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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 |
Use the expression below to answer the question that follows:
\( \large \dfrac{\left( 7,154 \right)\times \left( 896 \right)}{216}\)
Which of the following is the best estimate of the expression above?
2,000Hint: The answer is bigger than 7,000. | |
20,000Hint: Estimate 896/216 first. | |
3,000Hint: The answer is bigger than 7,000. | |
30,000Hint: \( \dfrac{896}{216} \approx 4\) and \(7154 \times 4\) is over 28,000, so this answer is closest. |
Question 3 |
Use the expression below to answer the question that follows.
\( \large \dfrac{\left( 4\times {{10}^{3}} \right)\times \left( 3\times {{10}^{4}} \right)}{6\times {{10}^{6}}}\)
Which of the following is equivalent to the expression above?
2Hint: \(10^3 \times 10^4=10^7\), and note that if you're guessing when the answers are so closely related, you're generally better off guessing one of the middle numbers. | |
20Hint: \( \dfrac{\left( 4\times {{10}^{3}} \right)\times \left( 3\times {{10}^{4}} \right)}{6\times {{10}^{6}}}=\dfrac {12 \times {{10}^{7}}}{6\times {{10}^{6}}}=\)\(2 \times {{10}^{1}}=20 \) | |
200Hint: \(10^3 \times 10^4=10^7\) | |
2000Hint: \(10^3 \times 10^4=10^7\), and note that if you're guessing when the answers are so closely related, you're generally better off guessing one of the middle numbers. |
Question 4 |
The following story situations model \( 12\div 3\):
I) Jack has 12 cookies, which he wants to share equally between himself and two friends. How many cookies does each person get?
II) Trent has 12 cookies, which he wants to put into bags of 3 cookies each. How many bags can he make?
III) Cicely has $12. Cookies cost $3 each. How many cookies can she buy?
Which of these questions illustrate the same model of division, either partitive (partioning) or measurement (quotative)?
I and II | |
I and III | |
II and IIIHint: Problem I is partitive (or partitioning or sharing) -- we put 12 objects into 3 groups. Problems II and III are quotative (or measurement) -- we put 12 objects in groups of 3. | |
All three problems model the same meaning of division |
Question 5 |
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 6 |
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. |
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 8 |
A biology class requires a lab fee, which is a whole number of dollars, and the same amount for all students. On Monday the instructor collected $70 in fees, on Tuesday she collected $126, and on Wednesday she collected $266. What is the largest possible amount the fee could be?
$2Hint: A possible fee, but not the largest possible fee. Check the other choices to see which are factors of all three numbers. | |
$7Hint: A possible fee, but not the largest possible fee. Check the other choices to see which are factors of all three numbers. | |
$14Hint: This is the greatest common factor of 70, 126, and 266. | |
$70Hint: Not a factor of 126 or 266, so couldn't be correct. |
Question 9 |
Here is a student's work on several multiplication problems:
For which of the following problems is this student most likely to get the correct solution, even though he is using an incorrect algorithm?
58 x 22Hint: This problem involves regrouping, which the student does not do correctly. | |
16 x 24Hint: This problem involves regrouping, which the student does not do correctly. | |
31 x 23Hint: There is no regrouping with this problem. | |
141 x 32Hint: This problem involves regrouping, which the student does not do correctly. |
Question 10 |
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 11 |
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?

Question 12 |
Which of the following is equivalent to
\( \large A-B+C\div D\times E\)?
\( \large A-B-\dfrac{C}{DE}
\) Hint: In the order of operations, multiplication and division have the same priority, so do them left to right; same with addition and subtraction. | |
\( \large A-B+\dfrac{CE}{D}\) 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. | |
\( \large \dfrac{AE-BE+CE}{D}\) Hint: Use order of operations, don't just compute left to right. | |
\( \large A-B+\dfrac{C}{DE}\) Hint: In the order of operations, multiplication and division have the same priority, so do them left to right |
Question 13 |
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 14 |
How many factors does 80 have?
\( \large8\) Hint: Don't forget 1 and 80. | |
\( \large9\) Hint: Only perfect squares have an odd number of factors -- otherwise factors come in pairs. | |
\( \large10\) Hint: 1,2,4,5,8,10,16,20,40,80 | |
\( \large12\) Hint: Did you count a number twice? Include a number that isn't a factor? |
Question 15 |
Below are front, side, and top views of a three-dimensional solid.
Which of the following could be the solid shown above?
A sphereHint: All views would be circles. | |
A cylinder | |
A coneHint: Two views would be triangles, not rectangles. | |
A pyramidHint: How would one view be a circle? |
Question 16 |
Solve for x: \(\large 4-\dfrac{2}{3}x=2x\)
\( \large x=3\) Hint: Try plugging x=3 into the equation. | |
\( \large x=-3\) Hint: Left side is positive, right side is negative when you plug this in for x. | |
\( \large x=\dfrac{3}{2}\) Hint: One way to solve: \(4=\dfrac{2}{3}x+2x\) \(=\dfrac{8}{3}x\).\(x=\dfrac{3 \times 4}{8}=\dfrac{3}{2}\). Another way is to just plug x=3/2 into the equation and see that each side equals 3 -- on a multiple choice test, you almost never have to actually solve for x. | |
\( \large x=-\dfrac{3}{2}\) Hint: Left side is positive, right side is negative when you plug this in for x. |
Question 17 |
Use the graph below to answer the question that follows.
Which of the following is a correct equation for the graph of the line depicted above?
\( \large y=-\dfrac{1}{2}x+2\) Hint: The slope is -1/2 and the y-intercept is 2. You can also try just plugging in points. For example, this is the only choice that gives y=1 when x=2. | |
\( \large 4x=2y\) Hint: This line goes through (0,0); the graph above does not. | |
\( \large y=x+2\) Hint: The line pictured has negative slope. | |
\( \large y=-x+2\) Hint: Try plugging x=4 into this equation and see if that point is on the graph above. |
Question 18 |
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 19 |
On a map the distance from Boston to Detroit is 6 cm, and these two cities are 702 miles away from each other. Assuming the scale of the map is the same throughout, which answer below is closest to the distance between Boston and San Francisco on the map, given that they are 2,708 miles away from each other?
21 cmHint: How many miles would correspond to 24 cm on the map? Try adjusting from there. | |
22 cmHint: How many miles would correspond to 24 cm on the map? Try adjusting from there. | |
23 cmHint: One way to solve this without a calculator is to note that 4 groups of 6 cm is 2808 miles, which is 100 miles too much. Then 100 miles would be about 1/7 th of 6 cm, or about 1 cm less than 24 cm. | |
24 cmHint: 4 groups of 6 cm is over 2800 miles on the map, which is too much. |
Question 20 |
The letters A, B, and C represent digits (possibly equal) in the twelve digit number x=111,111,111,ABC. For which values of A, B, and C is x divisible by 40?
\( \large A = 3, B = 2, C=0\) Hint: Note that it doesn't matter what the first 9 digits are, since 1000 is divisible by 40, so DEF,GHI,JKL,000 is divisible by 40 - we need to check the last 3. | |
\( \large A = 0, B = 0, C=4\) Hint: Not divisible by 10, since it doesn't end in 0. | |
\( \large A = 4, B = 2, C=0\) Hint: Divisible by 10 and by 4, but not by 40, as it's not divisible by 8. Look at 40 as the product of powers of primes -- 8 x 5, and check each. To check 8, either check whether 420 is divisible by 8, or take ones place + twice tens place + 4 * hundreds place = 18, which is not divisible by 8. | |
\( \large A =1, B=0, C=0\) Hint: Divisible by 10 and by 4, but not by 40, as it's not divisible by 8. Look at 40 as the product of powers of primes -- 8 x 5, and check each. To check 8, either check whether 100 is divisible by 8, or take ones place + twice tens place + 4 * hundreds place = 4, which is not divisible by 8. |
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