GMAT Quiz: Linear Equations And Inequalities

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.

Question 1

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

Question 2

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

1. $a = 3$
2. $a = 9$
3. $a = 6$
4. $a = 4$ (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: $2x + 3y = 6$ and $ax + 6y = 12$. For infinitely many solutions, the second equation must be a multiple of the first. Notice that the coefficient of $y$ in the second equation (6) is exactly twice the coefficient of $y$ 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: $2(2x + 3y) = 2(6)$, which gives us $4x + 6y = 12$. For this to match our second equation $ax + 6y = 12$, we need $a = 4$. Let's check why the other answers are wrong. Choice (A) $a = 3$ would give us $3x + 6y = 12$, which isn't a multiple of the first equation. Choice (B) $a = 9$ yields $9x + 6y = 12$, also not a multiple. Choice (C) $a = 6$ produces $6x + 6y = 12$, 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

1. $x \leq \frac{5}{2}$ and $x \neq 1$
2. $x \geq \frac{5}{2}$ or $x < 1$
3. $1 < x \leq \frac{5}{2}$ (correct answer)
4. $x > 1$ and $x \leq \frac{5}{2}$

Explanation: We need $\frac{x + 2}{x - 1} \geq 3$, which means $\frac{x + 2}{x - 1} - 3 \geq 0$. This becomes $\frac{x + 2 - 3(x - 1)}{x - 1} \geq 0$, so $\frac{x + 2 - 3x + 3}{x - 1} \geq 0$, which simplifies to $\frac{5 - 2x}{x - 1} \geq 0$. For this rational expression to be non-negative, we need the numerator and denominator to have the same sign. The numerator $5 - 2x \geq 0$ when $x \leq \frac{5}{2}$. The denominator $x - 1 > 0$ when $x > 1$ (note $x \neq 1$ since it makes the original expression undefined). For both to be positive: $x > 1$ and $x \leq \frac{5}{2}$, giving $1 < x \leq \frac{5}{2}$. We also need to check when both are negative: $x \geq \frac{5}{2}$ and $x < 1$, which is impossible. Therefore the answer is $1 < x \leq \frac{5}{2}$. 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

1. 0 (correct answer)
2. 2
3. 6
4. 12

Explanation: First solve for x: $3(2x - 5) + 4 = 2(x + 1) - 7$ becomes $6x - 15 + 4 = 2x + 2 - 7$, which simplifies to $6x - 11 = 2x - 5$. Solving: $4x = 6$, so $x = \frac{3}{2}$. Now substitute into $x^2 - 3x + 2$: $\left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 2 = \frac{9}{4} - \frac{9}{2} + 2 = \frac{9}{4} - \frac{18}{4} + \frac{8}{4} = \frac{-1}{4} + \frac{8}{4} = 0$. Choice B results from incorrectly solving the linear equation as $x = 2$. Choice C comes from finding $x = \frac{3}{2}$ correctly but then calculating $x^2 + 3x + 2$ instead. Choice D results from solving incorrectly to get $x = 3$.

Question 6

1. 2
2. 3 (correct answer)
3. 4
4. 5

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