20 Random Questions

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

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Question 1

Which of the lists below contains only irrational numbers?

A	$ \large\pi , \quad \sqrt{6},\quad \sqrt{\dfrac{1}{2}}$
B	$ \large\pi , \quad \sqrt{9}, \quad \pi +1$ Hint: $ \sqrt{9}=3$
C	$ \large\dfrac{1}{3},\quad \dfrac{5}{4},\quad \dfrac{2}{9}$ Hint: These are all rational.
D	$ \large-3,\quad 14,\quad 0$ Hint: These are all rational.

Question 1 Explanation:

Topic: Identifying rational and irrational numbers (Objective 0016).

Question 2

In each expression below N represents a negative integer. Which expression could have a negative value?

A	$ \large {{N}^{2}}$ Hint: Squaring always gives a non-negative value.
B	$ \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.
C	$ \large -N$ Hint: If N is negative, then -N is positive
D	$ \large 6+N$ Hint: For example, if $N=-10$, then $6+N = -4$

Question 2 Explanation:

If you are stuck on a question like this, try a few examples to eliminate some choices and to help you understand what the question means. Topic: Characteristics of integers (Objective 0016).

Question 3

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?

A	$ \large t=s+15$ Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation?
B	$ \large s=t+15$ Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation?
C	$ \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.
D	$ \large s=15t$

Question 3 Explanation:

Topic: Select the linear equation that best models a real-world situation (Objective 0022).

Question 4

Which of the following is not possible?

A	An equiangular triangle that is not equilateral. Hint: The AAA property of triangles states that all triangles with corresponding angles congruent are similar. Thus all triangles with three equal angles are similar, and are equilateral.
B	An equiangular quadrilateral that is not equilateral. Hint: A rectangle is equiangular (all angles the same measure), but if it's not a square, it's not equilateral (all sides the same length).
C	An equilateral quadrilateral that is not equiangular. Hint: This rhombus has equal sides, but it doesn't have equal angles:
D	An equiangular hexagon that is not equilateral. Hint: This hexagon has equal angles, but it doesn't have equal sides:

Question 4 Explanation:

Topic: Classify and analyze polygons using attributes of sides and angles (Objective 0024).

Question 5

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?

A	$ \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.
B	$ \large \dfrac{1}{8}$ Hint: Are you adding things that you should be multiplying?
C	$ \large \dfrac{1}{9}$ Hint: This would be the probability if the gumballs were returned to the bag.
D	$ \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 5 Explanation:

Topic: Calculate the probabilities of simple and compound events and of independent and dependent events (Objective 0026).

Question 6

A map has a scale of 3 inches = 100 miles. Cities A and B are 753 miles apart. Let d be the distance between the two cities on the map. Which of the following is not correct?

A	$ \large \dfrac{3}{100}=\dfrac{d}{753}$ Hint: Units on both side are inches/mile, and both numerators and denominators correspond -- this one is correct.
B	$ \large \dfrac{3}{100}=\dfrac{753}{d}$ Hint: Unit on the left is inches per mile, and on the right is miles per inch. The proportion is set up incorrectly (which is what we wanted). Another strategy is to notice that one of A or B has to be the answer because they cannot both be correct proportions. Then check that cross multiplying on A gives part D, so B is the one that is different from the other 3.
C	$ \large \dfrac{3}{d}=\dfrac{100}{753}$ Hint: Unitless on each side, as inches cancel on the left and miles on the right. Numerators correspond to the map, and denominators to the real life distances -- this one is correct.
D	$ \large 100d=3\cdot 753$ Hint: This is equivalent to part A.

Question 6 Explanation:

Topic: Analyze the relationships among proportions, constant rates, and linear functions (Objective 0022).

Question 7

Here is a student's work on several multiplication problems:

For which of the following problems is this student most likely to get the correct solution, even though he is using an incorrect algorithm?

A	58 x 22 Hint: This problem involves regrouping, which the student does not do correctly.
B	16 x 24 Hint: This problem involves regrouping, which the student does not do correctly.
C	31 x 23 Hint: There is no regrouping with this problem.
D	141 x 32 Hint: This problem involves regrouping, which the student does not do correctly.

Question 7 Explanation:

Topic: Analyze computational algorithms (Objective 0019).

Question 8

Which of the lines depicted below is a graph of $ \large y=2x-5$?

A	a Hint: The slope of line a is negative.
B	b Hint: Wrong slope and wrong intercept.
C	c Hint: The intercept of line c is positive.
D	d Hint: 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 8 Explanation:

Topic: Find a linear equation that represents a graph (Objective 0022).

Question 9

The picture below represents a board with pegs on it, where the closest distance between two pegs is 1 cm. What is the area of the pentagon shown?

A	$ \large 8\text{ c}{{\text{m}}^{2}} $ Hint: Don't just count the dots inside, that doesn't give the area. Try adding segments so that the slanted lines become the diagonals of rectangles.
B	$ \large 11\text{ c}{{\text{m}}^{2}}$ Hint: Try adding segments so that the slanted lines become the diagonals of rectangles.
C	$ \large 11.5\text{ c}{{\text{m}}^{2}}$ Hint: An easy way to do this problem is to use Pick's Theorem (of course, it's better if you understand why Pick's theorem works): area = # pegs inside + half # pegs on the border - 1. In this case 8+9/2-1=11.5. A more appropriate strategy for elementary classrooms is to add segments; here's one way. There are 20 1x1 squares enclosed, and the total area of the triangles that need to be subtracted is 8.5
D	$ \large 12.5\text{ c}{{\text{m}}^{2}}$ Hint: Try adding segments so that the slanted lines become the diagonals of rectangles.

Question 9 Explanation:

Topics: Calculate measurements and derive and use formulas for calculating the areas of geometric shapes and figures (Objective 0023).

Question 10

Use the expression below to answer the question that follows.

$ \large \dfrac{\left( 4\times {{10}^{3}} \right)\times \left( 3\times {{10}^{4}} \right)}{6\times {{10}^{6}}}$

Which of the following is equivalent to the expression above?

A	2 Hint: $10^3 \times 10^4=10^7$, and note that if you're guessing when the answers are so closely related, you're generally better off guessing one of the middle numbers.
B	20 Hint: $ \dfrac{\left( 4\times {{10}^{3}} \right)\times \left( 3\times {{10}^{4}} \right)}{6\times {{10}^{6}}}=\dfrac {12 \times {{10}^{7}}}{6\times {{10}^{6}}}=$$2 \times {{10}^{1}}=20 $
C	200 Hint: $10^3 \times 10^4=10^7$
D	2000 Hint: $10^3 \times 10^4=10^7$, and note that if you're guessing when the answers are so closely related, you're generally better off guessing one of the middle numbers.

Question 10 Explanation:

Topics: Scientific notation, exponents, simplifying fractions (Objective 0016, although overlaps with other objectives too).

Question 11

What is the least common multiple of 540 and 216?

A	$ \large{{2}^{5}}\cdot {{3}^{6}}\cdot 5$ Hint: This is the product of the numbers, not the LCM.
B	$ \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.
C	$ \large{{2}^{2}}\cdot {{3}^{3}}\cdot 5$ Hint: 216 is a multiple of 8.
D	$ \large{{2}^{2}}\cdot {{3}^{2}}\cdot {{5}^{2}}$ Hint: Not a multiple of 216 and not a multiple of 540.

Question 11 Explanation:

Topic: Find the least common multiple of a set of numbers (Objective 0018).

Question 12

The expression $ \large{{8}^{3}}\cdot {{2}^{-10}}$ is equal to which of the following?

A	$ \large 2$ Hint: Write $8^3$ as a power of 2.
B	$ \large \dfrac{1}{2}$ Hint: $8^3 \cdot {2}^{-10}={(2^3)}^3 \cdot {2}^{-10}$ =$2^9 \cdot {2}^{-10} =2^{-1}$
C	$ \large 16$ Hint: Write $8^3$ as a power of 2.
D	$ \large \dfrac{1}{16}$ Hint: Write $8^3$ as a power of 2.

Question 12 Explanation:

Topic: Laws of Exponents (Objective 0019).

Question 13

A homeowner is planning to tile the kitchen floor with tiles that measure 6 inches by 8 inches. The kitchen floor is a rectangle that measures 10 ft by 12 ft, and there are no gaps between the tiles. How many tiles does the homeowner need?

A	30 Hint: The floor is 120 sq feet, and the tiles are smaller than 1 sq foot. Also, remember that 1 sq foot is 12 $\times$ 12=144 sq inches.
B	120 Hint: The floor is 120 sq feet, and the tiles are smaller than 1 sq foot.
C	300 Hint: Recheck your calculations.
D	360 Hint: One way to do this is to note that 6 inches = 1/2 foot and 8 inches = 2/3 foot, so the area of each tile is 1/2 $\times$ 2/3=1/3 sq foot, or each square foot of floor requires 3 tiles. The area of the floor is 120 square feet. Note that the tiles would fit evenly oriented in either direction, parallel to the walls.

Question 13 Explanation:

Topic: Estimate and calculate measurements, use unit conversions to solve measurement problems, solve measurement problems in real-world situations (Objective 0023).

Question 14

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.

A	$ \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.
B	$ \large \dfrac{1}{14}$ Hint: What would be the sample space here? Ie, how would you list 14 things that you pick one from?
C	$ \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.
D	$ \large \dfrac{1}{49}$ Hint: This is the probability that they are both born on a particular day, e.g. Sunday.

Question 14 Explanation:

Topic: Calculate the probabilities of simple and compound events and of independent and dependent events (Objective 0026).

Question 15

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?

A	$ \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.
B	$ \large A = 0, B = 0, C=4$ Hint: Not divisible by 10, since it doesn't end in 0.
C	$ \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.
D	$ \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 Explanation:

Topic: Understand divisibility rules and why they work (Objective 018).

Question 16

The least common multiple of 60 and N is 1260. Which of the following could be the prime factorization of N?

A	$ \large2\cdot 5\cdot 7$ Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM.
B	$ \large{{2}^{3}}\cdot {{3}^{2}}\cdot 5 \cdot 7$ Hint: 1260 is not divisible by 8, so it isn't a multiple of this N.
C	$ \large3 \cdot 5 \cdot 7$ Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM.
D	$ \large{{3}^{2}}\cdot 5\cdot 7$ Hint: $1260=2^2 \cdot 3^2 \cdot 5 \cdot 7$ and $60=2^2 \cdot 3 \cdot 5$. In order for 1260 to be the LCM, N has to be a multiple of $3^2$ and of 7 (because 60 is not a multiple of either of these). N also cannot introduce a factor that would require the LCM to be larger (as in choice b).

Question 16 Explanation:

Topic: Least Common Multiple (Objective 0018)

Question 17

What is the perimeter of a right triangle with legs of lengths x and 2x?

A	$ \large 6x$ Hint: Use the Pythagorean Theorem.
B	$ \large 3x+5{{x}^{2}}$ Hint: Don't forget to take square roots when you use the Pythagorean Theorem.
C	$ \large 3x+\sqrt{5}{{x}^{2}}$ Hint: $\sqrt {5 x^2}$ is not $\sqrt {5}x^2$.
D	$ \large 3x+\sqrt{5}{{x}^{{}}}$ Hint: To find the hypotenuse, h, use the Pythagorean Theorem: $x^2+(2x)^2=h^2.$ $5x^2=h^2,h=\sqrt{5}x$. The perimeter is this plus x plus 2x.

Question 17 Explanation:

Topic: Recognize and apply connections between algebra and geometry (e.g., the use of coordinate systems, the Pythagorean theorem) (Objective 0024).

Question 18

Cell phone plan A charges $3 per month plus $0.10 per minute. Cell phone plan B charges $29.99 per month, with no fee for the first 400 minutes and then $0.20 for each additional minute.

Which equation can be used to solve for the number of minutes, m (with m>400) that a person would have to spend on the phone each month in order for the bills for plan A and plan B to be equal?

A	$ \large 3.10m=400+0.2m$ Hint: These are the numbers in the problem, but this equation doesn't make sense. If you don't know how to make an equation, try plugging in an easy number like m=500 minutes to see if each side equals what it should.
B	$ \large 3+0.1m=29.99+.20m$ Hint: Doesn't account for the 400 free minutes.
C	$ \large 3+0.1m=400+29.99+.20(m-400)$ Hint: Why would you add 400 minutes and $29.99? If you don't know how to make an equation, try plugging in an easy number like m=500 minutes to see if each side equals what it should.
D	$ \large 3+0.1m=29.99+.20(m-400)$ Hint: The left side is $3 plus $0.10 times the number of minutes. The right is $29.99 plus $0.20 times the number of minutes over 400.

Question 18 Explanation:

Identify variables and derive algebraic expressions that represent real-world situations (Objective 0020).

Question 19

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?

A	$ \large 65\text{ }{{\text{m}}^{3}}$ Hint: A bigger pool would hold more water.
B	$ \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.
C	$ \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.
D	$ \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 19 Explanation:

Topic: Determine how the characteristics (e.g., area, volume) of geometric figures and shapes are affected by changes in their dimensions (Objective 0023).

Question 20

Which of the following is equal to eleven billion four hundred thousand?

A	$ \large 11,400,000$ Hint: That's eleven million four hundred thousand.
B	$\large11,000,400,000$
C	$ \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).
D	$ \large 11,400,000,000$ Hint: That's eleven billion four hundred million

Question 20 Explanation:

Topic: Place Value (Objective 0016)

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A	\( \large\pi , \quad \sqrt{6},\quad \sqrt{\dfrac{1}{2}}\)
B	\( \large\pi , \quad \sqrt{9}, \quad \pi +1\) Hint: \( \sqrt{9}=3\)
C	\( \large\dfrac{1}{3},\quad \dfrac{5}{4},\quad \dfrac{2}{9}\) Hint: These are all rational.
D	\( \large-3,\quad 14,\quad 0\) Hint: These are all rational.

A	\( \large {{N}^{2}}\) Hint: Squaring always gives a non-negative value.
B	\( \large 6-N\) 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.
C	\( \large -N\) Hint: If N is negative, then -N is positive
D	\( \large 6+N\) Hint: For example, if \(N=-10\), then \(6+N = -4\)

A	\( \large t=s+15\) Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation?
B	\( \large s=t+15\) Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation?
C	\( \large t=15s\) 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.
D	\( \large s=15t\)

A	\( \large \dfrac{1}{3}\) Hint: This is the probability that the first child picks a red gumball, but not that the first two children pick red gumballs.
B	\( \large \dfrac{1}{8}\) Hint: Are you adding things that you should be multiplying?
C	\( \large \dfrac{1}{9}\) Hint: This would be the probability if the gumballs were returned to the bag.
D	\( \large \dfrac{1}{15}\) 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.

A	\( \large \dfrac{3}{100}=\dfrac{d}{753}\) Hint: Units on both side are inches/mile, and both numerators and denominators correspond -- this one is correct.
B	\( \large \dfrac{3}{100}=\dfrac{753}{d}\) Hint: Unit on the left is inches per mile, and on the right is miles per inch. The proportion is set up incorrectly (which is what we wanted). Another strategy is to notice that one of A or B has to be the answer because they cannot both be correct proportions. Then check that cross multiplying on A gives part D, so B is the one that is different from the other 3.
C	\( \large \dfrac{3}{d}=\dfrac{100}{753}\) Hint: Unitless on each side, as inches cancel on the left and miles on the right. Numerators correspond to the map, and denominators to the real life distances -- this one is correct.
D	\( \large 100d=3\cdot 753\) Hint: This is equivalent to part A.

A	\( \large 8\text{ c}{{\text{m}}^{2}} \) Hint: Don't just count the dots inside, that doesn't give the area. Try adding segments so that the slanted lines become the diagonals of rectangles.
B	\( \large 11\text{ c}{{\text{m}}^{2}}\) Hint: Try adding segments so that the slanted lines become the diagonals of rectangles.
C	\( \large 11.5\text{ c}{{\text{m}}^{2}}\) Hint: An easy way to do this problem is to use Pick's Theorem (of course, it's better if you understand why Pick's theorem works): area = # pegs inside + half # pegs on the border - 1. In this case 8+9/2-1=11.5. A more appropriate strategy for elementary classrooms is to add segments; here's one way. There are 20 1x1 squares enclosed, and the total area of the triangles that need to be subtracted is 8.5
D	\( \large 12.5\text{ c}{{\text{m}}^{2}}\) Hint: Try adding segments so that the slanted lines become the diagonals of rectangles.

A	2 Hint: \(10^3 \times 10^4=10^7\), and note that if you're guessing when the answers are so closely related, you're generally better off guessing one of the middle numbers.
B	20 Hint: \( \dfrac{\left( 4\times {{10}^{3}} \right)\times \left( 3\times {{10}^{4}} \right)}{6\times {{10}^{6}}}=\dfrac {12 \times {{10}^{7}}}{6\times {{10}^{6}}}=\)\(2 \times {{10}^{1}}=20 \)
C	200 Hint: \(10^3 \times 10^4=10^7\)
D	2000 Hint: \(10^3 \times 10^4=10^7\), and note that if you're guessing when the answers are so closely related, you're generally better off guessing one of the middle numbers.

A	\( \large{{2}^{5}}\cdot {{3}^{6}}\cdot 5\) Hint: This is the product of the numbers, not the LCM.
B	\( \large{{2}^{3}}\cdot {{3}^{3}}\cdot 5\) 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.
C	\( \large{{2}^{2}}\cdot {{3}^{3}}\cdot 5\) Hint: 216 is a multiple of 8.
D	\( \large{{2}^{2}}\cdot {{3}^{2}}\cdot {{5}^{2}}\) Hint: Not a multiple of 216 and not a multiple of 540.

A	\( \large 2\) Hint: Write \(8^3\) as a power of 2.
B	\( \large \dfrac{1}{2}\) Hint: \(8^3 \cdot {2}^{-10}={(2^3)}^3 \cdot {2}^{-10}\) =\(2^9 \cdot {2}^{-10} =2^{-1}\)
C	\( \large 16\) Hint: Write \(8^3\) as a power of 2.
D	\( \large \dfrac{1}{16}\) Hint: Write \(8^3\) as a power of 2.

A	\( \large \dfrac{1}{7}\) 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.
B	\( \large \dfrac{1}{14}\) Hint: What would be the sample space here? Ie, how would you list 14 things that you pick one from?
C	\( \large \dfrac{1}{42}\) 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.
D	\( \large \dfrac{1}{49}\) Hint: This is the probability that they are both born on a particular day, e.g. Sunday.

A	\( \large A = 3, B = 2, C=0\) 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.
B	\( \large A = 0, B = 0, C=4\) Hint: Not divisible by 10, since it doesn't end in 0.
C	\( \large A = 4, B = 2, C=0\) 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.
D	\( \large A =1, B=0, C=0\) 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.

A	\( \large2\cdot 5\cdot 7\) Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM.
B	\( \large{{2}^{3}}\cdot {{3}^{2}}\cdot 5 \cdot 7\) Hint: 1260 is not divisible by 8, so it isn't a multiple of this N.
C	\( \large3 \cdot 5 \cdot 7\) Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM.
D	\( \large{{3}^{2}}\cdot 5\cdot 7\) Hint: \(1260=2^2 \cdot 3^2 \cdot 5 \cdot 7\) and \(60=2^2 \cdot 3 \cdot 5\). In order for 1260 to be the LCM, N has to be a multiple of \(3^2\) and of 7 (because 60 is not a multiple of either of these). N also cannot introduce a factor that would require the LCM to be larger (as in choice b).

A	\( \large 6x\) Hint: Use the Pythagorean Theorem.
B	\( \large 3x+5{{x}^{2}}\) Hint: Don't forget to take square roots when you use the Pythagorean Theorem.
C	\( \large 3x+\sqrt{5}{{x}^{2}}\) Hint: \(\sqrt {5 x^2}\) is not \(\sqrt {5}x^2\).
D	\( \large 3x+\sqrt{5}{{x}^{{}}}\) Hint: To find the hypotenuse, h, use the Pythagorean Theorem: \(x^2+(2x)^2=h^2.\) \(5x^2=h^2,h=\sqrt{5}x\). The perimeter is this plus x plus 2x.

A	\( \large 3.10m=400+0.2m\) Hint: These are the numbers in the problem, but this equation doesn't make sense. If you don't know how to make an equation, try plugging in an easy number like m=500 minutes to see if each side equals what it should.
B	\( \large 3+0.1m=29.99+.20m\) Hint: Doesn't account for the 400 free minutes.
C	\( \large 3+0.1m=400+29.99+.20(m-400)\) Hint: Why would you add 400 minutes and $29.99? If you don't know how to make an equation, try plugging in an easy number like m=500 minutes to see if each side equals what it should.
D	\( \large 3+0.1m=29.99+.20(m-400)\) Hint: The left side is $3 plus $0.10 times the number of minutes. The right is $29.99 plus $0.20 times the number of minutes over 400.

A	\( \large 65\text{ }{{\text{m}}^{3}}\) Hint: A bigger pool would hold more water.
B	\( \large 65\cdot 2\text{ }{{\text{m}}^{3}}\) Hint: Try a simpler example, say doubling the sides of the base of a 1 x 1 x 1 cube.
C	\( \large 65\cdot 4\text{ }{{\text{m}}^{3}}\) 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.
D	\( \large 65\cdot 8\text{ }{{\text{m}}^{3}}\) Hint: Try a simpler example, say doubling the sides of the base of a 1 x 1 x 1 cube.

A	\( \large 11,400,000\) Hint: That's eleven million four hundred thousand.
B	\(\large11,000,400,000\)
C	\( \large11,000,000,400,000\) Hint: That's eleven trillion four hundred thousand (although with British conventions; this answer is correct, but in the US, it isn't).
D	\( \large 11,400,000,000\) Hint: That's eleven billion four hundred million

MTEL General Curriculum Mathematics Practice

Which of the lists below contains only irrational numbers?

In each expression below N represents a negative integer. Which expression could have a negative value?

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?

Which of the following is not possible?

An equiangular triangle that is not equilateral.

An equiangular quadrilateral that is not equilateral.

An equilateral quadrilateral that is not equiangular.

An equiangular hexagon that is not equilateral.

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?

A map has a scale of 3 inches = 100 miles. Cities A and B are 753 miles apart. Let d be the distance between the two cities on the map. Which of the following is not correct?

Here is a student's work on several multiplication problems:

For which of the following problems is this student most likely to get the correct solution, even though he is using an incorrect algorithm?

58 x 22

16 x 24

31 x 23

141 x 32

Which of the lines depicted below is a graph of \( \large y=2x-5\)?

a

b

c

d