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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
Your answers are highlighted below.
Question 1 |
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 1 Explanation:
Topics: Laws of exponents, order of operations, interpret absolute value (Objective 0019).
Question 2 |
Which of the following is closest to the height of a college student in centimeters?
1.6 cmHint: This is more the height of a Lego toy college student -- less than an inch! | |
16 cmHint: Less than knee high on most college students. | |
160 cmHint: Remember, a meter stick (a little bigger than a yard stick) is 100 cm. Also good to know is that 4 inches is approximately 10 cm. | |
1600 cmHint: This college student might be taller than some campus buildings! |
Question 2 Explanation:
Topic: Estimate and calculate measurements using customary, metric, and
nonstandard units of measurement (Objective 0023).
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 |
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 4 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 5 |
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 5 Explanation:
Topic: Apply geometric transformations (e.g., translations,
rotations, reflections, dilations); relate them to
similarity, ; and use these concepts to solve problems (Objective 0024) . Fits in other places too.
Question 6 |
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 6 Explanation:
Topic: Characteristics of Integers (Objective 0016)
Question 7 |
A family has four children. What is the probability that two children are girls and two are boys? Assume the the probability of having a boy (or a girl) is 50%.
\( \large \dfrac{1}{2}\) Hint: How many different configurations are there from oldest to youngest, e.g. BGGG? How many of them have 2 boys and 2 girls? | |
\( \large \dfrac{1}{4}\) Hint: How many different configurations are there from oldest to youngest, e.g. BGGG? How many of them have 2 boys and 2 girls? | |
\( \large \dfrac{1}{5}\) Hint: Some configurations are more probable than others -- i.e. it's more likely to have two boys and two girls than all boys. Be sure you are weighting properly. | |
\( \large \dfrac{3}{8}\) Hint: There are two possibilities for each child, so there are \(2 \times 2 \times 2 \times 2 =16\) different configurations, e.g. from oldest to youngest BBBG, BGGB, GBBB, etc. Of these configurations, there are 6 with two boys and two girls (this is the combination \(_{4}C_{2}\) or "4 choose 2"): BBGG, BGBG, BGGB, GGBB, GBGB, and GBBG. Thus the probability is 6/16=3/8. |
Question 7 Explanation:
Topic: Apply knowledge of combinations and permutations to
the computation of probabilities (Objective 0026).
Question 8 |
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 8 Explanation:
Topics: Estimation, simplifying fractions (Objective 0016).
Question 9 |
The expression \( \large {{7}^{-4}}\cdot {{8}^{-6}}\) is equal to which of the following?
\( \large \dfrac{8}{{{\left( 56 \right)}^{4}}}\) Hint: The bases are whole numbers, and the exponents are negative. How can the numerator be 8? | |
\( \large \dfrac{64}{{{\left( 56 \right)}^{4}}}\) Hint: The bases are whole numbers, and the exponents are negative. How can the numerator be 64? | |
\( \large \dfrac{1}{8\cdot {{\left( 56 \right)}^{4}}}\) Hint: \(8^{-6}=8^{-4} \times 8^{-2}\) | |
\( \large \dfrac{1}{64\cdot {{\left( 56 \right)}^{4}}}\) |
Question 9 Explanation:
Topics: Laws of exponents (Objective 0019).
Question 10 |
M is a multiple of 26. Which of the following cannot be true?
M is odd.Hint: All multiples of 26 are also multiples of 2, so they must be even. | |
M is a multiple of 3.Hint: 3 x 26 is a multiple of both 3 and 26. | |
M is 26.Hint: 1 x 26 is a multiple of 26. | |
M is 0.Hint: 0 x 26 is a multiple of 26. |
Question 10 Explanation:
Topic: Characteristics of composite numbers (Objective 0018).
Question 11 |
At a school fundraising event, people can buy a ticket to spin a spinner like the one below. The region that the spinner lands in tells which, if any, prize the person wins.
If 240 people buy tickets to spin the spinner, what is the best estimate of the number of keychains that will be given away?
40Hint: "Keychain" appears on the spinner twice. | |
80Hint: The probability of getting a keychain is 1/3, and so about 1/3 of the time the spinner will win. | |
100Hint: What is the probability of winning a keychain? | |
120Hint: That would be the answer for getting any prize, not a keychain specifically. |
Question 11 Explanation:
Topic: I would call this topic expected value, which is not listed on the objectives. This question is very similar to one on the sample test. It's not a good question in that it's oversimplified (a more difficult and interesting question would be something like, "The school bought 100 keychains for prizes, what is the probability that they will run out before 240 people play?"). In any case, I believe the objective this is meant for is, "Recognize the difference between experimentally and theoretically
determined probabilities in real-world situations. (Objective 0026)." This is not something easily assessed with multiple choice .
Question 12 |
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 12 Explanation:
Topic: Find the greatest common factor of a set of numbers (Objective 0018).
Question 13 |
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 13 Explanation:
Converting between fractions, decimals, and percents (Objective 0017).
Question 14 |
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 14 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 15 |
A cylindrical soup can has diameter 7 cm and height 11 cm. The can holds g grams of soup. How many grams of the same soup could a cylindrical can with diameter 14 cm and height 33 cm hold?
\( \large 6g\) Hint: You must scale in all three dimensions. | |
\( \large 12g\) Hint: Height is multiplied by 3, and diameter and radius are multiplied by 2. Since the radius is squared, final result is multiplied by \(2^2\times 3=12\). | |
\( \large 18g\) Hint: Don't square the height scale factor. | |
\( \large 36g\) Hint: Don't square the height scale factor. |
Question 15 Explanation:
Topic: Determine how the characteristics (e.g., area, volume) of geometric
figures and shapes are affected by changes in their dimensions (Objective 0023).
Question 16 |
The equation \( \large F=\frac{9}{5}C+32\) is used to convert a temperature measured in Celsius to the equivalent Farentheit temperature.
A patient's temperature increased by 1.5° Celcius. By how many degrees Fahrenheit did her temperature increase?
1.5°Hint: Celsius and Fahrenheit don't increase at the same rate. | |
1.8°Hint: That's how much the Fahrenheit temp increases when the Celsius temp goes up by 1 degree. | |
2.7°Hint: Each degree increase in Celsius corresponds to a \(\dfrac{9}{5}=1.8\) degree increase in Fahrenheit. Thus the increase is 1.8+0.9=2.7. | |
Not enough information.Hint: A linear equation has constant slope, which means that every increase of the same amount in one variable, gives a constant increase in the other variable. It doesn't matter what temperature the patient started out at. |
Question 16 Explanation:
Topic: Interpret the meaning of the slope and the intercepts of a linear equation
that models a real-world situation (Objective 0022).
Question 17 |
The least common multiple of 60 and N is 1260. Which of the following could be the prime factorization of N?
\( \large2\cdot 5\cdot 7\) Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM. | |
\( \large{{2}^{3}}\cdot {{3}^{2}}\cdot 5 \cdot 7\) Hint: 1260 is not divisible by 8, so it isn't a multiple of this N. | |
\( \large3 \cdot 5 \cdot 7\) Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM. | |
\( \large{{3}^{2}}\cdot 5\cdot 7\) 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 17 Explanation:
Topic: Least Common Multiple (Objective 0018)
Question 18 |
Which of the following is an irrational number?
\( \large \sqrt[3]{8}\) Hint: This answer is the cube root of 8. Since 2 x 2 x 2 =8, this is equal to 2, which is rational because 2 = 2/1. | |
\( \large \sqrt{8}\) Hint: It is not trivial to prove that this is irrational, but you can get this answer by eliminating the other choices. | |
\( \large \dfrac{1}{8}\) Hint: 1/8 is the RATIO of two integers, so it is rational. | |
\( \large -8\) Hint: Negative integers are also rational, -8 = -8/1, a ratio of integers. |
Question 18 Explanation:
Topic: Identifying rational and irrational numbers (Objective 0016).
Question 19 |
In March of 2012, 1 dollar was worth the same as 0.761 Euros, and 1 dollar was also worth the same as 83.03 Japanese Yen. Which of the expressions below gives the number of Yen that are worth 1 Euro?
\( \large {83}.0{3}\cdot 0.{761}\) Hint: This equation gives less than the number of yen per dollar, but 1 Euro is worth more than 1 dollar. | |
\( \large \dfrac{0.{761}}{{83}.0{3}}\) Hint: Number is way too small. | |
\( \large \dfrac{{83}.0{3}}{0.{761}}\) Hint: One strategy here is to use easier numbers, say 1 dollar = .5 Euros and 100 yen, then 1 Euro would be 200 Yen (change the numbers in the equations and see what works). Another is to use dimensional analysis: we want # yen per Euro, or yen/Euro = yen/dollar \(\times\) dollar/Euro = \(83.03 \times \dfrac {1}{0.761}\) | |
\( \large \dfrac{1}{0.{761}}\cdot \dfrac{1}{{83}.0{3}}\) Hint: Number is way too small. |
Question 19 Explanation:
Topic: Analyze the relationships among proportions, constant rates, and linear
functions (Objective 0022).
Question 20 |
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 20 Explanation:
Topics: Calculate measurements and derive and use formulas for calculating the areas of geometric shapes and figures (Objective 0023).
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