ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Linear Inequalities
20 questions · exam conditions
0:00
Linear InequalitiesQuestion 1 of 20

Solve the inequality 3(x−4)>2x+83(x - 4) > 2x + 8 and express the solution in interval notation.

(−∞,4)(-\infty, 4)
(−∞,20)(-\infty, 20)
(4,∞)(4, \infty)
(20,∞)(20, \infty)
← Back to quizzes

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Linear Inequalities

Practice Linear Inequalities in ACCUPLACER Quantitative Reasoning, Algebra & Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Linear Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Quantitative Reasoning, Algebra & Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Solve the inequality 3(x−4)>2x+83(x - 4) > 2x + 8 and express the solution in interval notation.

  1. (−∞,4)(-\infty, 4)
  2. (−∞,20)(-\infty, 20)
  3. (4,∞)(4, \infty)
  4. (20,∞)(20, \infty) (correct answer)
Explanation: When you encounter linear inequalities, your goal is to isolate the variable while carefully handling the inequality sign. The process mirrors solving equations, with one crucial exception regarding multiplication or division by negative numbers. Start by distributing and simplifying: 3(x−4)>2x+83(x - 4) > 2x + 8 becomes 3x−12>2x+83x - 12 > 2x + 8. Next, collect like terms by subtracting 2x2x from both sides: 3x−2x−12>83x - 2x - 12 > 8, which gives you x−12>8x - 12 > 8. Finally, add 12 to both sides: x>20x > 20. In interval notation, this is (20,∞)(20, \infty), making D correct. Let's examine why the other answers are wrong. Choice A, (−∞,4)(-\infty, 4), represents x<4x < 4, which would come from incorrectly flipping the inequality sign without justification. Choice B, (−∞,20)(-\infty, 20), represents x<20x < 20, suggesting you flipped the inequality sign when you shouldn't have—remember, you only flip when multiplying or dividing by a negative number. Choice C, (4,∞)(4, \infty), represents x>4x > 4, which appears to come from an algebraic error, possibly incorrectly combining the constants or making a sign error during distribution. The key strategy for inequality problems is to treat them exactly like equations until you need to multiply or divide by a negative number—only then do you flip the inequality sign. Always double-check your final answer by testing a value from your solution set in the original inequality.

Question 2

Solve the compound inequality −4<2x−6≤8-4 < 2x - 6 \leq 8 and express the solution in interval notation.

  1. [1,7)[1, 7)
  2. (1,7](1, 7] (correct answer)
  3. (1,7)(1, 7)
  4. [1,7][1, 7]
Explanation: When you encounter a compound inequality like this one, you're dealing with two inequalities joined together that must both be satisfied simultaneously. Think of it as finding values of x that fall within a specific range. To solve −4<2x−6≤8-4 < 2x - 6 \leq 8, you need to isolate x by performing the same operations on all three parts. First, add 6 to each section: −4+6<2x−6+6≤8+6-4 + 6 < 2x - 6 + 6 \leq 8 + 6, which gives you 2<2x≤142 < 2x \leq 14. Next, divide everything by 2: 1<x≤71 < x \leq 7. This means x is greater than 1 but less than or equal to 7. In interval notation, a parenthesis indicates the endpoint is not included, while a bracket means it is included. Since x > 1 (not x ≥ 1), you use a parenthesis at 1. Since x ≤ 7, you use a bracket at 7. Therefore, the answer is (1,7](1, 7], which is choice B. Choice A uses [1,7)[1, 7), incorrectly including 1 and excluding 7 — the opposite of what the inequality shows. Choice C uses (1,7)(1, 7), correctly excluding 1 but wrongly excluding 7 as well. Choice D uses [1,7][1, 7], incorrectly including both endpoints when only 7 should be included. Remember: when translating inequality symbols to interval notation, < and > become parentheses, while ≤ and ≥ become brackets. Always double-check which endpoints your solution actually includes by looking at the inequality symbols carefully.

Question 3

If kk is a negative constant, which of the following inequalities is equivalent to kx<10kx < 10?

  1. x<10kx < \frac{10}{k}
  2. x>10kx > \frac{10}{k} (correct answer)
  3. x<10−kx < 10 - k
  4. x>10−kx > 10 - k
Explanation: To solve the inequality kx<10kx < 10 for xx, you must divide both sides by kk. The problem states that kk is a negative constant. A fundamental rule of inequalities is that when you multiply or divide both sides by a negative number, you must reverse the direction of the inequality sign. Therefore, dividing by the negative constant kk changes the '<<' to a '>>'. The result is x>10kx > \frac{10}{k}.

Question 4

What is the solution set for the inequality 3(2x+5)−10<6x+73(2x + 5) - 10 < 6x + 7?

  1. x<2x < 2
  2. x>−1x > -1
  3. No solution
  4. All real numbers (correct answer)
Explanation: First, distribute the 3 on the left side: 6x+15−10<6x+76x + 15 - 10 < 6x + 7. Combine the constant terms on the left: 6x+5<6x+76x + 5 < 6x + 7. Next, subtract 6x6x from both sides: 5<75 < 7. The variable xx has been eliminated, and the resulting statement, 5<75 < 7, is always true. This means that the original inequality is true for any real number value of xx. Therefore, the solution set is all real numbers.

Question 5

What is the smallest integer nn for which 12−5n≤−1312 - 5n \le -13?

  1. -5
  2. 4
  3. 5 (correct answer)
  4. 6
Explanation: First, solve the inequality for nn. Subtract 12 from both sides: −5n≤−13−12-5n \le -13 - 12, which simplifies to −5n≤−25-5n \le -25. Next, divide both sides by -5. Remember to reverse the inequality sign when dividing by a negative number: n≥−25−5n \ge \frac{-25}{-5}, which simplifies to n≥5n \ge 5. The question asks for the smallest integer that satisfies this condition. The integers that are greater than or equal to 5 are 5, 6, 7, and so on. The smallest of these is 5.

Question 6

Find the solution set for −34x+5≥11-\frac{3}{4}x + 5 \ge 11.

  1. x≥−8x \ge -8
  2. x≤−8x \le -8 (correct answer)
  3. x≥8x \ge 8
  4. x≤−643x \le -\frac{64}{3}
Explanation: First, isolate the term with xx by subtracting 5 from both sides: −34x≥11−5-\frac{3}{4}x \ge 11 - 5, which simplifies to −34x≥6-\frac{3}{4}x \ge 6. To solve for xx, multiply both sides by the reciprocal of −34-\frac{3}{4}, which is −43-\frac{4}{3}. Because you are multiplying by a negative number, you must reverse the inequality sign: x≤6⋅(−43)x \le 6 \cdot (-\frac{4}{3}). This simplifies to x≤−243x \le -\frac{24}{3}, or x≤−8x \le -8.

Question 7

Given that a>ba > b, which of the following must be true for the inequality ax+5>bx+8ax + 5 > bx + 8?

  1. x<3a−bx < \frac{3}{a - b}
  2. x>3a−bx > \frac{3}{a - b} (correct answer)
  3. x<13a+bx < \frac{13}{a + b}
  4. x>13a+bx > \frac{13}{a + b}
Explanation: To solve for xx, first isolate the terms containing xx on one side. Subtract bxbx from both sides: ax−bx+5>8ax - bx + 5 > 8. Then, subtract 5 from both sides: ax−bx>3ax - bx > 3. Factor out xx on the left side: x(a−b)>3x(a - b) > 3. To isolate xx, divide both sides by (a−b)(a - b). Since it is given that a>ba > b, the quantity (a−b)(a - b) is positive. Therefore, you do not need to reverse the inequality sign. The result is x>3a−bx > \frac{3}{a - b}.

Question 8

A rental company charges a flat fee of $40 plus $0.15 per mile. A second company charges a flat fee of $25 plus $0.25 per mile. For how many miles, mm, is the cost of the first company less than or equal to the cost of the second company?

  1. m≤15m \le 15
  2. m≥15m \ge 15
  3. m≤150m \le 150
  4. m≥150m \ge 150 (correct answer)
Explanation: Let mm be the number of miles. The cost for the first company is 40+0.15m40 + 0.15m. The cost for the second company is 25+0.25m25 + 0.25m. We want to find when the first company's cost is less than or equal to the second's: 40+0.15m≤25+0.25m40 + 0.15m \le 25 + 0.25m. Subtract (0.15m) from both sides: 40≤25+0.10m40 \le 25 + 0.10m. Subtract 25 from both sides: 15≤0.10m15 \le 0.10m. Divide both sides by 0.10: 150≤m150 \le m, which is the same as m≥150m \ge 150.

Question 9

To earn a B in a course, a student needs a final average of at least 80. The student's grades on three exams are 75, 83, and 78. The final exam counts as two regular exams. What is the minimum score the student must get on the final exam to earn a B?

  1. 80
  2. 81
  3. 82 (correct answer)
  4. 84
Explanation: Let xx be the score on the final exam. Since the final exam counts as two exams, we can treat the total number of exams as 3+2=53 + 2 = 5. The sum of the scores is 75+83+78+2x75 + 83 + 78 + 2x. The average must be at least 80, so we set up the inequality: 75+83+78+2x5≥80\frac{75 + 83 + 78 + 2x}{5} \ge 80. Simplify the sum: 236+2x5≥80\frac{236 + 2x}{5} \ge 80. Multiply both sides by 5: 236+2x≥400236 + 2x \ge 400. Subtract 236 from both sides: 2x≥1642x \ge 164. Divide by 2: x≥82x \ge 82. The minimum score required is 82.

Question 10

What is the solution set for the inequality −1≤4−x3≤2-1 \le \frac{4-x}{3} \le 2?

  1. −7≤x≤2-7 \le x \le 2
  2. −5≤x≤1-5 \le x \le 1
  3. −2≤x≤7-2 \le x \le 7 (correct answer)
  4. x≤−2x \le -2 or x≥7x \ge 7
Explanation: To solve this compound inequality, apply each operation to all three parts. First, multiply all parts by 3 to clear the fraction: −3≤4−x≤6-3 \le 4 - x \le 6. Next, subtract 4 from all parts: −3−4≤4−x−4≤6−4-3 - 4 \le 4 - x - 4 \le 6 - 4, which simplifies to −7≤−x≤2-7 \le -x \le 2. Finally, multiply all parts by -1 to isolate xx. Remember to reverse the inequality signs when multiplying by a negative number: 7≥x≥−27 \ge x \ge -2. This is equivalent to −2≤x≤7-2 \le x \le 7.

Question 11

A company's monthly profit, PP, is given by the formula P=150x−45,000P = 150x - 45,000, where xx is the number of units sold. How many units must be sold for the company to have a profit of at least $15,000?

  1. x≤200x \le 200
  2. x≥200x \ge 200
  3. x≤400x \le 400
  4. x≥400x \ge 400 (correct answer)
Explanation: The phrase 'at least $15,000' means the profit PP must be greater than or equal to 15,000. So we set up the inequality: 150x−45,000≥15,000150x - 45,000 \ge 15,000. To solve for xx, first add 45,000 to both sides: 150x≥15,000+45,000150x \ge 15,000 + 45,000, which simplifies to 150x≥60,000150x \ge 60,000. Finally, divide both sides by 150: x≥60,000150x \ge \frac{60,000}{150}, which gives x≥400x \ge 400. The company must sell at least 400 units.

Question 12

A student has a budget of $100 to buy notebooks and pens. Notebooks cost $4 each and pens cost $2.50 each. The student must buy exactly 10 notebooks. What is the maximum number of pens the student can buy?

  1. 23
  2. 24 (correct answer)
  3. 25
  4. 40
Explanation: First, calculate the cost of the notebooks. The student buys 10 notebooks at $4 each, so the cost is 10 \times \4 = $40.Theremainingbudgetforpensis. The remaining budget for pens is $100 - $40 = $60.Let. Let pbethenumberofpens.Thecostofthepensisbe the number of pens. The cost of the pens is2.50p.Thiscostmustbelessthanorequaltotheremainingbudget:. This cost must be less than or equal to the remaining budget: 2.50p \le 60.Tofind. To find p,dividebothsidesby2.50:, divide both sides by 2.50: p \le \frac{60}{2.50},whichis, which is p \le 24$. The maximum number of pens the student can buy is 24.

Question 13

Which of the following is equivalent to the statement −2<2(x+1)<8-2 < 2(x + 1) < 8?

  1. −2<x<3-2 < x < 3 (correct answer)
  2. −1<x<4-1 < x < 4
  3. −4<x<6-4 < x < 6
  4. x<3x < 3
Explanation: There are two ways to solve this compound inequality. Method 1: Divide all three parts by 2 first: −22<2(x+1)2<82\frac{-2}{2} < \frac{2(x+1)}{2} < \frac{8}{2}, which simplifies to −1<x+1<4-1 < x + 1 < 4. Then, subtract 1 from all three parts: −1−1<x+1−1<4−1-1 - 1 < x + 1 - 1 < 4 - 1, which gives −2<x<3-2 < x < 3. Method 2: Distribute the 2 first: −2<2x+2<8-2 < 2x + 2 < 8. Then, subtract 2 from all three parts: −4<2x<6-4 < 2x < 6. Finally, divide all three parts by 2: −2<x<3-2 < x < 3.

Question 14

Given that x<yx < y. If both sides of the inequality are multiplied by a constant cc, the resulting inequality is cx>cycx > cy. Which of the following must be true about cc?

  1. c>1c > 1
  2. c=0c = 0
  3. c<0c < 0 (correct answer)
  4. cc is a non-integer
Explanation: The original inequality is x<yx < y. After multiplying by cc, the inequality becomes cx>cycx > cy. The inequality symbol has been reversed (from less than to greater than). The rule for inequalities states that the symbol is reversed only when multiplying or dividing both sides by a negative number. Therefore, the constant cc must be negative (c<0c < 0).

Question 15

A salesperson earns a base salary of $1,200 per month plus a 5% commission on all sales over $10,000. Which inequality represents the total sales, SS, needed for the salesperson to earn a total monthly income of at least $3,000?

  1. S≥36,000S \ge 36,000
  2. S≥46,000S \ge 46,000 (correct answer)
  3. S≥70,000S \ge 70,000
  4. S≥84,000S \ge 84,000
Explanation: Let SS be the total sales. The commission is 5% (or 0.05) of the sales amount that is over $10,000, which is represented by (S−10,000)(S - 10,000). The total income is the base salary plus the commission: 1200+0.05(S−10,000)1200 + 0.05(S - 10,000). The salesperson wants to earn at least $3,000, so we set up the inequality: 1200+0.05(S−10,000)≥30001200 + 0.05(S - 10,000) \ge 3000. Subtract 1200 from both sides: 0.05(S−10,000)≥18000.05(S - 10,000) \ge 1800. Divide both sides by 0.05: S−10,000≥36,000S - 10,000 \ge 36,000. Finally, add 10,000 to both sides: S≥46,000S \ge 46,000.

Question 16

The solution to the inequality ∣2x−6∣≤8|2x - 6| \leq 8 can be written as a≤x≤ba \leq x \leq b. What is the value of a+ba + b?

  1. 4
  2. 10
  3. 6 (correct answer)
  4. 8
Explanation: When you encounter absolute value inequalities, remember that the absolute value represents distance on a number line. The inequality ∣2x−6∣≤8|2x - 6| \leq 8 means the expression inside the absolute value bars must be within 8 units of zero. To solve this, you need to consider both cases of the absolute value. The inequality ∣2x−6∣≤8|2x - 6| \leq 8 is equivalent to the compound inequality: −8≤2x−6≤8-8 \leq 2x - 6 \leq 8 Now solve both parts simultaneously. Add 6 to all three parts: −8+6≤2x−6+6≤8+6-8 + 6 \leq 2x - 6 + 6 \leq 8 + 6 −2≤2x≤14-2 \leq 2x \leq 14 Divide everything by 2: −1≤x≤7-1 \leq x \leq 7 Therefore, a=−1a = -1 and b=7b = 7, so a+b=−1+7=6a + b = -1 + 7 = 6. Looking at the wrong answers: Choice A) 4 likely comes from forgetting the negative solution and only considering the positive boundary. Choice B) 10 might result from adding the absolute values of the boundaries (|-1| + |7| = 8, though this gives 8, not 10). Choice D) 8 could come from mistakenly using the original constraint value or adding the distances incorrectly. The key strategy here is to always convert absolute value inequalities into compound inequalities. Remember that ∣expression∣≤k|expression| \leq k becomes −k≤expression≤k-k \leq expression \leq k, while ∣expression∣≥k|expression| \geq k becomes two separate inequalities. This systematic approach prevents common sign errors and ensures you capture both boundaries of the solution.

Question 17

If 3x+2y≤123x + 2y \leq 12 and y≥0y \geq 0, what is the maximum value of xx when y=1y = 1?

  1. 44
  2. 103\frac{10}{3} (correct answer)
  3. 55
  4. 143\frac{14}{3}
Explanation: When you encounter a system of inequalities with constraints, you're looking for values that satisfy all given conditions simultaneously. Here, you need to find the maximum value of xx when y=1y = 1, subject to the constraints 3x+2y≤123x + 2y \leq 12 and y≥0y \geq 0. Since y=1y = 1 and y≥0y \geq 0, the second constraint is automatically satisfied. Now substitute y=1y = 1 into the first inequality: 3x+2(1)≤123x + 2(1) \leq 12, which simplifies to 3x+2≤123x + 2 \leq 12. Subtracting 2 from both sides gives 3x≤103x \leq 10, so x≤103x \leq \frac{10}{3}. The maximum value of xx occurs when this inequality becomes an equality, making x=103x = \frac{10}{3}. Looking at the wrong answers: Choice A (44) exceeds the maximum allowed value since 4>103≈3.334 > \frac{10}{3} \approx 3.33. If x=4x = 4, then 3(4)+2(1)=14>123(4) + 2(1) = 14 > 12, violating the constraint. Choice C (55) is even larger and would give 3(5)+2(1)=17>123(5) + 2(1) = 17 > 12. Choice D (143\frac{14}{3}) would yield 3(143)+2(1)=14+2=16>123(\frac{14}{3}) + 2(1) = 14 + 2 = 16 > 12, also violating the constraint. For inequality problems with constraints, always substitute the given values first, then solve for the remaining variable. Remember that "maximum value" in a constraint problem means finding the boundary where the inequality becomes an equality.

Question 18

A rental car company charges a base fee of $45 plus $0.25 per mile driven. If Marcus wants to spend at most $120 on his rental, what inequality represents the maximum number of miles $mm $ he can drive?

  1. 45+0.25m≤12045 + 0.25m \leq 120 (correct answer)
  2. 45+0.25m≥12045 + 0.25m \geq 120
  3. 0.25m−45≤1200.25m - 45 \leq 120
  4. 45m+0.25≤12045m + 0.25 \leq 120
Explanation: The total cost is the base fee (45)plusthecostpermile(45) plus the cost per mile (0.25m). Since Marcus wants to spend 'at most' $120, the total cost must be less than or equal to $120, giving us 45+0.25m≤12045 + 0.25m ≤ 120. Choice B uses the wrong inequality direction. Choice C incorrectly subtracts the base fee. Choice D reverses the coefficients.

Question 19

Which of the following describes the solution to the inequality 3(1−2x)≥2(x+13)3(1 - 2x) \ge 2(x + 13)?

  1. x≤−294x \le -\frac{29}{4}
  2. x≤−238x \le -\frac{23}{8} (correct answer)
  3. x≥−238x \ge -\frac{23}{8}
  4. x≥4x \ge 4
Explanation: First, distribute on both sides of the inequality: 3−6x≥2x+263 - 6x \ge 2x + 26. Next, gather the xx terms on one side and the constants on the other. Subtract 2x2x from both sides: 3−8x≥263 - 8x \ge 26. Subtract 3 from both sides: −8x≥23-8x \ge 23. Finally, divide both sides by -8. Because you are dividing by a negative number, you must reverse the inequality sign: x≤−238x \le -\frac{23}{8}.

Question 20

What is the smallest integer value of xx that satisfies the inequality x3−12<3x4+1\frac{x}{3} - \frac{1}{2} < \frac{3x}{4} + 1?

  1. -4
  2. -3 (correct answer)
  3. -2
  4. 4
Explanation: To clear the fractions, multiply the entire inequality by the least common denominator, which is 12: 12(x3−12)<12(3x4+1)12(\frac{x}{3} - \frac{1}{2}) < 12(\frac{3x}{4} + 1). This gives 4x−6<9x+124x - 6 < 9x + 12. Next, solve for xx. Subtract 4x4x from both sides: −6<5x+12-6 < 5x + 12. Subtract 12 from both sides: −18<5x-18 < 5x. Divide by 5: −185<x-\frac{18}{5} < x, or −3.6<x-3.6 < x. The question asks for the smallest integer value that satisfies this condition. The integers greater than -3.6 are -3, -2, -1, 0, ... The smallest of these is -3.