Free Algebra 1 Practice Test Questions

1. C. Let x represent the total value of last year’s sales, Set up an equation and solve it for x. Since the salesman’s sales increased by 20% since last year’s, his current sales 120% of x, or 1.2x. So,
1.2x = 60,000

Solve the equation for x by dividing both sides by 1.2.

Therefore, the salesman sold $50,000 worth of steak knives last year.

2. E. This equation is a proportion, so it can be solved by cross-multiplication. Form a new equation by multiplying the numerator of each fraction by the denominator of the fraction on the other side. Then, simplify the result and solve for x.

x/3 = (2x + 3)/7
7x = 3(2x + 3)
7x = 6x + 9
x = 9

3. E. To begin, simplify the right side of the equation by distributing the 3.

3(2y + 4) = 8y
6y + 12 = 8y

Then, solve the equation by isolating the variable and dividing both sides by the coefficient.

4. E. This equation involves an absolute value function. The absolute value of a number is its distance from zero on a number line. Since distances are never negative, the absolute value of a number is always positive (or equal to zero). In order to make the equation true, the expression inside the absolute value, x + 5, can equal either -3 or 3 since the absolute value of both values is 3. Write two equations and solve each.

5. D. To begin, solve the given equation for x.

3x + 8x + 4x = 6x + 63
15x = 6x + 63
9x = 63
x = 7
Next, substitute 7 for x in the expression 5x + 23 and simplify the result.

5(7) + 23 = 35 + 23 = 58

6. B. The product of a number and its reciprocal, or multiplicative inverse, is 1. For a fraction, the reciprocal can be found by inverting (or switching) the numerator and denominator. Since –3 can be written as ,
its reciprocal is
.

7. C. Since the square root of x is between 3 and 11, we know that the inequality 3

11 is true. To find the value of x, square each part of the inequality. The result is the inequality 9 < x < 121.

8. A. Write each piece of information as an equation using the variables A, B, and C for the current ages of Andrew, Brad, and Carol, respectively.

C = 3A
B = A + 2
(A + 6) + (B + 6) = C + 6

This is a system of equations. Since the first two equations are already solved for C and B, substitute the expressions on the right side into the third equation. Then, solve for A.

(A + 6) + (B + 6) = C + 6
(A + 6) + [(A + 2) + 6) = (3A) + 6
2A + 14 = 3A + 6
A = 8

Therefore, Andrew is 8 years old. To find Carol’s age, multiply Andrew’s age by three. Thus, Carol is currently 24 years old.

9. B. To begin, write an equation relating the cost C to the distance D. If one travels more than half a mile, the cost is $3.25 plus the $0.70 times the distance in miles, excluding the first half-mile. Because the first half-mile is excluded, ½, or 0.5 must be subtracted from the distance when multiplying by 0.70.

C = 3.25 + 0.70(D – 0.5)

To find how far someone can travel with $12, substitute 12 for C and solve for D.

12 = 3.25 + 0.70(D – 0.5)
12 = 3.25 + 0.7D – 0.35
9.1 = 0.7D
D = 13

Therefore, someone can travel 13 miles on $12.

10. D. First, simplify the left side of the equation.

13 – 2(2x + 1) = 1
13 – 4x – 2 = 1
–4x + 11 = 1

Then, isolate the variable and solve for x.

–4x = –10

Last Updated: June 4, 2019