CLEP College Algebra FREE Sample Practice Questions

3- A
Subtracting (2x) and adding 5 to both sides of (2x – 5 ≥ 3x – 1) gives (-4 ≥ x). Therefore, (x) is a solution to (2x – 5 ≥ 3x – 1) if and only if (x) is less than or equal to (-4) and x is NOT a solution to( 2x – 5 ≥ 3x – 1) if and only if (x) is greater than (-4). Of the choices given, only (-2) is greater than 9-4) and, therefore, cannot be a value of (x).

4- D
Given the two equations, substitute the numerical value of a into the second equation to solve for (x). (a=sqrt{3}, 4a=sqrt{4x})
Substituting the numerical value for an into the equation with x is as follows.
(4(sqrt{3})=sqrt{4x}),From here, distribute the 4. (4sqrt{3}=sqrt{4x})
Now square both sides of the equation. ((4sqrt{3})^2=(sqrt{4x})^2)
Remember to square both terms within the parentheses. Also, recall that squaring a square root sign cancels them out. (4^2 sqrt{3}^2=4x, 16(3)=4x, 48=4x, x=12)

5- C
First square both sides of the equation to get (4m-3=m^2)
Subtracting both sides by (4m-3) gives us the equation (m^2-4m+3=0)
Here you can solve the quadratic equation by factoring to get ((m-1)(m-3)=0)
For the phrase ((m-1)(m-3)) to equal zero, (m=1) or (m=3)

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6- C
Let (x) be the number. Write the equation and solve for (x. (28-x)÷x= 3)
Multiply both sides by (x). ((28-x)= 3x), then add x both sides. (28=4x), now divide both sides by 4.( x=7)

7- B
The sum of supplement angles is 180. Let (x) be that angle. Therefore, (x+9x=180)
(10x=180), divide both sides by 10: (x=18)

8- B
tan(θ=frac{opposite}{adjacent})
tan(θ=frac{5}{12}⇒) we have the following right triangle. Then
(c=sqrt{5^2+12^2 }=sqrt{25+144}=sqrt{169}=13)
cos(θ=frac{adjacent}{hypotenuse}=frac{12}{13})
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9- D
The amplitude in the graph of the equation (y=)acosb(x) is a. (a and b are constant)
In the equation (y=)cos(x), the amplitude is 2 and the period of the graph is (2π).
The only choice that has two times the amplitude of graph (y =) cos (x) is (y=2+2) cos (x)
They both have an amplitude of 2 and a period of (2π).

10- D
Substituting 6 for (x) and 14 for (y) in (y = nx+2) gives (14=(n)(6)+2),
which gives (n=2). Hence, (y=2x+2). Therefore, when (x = 10), the value of (y) is
(y=(2)(10)+2 = 22).

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Official source: the full list of CLEP exams, what each one covers and how colleges award credit for them are published by the College Board at clep.collegeboard.org.