TExES Mathematics 7-12 (235)
TExES Mathematics 7-12 (235) is a trademark of its owner, which is not affiliated with and does not endorse Test Treasure Publication or TTP Prep.
TExES Mathematics 7-12 (235) is a trademark of its owner, which is not affiliated with and does not endorse Test Treasure Publication or TTP Prep.
10 sample questions for the TExES Mathematics 7-12 (235), with the correct answer and an explanation for each. No account needed. Try each one before you open the answer.
Sections covered on the exam: Number Concepts, Patterns and Algebra, Geometry and Measurement, Probability and Statistics, Mathematics.
Question 1 · Number Concepts
Answer: B. 13/55
Let x = 0.2363636.... Then 1,000x = 236.3636... and 10x = 2.3636..., so 990x = 234 and x = 234/990 = 13/55. The fraction 59/250 equals the terminating decimal 0.236, and 118/495 comes from using 236/990, which counts the non-repeating digit twice.
Question 2 · Patterns and Algebra
Answer: D. f^(-1)(x) = (x + 3)/(x − 2)
Swap and solve: x = (2y + 3)/(y − 1) gives xy − x = 2y + 3, so y(x − 2) = x + 3 and y = (x + 3)/(x − 2). The reciprocal (x − 1)/(2x + 3) confuses the inverse function with 1/f, and (x + 3)/(2 − x) comes from a sign slip when collecting the y terms.
Question 3 · Patterns and Algebra
Answer: B. -12/5
The range of arccos is [0, π], so θ = arccos(-5/13) lies in Quadrant II with cos θ = -5/13 and sin θ = 12/13. Then tan θ = (12/13)/(-5/13) = -12/5. Placing θ in Quadrant I gives 12/5, and inverting the ratio gives -5/12.
Question 4 · Geometry and Measurement
A solid is built from identical unit cubes stacked in columns on a 2-by-2 base.
Top view: all four squares of the 2-by-2 base are covered.
Front view: the left column of the view is 3 cubes tall and the right column is 2 cubes tall.
Right-side view: the column nearer the front is 3 cubes tall and the column farther back is 2 cubes tall.
Answer: B. 9
Each stack's height is limited by both views that show it: the left-front stack can be min(3, 3) = 3, the left-back stack min(3, 2) = 2, the right-front stack min(2, 3) = 2, and the right-back stack min(2, 2) = 2. The maximum is 3 + 2 + 2 + 2 = 9. The value 7 is the least number of cubes consistent with the views, not the greatest.
Question 5 · Patterns and Algebra
Answer: D. If a quadrilateral's diagonals are not perpendicular, then it is not a square.
A conditional p → q is equivalent only to its contrapositive ¬q → ¬p, which is the statement about diagonals that are not perpendicular. The converse and the inverse are equivalent to each other but not to the original; a rhombus that is not a square has perpendicular diagonals, so the converse and inverse fail, and so does the biconditional.
Question 6 · Number Concepts
Answer: C. sqrt(12) + sqrt(27)
sqrt(12) + sqrt(27) = 2sqrt(3) + 3sqrt(3) = 5sqrt(3), which is irrational. The others simplify to rationals: sqrt(50)/sqrt(8) = sqrt(25/4) = 5/2, (2 − sqrt(3))(2 + sqrt(3)) = 4 − 3 = 1, and (sqrt(2) + 2sqrt(2))^2 = (3sqrt(2))^2 = 18.
Question 7 · Patterns and Algebra
Answer: B. 3
g(1) = 3 and f(3) = 1, so f(g(1)) = 1; f^(-1)(3) is the input with f(x) = 3, which is 2. Composing in the wrong order uses g(f(1)) = g(4) = 2 and gives a total of 4, and reading f^(-1)(3) as f(3) gives a total of 2. The correct total is 1 + 2 = 3.
Question 8 · Patterns and Algebra
Answer: D. 4 sqrt(5)/9
In Quadrant III cosine is negative, so cos θ = -sqrt(1 - 4/9) = -sqrt(5)/3. Then sin 2θ = 2 sin θ cos θ = 2(-2/3)(-sqrt(5)/3) = 4 sqrt(5)/9. Taking cos θ positive gives the negative value, and dropping the factor 2 gives 2 sqrt(5)/9.
Question 9 · Geometry and Measurement
Answer: C. 100 square cm
The volume ratio 250/54 = 125/27 = (5/3)^3, so the linear scale factor is 5/3 and areas scale by (5/3)^2 = 25/9. The larger lateral area is 36 × 25/9 = 100 square cm. Multiplying by 5/3 treats area as length (60), and multiplying by the volume ratio gives 166.7.
Question 10 · Patterns and Algebra
Answer: A. Some student in the class failed every quiz.
The statement has the form "for every student, there exists a quiz the student passed," so its negation is "there exists a student such that for every quiz, the student did not pass," that is, some student failed every quiz. Negating only the inner part gives "some student failed at least one quiz," which can be true at the same time as the original.
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