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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
Question 1 |
The polygon depicted below is drawn on dot paper, with the dots spaced 1 unit apart. What is the perimeter of the polygon?

\( \large 18+\sqrt{2} \text{ units}\) Hint: Be careful with the Pythagorean Theorem. | |
\( \large 18+2\sqrt{2}\text{ units}\) Hint: There are 13 horizontal or vertical 1 unit segments. The longer diagonal is the hypotenuse of a 3-4-5 right triangle, so its length is 5 units. The shorter diagonal is the hypotenuse of a 45-45-90 right triangle with side 2, so its hypotenuse has length \(2 \sqrt{2}\). | |
\( \large 18 \text{ units}
\) Hint: Use the Pythagorean Theorem to find the lengths of the diagonal segments. | |
\( \large 20 \text{ units}\) Hint: Use the Pythagorean Theorem to find the lengths of the diagonal segments. |
Question 2 |
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. |
Question 3 |
A sales companies pays its representatives $2 for each item sold, plus 40% of the price of the item. The rest of the money that the representatives collect goes to the company. All transactions are in cash, and all items cost $4 or more. If the price of an item in dollars is p, which expression represents the amount of money the company collects when the item is sold?
\( \large \dfrac{3}{5}p-2\) Hint: The company gets 3/5=60% of the price, minus the $2 per item. | |
\( \large \dfrac{3}{5}\left( p-2 \right)\) Hint: This is sensible, but not what the problem states. | |
\( \large \dfrac{2}{5}p+2\) Hint: The company pays the extra $2; it doesn't collect it. | |
\( \large \dfrac{2}{5}p-2\) Hint: This has the company getting 2/5 = 40% of the price of each item, but that's what the representative gets. |
Question 4 |
Which of the lists below is in order from least to greatest value?
\( \large -0.044,\quad -0.04,\quad 0.04,\quad 0.044\) Hint: These are easier to compare if you add trailing zeroes (this is finding a common denominator) -- all in thousandths, -0.044, -0.040,0 .040, 0.044. The middle two numbers, -0.040 and 0.040 can be modeled as owing 4 cents and having 4 cents. The outer two numbers are owing or having a bit more. | |
\( \large -0.04,\quad -0.044,\quad 0.044,\quad 0.04\) Hint: 0.04=0.040, which is less than 0.044. | |
\( \large -0.04,\quad -0.044,\quad 0.04,\quad 0.044\) Hint: -0.04=-0.040, which is greater than \(-0.044\). | |
\( \large -0.044,\quad -0.04,\quad 0.044,\quad 0.04\) Hint: 0.04=0.040, which is less than 0.044. |
Question 5 |
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 6 |
In which table below is y a function of x?
![]() Hint: If x=3, y can have two different values, so it's not a function. | |
![]() Hint: If x=3, y can have two different values, so it's not a function. | |
![]() Hint: If x=1, y can have different values, so it's not a function. | |
![]() Hint: Each value of x always corresponds to the same value of y. |
Question 7 |
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 8 |
The "houses" below are made of toothpicks and gum drops.
Which of the following does not represent the number of gumdrops in a row of h houses?
\( \large 2+3h\) Hint: Think of this as start with 2 gumdrops on the left wall, and then add 3 gumdrops for each house. | |
\( \large 5+3(h-1)\) Hint: Think of this as start with one house, and then add 3 gumdrops for each of the other h-1 houses. | |
\( \large h+(h+1)+(h+1)\) Hint: Look at the gumdrops in 3 rows: h gumdrops for the "rooftops," h+1 for the tops of the vertical walls, and h+1 for the floors. | |
\( \large 5+3h\) Hint: This one is not a correct equation (which makes it the correct answer!). Compare to choice A. One of them has to be wrong, as they differ by 3. |
Question 9 |
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 10 |
The column below consists of two cubes and a cylinder. The cylinder has diameter y, which is also the length of the sides of each cube. The total height of the column is 5y. Which of the formulas below gives the volume of the column?
\( \large 2{{y}^{3}}+\dfrac{3\pi {{y}^{3}}}{4}\) Hint: The cubes each have volume \(y^3\). The cylinder has radius \(\dfrac{y}{2}\) and height \(3y\). The volume of a cylinder is \(\pi r^2 h=\pi ({\dfrac{y}{2}})^2(3y)=\dfrac{3\pi {{y}^{3}}}{4}\). Note that the volume of a cylinder is analogous to that of a prism -- area of the base times height. | |
\( \large 2{{y}^{3}}+3\pi {{y}^{3}}\) Hint: y is the diameter of the circle, not the radius. | |
\( \large {{y}^{3}}+5\pi {{y}^{3}}\) Hint: Don't forget to count both cubes. | |
\( \large 2{{y}^{3}}+\dfrac{3\pi {{y}^{3}}}{8}\) Hint: Make sure you know how to find the volume of a cylinder. |
Question 11 |
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 12 |
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 13 |
What is the probability that two randomly selected people were born on the same day of the week? Assume that all days are equally probable.
\( \large \dfrac{1}{7}\) Hint: It doesn't matter what day the first person was born on. The probability that the second person will match is 1/7 (just designate one person the first and the other the second). Another way to look at it is that if you list the sample space of all possible pairs, e.g. (Wed, Sun), there are 49 such pairs, and 7 of them are repeats of the same day, and 7/49=1/7. | |
\( \large \dfrac{1}{14}\) Hint: What would be the sample space here? Ie, how would you list 14 things that you pick one from? | |
\( \large \dfrac{1}{42}\) Hint: If you wrote the seven days of the week on pieces of paper and put the papers in a jar, this would be the probability that the first person picked Sunday and the second picked Monday from the jar -- not the same situation. | |
\( \large \dfrac{1}{49}\) Hint: This is the probability that they are both born on a particular day, e.g. Sunday. |
Question 14 |
A car is traveling at 60 miles per hour. Which of the expressions below could be used to compute how many feet the car travels in 1 second? Note that 1 mile = 5,280 feet.
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot 5280\dfrac{\text{feet}}{\text{mile}}\cdot 60\dfrac{\text{minutes}}{\text{hour}}\cdot 60\dfrac{\text{seconds}}{\text{minute}}
\) Hint: This answer is not in feet/second. | |
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot 5280\dfrac{\text{feet}}{\text{mile}}\cdot \dfrac{1}{60}\dfrac{\text{hour}}{\text{minutes}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}
\) Hint: This is the only choice where the answer is in feet per second and the unit conversions are correct. | |
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot \dfrac{1}{5280}\dfrac{\text{foot}}{\text{miles}}\cdot 60\dfrac{\text{hours}}{\text{minute}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}\) Hint: Are there really 60 hours in a minute? | |
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot \dfrac{1}{5280}\dfrac{\text{mile}}{\text{feet}}\cdot 60\dfrac{\text{minutes}}{\text{hour}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}\) Hint: This answer is not in feet/second. |
Question 15 |
An above-ground swimming pool is in the shape of a regular hexagonal prism, is one meter high, and holds 65 cubic meters of water. A second pool has a base that is also a regular hexagon, but with sides twice as long as the sides in the first pool. This second pool is also one meter high. How much water will the second pool hold?
\( \large 65\text{ }{{\text{m}}^{3}}\) Hint: A bigger pool would hold more water. | |
\( \large 65\cdot 2\text{ }{{\text{m}}^{3}}\) Hint: Try a simpler example, say doubling the sides of the base of a 1 x 1 x 1 cube. | |
\( \large 65\cdot 4\text{ }{{\text{m}}^{3}}\) Hint: If we think of the pool as filled with 1 x 1 x 1 cubes (and some fractions of cubes), then scaling to the larger pool changes each 1 x 1 x 1 cube to a 2 x 2 x 1 prism, or multiplies volume by 4. | |
\( \large 65\cdot 8\text{ }{{\text{m}}^{3}}\) Hint: Try a simpler example, say doubling the sides of the base of a 1 x 1 x 1 cube. |
Question 16 |
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 17 |
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 18 |
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 19 |
A solution requires 4 ml of saline for every 7 ml of medicine. How much saline would be required for 50 ml of medicine?
\( \large 28 \dfrac{4}{7}\) ml Hint: 49 ml of medicine requires 28 ml of saline. The extra ml of saline requires 4 ml saline/ 7 ml medicine = 4/7 ml saline per 1 ml medicine. | |
\( \large 28 \dfrac{1}{4}\) ml Hint: 49 ml of medicine requires 28 ml of saline. How much saline does the extra ml require? | |
\( \large 28 \dfrac{1}{7}\) ml Hint: 49 ml of medicine requires 28 ml of saline. How much saline does the extra ml require? | |
\( \large 87.5\) ml Hint: 49 ml of medicine requires 28 ml of saline. How much saline does the extra ml require? |
Question 20 |
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. |
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