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 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 2 |
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 3 |
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 4 |
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 5 |
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 6 |
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 7 |
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 8 |
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 9 |
There are 15 students for every teacher. Let t represent the number of teachers and let s represent the number of students. Which of the following equations is correct?
\( \large t=s+15\) Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation? | |
\( \large s=t+15\) Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation? | |
\( \large t=15s\) Hint: This is a really easy mistake to make, which comes from transcribing directly from English, "1 teachers equals 15 students." To see that it's wrong, plug in s=2; do you really need 30 teachers for 2 students? To avoid this mistake, insert the word "number," "Number of teachers equals 15 times number of students" is more clearly problematic. | |
\( \large s=15t\) |
Question 10 |
The picture below shows identical circles drawn on a piece of paper. The rectangle represents an index card that is blocking your view of \( \dfrac{3}{5}\) of the circles on the paper. How many circles are covered by the rectangle?

4Hint: The card blocks more than half of the circles, so this number is too small. | |
5Hint: The card blocks more than half of the circles, so this number is too small. | |
8Hint: The card blocks more than half of the circles, so this number is too small. | |
12Hint: 2/5 of the circles or 8 circles are showing. Thus 4 circles represent 1/5 of the circles, and \(4 \times 5=20\) circles represent 5/5 or all the circles. Thus 12 circles are hidden. |
Question 11 |
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 12 |
How many lines of reflective symmetry and how many centers of rotational symmetry does the parallelogram depicted below have?
4 lines of reflective symmetry, 1 center of rotational symmetry.Hint: Try cutting out a shape like this one from paper, and fold where you think the lines of reflective symmetry are (or put a mirror there). Do things line up as you thought they would? | |
2 lines of reflective symmetry, 1 center of rotational symmetry.Hint: Try cutting out a shape like this one from paper, and fold where you think the lines of reflective symmetry are (or put a mirror there). Do things line up as you thought they would? | |
0 lines of reflective symmetry, 1 center of rotational symmetry.Hint: The intersection of the diagonals is a center of rotational symmetry. There are no lines of reflective symmetry, although many people get confused about this fact (best to play with hands on examples to get a feel). Just fyi, the letter S also has rotational, but not reflective symmetry, and it's one that kids often write backwards. | |
2 lines of reflective symmetry, 0 centers of rotational symmetry.Hint: Try cutting out a shape like this one from paper. Trace onto another sheet of paper. See if there's a way to rotate the cut out shape (less than a complete turn) so that it fits within the outlines again. |
Question 13 |
Which of the numbers below is not equivalent to 4%?
\( \large \dfrac{1}{25}\) Hint: 1/25=4/100, so this is equal to 4% (be sure you read the question correctly). | |
\( \large \dfrac{4}{100}\) Hint: 4/100=4% (be sure you read the question correctly). | |
\( \large 0.4\) Hint: 0.4=40% so this is not equal to 4% | |
\( \large 0.04\) Hint: 0.04=4/100, so this is equal to 4% (be sure you read the question correctly). |
Question 14 |
Which of the following is closest to the height of a college student in centimeters?
1.6 cmHint: This is more the height of a Lego toy college student -- less than an inch! | |
16 cmHint: Less than knee high on most college students. | |
160 cmHint: Remember, a meter stick (a little bigger than a yard stick) is 100 cm. Also good to know is that 4 inches is approximately 10 cm. | |
1600 cmHint: This college student might be taller than some campus buildings! |
Question 15 |
Which of the numbers below is the decimal equivalent of \( \dfrac{3}{8}?\)
0.38Hint: If you are just writing the numerator next to the denominator then your technique is way off, but by coincidence your answer is close; try with 2/3 and 0.23 is nowhere near correct. | |
0.125Hint: This is 1/8, not 3/8. | |
0.375 | |
0.83Hint: 3/8 is less than a half, and 0.83 is more than a half, so they can't be equal. |
Question 16 |
Which of the following is the equation of a linear function?
\( \large y={{x}^{2}}+2x+7\) Hint: This is a quadratic function. | |
\( \large y={{2}^{x}}\) Hint: This is an exponential function. | |
\( \large y=\dfrac{15}{x}\) Hint: This is an inverse function. | |
\( \large y=x+(x+4)\) Hint: This is a linear function, y=2x+4, it's graph is a straight line with slope 2 and y-intercept 4. |
Question 17 |
Kendra is trying to decide which fraction is greater, \( \dfrac{4}{7}\) or \( \dfrac{5}{8}\). Which of the following answers shows the best reasoning?
\( \dfrac{4}{7}\) is \( \dfrac{3}{7}\)away from 1, and \( \dfrac{5}{8}\) is \( \dfrac{3}{8}\)away from 1. Since eighth‘s are smaller than seventh‘s, \( \dfrac{5}{8}\) is closer to 1, and is the greater of the two fractions. | |
\( 7-4=3\) and \( 8-5=3\), so the fractions are equal.Hint: Not how to compare fractions. By this logic, 1/2 and 3/4 are equal, but 1/2 and 2/4 are not. | |
\( 4\times 8=32\) and \( 7\times 5=35\). Since \( 32<35\) , \( \dfrac{5}{8}<\dfrac{4}{7}\)Hint: Starts out as something that works, but the conclusion is wrong. 4/7 = 32/56 and 5/8 = 35/56. The cross multiplication gives the numerators, and 35/56 is bigger. | |
\( 4<5\) and \( 7<8\), so \( \dfrac{4}{7}<\dfrac{5}{8}\)Hint: Conclusion is correct, logic is wrong. With this reasoning, 1/2 would be less than 2/100,000. |
Question 18 |
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 19 |
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 20 |
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. |
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