,\quad -0.044,\quad 0.04,\quad 0.044\) Hint: -0.04=-0.040, which is greater than \(-0.044\). | |
\( \large -0.044,\quad -0.04,\quad 0.044,\quad 0.04\) Hint: 0.04=0.040, which is less than 0.044. |
Question 2 |
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 3 |
Below is a pictorial representation of \(2\dfrac{1}{2}\div \dfrac{2}{3}\):
Which of the following is the best description of how to find the quotient from the picture?
The quotient is \(3\dfrac{3}{4}\). There are 3 whole blocks each representing \(\dfrac{2}{3}\) and a partial block composed of 3 small rectangles. The 3 small rectangles represent \(\dfrac{3}{4}\) of \(\dfrac{2}{3}\). | |
The quotient is \(3\dfrac{1}{2}\). There are 3 whole blocks each representing \(\dfrac{2}{3}\) and a partial block composed of 3 small rectangles. The 3 small rectangles represent \(\dfrac{3}{6}\) of a whole, or \(\dfrac{1}{2}\).Hint: We are counting how many 2/3's are in 2 1/2: the unit becomes 2/3, not 1. | |
The quotient is \(\dfrac{4}{15}\). There are four whole blocks separated into a total of 15 small rectangles.Hint: This explanation doesn't make much sense. Probably you are doing "invert and multiply," but inverting the wrong thing. | |
This picture cannot be used to find the quotient because it does not show how to separate \(2\dfrac{1}{2}\) into equal sized groups.Hint: Study the measurement/quotative model of division. It's often very useful with fractions. |
Question 4 |
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 5 |
In January 2011, the national debt was about 14 trillion dollars and the US population was about 300 million people. Someone reading these figures estimated that the national debt was about $5,000 per person. Which of these statements best describes the reasonableness of this estimate?
It is too low by a factor of 10Hint: 14 trillion \( \approx 15 \times {{10}^{12}} \) and 300 million \( \approx 3 \times {{10}^{8}}\), so the true answer is about \( 5 \times {{10}^{4}} \) or $50,000. | |
It is too low by a factor of 100 | |
It is too high by a factor of 10 | |
It is too high by a factor of 100 |
Question 6 |
Aya and Kendra want to estimate the height of a tree. On a sunny day, Aya measures Kendra's shadow as 3 meters long, and Kendra measures the tree's shadow as 15 meters long. Kendra is 1.5 meters tall. How tall is the tree?
7.5 metersHint: Here is a picture, note that the large and small right triangles are similar: ![]() One way to do the problem is to note that there is a dilation (scale) factor of 5 on the shadows, so there must be that factor on the heights too. Another way is to note that the shadows are twice as long as the heights. | |
22.5 metersHint: Draw a picture. | |
30 metersHint: Draw a picture. | |
45 metersHint: Draw a picture. |
Question 7 |
The column below consists of two cubes and a cylinder. The cylinder has diameter y, which is also the length of the sides of each cube. The total height of the column is 5y. Which of the formulas below gives the volume of the column?
\( \large 2{{y}^{3}}+\dfrac{3\pi {{y}^{3}}}{4}\) Hint: The cubes each have volume \(y^3\). The cylinder has radius \(\dfrac{y}{2}\) and height \(3y\). The volume of a cylinder is \(\pi r^2 h=\pi ({\dfrac{y}{2}})^2(3y)=\dfrac{3\pi {{y}^{3}}}{4}\). Note that the volume of a cylinder is analogous to that of a prism -- area of the base times height. | |
\( \large 2{{y}^{3}}+3\pi {{y}^{3}}\) Hint: y is the diameter of the circle, not the radius. | |
\( \large {{y}^{3}}+5\pi {{y}^{3}}\) Hint: Don't forget to count both cubes. | |
\( \large 2{{y}^{3}}+\dfrac{3\pi {{y}^{3}}}{8}\) Hint: Make sure you know how to find the volume of a cylinder. |
Question 8 |
If x is an integer, which of the following must also be an integer?
\( \large \dfrac{x}{2}\) Hint: If x is odd, then \( \dfrac{x}{2} \) is not an integer, e.g. 3/2 = 1.5. | |
\( \large \dfrac{2}{x}\) Hint: Only an integer if x = -2, -1, 1, or 2. | |
\( \large-x\) Hint: -1 times any integer is still an integer. | |
\(\large\sqrt{x}\) Hint: Usually not an integer, e.g. \( \sqrt{2} \approx 1.414 \). |
Question 9 |
Elena is going to use a calculator to check whether or not 267 is prime. She will pick certain divisors, and then find 267 divided by each, and see if she gets a whole number. If she never gets a whole number, then she's found a prime. Which numbers does Elena NEED to check before she can stop checking and be sure she has a prime?
All natural numbers from 2 to 266.Hint: She only needs to check primes -- checking the prime factors of any composite is enough to look for divisors. As a test taking strategy, the other three choices involve primes, so worth thinking about. | |
All primes from 2 to 266 .Hint: Remember, factors come in pairs (except for square root factors), so she would first find the smaller of the pair and wouldn't need to check the larger. | |
All primes from 2 to 133 .Hint: She doesn't need to check this high. Factors come in pairs, and something over 100 is going to be paired with something less than 3, so she will find that earlier. | |
All primes from \( \large 2\) to \( \large \sqrt{267}\).Hint: \(\sqrt{267} \times \sqrt{267}=267\). Any other pair of factors will have one factor less than \( \sqrt{267}\) and one greater, so she only needs to check up to \( \sqrt{267}\). |
Question 10 |
The polygon depicted below is drawn on dot paper, with the dots spaced 1 unit apart. What is the perimeter of the polygon?

\( \large 18+\sqrt{2} \text{ units}\) Hint: Be careful with the Pythagorean Theorem. | |
\( \large 18+2\sqrt{2}\text{ units}\) Hint: There are 13 horizontal or vertical 1 unit segments. The longer diagonal is the hypotenuse of a 3-4-5 right triangle, so its length is 5 units. The shorter diagonal is the hypotenuse of a 45-45-90 right triangle with side 2, so its hypotenuse has length \(2 \sqrt{2}\). | |
\( \large 18 \text{ units}
\) Hint: Use the Pythagorean Theorem to find the lengths of the diagonal segments. | |
\( \large 20 \text{ units}\) Hint: Use the Pythagorean Theorem to find the lengths of the diagonal segments. |
Question 11 |
Use the expression below to answer the question that follows.
\( \large \dfrac{\left( 4\times {{10}^{3}} \right)\times \left( 3\times {{10}^{4}} \right)}{6\times {{10}^{6}}}\)
Which of the following is equivalent to the expression above?
2Hint: \(10^3 \times 10^4=10^7\), and note that if you're guessing when the answers are so closely related, you're generally better off guessing one of the middle numbers. | |
20Hint: \( \dfrac{\left( 4\times {{10}^{3}} \right)\times \left( 3\times {{10}^{4}} \right)}{6\times {{10}^{6}}}=\dfrac {12 \times {{10}^{7}}}{6\times {{10}^{6}}}=\)\(2 \times {{10}^{1}}=20 \) | |
200Hint: \(10^3 \times 10^4=10^7\) | |
2000Hint: \(10^3 \times 10^4=10^7\), and note that if you're guessing when the answers are so closely related, you're generally better off guessing one of the middle numbers. |
Question 12 |
The histogram below shows the number of pairs of footware owned by a group of college students.
Which of the following statements can be inferred from the graph above?
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 13 |
Which of the lists below contains only irrational numbers?
\( \large\pi , \quad \sqrt{6},\quad \sqrt{\dfrac{1}{2}}\) | |
\( \large\pi , \quad \sqrt{9}, \quad \pi +1\) Hint: \( \sqrt{9}=3\) | |
\( \large\dfrac{1}{3},\quad \dfrac{5}{4},\quad \dfrac{2}{9}\) Hint: These are all rational. | |
\( \large-3,\quad 14,\quad 0\) Hint: These are all rational. |
Question 14 |
Given that 10 cm is approximately equal to 4 inches, which of the following expressions models a way to find out approximately how many inches are equivalent to 350 cm?
\( \large 350\times \left( \dfrac{10}{4} \right)\) Hint: The final result should be smaller than 350, and this answer is bigger. | |
\( \large 350\times \left( \dfrac{4}{10} \right)\) Hint: Dimensional analysis can help here: \(350 \text{cm} \times \dfrac{4 \text{in}}{10 \text{cm}}\). The cm's cancel and the answer is in inches. | |
\( \large (10-4) \times 350
\) Hint: This answer doesn't make much sense. Try with a simpler example (e.g. 20 cm not 350 cm) to make sure that your logic makes sense. | |
\( \large (350-10) \times 4\) Hint: This answer doesn't make much sense. Try with a simpler example (e.g. 20 cm not 350 cm) to make sure that your logic makes sense. |
Question 15 |
Which of the following sets of polygons can be assembled to form a pentagonal pyramid?
2 pentagons and 5 rectangles.Hint: These can be assembled to form a pentagonal prism, not a pentagonal pyramid. | |
1 square and 5 equilateral triangles.Hint: You need a pentagon for a pentagonal pyramid. | |
1 pentagon and 5 isosceles triangles. | |
1 pentagon and 10 isosceles triangles. |
Question 16 |
Use the expression below to answer the question that follows.
\(\large \dfrac{\left( 155 \right)\times \left( 6,124 \right)}{977}\)
Which of the following is the best estimate of the expression above?
100Hint: 6124/977 is approximately 6. | |
200Hint: 6124/977 is approximately 6. | |
1,000Hint: 6124/977 is approximately 6. 155 is approximately 150, and \( 6 \times 150 = 3 \times 300 = 900\), so this answer is closest. | |
2,000Hint: 6124/977 is approximately 6. |
Question 17 |
A car is traveling at 60 miles per hour. Which of the expressions below could be used to compute how many feet the car travels in 1 second? Note that 1 mile = 5,280 feet.
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot 5280\dfrac{\text{feet}}{\text{mile}}\cdot 60\dfrac{\text{minutes}}{\text{hour}}\cdot 60\dfrac{\text{seconds}}{\text{minute}}
\) Hint: This answer is not in feet/second. | |
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot 5280\dfrac{\text{feet}}{\text{mile}}\cdot \dfrac{1}{60}\dfrac{\text{hour}}{\text{minutes}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}
\) Hint: This is the only choice where the answer is in feet per second and the unit conversions are correct. | |
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot \dfrac{1}{5280}\dfrac{\text{foot}}{\text{miles}}\cdot 60\dfrac{\text{hours}}{\text{minute}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}\) Hint: Are there really 60 hours in a minute? | |
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot \dfrac{1}{5280}\dfrac{\text{mile}}{\text{feet}}\cdot 60\dfrac{\text{minutes}}{\text{hour}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}\) Hint: This answer is not in feet/second. |
Question 18 |
The pattern below consists of a row of black squares surrounded by white squares.
How many white squares would surround a row of 157 black squares?
314Hint: Try your procedure on a smaller number that you can count to see where you made a mistake. | |
317Hint: Are there ever an odd number of white squares? | |
320Hint: One way to see this is that there are 6 tiles on the left and right ends, and the rest of the white tiles are twice the number of black tiles (there are many other ways to look at it too). | |
322Hint: Try your procedure on a smaller number that you can count to see where you made a mistake. |
Question 19 |
If two fair coins are flipped, what is the probability that one will come up heads and the other tails?
\( \large \dfrac{1}{4}\) Hint: Think of the coins as a penny and a dime, and list all possibilities. | |
\( \large \dfrac{1}{3} \) Hint: This is a very common misconception. There are three possible outcomes -- both heads, both tails, and one of each -- but they are not equally likely. Think of the coins as a penny and a dime, and list all possibilities. | |
\( \large \dfrac{1}{2}\) Hint: The possibilities are HH, HT, TH, TT, and all are equally likely. Two of the four have one of each coin, so the probability is 2/4=1/2. | |
\( \large \dfrac{3}{4}\) Hint: Think of the coins as a penny and a dime, and list all possibilities. |
Question 20 |
Below are four inputs and outputs for a function machine representing the function A:
Which of the following equations could also represent A for the values shown?
\( \large A(n)=n+4\) Hint: For a question like this, you don't have to find the equation yourself, you can just try plugging the function machine inputs into the equation, and see if any values come out wrong. With this equation n= -1 would output 3, not 0 as the machine does. | |
\( \large A(n)=n+2\) Hint: For a question like this, you don't have to find the equation yourself, you can just try plugging the function machine inputs into the equation, and see if any values come out wrong. With this equation n= 2 would output 4, not 6 as the machine does. | |
\( \large A(n)=2n+2\) Hint: Simply plug in each of the four function machine input values, and see that the equation produces the correct output, e.g. A(2)=6, A(-1)=0, etc. | |
\( \large A(n)=2\left( n+2 \right)\) Hint: For a question like this, you don't have to find the equation yourself, you can just try plugging the function machine inputs into the equation, and see if any values come out wrong. With this equation n= 2 would output 8, not 6 as the machine does. |
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