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 |
If two fair coins are flipped, what is the probability that one will come up heads and the other tails?
\( \large \dfrac{1}{4}\) Hint: Think of the coins as a penny and a dime, and list all possibilities. | |
\( \large \dfrac{1}{3} \) Hint: This is a very common misconception. There are three possible outcomes -- both heads, both tails, and one of each -- but they are not equally likely. Think of the coins as a penny and a dime, and list all possibilities. | |
\( \large \dfrac{1}{2}\) Hint: The possibilities are HH, HT, TH, TT, and all are equally likely. Two of the four have one of each coin, so the probability is 2/4=1/2. | |
\( \large \dfrac{3}{4}\) Hint: Think of the coins as a penny and a dime, and list all possibilities. |
Question 2 |
Which of the following is equivalent to
\( \large A-B+C\div D\times E\)?
\( \large A-B-\dfrac{C}{DE}
\) Hint: In the order of operations, multiplication and division have the same priority, so do them left to right; same with addition and subtraction. | |
\( \large A-B+\dfrac{CE}{D}\) Hint: In practice, you're better off using parentheses than writing an expression like the one in the question. The PEMDAS acronym that many people memorize is misleading. Multiplication and division have equal priority and are done left to right. They have higher priority than addition and subtraction. Addition and subtraction also have equal priority and are done left to right. | |
\( \large \dfrac{AE-BE+CE}{D}\) Hint: Use order of operations, don't just compute left to right. | |
\( \large A-B+\dfrac{C}{DE}\) Hint: In the order of operations, multiplication and division have the same priority, so do them left to right |
Question 3 |
The histogram below shows the frequency of a class's scores on a 4 question quiz.
What was the mean score on the quiz?
\( \large 2.75\) Hint: There were 20 students who took the quiz. Total points earned: \(2 \times 1+6 \times 2+ 7\times 3+5 \times 4=55\), and 55/20 = 2.75. | |
\( \large 2\) Hint: How many students are there total? Did you count them all? | |
\( \large 3\) Hint: How many students are there total? Did you count them all? Be sure you're finding the mean, not the median or the mode. | |
\( \large 2.5\) Hint: How many students are there total? Did you count them all? Don't just take the mean of 1, 2, 3, 4 -- you have to weight them properly. |
Question 4 |
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 5 |
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 6 |
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 7 |
The Americans with Disabilties Act (ADA) regulations state that the maximum slope for a wheelchair ramp in new construction is 1:12, although slopes between 1:16 and 1:20 are preferred. The maximum rise for any run is 30 inches. The graph below shows the rise and runs of four different wheelchair ramps. Which ramp is in compliance with the ADA regulations for new construction?

AHint: Rise is more than 30 inches. | |
BHint: Run is almost 24 feet, so rise can be almost 2 feet. | |
CHint: Run is 12 feet, so rise can be at most 1 foot. | |
DHint: Slope is 1:10 -- too steep. |
Question 8 |
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 9 |
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 10 |
Each individual cube that makes up the rectangular solid depicted below has 6 inch sides. What is the surface area of the solid in square feet?
\( \large 11\text{ f}{{\text{t}}^{2}}\) Hint: Check your units and make sure you're using feet and inches consistently. | |
\( \large 16.5\text{ f}{{\text{t}}^{2}}\) Hint: Each square has surface area \(\dfrac{1}{2} \times \dfrac {1}{2}=\dfrac {1}{4}\) sq feet. There are 9 squares on the top and bottom, and 12 on each of 4 sides, for a total of 66 squares. 66 squares \(\times \dfrac {1}{4}\) sq feet/square =16.5 sq feet. | |
\( \large 66\text{ f}{{\text{t}}^{2}}\) Hint: The area of each square is not 1. | |
\( \large 2376\text{ f}{{\text{t}}^{2}}\) Hint: Read the question more carefully -- the answer is supposed to be in sq feet, not sq inches.
|
Question 11 |
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 12 |
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 13 |
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 14 |
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 15 |
What is the least common multiple of 540 and 216?
\( \large{{2}^{5}}\cdot {{3}^{6}}\cdot 5\) Hint: This is the product of the numbers, not the LCM. | |
\( \large{{2}^{3}}\cdot {{3}^{3}}\cdot 5\) Hint: One way to solve this is to factor both numbers: \(540=2^2 \cdot 3^3 \cdot 5\) and \(216=2^3 \cdot 3^3\). Then for each prime that's a factor of either number, use the largest exponent that appears in one of the factorizations. You can also take the product of the two numbers divided by their GCD. | |
\( \large{{2}^{2}}\cdot {{3}^{3}}\cdot 5\) Hint: 216 is a multiple of 8. | |
\( \large{{2}^{2}}\cdot {{3}^{2}}\cdot {{5}^{2}}\) Hint: Not a multiple of 216 and not a multiple of 540. |
Question 16 |
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 17 |
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 18 |
Use the table below to answer the question that follows:
Each number in the table above represents a value W that is determined by the values of x and y. For example, when x=3 and y=1, W=5. What is the value of W when x=9 and y=14? Assume that the patterns in the table continue as shown.
\( \large W=-5\) Hint: When y is even, W is even. | |
\( \large W=4\) Hint: Note that when x increases by 1, W increases by 2, and when y increases by 1, W decreases by 1. At x=y=0, W=0, so at x=9, y=14, W has increased by \(9 \times 2\) and decreased by 14, or W=18-14=4. | |
\( \large W=6\) Hint: Try fixing x or y at 0, and start by finding W for x=0 y=14 or x=9, y=0. | |
\( \large W=32\) Hint: Try fixing x or y at 0, and start by finding W for x=0 y=14 or x=9, y=0. |
Question 19 |
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 20 |
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 21 |
Elena is going to use a calculator to check whether or not 267 is prime. She will pick certain divisors, and then find 267 divided by each, and see if she gets a whole number. If she never gets a whole number, then she's found a prime. Which numbers does Elena NEED to check before she can stop checking and be sure she has a prime?
All natural numbers from 2 to 266.Hint: She only needs to check primes -- checking the prime factors of any composite is enough to look for divisors. As a test taking strategy, the other three choices involve primes, so worth thinking about. | |
All primes from 2 to 266 .Hint: Remember, factors come in pairs (except for square root factors), so she would first find the smaller of the pair and wouldn't need to check the larger. | |
All primes from 2 to 133 .Hint: She doesn't need to check this high. Factors come in pairs, and something over 100 is going to be paired with something less than 3, so she will find that earlier. | |
All primes from \( \large 2\) to \( \large \sqrt{267}\).Hint: \(\sqrt{267} \times \sqrt{267}=267\). Any other pair of factors will have one factor less than \( \sqrt{267}\) and one greater, so she only needs to check up to \( \sqrt{267}\). |
Question 22 |
What is the greatest common factor of 540 and 216?
\( \large{{2}^{2}}\cdot {{3}^{3}}\) Hint: One way to solve this is to factor both numbers: \(540=2^2 \cdot 3^3 \cdot 5\) and \(216=2^3 \cdot 3^3\). Then take the smaller power for each prime that is a factor of both numbers. | |
\( \large2\cdot 3\) Hint: This is a common factor of both numbers, but it's not the greatest common factor. | |
\( \large{{2}^{3}}\cdot {{3}^{3}}\) Hint: \(2^3 = 8\) is not a factor of 540. | |
\( \large{{2}^{2}}\cdot {{3}^{2}}\) Hint: This is a common factor of both numbers, but it's not the greatest common factor. |
Question 23 |
The following story situations model \( 12\div 3\):
I) Jack has 12 cookies, which he wants to share equally between himself and two friends. How many cookies does each person get?
II) Trent has 12 cookies, which he wants to put into bags of 3 cookies each. How many bags can he make?
III) Cicely has $12. Cookies cost $3 each. How many cookies can she buy?
Which of these questions illustrate the same model of division, either partitive (partioning) or measurement (quotative)?
I and II | |
I and III | |
II and IIIHint: Problem I is partitive (or partitioning or sharing) -- we put 12 objects into 3 groups. Problems II and III are quotative (or measurement) -- we put 12 objects in groups of 3. | |
All three problems model the same meaning of division |
Question 24 |
What set of transformations will transform the leftmost image into the rightmost image?
A 90 degree clockwise rotation about (2,1) followed by a translation of two units to the right.Hint: Part of the figure would move below the x-axis with these transformations. | |
A translation 3 units up, followed by a reflection about the line y=x.Hint: See what happens to the point (5,1) under this set of transformations. | |
A 90 degree clockwise rotation about (5,1), followed by a translation of 2 units up. | |
A 90 degree clockwise rotation about (2,1) followed by a translation of 2 units to the right.Hint: See what happens to the point (3,3) under this set of transformations. |
Question 25 |
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 26 |
The Venn Diagram below gives data on the number of seniors, athletes, and vegetarians in the student body at a college:
How many students at the college are seniors who are not vegetarians?
\( \large 137\) Hint: Doesn't include the senior athletes who are not vegetarians. | |
\( \large 167\) | |
\( \large 197\) Hint: That's all seniors, including vegetarians. | |
\( \large 279\) Hint: Includes all athletes who are not vegetarians, some of whom are not seniors. |
Question 27 |
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 28 |
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 29 |
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 30 |
Use the expression below to answer the question that follows.
\(\large \dfrac{\left( 155 \right)\times \left( 6,124 \right)}{977}\)
Which of the following is the best estimate of the expression above?
100Hint: 6124/977 is approximately 6. | |
200Hint: 6124/977 is approximately 6. | |
1,000Hint: 6124/977 is approximately 6. 155 is approximately 150, and \( 6 \times 150 = 3 \times 300 = 900\), so this answer is closest. | |
2,000Hint: 6124/977 is approximately 6. |
Question 31 |
AHint: \(\frac{34}{135} \approx \frac{1}{4}\) and \( \frac{53}{86} \approx \frac {2}{3}\). \(\frac {1}{4}\) of \(\frac {2}{3}\) is small and closest to A. | |
BHint: Estimate with simpler fractions. | |
CHint: Estimate with simpler fractions. | |
DHint: Estimate with simpler fractions. |
Question 32 |
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 33 |
In January 2011, the national debt was about 14 trillion dollars and the US population was about 300 million people. Someone reading these figures estimated that the national debt was about $5,000 per person. Which of these statements best describes the reasonableness of this estimate?
It is too low by a factor of 10Hint: 14 trillion \( \approx 15 \times {{10}^{12}} \) and 300 million \( \approx 3 \times {{10}^{8}}\), so the true answer is about \( 5 \times {{10}^{4}} \) or $50,000. | |
It is too low by a factor of 100 | |
It is too high by a factor of 10 | |
It is too high by a factor of 100 |
Question 34 |
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 35 |
Which of the following nets will not fold into a cube?
![]() Hint: If you have trouble visualizing, cut them out and fold (during the test, you can tear paper to approximate). | |
![]() | |
![]() Hint: If you have trouble visualizing, cut them out and fold (during the test, you can tear paper to approximate). | |
![]() Hint: If you have trouble visualizing, cut them out and fold (during the test, you can tear paper to approximate). |
Question 36 |
Use the graph below to answer the question that follows:
The graph above best matches which of the following scenarios:
George left home at 10:00 and drove to work on a crooked path. He was stopped in traffic at 10:30 and 10:45. He drove 30 miles total.Hint: Just because he ended up 30 miles from home doesn't mean he drove 30 miles total. | |
George drove to work. On the way to work there is a little hill and a big hill. He slowed down for them. He made it to work at 11:15.Hint: The graph is not a picture of the roads. | |
George left home at 10:15. He drove 10 miles, then realized he‘d forgotten something at home. He turned back and got what he‘d forgotten. Then he drove in a straight line, at many different speeds, until he got to work around 11:15.Hint: A straight line on a distance versus time graph means constant speed. | |
George left home at 10:15. He drove 10 miles, then realized he‘d forgotten something at home. He turned back and got what he‘d forgotten. Then he drove at a constant speed until he got to work around 11:15. |
Question 37 |
Which of the following sets of polygons can be assembled to form a pentagonal pyramid?
2 pentagons and 5 rectangles.Hint: These can be assembled to form a pentagonal prism, not a pentagonal pyramid. | |
1 square and 5 equilateral triangles.Hint: You need a pentagon for a pentagonal pyramid. | |
1 pentagon and 5 isosceles triangles. | |
1 pentagon and 10 isosceles triangles. |
Question 38 |
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 39 |
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 40 |
P is a prime number that divides 240. Which of the following must be true?
P divides 30Hint: 2, 3, and 5 are the prime factors of 240, and all divide 30. | |
P divides 48Hint: P=5 doesn't work. | |
P divides 75Hint: P=2 doesn't work. | |
P divides 80Hint: P=3 doesn't work. |
Question 41 |
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 42 |
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. |
Question 43 |
A biology class requires a lab fee, which is a whole number of dollars, and the same amount for all students. On Monday the instructor collected $70 in fees, on Tuesday she collected $126, and on Wednesday she collected $266. What is the largest possible amount the fee could be?
$2Hint: A possible fee, but not the largest possible fee. Check the other choices to see which are factors of all three numbers. | |
$7Hint: A possible fee, but not the largest possible fee. Check the other choices to see which are factors of all three numbers. | |
$14Hint: This is the greatest common factor of 70, 126, and 266. | |
$70Hint: Not a factor of 126 or 266, so couldn't be correct. |
Question 44 |
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 45 |
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. |
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