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MTEL General Curriculum Mathematics Practice
Question 1 |
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 2 |
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 3 |
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 4 |
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 5 |
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 6 |
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 7 |
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 8 |
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 9 |
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 10 |
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 11 |
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 12 |
What is the perimeter of a right triangle with legs of lengths x and 2x?
\( \large 6x\) Hint: Use the Pythagorean Theorem. | |
\( \large 3x+5{{x}^{2}}\) Hint: Don't forget to take square roots when you use the Pythagorean Theorem. | |
\( \large 3x+\sqrt{5}{{x}^{2}}\) Hint: \(\sqrt {5 x^2}\) is not \(\sqrt {5}x^2\). | |
\( \large 3x+\sqrt{5}{{x}^{{}}}\) Hint: To find the hypotenuse, h, use the Pythagorean Theorem: \(x^2+(2x)^2=h^2.\) \(5x^2=h^2,h=\sqrt{5}x\). The perimeter is this plus x plus 2x. |
Question 13 |
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 14 |
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 15 |
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 16 |
The student used a method that worked for this problem and can be generalized to any subtraction problem.Hint: Note that this algorithm is taught as the "standard" algorithm in much of Europe (it's where the term "borrowing" came from -- you borrow on top and "pay back" on the bottom). | |
The student used a method that worked for this problem and that will work for any subtraction problem that only requires one regrouping; it will not work if more regrouping is required.Hint: Try some more examples. | |
The student used a method that worked for this problem and will work for all three-digit subtraction problems, but will not work for larger problems.Hint: Try some more examples. | |
The student used a method that does not work. The student made two mistakes that cancelled each other out and was lucky to get the right answer for this problem.Hint: Remember, there are many ways to do subtraction; there is no one "right" algorithm. |
Question 17 |
The histogram below shows the number of pairs of footware owned by a group of college students.
Which of the following statements can be inferred from the graph above?
The median number of pairs of footware owned is between 50 and 60 pairs.Hint: The same number of data points are less than the median as are greater than the median -- but on this histogram, clearly more than half the students own less than 50 pairs of shoes, so the median is less than 50. | |
The mode of the number of pairs of footware owned is 20.Hint: The mode is the most common number of pairs of footwear owned. We can't tell it from this histogram because each bar represents 10 different numbers-- perhaps 8 students each own each number from 10 to 19, but 40 students own exactly 6 pairs of shoes.... or perhaps not.... | |
The mean number of pairs of footware owned is less than the median number of pairs of footware owned.Hint: This is a right skewed distribution, and so the mean is bigger than the median -- the few large values on the right pull up the mean, but have little effect on the median. | |
The median number of pairs of footware owned is between 10 and 20.Hint: There are approximately 230 students represented in this survey, and the 41st through 120th lowest values are between 10 and 20 -- thus the middle value is in that range. |
Question 18 |
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 19 |
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 20 |
Here is a mental math strategy for computing 26 x 16:
Step 1: 100 x 16 = 1600
Step 2: 25 x 16 = 1600 ÷· 4 = 400
Step 3: 26 x 16 = 400 + 16 = 416
Which property best justifies Step 3 in this strategy?
Commutative Property.Hint: For addition, the commutative property is \(a+b=b+a\) and for multiplication it's \( a \times b = b \times a\). | |
Associative Property.Hint: For addition, the associative property is \((a+b)+c=a+(b+c)\) and for multiplication it's \((a \times b) \times c=a \times (b \times c)\) | |
Identity Property.Hint: 0 is the additive identity, because \( a+0=a\) and 1 is the multiplicative identity because \(a \times 1=a\). The phrase "identity property" is not standard. | |
Distributive Property.Hint: \( (25+1) \times 16 = 25 \times 16 + 1 \times 16 \). This is an example of the distributive property of multiplication over addition. |
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