Top 10 CLEP College Algebra Practice Questions

B. \( y = 5x\)

C. \( y≥x\)

D. \( y+4x=z\)

E. \( y>x\)

9- What is the solution to the following inequality?
\(|x-10|≤3\)

A. \(x≥13 ∪ x≤7\)

B. \( 7≤x≤13\)

C. \( x≥13 \)

D. \( x≤7\)

E. Set of real numbers

10- Two-third of 15 is equal to \(\frac{2}{5}\) of what number?

A. 12

B. 20

C. 25

D. 60

E. 90

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Answers:

2- D
Here we can substitute 8 for \(x\) in the equation. Thus, \(y-3=2(8+5), y-3=26\)
Adding 3 to both sides of the equation: \(y=26+3, y=29\)

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3- D
Use Pythagorean Theorem: \(a^2+b^2=c^2⇒5^2+12 ^2=c^2⇒169=c^2⇒c=13\)

4- E
The ratio of boys to girls is 2:3. Therefore, there are 2 boys out of 5 students. To find the answer, first, divide the total number of students by 5, then multiply the result by 2.
\(500÷5=100 ⇒100 × 2=200\)

5- E
Use FOIL (First, Out, In, Last).\((7x+2y)(5x+2y)= \)
\(35x^2+14xy+10xy+4y^2=\)
\(35x^2+ 24xy+4y^2\)

6- E
Use distributive property: \(5x(4+2y)=20x+10xy\)

7- E
\(y=5ab+3b^3\). Plug in the values of a and b in the equation: \(a=2\) and \(b=3\).
\(y=5 (2)(3)+3 (3)^3=30+3(27)=30+81=111\)

8- D
\(x\) and \(z\) are colinear. \(y\) and \(5x\) are colinear. Therefore,
\(x+z=y+5x\),subtract \(x\) from both sides,then,\(z=y+4x\)

9- B
\(|x-10|≤3→-3≤x-10≤3→-3+10≤x-10+10≤3+10→7≤x≤13\)

10- C
Let \(x\) be the number. Write the equation and solve for \(x\). \(\frac{2}{3} ×15= \frac{2}{5}\). \(x⇒\frac{2×15}{3}=\frac{2x}{5 }\), use cross multiplication to solve for \(x\). \(5×30=2x×3⇒150=6x⇒x=25\)

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