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MTEL General Curriculum Mathematics Practice
Question 1 |
Use the graph below to answer the question that follows:
The graph above best matches which of the following scenarios:
George left home at 10:00 and drove to work on a crooked path. He was stopped in traffic at 10:30 and 10:45. He drove 30 miles total.Hint: Just because he ended up 30 miles from home doesn't mean he drove 30 miles total. | |
George drove to work. On the way to work there is a little hill and a big hill. He slowed down for them. He made it to work at 11:15.Hint: The graph is not a picture of the roads. | |
George left home at 10:15. He drove 10 miles, then realized he‘d forgotten something at home. He turned back and got what he‘d forgotten. Then he drove in a straight line, at many different speeds, until he got to work around 11:15.Hint: A straight line on a distance versus time graph means constant speed. | |
George left home at 10:15. He drove 10 miles, then realized he‘d forgotten something at home. He turned back and got what he‘d forgotten. Then he drove at a constant speed until he got to work around 11:15. |
Question 2 |
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 3 |
Below are four inputs and outputs for a function machine representing the function A:
Which of the following equations could also represent A for the values shown?
\( \large A(n)=n+4\) Hint: For a question like this, you don't have to find the equation yourself, you can just try plugging the function machine inputs into the equation, and see if any values come out wrong. With this equation n= -1 would output 3, not 0 as the machine does. | |
\( \large A(n)=n+2\) Hint: For a question like this, you don't have to find the equation yourself, you can just try plugging the function machine inputs into the equation, and see if any values come out wrong. With this equation n= 2 would output 4, not 6 as the machine does. | |
\( \large A(n)=2n+2\) Hint: Simply plug in each of the four function machine input values, and see that the equation produces the correct output, e.g. A(2)=6, A(-1)=0, etc. | |
\( \large A(n)=2\left( n+2 \right)\) Hint: For a question like this, you don't have to find the equation yourself, you can just try plugging the function machine inputs into the equation, and see if any values come out wrong. With this equation n= 2 would output 8, not 6 as the machine does. |
Question 4 |
Use the samples of a student's work below to answer the question that follows:
This student divides fractions by first finding a common denominator, then dividing the numerators.
\( \large \dfrac{2}{3} \div \dfrac{3}{4} \longrightarrow \dfrac{8}{12} \div \dfrac{9}{12} \longrightarrow 8 \div 9 = \dfrac {8}{9}\) \( \large \dfrac{2}{5} \div \dfrac{7}{20} \longrightarrow \dfrac{8}{20} \div \dfrac{7}{20} \longrightarrow 8 \div 7 = \dfrac {8}{7}\) \( \large \dfrac{7}{6} \div \dfrac{3}{4} \longrightarrow \dfrac{14}{12} \div \dfrac{9}{12} \longrightarrow 14 \div 9 = \dfrac {14}{9}\)Which of the following best describes the mathematical validity of the algorithm the student is using?
It is not valid. Common denominators are for adding and subtracting fractions, not for dividing them.Hint: Don't be so rigid! Usually there's more than one way to do something in math. | |
It got the right answer in these three cases, but it isn‘t valid for all rational numbers.Hint: Did you try some other examples? What makes you say it's not valid? | |
It is valid if the rational numbers in the division problem are in lowest terms and the divisor is not zero.Hint: Lowest terms doesn't affect this problem at all. | |
It is valid for all rational numbers, as long as the divisor is not zero.Hint: When we have common denominators, the problem is in the form a/b divided by c/b, and the answer is a/c, as the student's algorithm predicts. |
Question 5 |
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 6 |
A publisher prints a series of books with covers made of identical material and using the same thickness of paper for each page. The covers of the book together are 0.4 cm thick, and 125 pieces of the paper used together are 1 cm thick.
The publisher uses a linear function to determine the total thickness, T(n) of a book made with n sheets of paper. What are the slope and intercept of T(n)?
Intercept = 0.4 cm, Slope = 125 cm/pageHint: This would mean that each page of the book was 125 cm thick. | |
Intercept =0.4 cm, Slope = \(\dfrac{1}{125}\)cm/pageHint: The intercept is how thick the book would be with no pages in it. The slope is how much 1 extra page adds to the thickness of the book. | |
Intercept = 125 cm, Slope = 0.4 cmHint: This would mean that with no pages in the book, it would be 125 cm thick. | |
Intercept = \(\dfrac{1}{125}\)cm, Slope = 0.4 pages/cmHint: This would mean that each new page of the book made it 0.4 cm thicker. |
Question 7 |
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 8 |
AHint: \(\frac{34}{135} \approx \frac{1}{4}\) and \( \frac{53}{86} \approx \frac {2}{3}\). \(\frac {1}{4}\) of \(\frac {2}{3}\) is small and closest to A. | |
BHint: Estimate with simpler fractions. | |
CHint: Estimate with simpler fractions. | |
DHint: Estimate with simpler fractions. |
Question 9 |
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?
\( \large \dfrac{3}{100}=\dfrac{d}{753}\) Hint: Units on both side are inches/mile, and both numerators and denominators correspond -- this one is correct. | |
\( \large \dfrac{3}{100}=\dfrac{753}{d}\) 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. | |
\( \large \dfrac{3}{d}=\dfrac{100}{753}\) 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. | |
\( \large 100d=3\cdot 753\) Hint: This is equivalent to part A. |
Question 10 |
Which of the following is equivalent to \( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{2}{8}\times \dfrac{1}{2}?\)
\( \large \dfrac{7}{16}\) Hint: Multiplication comes before addition and subtraction in the order of operations. | |
\( \large \dfrac{1}{2}\) Hint: Addition and subtraction are of equal priority in the order of operations -- do them left to right. | |
\( \large \dfrac{3}{4}\) Hint: \( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{2}{8}\times \dfrac{1}{2}\)=\( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{1}{8}\)=\( \dfrac{3}{4}+-\dfrac{1}{8}+\dfrac{1}{8}\)=\( \dfrac{3}{4}\) | |
\( \large \dfrac{3}{16}\) Hint: Multiplication comes before addition and subtraction in the order of operations. |
Question 11 |
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 12 |
The function d(x) gives the result when 12 is divided by x. Which of the following is a graph of d(x)?
![]() Hint: d(x) is 12 divided by x, not x divided by 12. | |
![]() Hint: When x=2, what should d(x) be? | |
![]() Hint: When x=2, what should d(x) be? | |
![]() |
Question 13 |
In each expression below N represents a negative integer. Which expression could have a negative value?
\( \large {{N}^{2}}\) Hint: Squaring always gives a non-negative value. | |
\( \large 6-N\) Hint: A story problem for this expression is, if it was 6 degrees out at noon and N degrees out at sunrise, by how many degrees did the temperature rise by noon? Since N is negative, the answer to this question has to be positive, and more than 6. | |
\( \large -N\) Hint: If N is negative, then -N is positive | |
\( \large 6+N\) Hint: For example, if \(N=-10\), then \(6+N = -4\) |
Question 14 |
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 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 16 |
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 17 |
A class is using base-ten block to represent numbers. A large cube represents 1000, a flat represents 100, a rod represents 10, and a little cube represents 1. Which of these is not a correct representation for 2,347?
23 flats, 4 rods, 7 little cubesHint: Be sure you read the question carefully: 2300+40+7=2347 | |
2 large cubes, 3 flats, 47 rodsHint: 2000+300+470 \( \neq\) 2347 | |
2 large cubes, 34 rods, 7 little cubesHint: Be sure you read the question carefully: 2000+340+7=2347 | |
2 large cubes, 3 flats, 4 rods, 7 little cubesHint: Be sure you read the question carefully: 2000+300+40+7=2347 |
Question 18 |
Use the solution procedure below to answer the question that follows:
\( \large {\left( x+3 \right)}^{2}=10\)
\( \large \left( x+3 \right)\left( x+3 \right)=10\)
\( \large {x}^{2}+9=10\)
\( \large {x}^{2}+9-9=10-9\)
\( \large {x}^{2}=1\)
\( \large x=1\text{ or }x=-1\)
Which of the following is incorrect in the procedure shown above?
The commutative property is used incorrectly.Hint: The commutative property is \(a+b=b+a\) or \(ab=ba\). | |
The associative property is used incorrectly.Hint: The associative property is \(a+(b+c)=(a+b)+c\) or
\(a \times (b \times c)=(a \times b) \times c\). | |
Order of operations is done incorrectly. | |
The distributive property is used incorrectly.Hint: \((x+3)(x+3)=x(x+3)+3(x+3)\)=\(x^2+3x+3x+9.\) |
Question 19 |
Which of the following is equal to eleven billion four hundred thousand?
\( \large 11,400,000\) Hint: That's eleven million four hundred thousand. | |
\(\large11,000,400,000\) | |
\( \large11,000,000,400,000\) Hint: That's eleven trillion four hundred thousand (although with British conventions; this answer is correct, but in the US, it isn't). | |
\( \large 11,400,000,000\) Hint: That's eleven billion four hundred million |
Question 20 |
Four children randomly line up, single file. What is the probability that they are in height order, with the shortest child in front? All of the children are different heights.
\( \large \dfrac{1}{4}\) Hint: Try a simpler question with 3 children -- call them big, medium, and small -- and list all the ways they could line up. Then see how to extend your logic to the problem with 4 children. | |
\( \large \dfrac{1}{256}
\) Hint: Try a simpler question with 3 children -- call them big, medium, and small -- and list all the ways they could line up. Then see how to extend your logic to the problem with 4 children. | |
\( \large \dfrac{1}{16}\) Hint: Try a simpler question with 3 children -- call them big, medium, and small -- and list all the ways they could line up. Then see how to extend your logic to the problem with 4 children. | |
\( \large \dfrac{1}{24}\) Hint: The number of ways for the children to line up is \(4!=4 \times 3 \times 2 \times 1 =24\) -- there are 4 choices for who is first in line, then 3 for who is second, etc. Only one of these lines has the children in the order specified. |
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