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MTEL General Curriculum Mathematics Practice
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Question 1 |
Which of the following inequalities describes all values of x with \(\large \dfrac{x}{2}\le \dfrac{x}{3}\)?
\( \large x < 0\) Hint: If x =0, then x/2 = x/3, so this answer can't be correct. | |
\( \large x \le 0\) | |
\( \large x > 0\) Hint: If x =0, then x/2 = x/3, so this answer can't be correct. | |
\( \large x \ge 0\) Hint: Try plugging in x = 6. |
Question 1 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 2 |
Which of the lines depicted below is a graph of \( \large y=2x-5\)?
aHint: The slope of line a is negative. | |
bHint: Wrong slope and wrong intercept. | |
cHint: The intercept of line c is positive. | |
dHint: Slope is 2 -- for every increase of 1 in x, y increases by 2. Intercept is -5 -- the point (0,-5) is on the line. |
Question 2 Explanation:
Topic: Find a linear equation that represents a graph (Objective 0022).
Question 3 |
Below is a portion of a number line:
Point B is halfway between two tick marks. What number is represented by Point B?
\( \large 0.645\) Hint: That point is marked on the line, to the right. | |
\( \large 0.6421\) Hint: That point is to the left of point B. | |
\( \large 0.6422\) Hint: That point is to the left of point B. | |
\( \large 0.6425\) |
Question 3 Explanation:
Topic: Using Number Lines (Objective 0017)
Question 4 |
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?
\( \large 8\text{ c}{{\text{m}}^{2}}
\) 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. | |
\( \large 11\text{ c}{{\text{m}}^{2}}\) Hint: Try adding segments so that the slanted lines become the diagonals of rectangles. | |
\( \large 11.5\text{ c}{{\text{m}}^{2}}\) 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 | |
\( \large 12.5\text{ c}{{\text{m}}^{2}}\) Hint: Try adding segments so that the slanted lines become the diagonals of rectangles. |
Question 4 Explanation:
Topics: Calculate measurements and derive and use formulas for calculating the areas of geometric shapes and figures (Objective 0023).
Question 5 |
Which of the following values of x satisfies the inequality \( \large \left| {{(x+2)}^{3}} \right|<3?\)
\( \large x=-3\) Hint: \( \left| {{(-3+2)}^{3}} \right|\)=\( \left | {(-1)}^3 \right | \)=\( \left | -1 \right |=1 \) . | |
\( \large x=0\) Hint: \( \left| {{(0+2)}^{3}} \right|\)=\( \left | {2}^3 \right | \)=\( \left | 8 \right | \) =\( 8\) | |
\( \large x=-4\) Hint: \( \left| {{(-4+2)}^{3}} \right|\)=\( \left | {(-2)}^3 \right | \)=\( \left | -8 \right | \) =\( 8\) | |
\( \large x=1\) Hint: \( \left| {{(1+2)}^{3}} \right|\)=\( \left | {3}^3 \right | \)=\( \left | 27 \right | \) = \(27\) |
Question 5 Explanation:
Topics: Laws of exponents, order of operations, interpret absolute value (Objective 0019).
Question 6 |
The chairs in a large room can be arranged in rows of 18, 25, or 60 with no chairs left over. If C is the smallest possible number of chairs in the room, which of the following inequalities does C satisfy?
\( \large C\le 300\) Hint: Find the LCM. | |
\( \large 300 < C \le 500 \) Hint: Find the LCM. | |
\( \large 500 < C \le 700 \) Hint: Find the LCM. | |
\( \large C>700\) Hint: The LCM is 900, which is the smallest number of chairs. |
Question 6 Explanation:
Topic: Apply LCM in "real-world" situations (according to standardized tests....) (Objective 0018).
Question 7 |
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 7 Explanation:
Topic: Demonstrate knowledge of divisibility rules (Objective 0018).
Question 8 |
Which of the lists below is in order from least to greatest value?
\( \large \dfrac{1}{2},\quad \dfrac{1}{3},\quad \dfrac{1}{4},\quad \dfrac{1}{5}\) Hint: This is ordered from greatest to least. | |
\( \large \dfrac{1}{3},\quad \dfrac{2}{7},\quad \dfrac{3}{8},\quad \dfrac{4}{11}\) Hint: 1/3 = 2/6 is bigger than 2/7. | |
\( \large \dfrac{1}{4},\quad \dfrac{2}{5},\quad \dfrac{2}{3},\quad \dfrac{4}{5}\) Hint: One way to look at this: 1/4 and 2/5 are both less than 1/2, and 2/3 and 4/5 are both greater than 1/2. 1/4 is 25% and 2/5 is 40%, so 2/5 is greater. The distance from 2/3 to 1 is 1/3 and from 4/5 to 1 is 1/5, and 1/5 is less than 1/3, so 4/5 is bigger. | |
\( \large \dfrac{7}{8},\quad \dfrac{6}{7},\quad \dfrac{5}{6},\quad \dfrac{4}{5}\) Hint: This is in order from greatest to least. |
Question 8 Explanation:
Topic: Ordering Fractions (Objective 0017)
Question 9 |
Below is a portion of a number line.
Point A is one-quarter of the distance from 0.26 to 0.28. What number is represented by point A?
\( \large0.26\) Hint: Please reread the question. | |
\( \large0.2625\) Hint: This is one-quarter of the distance between 0.26 and 0.27, which is not what the question asked. | |
\( \large0.265\) | |
\( \large0.27\) Hint: Please read the question more carefully. This answer would be correct if Point A were halfway between the tick marks, but it's not. |
Question 9 Explanation:
Topic: Using number lines (Objective 0017)
Question 10 |
Here are some statements:
I) 5 is an integer II)\( -5 \) is an integer III) \(0\) is an integer
Which of the statements are true?
I only | |
I and II only | |
I and III only | |
I, II, and IIIHint: The integers are ...-3, -2, -1, 0, 1, 2, 3, .... |
Question 10 Explanation:
Topic: Characteristics of Integers (Objective 0016)
Question 11 |
Which of the following is equal to one million three hundred thousand?
\(\large1.3\times {{10}^{6}}\)
| |
\(\large1.3\times {{10}^{9}}\)
Hint: That's one billion three hundred million. | |
\(\large1.03\times {{10}^{6}}\)
Hint: That's one million thirty thousand. | |
\(\large1.03\times {{10}^{9}}\) Hint: That's one billion thirty million |
Question 11 Explanation:
Topic: Scientific Notation (Objective 0016)
Question 12 |
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?
\( \large 3.10m=400+0.2m\) 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. | |
\( \large 3+0.1m=29.99+.20m\) Hint: Doesn't account for the 400 free minutes. | |
\( \large 3+0.1m=400+29.99+.20(m-400)\) 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. | |
\( \large 3+0.1m=29.99+.20(m-400)\) 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 12 Explanation:
Identify variables and derive algebraic expressions that represent real-world situations (Objective 0020).
Question 13 |
The table below gives data from various years on how many young girls drank milk.
Based on the data given above, what was the probability that a randomly chosen girl in 1990 drank milk?
\( \large \dfrac{502}{1222}\) Hint: This is the probability that a randomly chosen girl who drinks milk was in the 1989-1991 food survey. | |
\( \large \dfrac{502}{2149}\) Hint: This is the probability that a randomly chosen girl from the whole survey drank milk and was also surveyed in 1989-1991. | |
\( \large \dfrac{502}{837}\) | |
\( \large \dfrac{1222}{2149}\) Hint: This is the probability that a randomly chosen girl from any year of the survey drank milk. |
Question 13 Explanation:
Topic: Recognize and apply the concept of conditional probability (Objective 0026).
Question 14 |
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 14 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 15 |
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?
\( \large N*2+20*5-100\div 10=N\) Hint: Use parentheses or else order of operations is off. | |
\( \large \left( \left( 2*N+20 \right)*5-100 \right)\div 10=N\) | |
\( \large \left( N+N+20 \right)*5-100\div 10=N\) Hint: With this answer you would subtract 10, instead of subtracting 100 and then dividing by 10. | |
\( \large \left( \left( \left( N\div 10 \right)-100 \right)*5+20 \right)*2=N\) Hint: This answer is quite backwards. |
Question 15 Explanation:
Topic: Recognize and apply the concepts of variable, function, equality, and equation to express relationships algebraically (Objective 0020).
Question 16 |
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 16 Explanation:
Topic: Identify prime and composite numbers (Objective 0018).
Question 17 |
What is the length of side \(\overline{BD}\) in the triangle below, where \(\angle DBA\) is a right angle?
\( \large 1\) Hint: Use the Pythagorean Theorem. | |
\( \large \sqrt{5}\) Hint: \(2^2+e^2=3^2\) or \(4+e^2=9;e^2=5; e=\sqrt{5}\). | |
\( \large \sqrt{13}\) Hint: e is not the hypotenuse. | |
\( \large 5\) Hint: Use the Pythagorean Theorem. |
Question 17 Explanation:
Topic: Derive and use formulas for calculating the lengths, perimeters, areas,
volumes, and surface areas of geometric shapes and figures (Objective 0023), and recognize and apply connections between algebra and geometry
(e.g., the use of coordinate systems, the Pythagorean theorem) (Objective 0024).
Question 18 |
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 18 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 19 |
Which of the following is the equation of a linear function?
\( \large y={{x}^{2}}+2x+7\) Hint: This is a quadratic function. | |
\( \large y={{2}^{x}}\) Hint: This is an exponential function. | |
\( \large y=\dfrac{15}{x}\) Hint: This is an inverse function. | |
\( \large y=x+(x+4)\) Hint: This is a linear function, y=2x+4, it's graph is a straight line with slope 2 and y-intercept 4. |
Question 19 Explanation:
Topic: Distinguish between linear and nonlinear functions (Objective 0022).
Question 20 |
P is a prime number that divides 240. Which of the following must be true?
P divides 30Hint: 2, 3, and 5 are the prime factors of 240, and all divide 30. | |
P divides 48Hint: P=5 doesn't work. | |
P divides 75Hint: P=2 doesn't work. | |
P divides 80Hint: P=3 doesn't work. |
Question 20 Explanation:
Topic: Find the prime factorization of a number and recognize its uses (Objective 0018).
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