Full-Length TSI Math Practice Test-Answers and Explanations
Did you take the TSI Math Practice Test? If so, then it’s time to review your results to see where you went wrong and what areas you need to improve.
TSI Mathematical Reasoning Practice Test Answers and Explanations
1- Choice D is correct
If \(1.05 < x ≤ 3.04\), then \(x\) cannot be equal to 3.40. Because: \(3.04<3.40\)
2- Choice B is correct
\(a=8⇒\) area of triangle is \(=\frac{1}{2} (8×8)=\frac{64}{2}=32\) cm
3- Choice C is correct
\(66^\circ + 42^\circ = 108^\circ\)
\(180^\circ – 108^\circ = 72^\circ\)
The value of the third angle is \(72^\circ\).
4- Choice B is correct
Simplify:
\(10 – \frac{2}{3} x ≥ 12 ⇒ – \frac{2}{3} x ≥ 2 ⇒ – x ≥ 3 ⇒ x ≤ – 3\)
5- Choice A is correct
Factor each trinomial \(x^2 – 2x – 8\) and \(x^2 – 6x + 8\)
\(x^2 – 2x – 8 ⇒ (x – 4)(x + 2) \)
\(x^2 – 6x + 8 ⇒ (x – 2)(x – 4) \)
\((x – 4)\) is a factor of both trinomial.
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6- Choice B is correct
\(\frac{1+b}{6b^2}=\frac{1}{b^2} ⇒(b≠0) b^2+b^3=6b^2⇒b^3-5b^2=0⇒b^2 (b-5)=0⇒b-5=0⇒b=5\)
7- Choice D is correct
Use FOIL (First, Out, In, Last)
\((x + 7) (x + 5) = x^2 + 5x + 7x + 35 = x^2 + 12x + 35\)
8- Choice A is correct
\(\frac{54}{6}=\frac{27}{3}=9, \frac{48}{6}=\frac{24}{3}=8, \frac{36}{6}=\frac{18}{3}=6, \frac{59}{6}=\frac{59}{6} \)
59 is prime number
The answer is 54.
9- Choice B is correct
\(x^2 – 64 = 0 ⇒ x^2 = 64 ⇒ x = 8\)
10- Choice C is correct
If \(a = 8\) then \(b = \frac{8^2}{4} + 4 ⇒ b = \frac{64}{4} + 4 ⇒b = 16 + 4 = 20\)
11- Choice D is correct
tan \((– π/6) = – \frac{\sqrt{3}}{3}\)
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12- Choice A is correct
\(\frac{\sqrt{32a^5 b^3 }}{\sqrt{2ab^2}}= \frac{4a^2 b\sqrt{2ab}}{b\sqrt{2a}} = 4a^2\sqrt{b}\)
13- Choice D is correct
\(E = 4 + A\)
\(A = S – 3\)
14- Choice C is correct
\(\begin{cases}5x + y = 9\\10x-7y= -18\end{cases} ⇒\) Multiplication \((–2)\) in first equation \(⇒ \begin{cases}-10x- 2y = -18\\10x-7y= -18\end{cases}\) Add two equations together \(⇒ –9y = –36 ⇒ y = 4\) then: \(x = 1\)
15- Choice B is correct
\((x – h)^2 + (y – k)^2 = r^2\) ⇒ center: (h,k) and radius: r
\((x – 3)^2 + (y + 6)^2 = 12\) ⇒ center: \((3,-6)\) and radius: \(2\sqrt{3}\)
16- Choice D is correct
c\((3)=(3)^2+10(3)+30=9+30+30=69 \)
\(4×3=12⇒12-69=-57⇒57,000\) loss
17- Choice B is correct
sinB\(=\frac{the length of the side that is opposite that angle}{the length of the longest side of the triangle}=\frac{4}{5}\)
18- Choice C is correct
\(-7y=-6x-12⇒y=\frac{-6}{-7} x-\frac{12}{-7}⇒y=\frac{6}{7} x+\frac{12}{7}\)
19- Choice D is correct
\((g – f)(x) = g(x) – f(x) = (– x^2 – 1 – 2x) – (5 + x) – x^2 – 1 – 2x – 5 – x = – x^2 – 3x – 6\)
20- Choice B is correct
\(\frac{|3+x|}{7}≤5⇒|3+x|≤35⇒-35≤3+x≤35⇒-35-3≤x≤35-3⇒-38≤x≤32\)
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