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
Your answers are highlighted below.
Question 1 |
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.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. | |
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. | |
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. | |
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 1 Explanation:
Topic: Analyze and interpret various graphic and nongraphic data
representations (e.g., frequency distributions, percentiles) (Objective 0025).
Question 2 |
The table below gives the result of a survey at a college, asking students whether they were residents or commuters:
Based on the above data, what is the probability that a randomly chosen commuter student is a junior or a senior?
\( \large \dfrac{34}{43}\) | |
\( \large \dfrac{34}{71}\) Hint: This is the probability that a randomly chosen junior or senior is a commuter student. | |
\( \large \dfrac{34}{147}\) Hint: This is the probability that a randomly chosen student is a junior or senior who is a commuter. | |
\( \large \dfrac{71}{147}\) Hint: This is the probability that a randomly chosen student is a junior or a senior. |
Question 2 Explanation:
Topic: Recognize and apply the concept of conditional probability (Objective 0026).
Question 3 |
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 3 Explanation:
Topic: Converting between fraction and decimal representations (Objective 0017)
Question 4 |
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.Hint: Try plugging into the original solution. | |
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. | |
The student did not correctly use the distributive property.Hint: Distributive property is \(a(b+c)=ab+ac\). | |
The student did not correctly use the commutative property.Hint: Commutative property is \(a+b=b+a\) or \(ab=ba\). |
Question 4 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 5 |
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 5 Explanation:
Topic: Identifying rational and irrational numbers (Objective 0016).
Question 6 |
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 6 Explanation:
Topic: Classify and analyze polygons using attributes of sides and angles,
including real-world applications. (Objective 0024).
Question 7 |
| I. \(\large \dfrac{1}{2}+\dfrac{1}{3}\) | II. \( \large .400000\) | III. \(\large\dfrac{1}{5}+\dfrac{1}{5}\) |
| IV. \( \large 40\% \) | V. \( \large 0.25 \) | VI. \(\large\dfrac{14}{35}\) |
Which of the lists below includes all of the above expressions that are equivalent to \( \dfrac{2}{5}\)?
I, III, V, VIHint: I and V are not at all how fractions and decimals work. | |
III, VIHint: These are right, but there are more. | |
II, III, VIHint: These are right, but there are more. | |
II, III, IV, VI |
Question 7 Explanation:
Topic: Converting between fractions, decimals, and percents (Objective 0017)
Question 8 |
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 8 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 9 |
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 9 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 10 |
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 10 Explanation:
Topics: Estimation, simplifying fractions (Objective 0016).
Question 11 |
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 11 Explanation:
Topics: Estimation, Scientific Notation in the real world (Objective 0016).
Question 12 |
The prime factorization of n can be written as n=pqr, where p, q, and r are distinct prime numbers. How many factors does n have, including 1 and itself?
\( \large3\) Hint: 1, p, q, r, and pqr are already 5, so this isn't enough. You might try plugging in p=2, q=3, and r=5 to help with this problem. | |
\( \large5\) Hint: Don't forget pq, etc. You might try plugging in p=2, q=3, and r=5 to help with this problem. | |
\( \large6\) Hint: You might try plugging in p=2, q=3, and r=5 to help with this problem. | |
\( \large8\) Hint: 1, p, q, r, pq, pr, qr, pqr. |
Question 12 Explanation:
Topic: Recognize uses of prime factorization of a number (Objective 0018).
Question 13 |
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 13 Explanation:
Topics: Place value, scientific notation, estimation (Objective 0016)
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 14 Explanation:
Topic: Applying fractions to word problems (Objective 0017) This problem is similar to one on the official sample test for that objective, but it might fit better into unit conversion and dimensional analysis (Objective 0023: Measurement)
Question 15 |
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 15 Explanation:
Identify variables and derive algebraic expressions that represent real-world situations (Objective 0020).
Question 16 |
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 16 Explanation:
Topic: Least Common Multiple (Objective 0018)
Question 17 |
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 17 Explanation:
Topic:Classify and analyze three-dimensional figures using attributes of faces,
edges, and vertices (Objective 0024).
Question 18 |
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 18 Explanation:
Topic: Recognize and analyze pictorial representations of number operations. (Objective 0019).
Question 19 |
What is the mathematical name of the three-dimensional polyhedron depicted below?
TetrahedronHint: All the faces of a tetrahedron are triangles. | |
Triangular PrismHint: A prism has two congruent, parallel bases, connected by parallelograms (since this is a right prism, the parallelograms are rectangles). | |
Triangular PyramidHint: A pyramid has one base, not two. | |
TrigonHint: A trigon is a triangle (this is not a common term). |
Question 19 Explanation:
Topic: Classify and analyze three-dimensional figures using attributes of faces,
edges, and vertices (Objective 0024).
Question 20 |
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 20 Explanation:
Topic: Interpret the meaning of the slope and the intercepts of a linear equation
that models a real-world situation (Objective 0022).
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 20 questions to complete.
|
List |
If you found a mistake or have comments on a particular question, please contact me (please copy and paste at least part of the question into the form, as the numbers change depending on how quizzes are displayed). General comments can be left here.