20 Random Questions

Hints will display for most wrong answers; explanations for most right answers. You can attempt a question multiple times; it will only be scored correct if you get it right the first time.

I used the official objectives and sample test to construct these questions, but cannot promise that they accurately reflect what’s on the real test. Some of the sample questions were more convoluted than I could bear to write. See terms of use. See the MTEL Practice Test main page to view questions on a particular topic or to download paper practice tests.

MTEL General Curriculum Mathematics Practice

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Question 1

Here is a number trick:

1) Pick a whole number

2) Double your number.

3) Add 20 to the above result.

4) Multiply the above by 5

5) Subtract 100

6) Divide by 10

The result is always the number that you started with! Suppose you start by picking N. Which of the equations below best demonstrates that the result after Step 6 is also N?

A	$ \large N2+205-100\div 10=N$ Hint: Use parentheses or else order of operations is off.
B	$ \large \left( \left( 2N+20 \right)5-100 \right)\div 10=N$
C	$ \large \left( N+N+20 \right)*5-100\div 10=N$ Hint: With this answer you would subtract 10, instead of subtracting 100 and then dividing by 10.
D	$ \large \left( \left( \left( N\div 10 \right)-100 \right)5+20 \right)2=N$ Hint: This answer is quite backwards.

Question 1 Explanation:

Topic: Recognize and apply the concepts of variable, function, equality, and equation to express relationships algebraically (Objective 0020).

Question 2

The least common multiple of 60 and N is 1260. Which of the following could be the prime factorization of N?

A	$ \large2\cdot 5\cdot 7$ Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM.
B	$ \large{{2}^{3}}\cdot {{3}^{2}}\cdot 5 \cdot 7$ Hint: 1260 is not divisible by 8, so it isn't a multiple of this N.
C	$ \large3 \cdot 5 \cdot 7$ Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM.
D	$ \large{{3}^{2}}\cdot 5\cdot 7$ Hint: $1260=2^2 \cdot 3^2 \cdot 5 \cdot 7$ and $60=2^2 \cdot 3 \cdot 5$. In order for 1260 to be the LCM, N has to be a multiple of $3^2$ and of 7 (because 60 is not a multiple of either of these). N also cannot introduce a factor that would require the LCM to be larger (as in choice b).

Question 2 Explanation:

Topic: Least Common Multiple (Objective 0018)

Question 3

A map has a scale of 3 inches = 100 miles. Cities A and B are 753 miles apart. Let d be the distance between the two cities on the map. Which of the following is not correct?

A	$ \large \dfrac{3}{100}=\dfrac{d}{753}$ Hint: Units on both side are inches/mile, and both numerators and denominators correspond -- this one is correct.
B	$ \large \dfrac{3}{100}=\dfrac{753}{d}$ Hint: Unit on the left is inches per mile, and on the right is miles per inch. The proportion is set up incorrectly (which is what we wanted). Another strategy is to notice that one of A or B has to be the answer because they cannot both be correct proportions. Then check that cross multiplying on A gives part D, so B is the one that is different from the other 3.
C	$ \large \dfrac{3}{d}=\dfrac{100}{753}$ Hint: Unitless on each side, as inches cancel on the left and miles on the right. Numerators correspond to the map, and denominators to the real life distances -- this one is correct.
D	$ \large 100d=3\cdot 753$ Hint: This is equivalent to part A.

Question 3 Explanation:

Topic: Analyze the relationships among proportions, constant rates, and linear functions (Objective 0022).

Question 4

Which property is not shared by all rhombi?

A	4 congruent sides Hint: The most common definition of a rhombus is a quadrilateral with 4 congruent sides.
B	A center of rotational symmetry Hint: The diagonal of a rhombus separates it into two congruent isosceles triangles. The center of this line is a center of 180 degree rotational symmetry that switches the triangles.
C	4 congruent angles Hint: Unless the rhombus is a square, it does not have 4 congruent angles.
D	2 sets of parallel sides Hint: All rhombi are parallelograms.

Question 4 Explanation:

Topic: Classify and analyze polygons using attributes of sides and angles, and symmetry (Objective 0024).

Question 5

The chart below gives percentiles for the number of sit-ups that boys of various ages can do in 60 seconds (source , June 24, 2011)

Which of the following statements can be inferred from the above chart?

A	95% of 12 year old boys can do 56 sit-ups in 60 seconds. Hint: The 95th percentile means that 95% of scores are less than or equal to 56, and 5% are greater than or equal to 56.
B	At most 25% of 7 year old boys can do 19 or more sit-ups in 60 seconds. Hint: The 25th percentile means that 25% of scores are less than or equal to 19, and 75% are greater than or equal to 19.
C	Half of all 13 year old boys can do less than 41 sit-ups in 60 seconds and half can do more than 41 sit-ups in 60 seconds. Hint: Close, but not quite. There's no accounting for boys who can do exactly 41 sit ups. Look at these data: 10, 20, 41, 41, 41, 41, 50, 60, 90. The median is 41, but more than half can do 41 or more.
D	At least 75% of 16 year old boys can only do 51 or fewer sit-ups in 60 seconds. Hint: The "at least" is necessary due to duplicates. Suppose the data were 10, 20, 51, 51. The 75th percentile is 51, but 100% of the boys can only do 51 or fewer situps.

Question 5 Explanation:

Topic: Analyze and interpret various graphic and nongraphic data representations (e.g., frequency distributions, percentiles) (Objective 0025).

Question 6

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 6 Explanation:

Topics: Calculate measurements and derive and use formulas for calculating the areas of geometric shapes and figures (Objective 0023).

Question 7

Which of the following inequalities describes all values of x with $\large \dfrac{x}{2}\le \dfrac{x}{3}$?

A	$ \large x < 0$ Hint: If x =0, then x/2 = x/3, so this answer can't be correct.
B	$ \large x \le 0$
C	$ \large x > 0$ Hint: If x =0, then x/2 = x/3, so this answer can't be correct.
D	$ \large x \ge 0$ Hint: Try plugging in x = 6.

Question 7 Explanation:

Topics: Inequalities, operations (Objective 0019) (not exactly sure how to classify, but this is like one of the problems on the official sample test).

Question 8

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?

A	$ \large t=s+15$ Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation?
B	$ \large s=t+15$ Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation?
C	$ \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.
D	$ \large s=15t$

Question 8 Explanation:

Topic: Select the linear equation that best models a real-world situation (Objective 0022).

Question 9

How many lines of reflective symmetry and how many centers of rotational symmetry does the parallelogram depicted below have?

A	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?
B	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?
C	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.
D	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 9 Explanation:

Topic: Analyze geometric transformations (e.g., translations, rotations, reflections, dilations); relate them to concepts of symmetry (Objective 0024).

Question 10

Here is a student's work solving an equation:

$ x-4=-2x+6$

$ x-4+4=-2x+6+4$

$ x=-2x+10$

$ x-2x=10$

$ x=10$

Which of the following statements is true?

A	The student‘s solution is correct. Hint: Try plugging into the original solution.
B	The student did not correctly use properties of equality. Hint: After $ x=-2x+10$, the student subtracted 2x on the left and added 2x on the right.
C	The student did not correctly use the distributive property. Hint: Distributive property is $a(b+c)=ab+ac$.
D	The student did not correctly use the commutative property. Hint: Commutative property is $a+b=b+a$ or $ab=ba$.

Question 10 Explanation:

Topic: Justify algebraic manipulations by application of the properties of equality, the order of operations, the number properties, and the order properties (Objective 0020).

Question 11

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?

A	A Hint: Rise is more than 30 inches.
B	B Hint: Run is almost 24 feet, so rise can be almost 2 feet.
C	C Hint: Run is 12 feet, so rise can be at most 1 foot.
D	D Hint: Slope is 1:10 -- too steep.

Question 11 Explanation:

Topic: Interpret meaning of slope in a real world situation (Objective 0022).

Question 12

Which of the following is equivalent to

$ \large A-B+C\div D\times E$?

A	$ \large A-B-\dfrac{C}{DE} $ Hint: In the order of operations, multiplication and division have the same priority, so do them left to right; same with addition and subtraction.
B	$ \large A-B+\dfrac{CE}{D}$ Hint: In practice, you're better off using parentheses than writing an expression like the one in the question. The PEMDAS acronym that many people memorize is misleading. Multiplication and division have equal priority and are done left to right. They have higher priority than addition and subtraction. Addition and subtraction also have equal priority and are done left to right.
C	$ \large \dfrac{AE-BE+CE}{D}$ Hint: Use order of operations, don't just compute left to right.
D	$ \large A-B+\dfrac{C}{DE}$ Hint: In the order of operations, multiplication and division have the same priority, so do them left to right

Question 12 Explanation:

Topic: Justify algebraic manipulations by application of the properties of order of operations (Objective 0020).

Question 13

If two fair coins are flipped, what is the probability that one will come up heads and the other tails?

A	$ \large \dfrac{1}{4}$ Hint: Think of the coins as a penny and a dime, and list all possibilities.
B	$ \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.
C	$ \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.
D	$ \large \dfrac{3}{4}$ Hint: Think of the coins as a penny and a dime, and list all possibilities.

Question 13 Explanation:

Topic: Calculate the probabilities of simple and compound events and of independent and dependent events (Objective 0026).

Question 14

Which of the following points is closest to $ \dfrac{34}{135} \times \dfrac{53}{86}$?

A	A Hint: $\frac{34}{135} \approx \frac{1}{4}$ and $ \frac{53}{86} \approx \frac {2}{3}$. $\frac {1}{4}$ of $\frac {2}{3}$ is small and closest to A.
B	B Hint: Estimate with simpler fractions.
C	C Hint: Estimate with simpler fractions.
D	D Hint: Estimate with simpler fractions.

Question 14 Explanation:

Topic: Understand meaning and models of operations on fractions (Objective 0019).

Question 15

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?

A	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.
B	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.
C	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.
D	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 15 Explanation:

Topic: Identify prime and composite numbers (Objective 0018).

Question 16

Which of the following values of x satisfies the inequality $ \large \left| {{(x+2)}^{3}} \right|<3?$

A	$ \large x=-3$ Hint: $ \left\| {{(-3+2)}^{3}} \right\|$=$ \left \| {(-1)}^3 \right \| $=$ \left \| -1 \right \|=1 $ .
B	$ \large x=0$ Hint: $ \left\| {{(0+2)}^{3}} \right\|$=$ \left \| {2}^3 \right \| $=$ \left \| 8 \right \| $ =$ 8$
C	$ \large x=-4$ Hint: $ \left\| {{(-4+2)}^{3}} \right\|$=$ \left \| {(-2)}^3 \right \| $=$ \left \| -8 \right \| $ =$ 8$
D	$ \large x=1$ Hint: $ \left\| {{(1+2)}^{3}} \right\|$=$ \left \| {3}^3 \right \| $=$ \left \| 27 \right \| $ = $27$

Question 16 Explanation:

Topics: Laws of exponents, order of operations, interpret absolute value (Objective 0019).

Question 17

Which of the numbers below is a fraction equivalent to $ 0.\bar{6}$?

A	$ \large \dfrac{4}{6}$ Hint: $ 0.\bar{6}=\dfrac{2}{3}=\dfrac{4}{6}$
B	$ \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.
C	$ \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.
D	$ \large \dfrac{1}{6}$ Hint: This is less than a half, and $ 0.\bar{6}$ is greater than a half.

Question 17 Explanation:

Topic: Converting between fraction and decimal representations (Objective 0017)

Question 18

A family went on a long car trip. Below is a graph of how far they had driven at each hour.

Which of the following is closest to their average speed driving on the trip?

A	$ \large d=20t$ Hint: Try plugging t=7 into the equation, and see how it matches the graph.
B	$ \large d=30t$ Hint: Try plugging t=7 into the equation, and see how it matches the graph.
C	$ \large d=40t$
D	$ \large d=50t$ Hint: Try plugging t=7 into the equation, and see how it matches the graph.

Question 18 Explanation:

Topic: Select the linear equation that best models a real-world situation (Objective 0022).

Question 19

What is the greatest common factor of 540 and 216?

A	$ \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.
B	$ \large2\cdot 3$ Hint: This is a common factor of both numbers, but it's not the greatest common factor.
C	$ \large{{2}^{3}}\cdot {{3}^{3}}$ Hint: $2^3 = 8$ is not a factor of 540.
D	$ \large{{2}^{2}}\cdot {{3}^{2}}$ Hint: This is a common factor of both numbers, but it's not the greatest common factor.

Question 19 Explanation:

Topic: Find the greatest common factor of a set of numbers (Objective 0018).

Question 20

Cell phone plan A charges $3 per month plus $0.10 per minute. Cell phone plan B charges $29.99 per month, with no fee for the first 400 minutes and then $0.20 for each additional minute.

Which equation can be used to solve for the number of minutes, m (with m>400) that a person would have to spend on the phone each month in order for the bills for plan A and plan B to be equal?

A	$ \large 3.10m=400+0.2m$ Hint: These are the numbers in the problem, but this equation doesn't make sense. If you don't know how to make an equation, try plugging in an easy number like m=500 minutes to see if each side equals what it should.
B	$ \large 3+0.1m=29.99+.20m$ Hint: Doesn't account for the 400 free minutes.
C	$ \large 3+0.1m=400+29.99+.20(m-400)$ Hint: Why would you add 400 minutes and $29.99? If you don't know how to make an equation, try plugging in an easy number like m=500 minutes to see if each side equals what it should.
D	$ \large 3+0.1m=29.99+.20(m-400)$ Hint: The left side is $3 plus $0.10 times the number of minutes. The right is $29.99 plus $0.20 times the number of minutes over 400.

Question 20 Explanation:

Identify variables and derive algebraic expressions that represent real-world situations (Objective 0020).

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A	\( \large N2+205-100\div 10=N\) Hint: Use parentheses or else order of operations is off.
B	\( \large \left( \left( 2N+20 \right)5-100 \right)\div 10=N\)
C	\( \large \left( N+N+20 \right)*5-100\div 10=N\) Hint: With this answer you would subtract 10, instead of subtracting 100 and then dividing by 10.
D	\( \large \left( \left( \left( N\div 10 \right)-100 \right)5+20 \right)2=N\) Hint: This answer is quite backwards.

A	\( \large2\cdot 5\cdot 7\) Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM.
B	\( \large{{2}^{3}}\cdot {{3}^{2}}\cdot 5 \cdot 7\) Hint: 1260 is not divisible by 8, so it isn't a multiple of this N.
C	\( \large3 \cdot 5 \cdot 7\) Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM.
D	\( \large{{3}^{2}}\cdot 5\cdot 7\) Hint: \(1260=2^2 \cdot 3^2 \cdot 5 \cdot 7\) and \(60=2^2 \cdot 3 \cdot 5\). In order for 1260 to be the LCM, N has to be a multiple of \(3^2\) and of 7 (because 60 is not a multiple of either of these). N also cannot introduce a factor that would require the LCM to be larger (as in choice b).

A	\( \large \dfrac{3}{100}=\dfrac{d}{753}\) Hint: Units on both side are inches/mile, and both numerators and denominators correspond -- this one is correct.
B	\( \large \dfrac{3}{100}=\dfrac{753}{d}\) Hint: Unit on the left is inches per mile, and on the right is miles per inch. The proportion is set up incorrectly (which is what we wanted). Another strategy is to notice that one of A or B has to be the answer because they cannot both be correct proportions. Then check that cross multiplying on A gives part D, so B is the one that is different from the other 3.
C	\( \large \dfrac{3}{d}=\dfrac{100}{753}\) Hint: Unitless on each side, as inches cancel on the left and miles on the right. Numerators correspond to the map, and denominators to the real life distances -- this one is correct.
D	\( \large 100d=3\cdot 753\) Hint: This is equivalent to part A.

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.

A	\( \large x < 0\) Hint: If x =0, then x/2 = x/3, so this answer can't be correct.
B	\( \large x \le 0\)
C	\( \large x > 0\) Hint: If x =0, then x/2 = x/3, so this answer can't be correct.
D	\( \large x \ge 0\) Hint: Try plugging in x = 6.

A	\( \large t=s+15\) Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation?
B	\( \large s=t+15\) Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation?
C	\( \large t=15s\) 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.
D	\( \large s=15t\)

A	\( \large A-B-\dfrac{C}{DE} \) Hint: In the order of operations, multiplication and division have the same priority, so do them left to right; same with addition and subtraction.
B	\( \large A-B+\dfrac{CE}{D}\) Hint: In practice, you're better off using parentheses than writing an expression like the one in the question. The PEMDAS acronym that many people memorize is misleading. Multiplication and division have equal priority and are done left to right. They have higher priority than addition and subtraction. Addition and subtraction also have equal priority and are done left to right.
C	\( \large \dfrac{AE-BE+CE}{D}\) Hint: Use order of operations, don't just compute left to right.
D	\( \large A-B+\dfrac{C}{DE}\) Hint: In the order of operations, multiplication and division have the same priority, so do them left to right

A	\( \large \dfrac{1}{4}\) Hint: Think of the coins as a penny and a dime, and list all possibilities.
B	\( \large \dfrac{1}{3} \) 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.
C	\( \large \dfrac{1}{2}\) 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.
D	\( \large \dfrac{3}{4}\) Hint: Think of the coins as a penny and a dime, and list all possibilities.

A	A Hint: \(\frac{34}{135} \approx \frac{1}{4}\) and \( \frac{53}{86} \approx \frac {2}{3}\). \(\frac {1}{4}\) of \(\frac {2}{3}\) is small and closest to A.
B	B Hint: Estimate with simpler fractions.
C	C Hint: Estimate with simpler fractions.
D	D Hint: Estimate with simpler fractions.

A	\( \large x=-3\) Hint: \( \left\| {{(-3+2)}^{3}} \right\|\)=\( \left \| {(-1)}^3 \right \| \)=\( \left \| -1 \right \|=1 \) .
B	\( \large x=0\) Hint: \( \left\| {{(0+2)}^{3}} \right\|\)=\( \left \| {2}^3 \right \| \)=\( \left \| 8 \right \| \) =\( 8\)
C	\( \large x=-4\) Hint: \( \left\| {{(-4+2)}^{3}} \right\|\)=\( \left \| {(-2)}^3 \right \| \)=\( \left \| -8 \right \| \) =\( 8\)
D	\( \large x=1\) Hint: \( \left\| {{(1+2)}^{3}} \right\|\)=\( \left \| {3}^3 \right \| \)=\( \left \| 27 \right \| \) = \(27\)

A	\( \large \dfrac{4}{6}\) Hint: \( 0.\bar{6}=\dfrac{2}{3}=\dfrac{4}{6}\)
B	\( \large \dfrac{3}{5}\) 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.
C	\( \large \dfrac{6}{10}\) 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.
D	\( \large \dfrac{1}{6}\) Hint: This is less than a half, and \( 0.\bar{6}\) is greater than a half.

A	\( \large d=20t\) Hint: Try plugging t=7 into the equation, and see how it matches the graph.
B	\( \large d=30t\) Hint: Try plugging t=7 into the equation, and see how it matches the graph.
C	\( \large d=40t\)
D	\( \large d=50t\) Hint: Try plugging t=7 into the equation, and see how it matches the graph.

A	\( \large{{2}^{2}}\cdot {{3}^{3}}\) 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.
B	\( \large2\cdot 3\) Hint: This is a common factor of both numbers, but it's not the greatest common factor.
C	\( \large{{2}^{3}}\cdot {{3}^{3}}\) Hint: \(2^3 = 8\) is not a factor of 540.
D	\( \large{{2}^{2}}\cdot {{3}^{2}}\) Hint: This is a common factor of both numbers, but it's not the greatest common factor.

A	\( \large 3.10m=400+0.2m\) Hint: These are the numbers in the problem, but this equation doesn't make sense. If you don't know how to make an equation, try plugging in an easy number like m=500 minutes to see if each side equals what it should.
B	\( \large 3+0.1m=29.99+.20m\) Hint: Doesn't account for the 400 free minutes.
C	\( \large 3+0.1m=400+29.99+.20(m-400)\) Hint: Why would you add 400 minutes and $29.99? If you don't know how to make an equation, try plugging in an easy number like m=500 minutes to see if each side equals what it should.
D	\( \large 3+0.1m=29.99+.20(m-400)\) Hint: The left side is $3 plus $0.10 times the number of minutes. The right is $29.99 plus $0.20 times the number of minutes over 400.

MTEL General Curriculum Mathematics Practice

Here is a number trick:

1) Pick a whole number

2) Double your number.

3) Add 20 to the above result.

4) Multiply the above by 5

5) Subtract 100

6) Divide by 10

The result is always the number that you started with! Suppose you start by picking N. Which of the equations below best demonstrates that the result after Step 6 is also N?

The least common multiple of 60 and N is 1260. Which of the following could be the prime factorization of N?

A map has a scale of 3 inches = 100 miles. Cities A and B are 753 miles apart. Let d be the distance between the two cities on the map. Which of the following is not correct?

Which property is not shared by all rhombi?

4 congruent sides

A center of rotational symmetry

4 congruent angles

2 sets of parallel sides

The chart below gives percentiles for the number of sit-ups that boys of various ages can do in 60 seconds (source , June 24, 2011)

Which of the following statements can be inferred from the above chart?

95% of 12 year old boys can do 56 sit-ups in 60 seconds.

At most 25% of 7 year old boys can do 19 or more sit-ups in 60 seconds.

Half of all 13 year old boys can do less than 41 sit-ups in 60 seconds and half can do more than 41 sit-ups in 60 seconds.

At least 75% of 16 year old boys can only do 51 or fewer sit-ups in 60 seconds.

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?

Which of the following inequalities describes all values of x with \(\large \dfrac{x}{2}\le \dfrac{x}{3}\)?

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?

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.

2 lines of reflective symmetry, 1 center of rotational symmetry.

0 lines of reflective symmetry, 1 center of rotational symmetry.

2 lines of reflective symmetry, 0 centers of rotational symmetry.

Here is a student's work solving an equation:

\( x-4=-2x+6\)

\( x-4+4=-2x+6+4\)

\( x=-2x+10\)

\( x-2x=10\)

\( x=10\)

Which of the following statements is true?

The student‘s solution is correct.

The student did not correctly use properties of equality.

The student did not correctly use the distributive property.

The student did not correctly use the commutative property.

A

B

C

D

Which of the following is equivalent to

\( \large A-B+C\div D\times E\)?

If two fair coins are flipped, what is the probability that one will come up heads and the other tails?

Which of the following points is closest to \( \dfrac{34}{135} \times \dfrac{53}{86}\)?

A

B

C

D

All natural numbers from 2 to 266.

All primes from 2 to 266 .

All primes from 2 to 133 .

All primes from \( \large 2\) to \( \large \sqrt{267}\).

Which of the following values of x satisfies the inequality \( \large \left| {{(x+2)}^{3}} \right|<3?\)

Which of the numbers below is a fraction equivalent to \( 0.\bar{6}\)?

A family went on a long car trip. Below is a graph of how far they had driven at each hour.

Which of the following is closest to their average speed driving on the trip?

What is the greatest common factor of 540 and 216?

Cell phone plan A charges $3 per month plus $0.10 per minute. Cell phone plan B charges $29.99 per month, with no fee for the first 400 minutes and then $0.20 for each additional minute.

Which equation can be used to solve for the number of minutes, m (with m>400) that a person would have to spend on the phone each month in order for the bills for plan A and plan B to be equal?