10 Most Common CLEP College Algebra Math Questions
B. 78
C. 84
D. 86
E. 92
10- The score of Emma was half as that of Ava and the score of Mia was twice that of Ava. If the score of Mia was 40, what is the score, Emma?
A. 10
B. 15
C. 20
D. 30
E. 40
Best CLEP College Algebra Math Prep Resource for 2026
Answers:
1- B
To find the discount, multiply the number by \((100\%\)-rate of discount). Therefore, for the first discount we get: (D) \((100\%-25\%)\)=(D)(0.75)= 0.75. For increase of \(10\%\): (0.75 D)\((100\%+10\%)\)=(0.75 D)(1.10)=0.82 D=\(82\%\) of D or 0.82 D.
2- D
The union of A and B is: \(A∪B\) = {1,2,3,4,5,6,11,15}
The intersection of \((A∪B)\) and C is: \((A∪B)∩C\)={5,11}
3- E
average\(=\frac{sum \ of \ terms}{number \ of \ terms} ⇒20=\frac{13+15+20+x}{4}⇒80=48+x ⇒ x=32\)
4- B
Since \(f(x)\) is linear function with a negative slop, then when \(x=-2,f(x)\) is maximum and when \(x=3,f(x)\) is minimum. Then the ratio of the minimum value to the maximum value of the function is: \(\frac{f(3)}{f(-2)}=\frac{-3(3)+1}{-3(-2)+1}=\frac{-8}{7}=-\frac{8}{7}\)
5- C
There can be 0, 1, or 2 solutions to a quadratic equation. In standard form, a quadratic equation is written as: \(ax^2+bx+c=0\)
For the quadratic equation, the phrase \(b^2-4ac\) is called the discriminant. If the discriminant is positive, there are 2 distinct solutions for the quadratic equation. If the discriminant is 0, there is one solution for the quadratic equation and if it is negative the equation does not have any solutions.
To find number of solutions for \(x^2=4x-3\), first, rewrite it as \(x^2-4x+3=0\).
Find the value of the discriminant. \(b^2-4ac=(-4)^2-4(1)(3)=16-12=4\)
Since the discriminant is positive, the quadratic equation has two distinct solutions.
6- C
If the value of \(|x-3|+3\) is equal to 0, then \(|x-3|+3=0\). Subtracting 3 from both sides of this equation gives \(|x-3|\)=\(-3\). The expression \(|x – 3|\) on the left side of the equation is the absolute value of \(x \)\(- 3\), and the absolute value can never be a negative number.
Thus \(|x-3|\)=\(-3\) has no solution. Therefore, there are no values for \(x\) for which the value of \(|x(-3)|+3\) is equal to 0.
7- A
Let \(x\) be the number of years. Therefore, $2,000 per year equals 2000\(x\). starting from $26,000 annual salary means you should add that amount to 2000\(x\). Income more than that is:
\(I>2000 x + 26000\)
8- E
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: \(a^2+b^2=c^2\)
\(60^2+80^2= c^2⇒ 3600 +6400 =c^2 ⇒ 10000 =c^2 ⇒ c=100\)
9- D
In the figure angle A is labeled \((3x-2)\) and it measures 37. Thus, 3\(x\)\(-2\)=37 and 3\(x\)=39 or \(x\)=13. That means that angle B, which is labeled \((5x)\), must measure 5×13=65.
Since the three angles of a triangle must add up to 180, 37+65+y\(-8\)=180, then:
y+94=108→y=180\(-94\)=86
10- A
If the score of Mia was 40, therefore the score of Ava is 20. Since the score of Emma was half as that of Ava, therefore, the score of Emma is 10.
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