GMAT Quiz: Linear Equations And Inequalities
6 questions · exam conditions
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Linear Equations And InequalitiesQuestion 1 of 6

A company's profit PP (in thousands of dollars) is given by P=3n−12P = 3n - 12, where nn is the number of units sold (in hundreds). If the company needs a profit of at least $18,000 and can sell at most 1,500 units, what is the range of units they can sell?

Between 1,000 and 1,400 units inclusive
Between 1,000 and 1,500 units inclusive
Between 1,200 and 1,500 units inclusive
Between 1,400 and 1,500 units inclusive
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GMAT Quiz

GMAT Quiz: Linear Equations And Inequalities

Practice Linear Equations And Inequalities in GMAT 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 Equations And Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT.

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

A company's profit PP (in thousands of dollars) is given by P=3n−12P = 3n - 12, where nn is the number of units sold (in hundreds). If the company needs a profit of at least $18,000 and can sell at most 1,500 units, what is the range of units they can sell?

  1. Between 1,000 and 1,400 units inclusive
  2. Between 1,000 and 1,500 units inclusive (correct answer)
  3. Between 1,200 and 1,500 units inclusive
  4. Between 1,400 and 1,500 units inclusive
Explanation: Since nn is in hundreds of units and PP is in thousands of dollars, the profit constraint is P≥18P \geq 18 (thousand dollars). So 3n−12≥183n - 12 \geq 18, which gives 3n≥303n \geq 30, so n≥10n \geq 10. Since nn represents hundreds of units, this means at least 1,000 units. The upper constraint is 1,500 units, which is n≤15n \leq 15. Therefore 10≤n≤1510 \leq n \leq 15, corresponding to 1,000 to 1,500 units inclusive. Choice A incorrectly uses 1,400 as the upper bound (n=14n = 14). Choice C uses n≥12n \geq 12 (1,200 units), possibly from solving 3n−12>183n - 12 > 18. Choice D uses both errors from A and C.

Question 2

A manufacturer produces widgets with both variable and fixed daily costs. The total daily cost C (in dollars) for producing n widgets is given by a linear function.

If the total cost for producing 50 widgets is $550 and the total cost for producing 80 widgets is $790, what is the fixed daily cost?

  1. $150 (correct answer)
  2. $200
  3. $250
  4. $300
Explanation: Let the linear cost function be C = an + b where a is the cost per unit and b is the fixed cost. We have two points: (50, 550) and (80, 790). The slope is a = (790 - 550)/(80 - 50) = 240/30 = 8. Using point-slope form with (50, 550): C - 550 = 8(n - 50), so C = 8n - 400 + 550 = 8n + 150. Therefore, the fixed daily cost is $150. Choice B results from incorrectly calculating the slope. Choice C comes from slope calculation errors. Choice D results from errors in the point-slope calculation.

Question 3

For which values of aa does the system $$ \begin{cases} 2x + 3y = 6 \ ax + 6y = 12 \end{cases}

  1. a=3a = 3
  2. a=9a = 9
  3. a=6a = 6
  4. a=4a = 4 (correct answer)
Explanation: When you encounter a system of linear equations and need to determine when it has infinitely many solutions, you're looking for when the equations represent the same line. This happens when one equation is a scalar multiple of the other. Let's examine the given system: 2x+3y=62x + 3y = 6 and ax+6y=12ax + 6y = 12. For infinitely many solutions, the second equation must be a multiple of the first. Notice that the coefficient of yy in the second equation (6) is exactly twice the coefficient of yy in the first equation (3). Similarly, the constant term in the second equation (12) is twice the constant term in the first equation (6). This means the second equation should be exactly 2 times the first equation. Multiplying the first equation by 2: 2(2x+3y)=2(6)2(2x + 3y) = 2(6), which gives us 4x+6y=124x + 6y = 12. For this to match our second equation ax+6y=12ax + 6y = 12, we need a=4a = 4. Let's check why the other answers are wrong. Choice (A) a=3a = 3 would give us 3x+6y=123x + 6y = 12, which isn't a multiple of the first equation. Choice (B) a=9a = 9 yields 9x+6y=129x + 6y = 12, also not a multiple. Choice (C) a=6a = 6 produces 6x+6y=126x + 6y = 12, which again isn't proportional to the first equation. Strategy tip: When checking for infinitely many solutions in linear systems, look for proportional coefficients across all terms. If the ratios between corresponding coefficients (including the constants) are equal, the system has infinitely many solutions.

Question 4

The inequality x+2x−1≥3\frac{x + 2}{x - 1} \geq 3 is equivalent to which of the following?

  1. x≤52x \leq \frac{5}{2} and x≠1x \neq 1
  2. x≥52x \geq \frac{5}{2} or x<1x < 1
  3. 1<x≤521 < x \leq \frac{5}{2} (correct answer)
  4. x>1x > 1 and x≤52x \leq \frac{5}{2}
Explanation: We need x+2x−1≥3\frac{x + 2}{x - 1} \geq 3, which means x+2x−1−3≥0\frac{x + 2}{x - 1} - 3 \geq 0. This becomes x+2−3(x−1)x−1≥0\frac{x + 2 - 3(x - 1)}{x - 1} \geq 0, so x+2−3x+3x−1≥0\frac{x + 2 - 3x + 3}{x - 1} \geq 0, which simplifies to 5−2xx−1≥0\frac{5 - 2x}{x - 1} \geq 0. For this rational expression to be non-negative, we need the numerator and denominator to have the same sign. The numerator 5−2x≥05 - 2x \geq 0 when x≤52x \leq \frac{5}{2}. The denominator x−1>0x - 1 > 0 when x>1x > 1 (note x≠1x \neq 1 since it makes the original expression undefined). For both to be positive: x>1x > 1 and x≤52x \leq \frac{5}{2}, giving 1<x≤521 < x \leq \frac{5}{2}. We also need to check when both are negative: x≥52x \geq \frac{5}{2} and x<1x < 1, which is impossible. Therefore the answer is 1<x≤521 < x \leq \frac{5}{2}. Choice A ignores the sign change when multiplying by a potentially negative denominator. Choice B incorrectly combines the cases. Choice D has the wrong inequality direction for one bound.

Question 5

If 3(2x−5)+4=2(x+1)−73(2x - 5) + 4 = 2(x + 1) - 7, what is the value of x2−3x+2x^2 - 3x + 2?

  1. 0 (correct answer)
  2. 2
  3. 6
  4. 12
Explanation: First solve for x: 3(2x−5)+4=2(x+1)−73(2x - 5) + 4 = 2(x + 1) - 7 becomes 6x−15+4=2x+2−76x - 15 + 4 = 2x + 2 - 7, which simplifies to 6x−11=2x−56x - 11 = 2x - 5. Solving: 4x=64x = 6, so x=32x = \frac{3}{2}. Now substitute into x2−3x+2x^2 - 3x + 2: (32)2−3(32)+2=94−92+2=94−184+84=−14+84=0\left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 2 = \frac{9}{4} - \frac{9}{2} + 2 = \frac{9}{4} - \frac{18}{4} + \frac{8}{4} = \frac{-1}{4} + \frac{8}{4} = 0. Choice B results from incorrectly solving the linear equation as x=2x = 2. Choice C comes from finding x=32x = \frac{3}{2} correctly but then calculating x2+3x+2x^2 + 3x + 2 instead. Choice D results from solving incorrectly to get x=3x = 3.

Question 6

If 3≤2x−1<93 \leq 2x - 1 < 9 and xx is an integer, how many possible values of xx satisfy the inequality?

  1. 2
  2. 3 (correct answer)
  3. 4
  4. 5
Explanation: From 3≤2x−1<93 \leq 2x - 1 < 9, we add 1 to all parts: 4≤2x<104 \leq 2x < 10. Dividing by 2: 2≤x<52 \leq x < 5. Since xx must be an integer, the possible values are x=2,3,4x = 2, 3, 4. That's 3 values. Choice A results from excluding one endpoint incorrectly. Choice C includes x=5x = 5, which fails the strict inequality 2x−1<92x - 1 < 9. Choice D includes both x=1x = 1 and x=5x = 5, representing errors with both bounds.