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MTEL General Curriculum Mathematics Practice
Question 1 |
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 2 |
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 3 |
Aya and Kendra want to estimate the height of a tree. On a sunny day, Aya measures Kendra's shadow as 3 meters long, and Kendra measures the tree's shadow as 15 meters long. Kendra is 1.5 meters tall. How tall is the tree?
7.5 metersHint: Here is a picture, note that the large and small right triangles are similar: ![]() One way to do the problem is to note that there is a dilation (scale) factor of 5 on the shadows, so there must be that factor on the heights too. Another way is to note that the shadows are twice as long as the heights. | |
22.5 metersHint: Draw a picture. | |
30 metersHint: Draw a picture. | |
45 metersHint: Draw a picture. |
Question 4 |
The prime factorization of n can be written as n=pqr, where p, q, and r are distinct prime numbers. How many factors does n have, including 1 and itself?
\( \large3\) Hint: 1, p, q, r, and pqr are already 5, so this isn't enough. You might try plugging in p=2, q=3, and r=5 to help with this problem. | |
\( \large5\) Hint: Don't forget pq, etc. You might try plugging in p=2, q=3, and r=5 to help with this problem. | |
\( \large6\) Hint: You might try plugging in p=2, q=3, and r=5 to help with this problem. | |
\( \large8\) Hint: 1, p, q, r, pq, pr, qr, pqr. |
Question 5 |
In the triangle below, \(\overline{AC}\cong \overline{AD}\cong \overline{DE}\) and \(m\angle CAD=100{}^\circ \). What is \(m\angle DAE\)?

\( \large 20{}^\circ \) Hint: Angles ACD and ADC are congruent since they are base angles of an isosceles triangle. Since the angles of a triangle sum to 180, they sum to 80, and they are 40 deg each. Thus angle ADE is 140 deg, since it makes a straight line with angle ADC. Angles DAE and DEA are base angles of an isosceles triangle and thus congruent-- they sum to 40 deg, so are 20 deg each. | |
\( \large 25{}^\circ \) Hint: If two sides of a triangle are congruent, then it's isosceles, and the base angles of an isosceles triangle are equal. | |
\( \large 30{}^\circ \) Hint: If two sides of a triangle are congruent, then it's isosceles, and the base angles of an isosceles triangle are equal. | |
\( \large 40{}^\circ \) Hint: Make sure you're calculating the correct angle. |
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 |
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 8 |
Which of the following is not possible?
An equiangular triangle that is not equilateral.Hint: The AAA property of triangles states that all triangles with corresponding angles congruent are similar. Thus all triangles with three equal angles are similar, and are equilateral. | |
An equiangular quadrilateral that is not equilateral.Hint: A rectangle is equiangular (all angles the same measure), but if it's not a square, it's not equilateral (all sides the same length). | |
An equilateral quadrilateral that is not equiangular.Hint: This rhombus has equal sides, but it doesn't have equal angles: ![]() | |
An equiangular hexagon that is not equilateral.Hint: This hexagon has equal angles, but it doesn't have equal sides: ![]() |
Question 9 |
A teacher has a list of all the countries in the world and their populations in March 2012. She is going to have her students use technology to compute the mean and median of the numbers on the list. Which of the following statements is true?
The teacher can be sure that the mean and median will be the same without doing any computation.Hint: Does this make sense? How likely is it that the mean and median of any large data set will be the same? | |
The teacher can be sure that the mean is bigger than the median without doing any computation.Hint: This is a skewed distribution, and very large countries like China and India contribute huge numbers to the mean, but are counted the same as small countries like Luxembourg in the median (the same thing happens w/data on salaries, where a few very high income people tilt the mean -- that's why such data is usually reported as medians). | |
The teacher can be sure that the median is bigger than the mean without doing any computation.Hint: Think about a set of numbers like 1, 2, 3, 4, 10,000 -- how do the mean/median compare? How might that relate to countries of the world? | |
There is no way for the teacher to know the relative size of the mean and median without computing them.Hint: Knowing the shape of the distribution of populations does give us enough info to know the relative size of the mean and median, even without computing them. |
Question 10 |
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 11 |
Use the four figures below to answer the question that follows:
How many of the figures pictured above have at least one line of reflective symmetry?
\( \large 1\) | |
\( \large 2\) Hint: The ellipse has 2 lines of reflective symmetry (horizontal and vertical, through the center) and the triangle has 3. The other two figures have rotational symmetry, but not reflective symmetry. | |
\( \large 3\) | |
\( \large 4\) Hint: All four have rotational symmetry, but not reflective symmetry. |
Question 12 |
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 13 |
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 14 |
Which of the following values of x satisfies the inequality \( \large \left| {{(x+2)}^{3}} \right|<3?\)
\( \large x=-3\) Hint: \( \left| {{(-3+2)}^{3}} \right|\)=\( \left | {(-1)}^3 \right | \)=\( \left | -1 \right |=1 \) . | |
\( \large x=0\) Hint: \( \left| {{(0+2)}^{3}} \right|\)=\( \left | {2}^3 \right | \)=\( \left | 8 \right | \) =\( 8\) | |
\( \large x=-4\) Hint: \( \left| {{(-4+2)}^{3}} \right|\)=\( \left | {(-2)}^3 \right | \)=\( \left | -8 \right | \) =\( 8\) | |
\( \large x=1\) Hint: \( \left| {{(1+2)}^{3}} \right|\)=\( \left | {3}^3 \right | \)=\( \left | 27 \right | \) = \(27\) |
Question 15 |
In each expression below N represents a negative integer. Which expression could have a negative value?
\( \large {{N}^{2}}\) Hint: Squaring always gives a non-negative value. | |
\( \large 6-N\) Hint: A story problem for this expression is, if it was 6 degrees out at noon and N degrees out at sunrise, by how many degrees did the temperature rise by noon? Since N is negative, the answer to this question has to be positive, and more than 6. | |
\( \large -N\) Hint: If N is negative, then -N is positive | |
\( \large 6+N\) Hint: For example, if \(N=-10\), then \(6+N = -4\) |
Question 16 |
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 17 |
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 18 |
The window glass below has the shape of a semi-circle on top of a square, where the side of the square has length x. It was cut from one piece of glass.

What is the perimeter of the window glass?
\( \large 3x+\dfrac{\pi x}{2}\) Hint: By definition, \(\pi\) is the ratio of the circumference of a circle to its diameter; thus the circumference is \(\pi d\). Since we have a semi-circle, its perimeter is \( \dfrac{1}{2} \pi x\). Only 3 sides of the square contribute to the perimeter. | |
\( \large 3x+2\pi x\) Hint: Make sure you know how to find the circumference of a circle. | |
\( \large 3x+\pi x\) Hint: Remember it's a semi-circle, not a circle. | |
\( \large 4x+2\pi x\) Hint: Only 3 sides of the square contribute to the perimeter. |
Question 19 |
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 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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