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
Question 1 |
The "houses" below are made of toothpicks and gum drops.
How many toothpicks are there in a row of 53 houses?
212Hint: Can the number of toothpicks be even? | |
213Hint: One way to see this is that every new "house" adds 4 toothpicks to the leftmost vertical toothpick -- so the total number is 1 plus 4 times the number of "houses." There are many other ways to look at the problem too. | |
217Hint: Try your strategy with a smaller number of "houses" so you can count and find your mistake. | |
265Hint: Remember that the "houses" overlap some walls. |
Question 2 |
The pattern below consists of a row of black squares surrounded by white squares.
How many white squares would surround a row of 157 black squares?
314Hint: Try your procedure on a smaller number that you can count to see where you made a mistake. | |
317Hint: Are there ever an odd number of white squares? | |
320Hint: One way to see this is that there are 6 tiles on the left and right ends, and the rest of the white tiles are twice the number of black tiles (there are many other ways to look at it too). | |
322Hint: Try your procedure on a smaller number that you can count to see where you made a mistake. |
Question 3 |
Exactly one of the numbers below is a prime number. Which one is it?
\( \large511 \) Hint: Divisible by 7. | |
\( \large517\) Hint: Divisible by 11. | |
\( \large519\) Hint: Divisible by 3. | |
\( \large521\) |
Question 4 |
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 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 |
Use the samples of a student's work below to answer the question that follows:
\( \large \dfrac{2}{3}\times \dfrac{3}{4}=\dfrac{4\times 2}{3\times 3}=\dfrac{8}{9}\) \( \large \dfrac{2}{5}\times \dfrac{7}{7}=\dfrac{7\times 2}{5\times 7}=\dfrac{2}{5}\) \( \large \dfrac{7}{6}\times \dfrac{3}{4}=\dfrac{4\times 7}{6\times 3}=\dfrac{28}{18}=\dfrac{14}{9}\)Which of the following best describes the mathematical validity of the algorithm the student is using?
It is not valid. It never produces the correct answer.Hint: In the middle example,the answer is correct. | |
It is not valid. It produces the correct answer in a few special cases, but it‘s still not a valid algorithm.Hint: Note that this algorithm gives a/b divided by c/d, not a/b x c/d, but some students confuse multiplication and cross-multiplication. If a=0 or if c/d =1, division and multiplication give the same answer. | |
It is valid if the rational numbers in the multiplication problem are in lowest terms.Hint: Lowest terms is irrelevant. | |
It is valid for all rational numbers.Hint: Can't be correct as the first and last examples have the wrong answers. |
Question 7 |
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 8 |
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 9 |
A homeowner is planning to tile the kitchen floor with tiles that measure 6 inches by 8 inches. The kitchen floor is a rectangle that measures 10 ft by 12 ft, and there are no gaps between the tiles. How many tiles does the homeowner need?
30Hint: The floor is 120 sq feet, and the tiles are smaller than 1 sq foot. Also, remember that 1 sq foot is 12 \(\times\) 12=144 sq inches. | |
120Hint: The floor is 120 sq feet, and the tiles are smaller than 1 sq foot. | |
300Hint: Recheck your calculations. | |
360Hint: One way to do this is to note that 6 inches = 1/2 foot and 8 inches = 2/3 foot, so the area of each tile is 1/2 \(\times\) 2/3=1/3 sq foot, or each square foot of floor requires 3 tiles. The area of the floor is 120 square feet. Note that the tiles would fit evenly oriented in either direction, parallel to the walls. |
Question 10 |
At a school fundraising event, people can buy a ticket to spin a spinner like the one below. The region that the spinner lands in tells which, if any, prize the person wins.
If 240 people buy tickets to spin the spinner, what is the best estimate of the number of keychains that will be given away?
40Hint: "Keychain" appears on the spinner twice. | |
80Hint: The probability of getting a keychain is 1/3, and so about 1/3 of the time the spinner will win. | |
100Hint: What is the probability of winning a keychain? | |
120Hint: That would be the answer for getting any prize, not a keychain specifically. |
Question 11 |
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 12 |
The speed of sound in dry air at 68 degrees F is 343.2 meters per second. Which of the expressions below could be used to compute the number of kilometers that a sound wave travels in 10 minutes (in dry air at 68 degrees F)?
\( \large 343.2\times 60\times 10\) Hint: In kilometers, not meters. | |
\( \large 343.2\times 60\times 10\times \dfrac{1}{1000}\) Hint: Units are meters/sec \(\times\) seconds/minute \(\times\) minutes \(\times\) kilometers/meter, and the answer is in kilometers. | |
\( \large 343.2\times \dfrac{1}{60}\times 10\) Hint: Include units and make sure answer is in kilometers. | |
\( \large 343.2\times \dfrac{1}{60}\times 10\times \dfrac{1}{1000}\) Hint: Include units and make sure answer is in kilometers. |
Question 13 |
Use the graph below to answer the question that follows.
If the polygon shown above is reflected about the y axis and then rotated 90 degrees clockwise about the origin, which of the following graphs is the result?
![]() Hint: Try following the point (1,4) to see where it goes after each transformation. | |
![]() | |
Hint: Make sure you're reflecting in the correct axis. | |
![]() Hint: Make sure you're rotating the correct direction. |
Question 14 |
A cylindrical soup can has diameter 7 cm and height 11 cm. The can holds g grams of soup. How many grams of the same soup could a cylindrical can with diameter 14 cm and height 33 cm hold?
\( \large 6g\) Hint: You must scale in all three dimensions. | |
\( \large 12g\) Hint: Height is multiplied by 3, and diameter and radius are multiplied by 2. Since the radius is squared, final result is multiplied by \(2^2\times 3=12\). | |
\( \large 18g\) Hint: Don't square the height scale factor. | |
\( \large 36g\) Hint: Don't square the height scale factor. |
Question 15 |
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 16 |
A family went on a long car trip. Below is a graph of how far they had driven at each hour.
Which of the following is closest to their average speed driving on the trip?
\( \large d=20t\) Hint: Try plugging t=7 into the equation, and see how it matches the graph. | |
\( \large d=30t\) Hint: Try plugging t=7 into the equation, and see how it matches the graph. | |
\( \large d=40t\) | |
\( \large d=50t\) Hint: Try plugging t=7 into the equation, and see how it matches the graph. |
Question 17 |
A family on vacation drove the first 200 miles in 4 hours and the second 200 miles in 5 hours. Which expression below gives their average speed for the entire trip?
\( \large \dfrac{200+200}{4+5}\) Hint: Average speed is total distance divided by total time. | |
\( \large \left( \dfrac{200}{4}+\dfrac{200}{5} \right)\div 2\) Hint: This seems logical, but the problem is that it weights the first 4 hours and the second 5 hours equally, when each hour should get the same weight in computing the average speed. | |
\( \large \dfrac{200}{4}+\dfrac{200}{5} \) Hint: This would be an average of 90 miles per hour! | |
\( \large \dfrac{400}{4}+\dfrac{400}{5} \) Hint: This would be an average of 180 miles per hour! Even a family of race car drivers probably doesn't have that average speed on a vacation! |
Question 18 |
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 19 |
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 20 |
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. |
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