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 |
At a school fundraising event, people can buy a ticket to spin a spinner like the one below. The region that the spinner lands in tells which, if any, prize the person wins.
If 240 people buy tickets to spin the spinner, what is the best estimate of the number of keychains that will be given away?
40Hint: "Keychain" appears on the spinner twice. | |
80Hint: The probability of getting a keychain is 1/3, and so about 1/3 of the time the spinner will win. | |
100Hint: What is the probability of winning a keychain? | |
120Hint: That would be the answer for getting any prize, not a keychain specifically. |
Question 2 |
A car is traveling at 60 miles per hour. Which of the expressions below could be used to compute how many feet the car travels in 1 second? Note that 1 mile = 5,280 feet.
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot 5280\dfrac{\text{feet}}{\text{mile}}\cdot 60\dfrac{\text{minutes}}{\text{hour}}\cdot 60\dfrac{\text{seconds}}{\text{minute}}
\) Hint: This answer is not in feet/second. | |
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot 5280\dfrac{\text{feet}}{\text{mile}}\cdot \dfrac{1}{60}\dfrac{\text{hour}}{\text{minutes}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}
\) Hint: This is the only choice where the answer is in feet per second and the unit conversions are correct. | |
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot \dfrac{1}{5280}\dfrac{\text{foot}}{\text{miles}}\cdot 60\dfrac{\text{hours}}{\text{minute}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}\) Hint: Are there really 60 hours in a minute? | |
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot \dfrac{1}{5280}\dfrac{\text{mile}}{\text{feet}}\cdot 60\dfrac{\text{minutes}}{\text{hour}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}\) Hint: This answer is not in feet/second. |
Question 3 |
Which of the following is equal to eleven billion four hundred thousand?
\( \large 11,400,000\) Hint: That's eleven million four hundred thousand. | |
\(\large11,000,400,000\) | |
\( \large11,000,000,400,000\) Hint: That's eleven trillion four hundred thousand (although with British conventions; this answer is correct, but in the US, it isn't). | |
\( \large 11,400,000,000\) Hint: That's eleven billion four hundred million |
Question 4 |
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 5 |
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 6 |
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 7 |
Solve for x: \(\large 4-\dfrac{2}{3}x=2x\)
\( \large x=3\) Hint: Try plugging x=3 into the equation. | |
\( \large x=-3\) Hint: Left side is positive, right side is negative when you plug this in for x. | |
\( \large x=\dfrac{3}{2}\) Hint: One way to solve: \(4=\dfrac{2}{3}x+2x\) \(=\dfrac{8}{3}x\).\(x=\dfrac{3 \times 4}{8}=\dfrac{3}{2}\). Another way is to just plug x=3/2 into the equation and see that each side equals 3 -- on a multiple choice test, you almost never have to actually solve for x. | |
\( \large x=-\dfrac{3}{2}\) Hint: Left side is positive, right side is negative when you plug this in for x. |
Question 8 |
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 9 |
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 10 |
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 11 |
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 12 |
Use the expression below to answer the question that follows.
\( \large 3\times {{10}^{4}}+2.2\times {{10}^{2}}\)
Which of the following is closest to the expression above?
Five millionHint: Pay attention to the exponents. Adding 3 and 2 doesn't work because they have different place values. | |
Fifty thousandHint: Pay attention to the exponents. Adding 3 and 2 doesn't work because they have different place values. | |
Three millionHint: Don't add the exponents. | |
Thirty thousandHint: \( 3\times {{10}^{4}} = 30,000;\) the other term is much smaller and doesn't change the estimate. |
Question 13 |
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 14 |
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 15 |
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 16 |
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 17 |
A family on vacation drove the first 200 miles in 4 hours and the second 200 miles in 5 hours. Which expression below gives their average speed for the entire trip?
\( \large \dfrac{200+200}{4+5}\) Hint: Average speed is total distance divided by total time. | |
\( \large \left( \dfrac{200}{4}+\dfrac{200}{5} \right)\div 2\) Hint: This seems logical, but the problem is that it weights the first 4 hours and the second 5 hours equally, when each hour should get the same weight in computing the average speed. | |
\( \large \dfrac{200}{4}+\dfrac{200}{5} \) Hint: This would be an average of 90 miles per hour! | |
\( \large \dfrac{400}{4}+\dfrac{400}{5} \) Hint: This would be an average of 180 miles per hour! Even a family of race car drivers probably doesn't have that average speed on a vacation! |
Question 18 |
Which of the numbers below is 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 19 |
A class is using base-ten block to represent numbers. A large cube represents 1000, a flat represents 100, a rod represents 10, and a little cube represents 1. Which of these is not a correct representation for 2,347?
23 flats, 4 rods, 7 little cubesHint: Be sure you read the question carefully: 2300+40+7=2347 | |
2 large cubes, 3 flats, 47 rodsHint: 2000+300+470 \( \neq\) 2347 | |
2 large cubes, 34 rods, 7 little cubesHint: Be sure you read the question carefully: 2000+340+7=2347 | |
2 large cubes, 3 flats, 4 rods, 7 little cubesHint: Be sure you read the question carefully: 2000+300+40+7=2347 |
Question 20 |
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 21 |
| 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 22 |
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 23 |
Use the graph below to answer the question that follows.
If the polygon shown above is reflected about the y axis and then rotated 90 degrees clockwise about the origin, which of the following graphs is the result?
![]() Hint: Try following the point (1,4) to see where it goes after each transformation. | |
![]() | |
Hint: Make sure you're reflecting in the correct axis. | |
![]() Hint: Make sure you're rotating the correct direction. |
Question 24 |
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 25 |
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 26 |
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 27 |
Which of the graphs below represent functions?
I.
II.
III.
IV. 
I and IV only.Hint: There are vertical lines that go through 2 points in IV . | |
I and III only.Hint: Even though III is not continuous, it's still a function (assuming that vertical lines between the "steps" do not go through 2 points). | |
II and III only.Hint: Learn about the vertical line test. | |
I, II, and IV only.Hint: There are vertical lines that go through 2 points in II. |
Question 28 |
Use the samples of a student's work below to answer the question that follows:
This student divides fractions by first finding a common denominator, then dividing the numerators.
\( \large \dfrac{2}{3} \div \dfrac{3}{4} \longrightarrow \dfrac{8}{12} \div \dfrac{9}{12} \longrightarrow 8 \div 9 = \dfrac {8}{9}\) \( \large \dfrac{2}{5} \div \dfrac{7}{20} \longrightarrow \dfrac{8}{20} \div \dfrac{7}{20} \longrightarrow 8 \div 7 = \dfrac {8}{7}\) \( \large \dfrac{7}{6} \div \dfrac{3}{4} \longrightarrow \dfrac{14}{12} \div \dfrac{9}{12} \longrightarrow 14 \div 9 = \dfrac {14}{9}\)Which of the following best describes the mathematical validity of the algorithm the student is using?
It is not valid. Common denominators are for adding and subtracting fractions, not for dividing them.Hint: Don't be so rigid! Usually there's more than one way to do something in math. | |
It got the right answer in these three cases, but it isn‘t valid for all rational numbers.Hint: Did you try some other examples? What makes you say it's not valid? | |
It is valid if the rational numbers in the division problem are in lowest terms and the divisor is not zero.Hint: Lowest terms doesn't affect this problem at all. | |
It is valid for all rational numbers, as long as the divisor is not zero.Hint: When we have common denominators, the problem is in the form a/b divided by c/b, and the answer is a/c, as the student's algorithm predicts. |
Question 29 |
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 30 |
The first histogram shows the average life expectancies for women in different countries in Africa in 1998; the second histogram gives similar data for Europe:
How much bigger is the range of the data for Africa than the range of the data for Europe?
0 yearsHint: Range is the maximum life expectancy minus the minimum life expectancy. | |
12 yearsHint: Are you subtracting frequencies? Range is about values of the data, not frequency. | |
18 yearsHint: It's a little hard to read the graph, but it doesn't matter if you're consistent. It looks like the range for Africa is 80-38= 42 years and for Europe is 88-64 = 24; 42-24=18. | |
42 yearsHint: Read the question more carefully. |
Question 31 |
Which property is not shared by all rhombi?
4 congruent sidesHint: The most common definition of a rhombus is a quadrilateral with 4 congruent sides. | |
A center of rotational symmetryHint: The diagonal of a rhombus separates it into two congruent isosceles triangles. The center of this line is a center of 180 degree rotational symmetry that switches the triangles. | |
4 congruent anglesHint: Unless the rhombus is a square, it does not have 4 congruent angles. | |
2 sets of parallel sidesHint: All rhombi are parallelograms. |
Question 32 |
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 33 |
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 34 |
Below is a pictorial representation of \(2\dfrac{1}{2}\div \dfrac{2}{3}\):
Which of the following is the best description of how to find the quotient from the picture?
The quotient is \(3\dfrac{3}{4}\). There are 3 whole blocks each representing \(\dfrac{2}{3}\) and a partial block composed of 3 small rectangles. The 3 small rectangles represent \(\dfrac{3}{4}\) of \(\dfrac{2}{3}\). | |
The quotient is \(3\dfrac{1}{2}\). There are 3 whole blocks each representing \(\dfrac{2}{3}\) and a partial block composed of 3 small rectangles. The 3 small rectangles represent \(\dfrac{3}{6}\) of a whole, or \(\dfrac{1}{2}\).Hint: We are counting how many 2/3's are in 2 1/2: the unit becomes 2/3, not 1. | |
The quotient is \(\dfrac{4}{15}\). There are four whole blocks separated into a total of 15 small rectangles.Hint: This explanation doesn't make much sense. Probably you are doing "invert and multiply," but inverting the wrong thing. | |
This picture cannot be used to find the quotient because it does not show how to separate \(2\dfrac{1}{2}\) into equal sized groups.Hint: Study the measurement/quotative model of division. It's often very useful with fractions. |
Question 35 |
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 36 |
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 37 |
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 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 |
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 40 |
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 41 |
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 42 |
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 43 |
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 44 |
Below are front, side, and top views of a three-dimensional solid.
Which of the following could be the solid shown above?
A sphereHint: All views would be circles. | |
A cylinder | |
A coneHint: Two views would be triangles, not rectangles. | |
A pyramidHint: How would one view be a circle? |
Question 45 |
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. |
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