TSI Math FREE Sample Practice Questions

TSI Math FREE Sample Practice Questions: work through realistic questions in the same format and difficulty as the real test, then check every answer against a full step-by-step solution so you can see where you lost marks and what to review next.

The TSI test has officially been the TSI Assessment 2.0 (TSIA2) since January 2021. Its math section starts with a 20-question College Readiness Classification test scored from 910 to 990, and a score of 950 or higher meets the Texas college readiness standard. If you score below 950, a diagnostic test follows, and reaching Diagnostic Level 6 also meets the standard.

10 Sample TSI Math Practice Questions

1- Solve the equation: \(\log_4(x+2) – \log_4(x-2) = 1\)

A. \(10\)

B. \(\frac{3}{10}\)

C. \(\frac{10}{3}\)

D. \(3\)

2- Solve: \(e^{5x + 1} = 10\)

A. \(\frac{\ln(10) + 1}{5}\)

B. \(\frac{\ln(10) – 1}{5}\)

C. \(5\ln(10) + 2\)

D. \(5\ln(10) – 2\)

3- If \(f(x) = x – \frac{5}{3}\) and \(f^{-1}\) is the inverse of \(f(x)\), what is the value of \(f^{-1}(5)\)?

A. \(\frac{10}{3}\)

B. \(\frac{3}{20}\)

C. \(\frac{20}{3}\)

D. \(\frac{3}{10}\)

4- What is \(\cos 30^\circ\)?

A. \(\frac{1}{2}\)

B. \(\frac{\sqrt{2}}{2}\)

C. \(\frac{\sqrt{3}}{2}\)

D. \(\sqrt{3}\)

5- If \(\theta\) is an acute angle and \(\sin\theta = \frac{3}{5}\), then \(\cos\theta = \space ?\)

A. \(-1\)

B. \(0\)

C. \(\frac{4}{5}\)

D. \(\frac{5}{4}\)

6- What is the solution of the following system of equations?
\(-2x – y = -9\)
\(5x – 2y = 18\)

A. \((-1, 2)\)

B. \((4, 1)\)

C. \((1, 4)\)

D. \((4, -2)\)

7- Solve.
\(|9 – (12 ÷ |2 – 5|)| = \space ?\)

A. \(9\)

B. \(-6\)

C. \(5\)

D. \(-5\)

8- If \(\log_2 x = 5\), then \(x = \space ?\)

A. \(2^{10}\)

B. \(\frac{5}{2}\)

C. \(2^{6}\)

D. \(32\)

9- What’s the reciprocal of \(\frac{x^3}{16}\)?

A. \(\frac{16}{x^3}-1\)

B. \(\frac{48}{x^3}\)

C. \(\frac{16}{x^3}+1\)

D. \(\frac{16}{x^3}\)

10- Find the inverse function for \(f(x) = \ln(2x + 1)\).

A. \(\frac{1}{2}(e^{x} – 1)\)

B. \(e^{x} + 1\)

C. \(\frac{1}{2}(e^{x} + 1)\)

D. \(e^{x} – 1\)

Answers:

1- C
We know that \(\log_a b – \log_a c = \log_a \frac{b}{c}\) and \(\log_a b = c ⇒ b = a^c\).
Then: \(\log_4(x+2) – \log_4(x-2) = \log_4 \frac{x+2}{x-2} = 1 ⇒ \frac{x+2}{x-2} = 4^1 = 4\)
\(x + 2 = 4(x – 2) ⇒ x + 2 = 4x – 8 ⇒ 3x = 10 ⇒ x = \frac{10}{3}\)
Check: \(x – 2 = \frac{4}{3} > 0\), so the answer works in the original equation.

2- B
Take the natural log of both sides:
\(\ln(e^{5x + 1}) = \ln(10)\)
Since \(\ln(e^{a}) = a\): \(5x + 1 = \ln(10)\)
\(5x = \ln(10) – 1 ⇒ x = \frac{\ln(10) – 1}{5}\)

3- C
\(f(x) = x – \frac{5}{3} ⇒ y = x – \frac{5}{3} ⇒ x = y + \frac{5}{3}\)
\(f^{-1}(x) = x + \frac{5}{3}\)
\(f^{-1}(5) = 5 + \frac{5}{3} = \frac{20}{3}\)

4- C
\(\cos 30^\circ = \frac{\sqrt{3}}{2}\)

5- C

sin\(θ=\frac{3}{5}\)⇒ we have following triangle, then:
\(c=\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4\)
\(\cos θ=\frac{4}{5}\)

6- B
\(-2x – y = -9\)
\(5x – 2y = 18\)
⇒ Multiply the first equation by \(-2\):
\(4x + 2y = 18\)
\(5x – 2y = 18\)
Add the two equations together ⇒ \(9x = 36 ⇒ x = 4\), then: \(-2(4) – y = -9 ⇒ y = 1\)

7- C
\(|9 – (12 ÷ |2 – 5|)| = |9 – (12 ÷ |-3|)| = |9 – (12 ÷ 3)| = |9 – 4| = |5| = 5\)

8- D
METHOD ONE
\(\log_2x = 5\)
Apply logarithm rule:\(a = \log_b(b^a)\)
\(5 = \log_2(2^5) = \log_2(32)\)
\(\log_2x = \log_2(32)\)
When the logs have the same base: \(\log_b(f(x)) = \log_b(g(x))⇒ f(x) = g(x)\)
then: \(x = 32\)
METHOD TWO
We know that:
\(\log_ab=c⇒b=a^c\)
\(\log_2x=5⇒x=2^5=32\)

9- D
\(\frac{x^3}{16}\)
⇒ reciprocal is : \(\frac{16}{x^3}\)

10- A
\(f(x) = \ln(2x + 1)\)
\(y = \ln(2x + 1)\)
Change variables \(x\) and \(y\): \(x = \ln(2y + 1)\)
Solve for \(y\): \(e^{x} = 2y + 1 ⇒ y = \frac{e^{x}-1}{2}=\frac{1}{2}(e^{x} – 1)\)

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