20 medium SAT Linear inequalities questions

Medium is where most scores are actually won and lost. These questions are not tricky for the sake of it, but every one of them has a wrong answer built to catch a specific shortcut.

Every question below is a real item from the SAT Climb bank, tagged medium by the same difficulty model the app uses to build your practice. Pick an answer before you open the explanation.

Math · Algebra~2 per testMedium tier
Question 1Medium

A caterer is preparing gift baskets containing cookies and brownies. Each basket with cookies weighs 2 pounds, and each basket with brownies weighs 3 pounds. The caterer's delivery vehicle can carry at most 180 pounds. The caterer must prepare at least 25 baskets with cookies to fulfill orders. What is the maximum number of baskets with brownies the caterer can prepare for this delivery?

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Why B is right

Let c represent cookie baskets and b represent brownie baskets. The constraints are 2c+3b≤1802c + 3b \le 180 and c≥25c \ge 25. To maximize b, set c=25c = 25. Substituting: 2(25)+3b≤1802(25) + 3b \le 180, so 50+3b≤18050 + 3b \le 180, giving 3b≤1303b \le 130 and b≤43.33b \le 43.33. The maximum whole number of brownie baskets is 43.

Why the others are wrong

  • AThis is the result of solving for the maximum number of cookie baskets when preparing zero brownie baskets (180 ÷ 3 = 60), which is the wrong quantity.
  • CThis results from an arithmetic error, such as computing 130 ÷ 3 as 45 instead of 43.33.
  • DThis results from dividing the remaining weight capacity by an incorrect divisor or using the wrong operation to combine constraints.
Question 2Medium

A number y is at most 10 less than 4 times the value of z. If z is 11, what is the greatest possible value of y?

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Why B is right

The phrase 'at most 10 less than 4 times the value of z' translates to y≤4z−10y \le 4z - 10. Substituting z=11z = 11 gives y≤4y \le 4(11) - 10 = 44 - 10 = 34. The greatest possible value of y is 34.

Why the others are wrong

  • AThis results from computing 11 - 10 = 1, failing to multiply z by 4 first.
  • CThis is 4(11) = 44, the value before subtracting 10.
  • DThis results from computing 4(11) + 10 = 54, incorrectly adding 10 instead of subtracting it.
Question 3Medium

In the xy-plane, a line with slope 12\displaystyle \frac{1}{2} passes through the point (4, 3). The region below the line, including the line itself, is shaded. Which inequality represents the graph?

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Why D is right

Using point-slope form with slope 1/2 and point (4,3): y−3=12(x−4)\displaystyle y - 3 = \frac{1}{2} (x - 4), which simplifies to y=12x+1\displaystyle y = \frac{1}{2} x + 1. Since the region below is shaded and the line is solid, the inequality is y≤12x+1\displaystyle y \le \frac{1}{2} x + 1.

Why the others are wrong

  • AThis makes an arithmetic error in calculating the y-intercept, getting 5 instead of 1.
  • BThis uses the correct slope and y-intercept but treats the boundary as dashed rather than solid.
  • CThis solves for x instead of y and uses the reciprocal slope.
Question 4Medium

Which inequality is equivalent to 15−5(x+1)≤015 - 5(x + 1) \le 0?

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Why B is right

Distributing -5 gives 15−5x−5≤015 - 5x - 5 \le 0. Simplifying gives 10−5x≤010 - 5x \le 0. Subtracting 10 from both sides gives −5x≤−10-5x \le -10. Dividing both sides by -5 and flipping the inequality sign gives x≥2x \ge 2.

Why the others are wrong

  • AThis results from correctly computing the boundary as 2 but failing to flip the inequality sign when dividing by -5.
  • CThis results from both making an error in simplifying 15 - 5 as 15 instead of 10 and incorrectly handling the inequality direction.
  • DThis results from correctly flipping the inequality sign but incorrectly computing -10 divided by -5 as 3 instead of 2.
Question 5Medium

Which inequality is equivalent to 7x−12≤3(x+4)7x - 12 \le 3(x + 4)?

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Why D is right

Distributing the 3 gives 7x−12≤3x+127x - 12 \le 3x + 12. Subtracting 3x from both sides gives 4x−12≤124x - 12 \le 12. Adding 12 to both sides gives 4x≤244x \le 24. Dividing both sides by 4 gives x≤6x \le 6.

Why the others are wrong

  • AThis results from incorrectly flipping the inequality sign when no sign flip is needed.
  • BThis results from incorrectly computing 12 + 12 as 12 instead of 24.
  • CThis results from both sign and computational errors combined.
Question 6Medium

A factory produces chairs and tables. Each chair requires 3 hours of labor and each table requires 5 hours of labor. The factory has at most 240 hours of labor available per week. Additionally, the factory must produce at least 20 chairs per week to meet customer demand. What is the maximum number of tables the factory can produce in one week?

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Why C is right

Let c represent chairs and t represent tables. The constraints are 3c+5t≤2403c + 5t \le 240 and c≥20c \ge 20. To maximize t, minimize c by setting c=20c = 20. Substituting: 3(20)+5t≤2403(20) + 5t \le 240, so 60+5t≤24060 + 5t \le 240, giving 5t≤1805t \le 180 and t≤36t \le 36. The maximum number of tables is 36.

Why the others are wrong

  • AThis is the result of solving for the maximum number of chairs when producing zero tables (240 ÷ 5 = 48), which is the wrong quantity.
  • BThis results from an arithmetic error, such as computing (240 - 60) ÷ 5 incorrectly as 40 instead of 36.
  • DThis results from dividing the total hours by the table requirement without accounting for the minimum chair constraint (240 ÷ 4 = 60, using the wrong operation).
Question 7Medium

A number h is at most 11 less than 6 times the value of j. If j is 10, what is the greatest possible value of h?

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Why B is right

The phrase 'at most 11 less than 6 times the value of j' translates to h≤6j−11h \le 6j - 11. Substituting j=10j = 10 gives h≤6h \le 6(10) - 11 = 60 - 11 = 49. The greatest possible value of h is 49.

Why the others are wrong

  • AThis is 6(10) = 60, the value before subtracting 11.
  • CThis results from computing 6(10) + 11 = 71, incorrectly adding 11 instead of subtracting it.
  • DThis results from computing 10 - 5 = 5, using incorrect operations.
Question 8Medium

In the xy-plane, a line has x-intercept 4 and y-intercept -6. The region below the line, not including the line itself, is shaded. Which inequality represents the graph?

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Why A is right

The line passes through (4,0) and (0,-6), giving slope (0-(-6))/(4-0) = 6/4 = 3/2. The equation is y=32x−6\displaystyle y = \frac{3}{2} x - 6. Since the region below is shaded and the line is not included, the inequality is y<32x−6\displaystyle y < \frac{3}{2} x - 6.

Why the others are wrong

  • BThis uses the correct slope and direction but treats the boundary as solid when it should be dashed.
  • CThis incorrectly negates the slope while keeping the y-intercept.
  • DThis solves for x instead of y and uses the reciprocal slope.
Question 9Medium

Which inequality is equivalent to 6(x+2)−4x≤366(x + 2) - 4x \le 36?

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Why D is right

Distributing gives 6x+12−4x≤366x + 12 - 4x \le 36, which simplifies to 2x+12≤362x + 12 \le 36. Subtracting 12 from both sides gives 2x≤242x \le 24. Dividing both sides by 2 gives x≤12x \le 12.

Why the others are wrong

  • AThis choice results from incorrectly computing 24 divided by 2 or failing to properly combine like terms after distributing.
  • BThis choice results from an error in simplification and incorrectly flipping the inequality sign when no negative division occurred.
  • CThis choice results from correctly finding the boundary value but reversing the direction of the inequality.
Question 10Medium

Which inequality is equivalent to 15−3x≥6x+2415 - 3x \ge 6x + 24?

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Why D is right

Adding 3x to both sides gives 15≥9x+2415 \ge 9x + 24. Subtracting 24 from both sides gives −9≥9x-9 \ge 9x. Dividing both sides by 9 gives −1≥x-1 \ge x, which is equivalent to x≤−1x \le -1.

Why the others are wrong

  • AThis results from incorrectly flipping the inequality sign when no sign flip is needed.
  • BThis results from incorrectly computing 15 - 24 as -117 instead of -9.
  • CThis results from both sign handling and arithmetic errors.
Question 11Medium

A community center offers yoga classes and dance classes. Each yoga class can accommodate 8 participants, and each dance class can accommodate 12 participants. The center has space for at most 192 total participants across all classes in a session. The center must offer at least 6 yoga classes to meet member preferences. What is the maximum number of dance classes the center can offer in one session?

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Why C is right

Let y represent yoga classes and d represent dance classes. The constraints are 8y+12d≤1928y + 12d \le 192 and y≥6y \ge 6. To maximize d, set y=6y = 6. Substituting: 8(6)+12d≤1928(6) + 12d \le 192, so 48+12d≤19248 + 12d \le 192, giving 12d≤14412d \le 144 and d≤12d \le 12. The maximum number of dance classes is 12.

Why the others are wrong

  • AThis is the result of solving for the maximum number of yoga classes when offering zero dance classes (192 ÷ 12 = 16), which is the wrong quantity.
  • BThis results from an arithmetic error, such as computing 144 ÷ 12 incorrectly or mishandling the constraint.
  • DThis results from dividing the total capacity by the yoga capacity instead of properly accounting for constraints (192 ÷ 8 = 24).
Question 12Medium

A number w is at most 9 less than 2 times the value of t. If t is 15, what is the greatest possible value of w?

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Why A is right

The phrase 'at most 9 less than 2 times the value of t' translates to w≤2t−9w \le 2t - 9. Substituting t=15t = 15 gives w≤2w \le 2(15) - 9 = 30 - 9 = 21. The greatest possible value of w is 21.

Why the others are wrong

  • BThis results from computing 2(15) + 9 = 39, incorrectly adding 9 instead of subtracting it.
  • CThis results from computing 15 - 9 = 6, failing to multiply t by 2 first.
  • DThis is 2(15) = 30, the value before subtracting 9.
Question 13Medium

In the xy-plane, a line passes through the origin and the point (3, 2). The region above the line, including the line itself, is shaded. Which inequality represents the graph?

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Why B is right

The line passes through (0,0) and (3,2), so its slope is 2/3. Since the region above the line is shaded and the line is solid (included), the inequality is y≥23\displaystyle y \ge \frac{2}{3}x.

Why the others are wrong

  • AThis reverses the slope by using x-coordinate over y-coordinate instead of rise over run.
  • CThis uses the correct slope and direction but treats the boundary as dashed rather than solid.
  • DThis reverses the inequality sign, shading below instead of above the line.
Question 14Medium

Which inequality is equivalent to 3(x−5)+2x≥203(x - 5) + 2x \ge 20?

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Why C is right

Distributing gives 3x−15+2x≥203x - 15 + 2x \ge 20, which simplifies to 5x−15≥205x - 15 \ge 20. Adding 15 to both sides gives 5x≥355x \ge 35. Dividing both sides by 5 gives x≥7x \ge 7.

Why the others are wrong

  • AThis choice results from incorrectly computing 35 divided by 5 or failing to add 15 properly.
  • BThis choice results from an error in combining like terms and incorrectly flipping the inequality sign.
  • DThis choice results from an arithmetic error and reversing the direction of the inequality.
Question 15Medium

Which inequality is equivalent to -2(x−8)<202(x - 8) < 20?

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Why C is right

Distributing the -2 gives −2x+16<20-2x + 16 < 20. Subtracting 16 from both sides gives −2x<4-2x < 4. Dividing both sides by -2 and flipping the inequality sign gives x>−2x > -2.

Why the others are wrong

  • AThis results from failing to flip the inequality sign when dividing by -2.
  • BThis results from incorrectly computing 20 - 16 as 36 instead of 4.
  • DThis results from both failing to flip the sign and making arithmetic errors.
Question 16Medium

A warehouse stores small boxes and large boxes. Each small box occupies 6 cubic feet of space, and each large box occupies 10 cubic feet of space. The warehouse has at most 480 cubic feet of available storage space. The warehouse must store at least 15 small boxes for inventory requirements. What is the maximum number of large boxes the warehouse can store?

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Why A is right

Let s represent small boxes and L represent large boxes. The constraints are 6s+10L≤4806s + 10L \le 480 and s≥15s \ge 15. To maximize L, set s=15s = 15. Substituting: 6(15)+10L≤4806(15) + 10L \le 480, so 90+10L≤48090 + 10L \le 480, giving 10L≤39010L \le 390 and L≤39L \le 39. The maximum number of large boxes is 39.

Why the others are wrong

  • BThis is the result of solving for the maximum number of small boxes when storing zero large boxes (480 ÷ 10 = 48), which is the wrong quantity.
  • CThis results from an arithmetic error in computing 390 ÷ 10 or mishandling the remaining space calculation.
  • DThis results from dividing the total space by the small box requirement instead of properly accounting for constraints (480 ÷ 6 = 80).
Question 17Medium

A number a is at most 8 less than 3 times the value of b. If b is 14, what is the greatest possible value of a?

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Why C is right

The phrase 'at most 8 less than 3 times the value of b' translates to a≤3b−8a \le 3b - 8. Substituting b=14b = 14 gives a≤3a \le 3(14) - 8 = 42 - 8 = 34. The greatest possible value of a is 34.

Why the others are wrong

  • AThis is 3(14) = 42, the value before subtracting 8.
  • BThis results from computing 14 - 8 = 6, failing to multiply b by 3 first.
  • DThis results from computing 3(14) + 8 = 50, incorrectly adding 8 instead of subtracting it.
Question 18Medium

In the xy-plane, a line passes through the origin and has slope−34\displaystyle slope -\frac{3}{4}. The region above the line, not including the line itself, is shaded. Which inequality represents the graph?

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Why D is right

A line through the origin with slope -3/4 has equation y=(−3/4)y = (-3/4)x. Since the region above is shaded and the line is not included, the inequality is y>(−3/4)y > (-3/4)x.

Why the others are wrong

  • AThis uses the reciprocal of the slope instead of the given slope.
  • BThis reverses the inequality direction, shading below instead of above the line.
  • CThis uses the correct slope and direction but treats the boundary as solid when it should be dashed.
Question 19Medium

Which inequality is equivalent to 7−3x≤2(x−4)7 - 3x \le 2(x - 4)?

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Why B is right

Distributing the right side gives 7−3x≤2x−87 - 3x \le 2x - 8. Adding 3x to both sides gives 7≤5x−87 \le 5x - 8. Adding 8 to both sides gives 15≤5x15 \le 5x, so x≥3x \ge 3.

Why the others are wrong

  • AThis results from incorrectly reversing the inequality during algebraic manipulation.
  • CThis results from incorrectly computing the boundary, treating 15 ÷ 5 as related to a different value.
  • DThis results from both an incorrect boundary computation and reversing the inequality sign.
Question 20Medium

Which inequality is equivalent to−6(x−3)≥42to -6(x - 3) \ge 42?

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Why C is right

Distributing the -6 gives −6x+18≥42-6x + 18 \ge 42. Subtracting 18 from both sides gives −6x≥24-6x \ge 24. Dividing both sides by -6 and flipping the inequality sign gives x≤−4x \le -4.

Why the others are wrong

  • AThis results from failing to flip the inequality sign when dividing by -6.
  • BThis results from incorrectly computing 42 - 18 as 60 instead of 24.
  • DThis results from both sign and computational errors combined.

What to do after medium

Medium is the tier that decides most scores. If these are landing, the hard set is where the remaining points are.

Questions are written by SAT Climb and drawn from its own item bank. SAT® is a registered trademark of College Board, which is not affiliated with and does not endorse SAT Climb.