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MTEL General Curriculum Mathematics Practice
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Question 1 |
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 1 Explanation:
Topic: Converting between fraction and decimal representations (Objective 0017)
Question 2 |
What is the probability that two randomly selected people were born on the same day of the week? Assume that all days are equally probable.
\( \large \dfrac{1}{7}\) Hint: It doesn't matter what day the first person was born on. The probability that the second person will match is 1/7 (just designate one person the first and the other the second). Another way to look at it is that if you list the sample space of all possible pairs, e.g. (Wed, Sun), there are 49 such pairs, and 7 of them are repeats of the same day, and 7/49=1/7. | |
\( \large \dfrac{1}{14}\) Hint: What would be the sample space here? Ie, how would you list 14 things that you pick one from? | |
\( \large \dfrac{1}{42}\) Hint: If you wrote the seven days of the week on pieces of paper and put the papers in a jar, this would be the probability that the first person picked Sunday and the second picked Monday from the jar -- not the same situation. | |
\( \large \dfrac{1}{49}\) Hint: This is the probability that they are both born on a particular day, e.g. Sunday. |
Question 2 Explanation:
Topic: Calculate the probabilities of simple and compound events and of
independent and dependent events (Objective 0026).
Question 3 |
Taxicab fares in Boston (Spring 2012) are $2.60 for the first \(\dfrac{1}{7}\) of a mile or less and $0.40 for each \(\dfrac{1}{7}\) of a mile after that.
Let d represent the distance a passenger travels in miles (with \(d>\dfrac{1}{7}\)). Which of the following expressions represents the total fare?
\( \large \$2.60+\$0.40d\) Hint: It's 40 cents for 1/7 of a mile, not per mile. | |
\( \large \$2.60+\$0.40\dfrac{d}{7}\) Hint: According to this equation, going 7 miles would cost $3; does that make sense? | |
\( \large \$2.20+\$2.80d\) Hint: You can think of the fare as $2.20 to enter the cab, and then $0.40 for each 1/7 of a mile, including the first 1/7 of a mile (or $2.80 per mile).
Alternatively, you pay $2.60 for the first 1/7 of a mile, and then $2.80 per mile for d-1/7 miles. The total is 2.60+2.80(d-1/7) = 2.60+ 2.80d -.40 = 2.20+2.80d. | |
\( \large \$2.60+\$2.80d\) Hint: Don't count the first 1/7 of a mile twice. |
Question 3 Explanation:
Topic: Identify variables and derive algebraic expressions that represent real-world situations (Objective 0020), and select the linear equation that best models a real-world situation (Objective 0022).
Question 4 |
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 4 Explanation:
Topic: Recognize and apply connections between algebra and geometry
(e.g., the use of coordinate systems, the Pythagorean theorem) (Objective 0024).
Question 5 |
Use the graph below to answer the question that follows.
If the polygon shown above is reflected about the y axis and then rotated 90 degrees clockwise about the origin, which of the following graphs is the result?
Hint: Try following the point (1,4) to see where it goes after each transformation. | |
 | |
Hint: Make sure you're reflecting in the correct axis. | |
 Hint: Make sure you're rotating the correct direction. |
Question 5 Explanation:
Topic: Analyze and apply geometric transformations (e.g., translations,
rotations, reflections, dilations); relate them to concepts of symmetry,
similarity, and congruence; and use these concepts to solve problems (Objective 0024).
Question 6 |
Which of the numbers below is not equivalent to 4%?
\( \large \dfrac{1}{25}\) Hint: 1/25=4/100, so this is equal to 4% (be sure you read the question correctly). | |
\( \large \dfrac{4}{100}\) Hint: 4/100=4% (be sure you read the question correctly). | |
\( \large 0.4\) Hint: 0.4=40% so this is not equal to 4% | |
\( \large 0.04\) Hint: 0.04=4/100, so this is equal to 4% (be sure you read the question correctly). |
Question 6 Explanation:
Converting between fractions, decimals, and percents (Objective 0017).
Question 7 |
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.
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Question 7 Explanation:
Topics: Use unit conversions to solve measurement
problems, and derive and use formulas for calculating surface areas of geometric shapes and figures (Objective 0023).
Question 8 |
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 8 Explanation:
Topic: Ordering decimals and integers (Objective 0017).
Question 9 |
Solve for x: \(\large 4-\dfrac{2}{3}x=2x\)
\( \large x=3\) Hint: Try plugging x=3 into the equation. | |
\( \large x=-3\) Hint: Left side is positive, right side is negative when you plug this in for x. | |
\( \large x=\dfrac{3}{2}\) Hint: One way to solve: \(4=\dfrac{2}{3}x+2x\) \(=\dfrac{8}{3}x\).\(x=\dfrac{3 \times 4}{8}=\dfrac{3}{2}\). Another way is to just plug x=3/2 into the equation and see that each side equals 3 -- on a multiple choice test, you almost never have to actually solve for x. | |
\( \large x=-\dfrac{3}{2}\) Hint: Left side is positive, right side is negative when you plug this in for x. |
Question 9 Explanation:
Topic: Solve linear equations (Objective 0020).
Question 10 |
In the triangle below, \(\overline{AC}\cong \overline{AD}\cong \overline{DE}\) and \(m\angle CAD=100{}^\circ \). What is \(m\angle DAE\)?
\( \large 20{}^\circ \) Hint: Angles ACD and ADC are congruent since they are base angles of an isosceles triangle. Since the angles of a triangle sum to 180, they sum to 80, and they are 40 deg each. Thus angle ADE is 140 deg, since it makes a straight line with angle ADC. Angles DAE and DEA are base angles of an isosceles triangle and thus congruent-- they sum to 40 deg, so are 20 deg each. | |
\( \large 25{}^\circ \) Hint: If two sides of a triangle are congruent, then it's isosceles, and the base angles of an isosceles triangle are equal. | |
\( \large 30{}^\circ \) Hint: If two sides of a triangle are congruent, then it's isosceles, and the base angles of an isosceles triangle are equal. | |
\( \large 40{}^\circ \) Hint: Make sure you're calculating the correct angle. |
Question 10 Explanation:
Topic: Classify and analyze polygons using attributes of sides and angles,
including real-world applications. (Objective 0024).
Question 11 |
A sales companies pays its representatives $2 for each item sold, plus 40% of the price of the item. The rest of the money that the representatives collect goes to the company. All transactions are in cash, and all items cost $4 or more. If the price of an item in dollars is p, which expression represents the amount of money the company collects when the item is sold?
\( \large \dfrac{3}{5}p-2\) Hint: The company gets 3/5=60% of the price, minus the $2 per item. | |
\( \large \dfrac{3}{5}\left( p-2 \right)\) Hint: This is sensible, but not what the problem states. | |
\( \large \dfrac{2}{5}p+2\) Hint: The company pays the extra $2; it doesn't collect it. | |
\( \large \dfrac{2}{5}p-2\) Hint: This has the company getting 2/5 = 40% of the price of each item, but that's what the representative gets. |
Question 11 Explanation:
Topic: Use algebra to solve word problems involving fractions, ratios, proportions, and percents (Objective 0020).
Question 12 |
What is the perimeter of a right triangle with legs of lengths x and 2x?
\( \large 6x\) Hint: Use the Pythagorean Theorem. | |
\( \large 3x+5{{x}^{2}}\) Hint: Don't forget to take square roots when you use the Pythagorean Theorem. | |
\( \large 3x+\sqrt{5}{{x}^{2}}\) Hint: \(\sqrt {5 x^2}\) is not \(\sqrt {5}x^2\). | |
\( \large 3x+\sqrt{5}{{x}^{{}}}\) Hint: To find the hypotenuse, h, use the Pythagorean Theorem: \(x^2+(2x)^2=h^2.\) \(5x^2=h^2,h=\sqrt{5}x\). The perimeter is this plus x plus 2x. |
Question 12 Explanation:
Topic: Recognize and apply connections between algebra and geometry
(e.g., the use of coordinate systems, the Pythagorean theorem) (Objective 0024).
Question 13 |
There are 15 students for every teacher. Let t represent the number of teachers and let s represent the number of students. Which of the following equations is correct?
\( \large t=s+15\) Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation? | |
\( \large s=t+15\) Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation? | |
\( \large t=15s\) Hint: This is a really easy mistake to make, which comes from transcribing directly from English, "1 teachers equals 15 students." To see that it's wrong, plug in s=2; do you really need 30 teachers for 2 students? To avoid this mistake, insert the word "number," "Number of teachers equals 15 times number of students" is more clearly problematic. | |
\( \large s=15t\) |
Question 13 Explanation:
Topic: Select the linear equation that best models a real-world situation (Objective 0022).
Question 14 |
Use the problem below to answer the question that follows:
T shirts are on sale for 20% off. Tasha paid $8.73 for a shirt. What is the regular price of the shirt? There is no tax on clothing purchases under $175.
Let p represent the regular price of these t-shirt. Which of the following equations is correct?
\( \large 0.8p=\$8.73\) Hint: 80% of the regular price = $8.73. | |
\( \large \$8.73+0.2*\$8.73=p\) Hint: The 20% off was off of the ORIGINAL price, not off the $8.73 (a lot of people make this mistake). Plus this is the same equation as in choice c. | |
\( \large 1.2*\$8.73=p\) Hint: The 20% off was off of the ORIGINAL price, not off the $8.73 (a lot of people make this mistake). Plus this is the same equation as in choice b. | |
\( \large p-0.2*\$8.73=p\) Hint: Subtract p from both sides of this equation, and you have -.2 x 8.73 =0. |
Question 14 Explanation:
Topics: Use algebra to solve word problems involving percents and identify variables, and derive algebraic expressions that represent real-world situations (Objective 0020).
Question 15 |
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 15 Explanation:
Topic: Derive and use formulas for calculating the lengths, perimeters, areas,
volumes, and surface areas of geometric shapes and figures (Objective 0023).
Question 16 |
The expression \( \large{{8}^{3}}\cdot {{2}^{-10}}\) is equal to which of the following?
\( \large 2\) Hint: Write \(8^3\) as a power of 2. | |
\( \large \dfrac{1}{2}\) Hint: \(8^3 \cdot {2}^{-10}={(2^3)}^3 \cdot {2}^{-10}\) =\(2^9 \cdot {2}^{-10} =2^{-1}\) | |
\( \large 16\) Hint: Write \(8^3\) as a power of 2. | |
\( \large \dfrac{1}{16}\) Hint: Write \(8^3\) as a power of 2. |
Question 16 Explanation:
Topic: Laws of Exponents (Objective 0019).
Question 17 |
Use the expression below to answer the question that follows.
\( \large 3\times {{10}^{4}}+2.2\times {{10}^{2}}\)
Which of the following is closest to the expression above?
Five millionHint: Pay attention to the exponents. Adding 3 and 2 doesn't work because they have different place values. | |
Fifty thousandHint: Pay attention to the exponents. Adding 3 and 2 doesn't work because they have different place values. | |
Three millionHint: Don't add the exponents. | |
Thirty thousandHint: \( 3\times {{10}^{4}} = 30,000;\) the other term is much smaller and doesn't change the estimate. |
Question 17 Explanation:
Topics: Place value, scientific notation, estimation (Objective 0016)
Question 18 |
How many lines of reflective symmetry and how many centers of rotational symmetry does the parallelogram depicted below have?
4 lines of reflective symmetry, 1 center of rotational symmetry.Hint: Try cutting out a shape like this one from paper, and fold where you think the lines of reflective symmetry are (or put a mirror there). Do things line up as you thought they would? | |
2 lines of reflective symmetry, 1 center of rotational symmetry.Hint: Try cutting out a shape like this one from paper, and fold where you think the lines of reflective symmetry are (or put a mirror there). Do things line up as you thought they would? | |
0 lines of reflective symmetry, 1 center of rotational symmetry.Hint: The intersection of the diagonals is a center of rotational symmetry. There are no lines of reflective symmetry, although many people get confused about this fact (best to play with hands on examples to get a feel). Just fyi, the letter S also has rotational, but not reflective symmetry, and it's one that kids often write backwards. | |
2 lines of reflective symmetry, 0 centers of rotational symmetry.Hint: Try cutting out a shape like this one from paper. Trace onto another sheet of paper. See if there's a way to rotate the cut out shape (less than a complete turn) so that it fits within the outlines again. |
Question 18 Explanation:
Topic: Analyze geometric transformations (e.g., translations,
rotations, reflections, dilations); relate them to concepts of symmetry (Objective 0024).
Question 19 |
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 19 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 20 |
The letters A, and B represent digits (possibly equal) in the ten digit number x=1,438,152,A3B. For which values of A and B is x divisible by 12, but not by 9?
\( \large A = 0, B = 4\) Hint: Digits add to 31, so not divisible by 3, so not divisible by 12. | |
\( \large A = 7, B = 2\) Hint: Digits add to 36, so divisible by 9. | |
\( \large A = 0, B = 6\) Hint: Digits add to 33, divisible by 3, not 9. Last digits are 36, so divisible by 4, and hence by 12. | |
\( \large A = 4, B = 8\) Hint: Digits add to 39, divisible by 3, not 9. Last digits are 38, so not divisible by 4, so not divisible by 12. |
Question 20 Explanation:
Topic: Demonstrate knowledge of divisibility rules (Objective 0018).
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