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 |
There are six gumballs in a bag — two red and four green. Six children take turns picking a gumball out of the bag without looking. They do not return any gumballs to the bag. What is the probability that the first two children to pick from the bag pick the red gumballs?
\( \large \dfrac{1}{3}\) Hint: This is the probability that the first child picks a red gumball, but not that the first two children pick red gumballs. | |
\( \large \dfrac{1}{8}\) Hint: Are you adding things that you should be multiplying? | |
\( \large \dfrac{1}{9}\) Hint: This would be the probability if the gumballs were returned to the bag. | |
\( \large \dfrac{1}{15}\) Hint: The probability that the first child picks red is 2/6 = 1/3. Then there are 5 gumballs in the bag, one red, so the probability that the second child picks red is 1/5. Thus 1/5 of the time, after the first child picks red, the second does too, so the probability is 1/5 x 1/3 = 1/15. |
Question 2 |
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 3 |
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 4 |
On a map the distance from Boston to Detroit is 6 cm, and these two cities are 702 miles away from each other. Assuming the scale of the map is the same throughout, which answer below is closest to the distance between Boston and San Francisco on the map, given that they are 2,708 miles away from each other?
21 cmHint: How many miles would correspond to 24 cm on the map? Try adjusting from there. | |
22 cmHint: How many miles would correspond to 24 cm on the map? Try adjusting from there. | |
23 cmHint: One way to solve this without a calculator is to note that 4 groups of 6 cm is 2808 miles, which is 100 miles too much. Then 100 miles would be about 1/7 th of 6 cm, or about 1 cm less than 24 cm. | |
24 cmHint: 4 groups of 6 cm is over 2800 miles on the map, which is too much. |
Question 5 |
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 6 |
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 7 |
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 8 |
Use the table below to answer the question that follows:
Gordon wants to buy three pounds of nuts. Each of the stores above ordinarily sells the nuts for $4.99 a pound, but is offering a discount this week. At which store can he buy the nuts for the least amount of money?
Store AHint: This would save about $2.50. You can quickly see that D saves more. | |
Store BHint: This saves 15% and C saves 25%. | |
Store C | |
Store DHint: This is about 20% off, which is less of a discount than C. |
Question 9 |
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 10 |
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 11 |
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 12 |
Which of the following is equivalent to \( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{2}{8}\times \dfrac{1}{2}?\)
\( \large \dfrac{7}{16}\) Hint: Multiplication comes before addition and subtraction in the order of operations. | |
\( \large \dfrac{1}{2}\) Hint: Addition and subtraction are of equal priority in the order of operations -- do them left to right. | |
\( \large \dfrac{3}{4}\) Hint: \( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{2}{8}\times \dfrac{1}{2}\)=\( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{1}{8}\)=\( \dfrac{3}{4}+-\dfrac{1}{8}+\dfrac{1}{8}\)=\( \dfrac{3}{4}\) | |
\( \large \dfrac{3}{16}\) Hint: Multiplication comes before addition and subtraction in the order of operations. |
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 |
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 15 |
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 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 chart below gives percentiles for the number of sit-ups that boys of various ages can do in 60 seconds (source , June 24, 2011)
Which of the following statements can be inferred from the above chart?
95% of 12 year old boys can do 56 sit-ups in 60 seconds.Hint: The 95th percentile means that 95% of scores are less than or equal to 56, and 5% are greater than or equal to 56. | |
At most 25% of 7 year old boys can do 19 or more sit-ups in 60 seconds.Hint: The 25th percentile means that 25% of scores are less than or equal to 19, and 75% are greater than or equal to 19. | |
Half of all 13 year old boys can do less than 41 sit-ups in 60 seconds and half can do more than 41 sit-ups in 60 seconds.Hint: Close, but not quite. There's no accounting for boys who can do exactly 41 sit ups. Look at these data: 10, 20, 41, 41, 41, 41, 50, 60, 90. The median is 41, but more than half can do 41 or more. | |
At least 75% of 16 year old boys can only do 51 or fewer sit-ups in 60 seconds.Hint: The "at least" is necessary due to duplicates. Suppose the data were 10, 20, 51, 51. The 75th percentile is 51, but 100% of the boys can only do 51 or fewer situps. |
Question 18 |
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 19 |
Which of the following is an irrational number?
\( \large \sqrt[3]{8}\) Hint: This answer is the cube root of 8. Since 2 x 2 x 2 =8, this is equal to 2, which is rational because 2 = 2/1. | |
\( \large \sqrt{8}\) Hint: It is not trivial to prove that this is irrational, but you can get this answer by eliminating the other choices. | |
\( \large \dfrac{1}{8}\) Hint: 1/8 is the RATIO of two integers, so it is rational. | |
\( \large -8\) Hint: Negative integers are also rational, -8 = -8/1, a ratio of integers. |
Question 20 |
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 21 |
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 22 |
Use the solution procedure below to answer the question that follows:
\( \large {\left( x+3 \right)}^{2}=10\)
\( \large \left( x+3 \right)\left( x+3 \right)=10\)
\( \large {x}^{2}+9=10\)
\( \large {x}^{2}+9-9=10-9\)
\( \large {x}^{2}=1\)
\( \large x=1\text{ or }x=-1\)
Which of the following is incorrect in the procedure shown above?
The commutative property is used incorrectly.Hint: The commutative property is \(a+b=b+a\) or \(ab=ba\). | |
The associative property is used incorrectly.Hint: The associative property is \(a+(b+c)=(a+b)+c\) or
\(a \times (b \times c)=(a \times b) \times c\). | |
Order of operations is done incorrectly. | |
The distributive property is used incorrectly.Hint: \((x+3)(x+3)=x(x+3)+3(x+3)\)=\(x^2+3x+3x+9.\) |
Question 23 |
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 24 |
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 25 |
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 26 |
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 27 |
Here are some statements:
I) 5 is an integer II)\( -5 \) is an integer III) \(0\) is an integer
Which of the statements are true?
I only | |
I and II only | |
I and III only | |
I, II, and IIIHint: The integers are ...-3, -2, -1, 0, 1, 2, 3, .... |
Question 28 |
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 29 |
If x is an integer, which of the following must also be an integer?
\( \large \dfrac{x}{2}\) Hint: If x is odd, then \( \dfrac{x}{2} \) is not an integer, e.g. 3/2 = 1.5. | |
\( \large \dfrac{2}{x}\) Hint: Only an integer if x = -2, -1, 1, or 2. | |
\( \large-x\) Hint: -1 times any integer is still an integer. | |
\(\large\sqrt{x}\) Hint: Usually not an integer, e.g. \( \sqrt{2} \approx 1.414 \). |
Question 30 |
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 31 |
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 32 |
Use the problem below to answer the question that follows:
T shirts are on sale for 20% off. Tasha paid $8.73 for a shirt. What is the regular price of the shirt? There is no tax on clothing purchases under $175.
Let p represent the regular price of these t-shirt. Which of the following equations is correct?
\( \large 0.8p=\$8.73\) Hint: 80% of the regular price = $8.73. | |
\( \large \$8.73+0.2*\$8.73=p\) Hint: The 20% off was off of the ORIGINAL price, not off the $8.73 (a lot of people make this mistake). Plus this is the same equation as in choice c. | |
\( \large 1.2*\$8.73=p\) Hint: The 20% off was off of the ORIGINAL price, not off the $8.73 (a lot of people make this mistake). Plus this is the same equation as in choice b. | |
\( \large p-0.2*\$8.73=p\) Hint: Subtract p from both sides of this equation, and you have -.2 x 8.73 =0. |
Question 33 |
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 34 |
Taxicab fares in Boston (Spring 2012) are $2.60 for the first \(\dfrac{1}{7}\) of a mile or less and $0.40 for each \(\dfrac{1}{7}\) of a mile after that.
Let d represent the distance a passenger travels in miles (with \(d>\dfrac{1}{7}\)). Which of the following expressions represents the total fare?
\( \large \$2.60+\$0.40d\) Hint: It's 40 cents for 1/7 of a mile, not per mile. | |
\( \large \$2.60+\$0.40\dfrac{d}{7}\) Hint: According to this equation, going 7 miles would cost $3; does that make sense? | |
\( \large \$2.20+\$2.80d\) Hint: You can think of the fare as $2.20 to enter the cab, and then $0.40 for each 1/7 of a mile, including the first 1/7 of a mile (or $2.80 per mile).
Alternatively, you pay $2.60 for the first 1/7 of a mile, and then $2.80 per mile for d-1/7 miles. The total is 2.60+2.80(d-1/7) = 2.60+ 2.80d -.40 = 2.20+2.80d. | |
\( \large \$2.60+\$2.80d\) Hint: Don't count the first 1/7 of a mile twice. |
Question 35 |
The table below gives the result of a survey at a college, asking students whether they were residents or commuters:
Based on the above data, what is the probability that a randomly chosen commuter student is a junior or a senior?
\( \large \dfrac{34}{43}\) | |
\( \large \dfrac{34}{71}\) Hint: This is the probability that a randomly chosen junior or senior is a commuter student. | |
\( \large \dfrac{34}{147}\) Hint: This is the probability that a randomly chosen student is a junior or senior who is a commuter. | |
\( \large \dfrac{71}{147}\) Hint: This is the probability that a randomly chosen student is a junior or a senior. |
Question 36 |
The letters A, and B represent digits (possibly equal) in the ten digit number x=1,438,152,A3B. For which values of A and B is x divisible by 12, but not by 9?
\( \large A = 0, B = 4\) Hint: Digits add to 31, so not divisible by 3, so not divisible by 12. | |
\( \large A = 7, B = 2\) Hint: Digits add to 36, so divisible by 9. | |
\( \large A = 0, B = 6\) Hint: Digits add to 33, divisible by 3, not 9. Last digits are 36, so divisible by 4, and hence by 12. | |
\( \large A = 4, B = 8\) Hint: Digits add to 39, divisible by 3, not 9. Last digits are 38, so not divisible by 4, so not divisible by 12. |
Question 37 |
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 38 |
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 39 |
Here is a mental math strategy for computing 26 x 16:
Step 1: 100 x 16 = 1600
Step 2: 25 x 16 = 1600 ÷· 4 = 400
Step 3: 26 x 16 = 400 + 16 = 416
Which property best justifies Step 3 in this strategy?
Commutative Property.Hint: For addition, the commutative property is \(a+b=b+a\) and for multiplication it's \( a \times b = b \times a\). | |
Associative Property.Hint: For addition, the associative property is \((a+b)+c=a+(b+c)\) and for multiplication it's \((a \times b) \times c=a \times (b \times c)\) | |
Identity Property.Hint: 0 is the additive identity, because \( a+0=a\) and 1 is the multiplicative identity because \(a \times 1=a\). The phrase "identity property" is not standard. | |
Distributive Property.Hint: \( (25+1) \times 16 = 25 \times 16 + 1 \times 16 \). This is an example of the distributive property of multiplication over addition. |
Question 40 |
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 41 |
Use the graph below to answer the question that follows:
The graph above represents the equation \( \large 3x+Ay=B\), where A and B are integers. What are the values of A and B?
\( \large A = -2, B= 6\) Hint: Plug in (2,0) to get B=6, then plug in (0,-3) to get A=-2. | |
\( \large A = 2, B = 6\) Hint: Try plugging (0,-3) into this equation. | |
\( \large A = -1.5, B=-3\) Hint: The problem said that A and B were integers and -1.5 is not an integer. Don't try to use slope-intercept form. | |
\( \large A = 2, B = -3\) Hint: Try plugging (2,0) into this equation. |
Question 42 |
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 43 |
| I. \(\large \dfrac{1}{2}+\dfrac{1}{3}\) | II. \( \large .400000\) | III. \(\large\dfrac{1}{5}+\dfrac{1}{5}\) |
| IV. \( \large 40\% \) | V. \( \large 0.25 \) | VI. \(\large\dfrac{14}{35}\) |
Which of the lists below includes all of the above expressions that are equivalent to \( \dfrac{2}{5}\)?
I, III, V, VIHint: I and V are not at all how fractions and decimals work. | |
III, VIHint: These are right, but there are more. | |
II, III, VIHint: These are right, but there are more. | |
II, III, IV, VI |
Question 44 |
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 45 |
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). |
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