SHSAT Math Quiz: Solving Inequalities
20 questions · exam conditions
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Solving InequalitiesQuestion 1 of 20

Solve for nn: −5n≥40-5n \ge 40.

n≥−8n \ge -8
n≤−8n \le -8
n≥8n \ge 8
n≤8n \le 8
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SHSAT Math Quiz

SHSAT Math Quiz: Solving Inequalities

Practice Solving Inequalities in SHSAT Math 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 Solving Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for SHSAT Math.

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 for nn: −5n≥40-5n \ge 40.

  1. n≥−8n \ge -8
  2. n≤−8n \le -8 (correct answer)
  3. n≥8n \ge 8
  4. n≤8n \le 8
Explanation: When solving inequalities involving negative coefficients, the key concept to remember is that multiplying or dividing both sides by a negative number flips the inequality sign. Starting with −5n≥40-5n \ge 40, you need to isolate nn by dividing both sides by −5-5. Since you're dividing by a negative number, the inequality sign must flip from ≥\ge to ≤\le: −5n−5≤40−5\frac{-5n}{-5} \le \frac{40}{-5} This simplifies to n≤−8n \le -8. Looking at the wrong answers: Choice A (n≥−8n \ge -8) represents the most common error—forgetting to flip the inequality sign when dividing by −5-5. This would give you the opposite relationship. Choice C (n≥8n \ge 8) makes two mistakes: not flipping the sign AND getting the wrong sign on the 8, likely from incorrectly handling the negative division. Choice D (n≤8n \le 8) correctly flips the inequality sign but incorrectly calculates 40−5\frac{40}{-5} as positive 8 instead of negative 8. You can verify the answer by testing a value: if n=−8n = -8, then −5(−8)=40-5(-8) = 40, which satisfies −5n≥40-5n \ge 40. If n=−9n = -9 (which is less than −8-8), then −5(−9)=45>40-5(-9) = 45 > 40, confirming our solution. Remember: whenever you multiply or divide an inequality by a negative number, always flip the inequality sign. This is one of the most tested concepts with inequalities on standardized exams.

Question 2

A rectangle has a length that is 4 units more than twice its width. If the perimeter must be less than 32 units, what is the range of possible values for the width ww?

  1. w<8w < 8
  2. 0<w<80 < w < 8
  3. w<4w < 4
  4. 0<w<40 < w < 4 (correct answer)
Explanation: When you encounter word problems involving geometric constraints, you need to translate the verbal relationships into mathematical expressions and then apply the given conditions. Start by defining the width as ww. Since the length is "4 units more than twice the width," you can express the length as 2w+42w + 4. The perimeter of a rectangle is P=2(length+width)P = 2(\text{length} + \text{width}), so: P=2(2w+4+w)=2(3w+4)=6w+8P = 2(2w + 4 + w) = 2(3w + 4) = 6w + 8 Since the perimeter must be less than 32 units: 6w+8<326w + 8 < 32. Solving this inequality: 6w<246w < 24, so w<4w < 4. However, since width represents a physical measurement, it must be positive, giving us 0<w<40 < w < 4. Choice A (w<8w < 8) incorrectly solves the inequality. This likely comes from making an error in the perimeter formula or algebraic manipulation. Choice B (0<w<80 < w < 8) makes the same algebraic error as A but correctly recognizes that width must be positive. Choice C (w<4w < 4) correctly solves the inequality but fails to account for the physical constraint that width must be positive—a rectangle cannot have zero or negative width. Choice D correctly incorporates both the mathematical constraint from the inequality and the real-world constraint that geometric measurements must be positive. Remember that word problems involving measurements require two checks: solving the mathematical relationships correctly and ensuring your answer makes sense in the physical context. Always verify that geometric measurements are positive values.

Question 3

A taxi company charges a flat fee of $2 plus $0.50 per mile. A customer has at most $10 to spend.

Which inequality represents the possible numbers of miles, mm, the customer can ride?

  1. 0.5m+2>100.5m+2>10
  2. 0.5m+2<100.5m+2<10
  3. 0.5m+2≤100.5m+2\le10 (correct answer)
  4. 0.5m+2≥100.5m+2\ge10
Explanation: When you encounter word problems involving costs and spending limits, you're working with inequalities that represent real-world constraints. The key is translating the language carefully into mathematical symbols. The taxi charges $2 plus $0.50 per mile, so the total cost for $mm milesismiles is 0.5m+20.5m + 2 .Thecustomerhas"atmost. The customer has "at most 10 to spend," which means they can spend 10orless.Thisgivesustheinequality10 or less. This gives us the inequality 0.5m+2≤100.5m + 2 \leq 10 $, which is answer choice C. Let's examine why the other choices are wrong. Choice A ( 0.5m + 2 > 10 ) represents situations where the cost exceeds 10, meaning the customer can't afford the ride. Choice B ($$0.5m + 2 < 10$$) only includes rides that cost strictly less than 10, excluding the possibility of spending exactly 10, which the customer can afford. Choice D ($$0.5m + 2 \geq 10$$) represents costs of 10 or more, which goes beyond what the customer can afford. The critical phrase here is "at most," which always translates to "less than or equal to" ( \leq ). You can verify this makes sense: if the customer has exactly 10,theyshouldbeabletoaffordaridethatcostsexactly10, they should be able to afford a ride that costs exactly 10. Remember this pattern: "at most" means \leq , "at least" means \geq , "more than" means > , and "less than" means < . Pay close attention to whether the limit value should be included in your solution set.

Question 4

Use the table showing four students' work on the inequality −23x+5>9-\frac{2}{3}x + 5 > 9. Which student's reasoning is entirely correct?

  1. Aisha
  2. Brandon
  3. Carmen (correct answer)
  4. Dion
Explanation: Correct solution: −23x>4⇒x<−6-\frac{2}{3}x > 4 \Rightarrow x < -6 (flip inequality when multiplying by −32-\frac{3}{2}). Aisha forgot to flip: x>−6x > -6. Brandon subtracted wrong: got −23x>14-\frac{2}{3}x > 14. Carmen correctly got x<−6x < -6. Dion flipped twice, ending with x>6x > 6.

Question 5

The table shows values of f(x)=3x−7f(x) = 3x - 7 and g(x)=−x+9g(x) = -x + 9. Based on the table, what is the smallest integer xx for which f(x)>g(x)f(x) > g(x)?

  1. 33
  2. 44
  3. 55 (correct answer)
  4. 66
Explanation: Solve 3x−7>−x+9⇒4x>16⇒x>43x - 7 > -x + 9 \Rightarrow 4x > 16 \Rightarrow x > 4. Smallest integer is 5. From the table, at x=4x=4, f(4)=5f(4)=5 and g(4)=5g(4)=5 (equal, not greater). At x=5x=5, f(5)=8>g(5)=4f(5)=8 > g(5)=4 ✓. (A) and (B) misread 'strictly greater'. (D) is one too large.

Question 6

Based on the diagram below, a phone plan charges a flat $12 monthly fee plus $0.08 per minute. A competing plan charges $0.20 per minute with no monthly fee. For how many whole minutes $mm $ per month is the first plan strictly less expensive than the second plan?

  1. m≥100m \geq 100
  2. m>100m > 100
  3. m≥101m \geq 101 (correct answer)
  4. m>150m > 150
Explanation: Need 12+0.08m<0.20m⇒12<0.12m⇒m>10012 + 0.08m < 0.20m \Rightarrow 12 < 0.12m \Rightarrow m > 100. Since mm must be a whole number and strictly greater than 100, the smallest value is 101, so m≥101m \geq 101. (A) includes m=100m=100 where costs are equal. (B) ignores 'whole minutes'. (D) uses wrong arithmetic (12/0.0812/0.08).

Question 7

The figure shows a triangle with sides of length xx, x+4x + 4, and 1111. Using the triangle inequality theorem, what is the complete set of possible values of xx?

  1. x>3.5x > 3.5
  2. 3.5<x<73.5 < x < 7 (correct answer)
  3. 3.5<x<153.5 < x < 15
  4. 0<x<150 < x < 15
Explanation: Triangle inequality requires the sum of any two sides to exceed the third side. We need: (1) x+(x+4)>11⇒x>3.5x + (x+4) > 11 \Rightarrow x > 3.5; (2) x+11>x+4⇒7>0x + 11 > x+4 \Rightarrow 7 > 0 (always true); (3) (x+4)+11>x⇒x<15(x+4) + 11 > x \Rightarrow x < 15. Also need x>0x > 0. However, since x+4x+4 is the longest side when x<7x < 7, we also need x+11>x+4x + 11 > x + 4, which gives x<7x < 7. Combined: 3.5<x<73.5 < x < 7.

Question 8

Refer to the figure. A rectangle has length (2x+3)(2x + 3) cm and width (x−1)(x - 1) cm, where x>1x > 1. The perimeter of the rectangle must be less than 40 cm but at least 16 cm. What is the complete set of integer values of xx that satisfies these conditions?

  1. {3,4,5,6}\{3, 4, 5, 6\}
  2. {2,3,4,5}\{2, 3, 4, 5\} (correct answer)
  3. {4,5,6}\{4, 5, 6\}
  4. {3,4,5}\{3, 4, 5\}
Explanation: Perimeter = 2[(2x+3)+(x−1)]=6x+42[(2x+3)+(x-1)] = 6x + 4. Need 16≤6x+4<4016 \leq 6x + 4 < 40, so 12≤6x<3612 \leq 6x < 36, giving 2≤x<62 \leq x < 6. Combined with x>1x > 1, we get 2≤x<62 \leq x < 6. Integer values are {2,3,4,5}\{2, 3, 4, 5\}.

Question 9

Refer to the figure. A trapezoidal garden has parallel sides of length xx and x+6x + 6 meters, and height 4 meters. The area must be greater than 40 square meters but not more than 60 square meters. Which inequality gives all valid values of xx (in meters)?

  1. 7<x≤127 < x \leq 12 (correct answer)
  2. 7≤x<127 \leq x < 12
  3. 10<x≤1510 < x \leq 15
  4. 13<x≤1813 < x \leq 18
Explanation: Area =12(x+x+6)(4)=2(2x+6)=4x+12= \frac{1}{2}(x + x+6)(4) = 2(2x+6) = 4x + 12. Need 40<4x+12≤60⇒28<4x≤48⇒7<x≤1240 < 4x + 12 \leq 60 \Rightarrow 28 < 4x \leq 48 \Rightarrow 7 < x \leq 12. (B) has inequality directions reversed. (C) forgets the +12+12 term. (D) computes as if base is just x+6x+6.

Question 10

If −4≤2x+6<10-4 \leq 2x + 6 < 10, what is the range of values for xx?

  1. −5<x≤2-5 < x \leq 2
  2. −5≤x<2-5 \leq x < 2 (correct answer)
  3. −10≤x<4-10 \leq x < 4
  4. 1≤x<81 \leq x < 8
Explanation: When you encounter compound inequalities like this one, you're solving for a variable that must satisfy two conditions simultaneously. The key is to isolate the variable by performing the same operations on all three parts of the inequality. Starting with −4≤2x+6<10-4 \leq 2x + 6 < 10, you need to isolate xx by first subtracting 6 from all three parts: −4−6≤2x+6−6<10−6-4 - 6 \leq 2x + 6 - 6 < 10 - 6, which gives you −10≤2x<4-10 \leq 2x < 4. Then divide all parts by 2: −5≤x<2-5 \leq x < 2. This means xx can equal -5 but must be strictly less than 2. Choice A (−5<x≤2-5 < x \leq 2) reverses both inequality symbols. This would mean xx cannot equal -5 but can equal 2, which contradicts our solution. Choice C (−10≤x<4-10 \leq x < 4) represents the inequality before dividing by 2 - a common error when students forget this final step. Choice D (1≤x<81 \leq x < 8) appears to come from incorrectly adding instead of subtracting 6, then making additional errors in the division step. Choice B (−5≤x<2-5 \leq x < 2) correctly shows that xx includes -5 (closed circle) and excludes 2 (open circle). Strategy tip: When solving compound inequalities, always perform identical operations on all three parts simultaneously, and pay careful attention to whether endpoints are included (≤, ≥) or excluded (<, >). Double-check by substituting boundary values back into the original inequality.

Question 11

A student needs to score at least 85% on a test with 40 questions to pass. If each question is worth the same number of points, what is the minimum number of questions the student must answer correctly?

  1. 33 questions
  2. 34 questions (correct answer)
  3. 35 questions
  4. 32 questions
Explanation: This problem tests percentage calculations and the critical skill of rounding up when dealing with "at least" requirements in real-world contexts. To find the minimum correct answers needed, you must calculate 85% of 40 questions: 0.85×40=340.85 \times 40 = 34 questions exactly. Since the student needs to score "at least 85%," and 34 questions gives exactly 85%, the minimum number of correct answers is 34. Let's examine why each answer choice works or fails. Choice A (33 questions) gives you 3340=0.825=82.5%\frac{33}{40} = 0.825 = 82.5\%, which falls short of the required 85%. Choice C (35 questions) would work since 3540=87.5%\frac{35}{40} = 87.5\%, but the question asks for the minimum number, making this more than necessary. Choice D (32 questions) gives only 3240=80%\frac{32}{40} = 80\%, clearly insufficient. Choice B (34 questions) is correct because 3440=85%\frac{34}{40} = 85\% exactly, meeting the minimum requirement. The key insight here is recognizing that when a percentage calculation yields a whole number, that's your answer for "at least" problems. If the calculation had produced 34.2, you'd round up to 35 since you can't answer a fraction of a question correctly. Strategy tip: When you see "at least" or "minimum" with percentages, calculate the exact percentage first, then remember that partial questions always round up in real-world scenarios. However, when the calculation is exact (like here), that exact number is your minimum.

Question 12

A taxi charges $3.50 for the first mile and $1.25 for each additional mile. If someone has $20 to spend, what is the maximum distance they can travel?

  1. 1616 miles
  2. 13.213.2 miles
  3. 14.214.2 miles (correct answer)
  4. 1717 miles
Explanation: The cost is 3.50+1.25(x−1)≤203.50 + 1.25(x-1) \leq 20 for xx miles. Simplifying: 3.50+1.25x−1.25≤203.50 + 1.25x - 1.25 \leq 20, so 2.25+1.25x≤202.25 + 1.25x \leq 20. Thus 1.25x≤17.751.25x \leq 17.75, giving x≤14.2x \leq 14.2 miles. Choice A uses 20÷1.2520 \div 1.25. Choice B uses (20−3.50)÷1.25(20-3.50) \div 1.25. Choice D incorrectly adds instead of subtracts the first mile adjustment.

Question 13

Refer to the graph shown. A student wants to buy notebooks at $3 each and pens at $1.50 each, spending no more than $30 total. The shaded region in the graph represents all valid combinations $(n,p)(n, p) wherewhere nn =numberofnotebooksand= number of notebooks and pp =numberofpens.Whichinequality,combinedwith= number of pens. Which inequality, combined with n≥0n \geq 0 andand p≥0p \geq 0 $, produces this shaded region?

  1. 3n+1.5p<303n + 1.5p < 30
  2. 3n+1.5p≤303n + 1.5p \leq 30 (correct answer)
  3. 1.5n+3p≤301.5n + 3p \leq 30
  4. 3n+1.5p≥303n + 1.5p \geq 30
Explanation: Total cost is 3n+1.5p3n + 1.5p and cannot exceed 30. Spending 'no more than' means ≤\leq, so 3n+1.5p≤303n + 1.5p \leq 30. The shaded region is bounded by a solid line (including the boundary). (A) uses strict inequality with dashed boundary. (C) swaps the prices. (D) reverses direction.

Question 14

The graph below shows two lines y1=12x+1y_1 = \frac{1}{2}x + 1 and y2=−x+7y_2 = -x + 7. For what values of xx is y1≥y2y_1 \geq y_2?

  1. x≥4x \geq 4 (correct answer)
  2. x≤4x \leq 4
  3. x≥3x \geq 3
  4. x≤3x \leq 3
Explanation: Set 12x+1≥−x+7⇒32x≥6⇒x≥4\frac{1}{2}x + 1 \geq -x + 7 \Rightarrow \frac{3}{2}x \geq 6 \Rightarrow x \geq 4. On the graph, y1y_1 is above (or equal to) y2y_2 to the right of the intersection point at (4,3)(4, 3). (B) reverses direction. (C) and (D) use yy-coordinate of intersection.

Question 15

The number line shown represents the solution set to which of the following inequalities?

  1. −3(2x−4)≥18-3(2x - 4) \geq 18 (correct answer)
  2. −3(2x−4)≤18-3(2x - 4) \leq 18
  3. 5−2x≥115 - 2x \geq 11
  4. 2x−4≥−62x - 4 \geq -6
Explanation: The number line shows x≤−1x \leq -1 with a closed circle at −1-1. Solving (A): −3(2x−4)≥18⇒−6x+12≥18⇒−6x≥6⇒x≤−1-3(2x-4) \geq 18 \Rightarrow -6x + 12 \geq 18 \Rightarrow -6x \geq 6 \Rightarrow x \leq -1 ✓. (B) gives x≥−1x \geq -1 (wrong direction). (C) gives −2x≥6⇒x≤−3-2x \geq 6 \Rightarrow x \leq -3 (wrong endpoint). (D) gives x≥−1x \geq -1 (wrong direction).

Question 16

The graph below shows the line y=−2x+6y = -2x + 6. For what values of xx is −2x+6<4-2x + 6 < 4? Use the graph shown to determine the answer.

  1. x<1x < 1
  2. x>1x > 1 (correct answer)
  3. x<−1x < -1
  4. x>5x > 5
Explanation: We need where the line is below y=4y=4. Setting −2x+6=4-2x + 6 = 4 gives x=1x = 1. Since the line has negative slope, it is below 4 when x>1x > 1. Algebraically: −2x<−2⇒x>1-2x < -2 \Rightarrow x > 1. (A) forgets to flip the inequality. (C) and (D) use wrong endpoint.

Question 17

Solve for xx: x+7<4x + 7 < 4.

  1. x<−3x < -3 (correct answer)
  2. x<11x < 11
  3. x>−3x > -3
  4. x>11x > 11
Explanation: When you encounter an inequality like this, you're solving it just like a regular equation, but you need to pay attention to the inequality sign throughout your work. To solve x+7<4x + 7 < 4, you need to isolate xx on one side. Subtract 7 from both sides: x+7−7<4−7x + 7 - 7 < 4 - 7, which simplifies to x<−3x < -3. This matches choice A. Let's examine why the other options are incorrect. Choice B (x<11x < 11) represents the mistake of adding 7 to both sides instead of subtracting it. If you mistakenly thought x+7<4x + 7 < 4 becomes x<4+7x < 4 + 7, you'd get this wrong answer. Choice C (x>−3x > -3) shows the correct number but the wrong inequality direction - this would happen if you forgot that the original inequality was "less than" rather than "greater than." Choice D (x>11x > 11) combines both errors: adding instead of subtracting AND flipping the inequality sign incorrectly. You can verify your answer by testing a value. Since x<−3x < -3, let's try x=−4x = -4. Substituting: −4+7=3-4 + 7 = 3, and indeed 3<43 < 4 ✓. If you tried x=0x = 0 (which doesn't satisfy x<−3x < -3), you'd get 0+7=70 + 7 = 7, and 7<47 < 4 is false. Key strategy: Unlike equations, inequalities keep their direction when you add or subtract the same number from both sides. Only when multiplying or dividing by negative numbers do you flip the inequality sign.

Question 18

A phone plan charges a monthly fee of $25 plus $0.15 per text message. If Maria wants to spend no more than $40 per month, what is the maximum number of text messages she can send?

  1. 100100 text messages (correct answer)
  2. 9999 text messages
  3. 101101 text messages
  4. 266266 text messages
Explanation: Set up the inequality: 25+0.15x≤4025 + 0.15x \leq 40, where xx is the number of text messages. Solving: 0.15x≤150.15x \leq 15, so x≤100x \leq 100. Therefore, the maximum is 100 text messages. Choice B (99) assumes strict inequality. Choice C (101) would exceed the budget. Choice D (266) incorrectly uses 40÷0.1540 \div 0.15.

Question 19

The solution to the inequality 3x−7>2x+53x - 7 > 2x + 5 is equivalent to which of the following?

  1. x>12x > 12 (correct answer)
  2. x<12x < 12
  3. x>−2x > -2
  4. x<−2x < -2
Explanation: Solving 3x−7>2x+53x - 7 > 2x + 5: subtract 2x2x from both sides to get x−7>5x - 7 > 5, then add 7 to get x>12x > 12. Choice B reverses the inequality sign. Choice C results from the error −7+5=−2-7 + 5 = -2. Choice D combines both the sign error and the arithmetic error.

Question 20

The inequality 5−3x<2x−105 - 3x < 2x - 10 has the same solution set as which of the following?

  1. x<−3x < -3
  2. x<3x < 3
  3. x>−3x > -3
  4. x>3x > 3 (correct answer)
Explanation: When you encounter an inequality that needs to be solved, you're working with the same algebraic principles as equations, but with one crucial difference: flipping the inequality sign when multiplying or dividing by a negative number. To solve 5−3x<2x−105 - 3x < 2x - 10, start by collecting like terms. Add 3x3x to both sides: 5<5x−105 < 5x - 10. Then add 1010 to both sides: 15<5x15 < 5x. Finally, divide both sides by 55: 3<x3 < x, which is equivalent to x>3x > 3. Let's check why the other options are incorrect. Choice A (x<−3x < -3) represents the complete opposite direction - it suggests xx values that are much smaller than what our solution requires. Choice B (x<3x < 3) has the wrong direction of the inequality; this would include values like x=0x = 0 or x=1x = 1, which don't satisfy our original inequality. Choice C (x>−3x > -3) has the right direction but the wrong boundary value - this would incorrectly include values between −3-3 and 33, such as x=0x = 0, which doesn't work in the original inequality. You can verify choice D is correct by testing a value like x=4x = 4: 5−3(4)=−75 - 3(4) = -7 and 2(4)−10=−22(4) - 10 = -2, so −7<−2-7 < -2 ✓. Remember: when solving inequalities, be extra careful with your algebraic steps and always consider testing a value from your solution set to verify your answer. The SHSAT often includes trap answers that reverse the inequality direction.