GRE Quantitative Quiz: Linear Equations Inequalities

What this quiz covers

This quiz focuses on Linear Equations Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for GRE Quantitative.

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. $0.25F+m=37.50$
2. $F+0.25m=37.50$ (correct answer)
3. $0.25(F+m)=37.50$
4. $F+\dfrac{m}{0.25}=37.50$
5. $F-0.25m=37.50$

Explanation: This question tests translating a word problem into a linear equation. The total charge consists of a flat fee F plus a per-mile charge of $0.25 times the number of miles m. The equation representing the total charge is F + 0.25m = 37.50, where F is the flat fee and 0.25m is the mileage charge. This matches choice B exactly. Choice A incorrectly multiplies the flat fee by 0.25, while choice C incorrectly applies the rate to the sum of F and m rather than just m.

Question 2

1. $x\le 11$
2. $x\ge 11$
3. $x<11$
4. $x>11$ (correct answer)
5. $x>7$

Explanation: This question tests solving linear inequalities involving fractions. Starting with (x - 5)/3 > 2, we multiply both sides by 3 to get x - 5 > 6. Adding 5 to both sides yields x > 11. Since this is a strict inequality (>) rather than ≥, the solution is x > 11, not x ≥ 11. A common error would be to incorrectly simplify the right side after multiplying, getting x - 5 > 2 instead of x - 5 > 6.

Question 3

1. The system has no solution.
2. The system has exactly one solution.
3. The system has infinitely many solutions. (correct answer)
4. The value of $x$ must be 9.
5. The value of $y$ must be 0.

Explanation: This question tests recognizing dependent equations in a system. Notice that the second equation 2x + 2y = 18 is exactly twice the first equation x + y = 9. When we multiply the first equation by 2, we get 2x + 2y = 18, which is identical to the second equation. This means the two equations represent the same line, so the system has infinitely many solutions - any point (x, y) satisfying x + y = 9 is a solution. The system is not inconsistent (which would have no solutions) nor does it have a unique solution.

Question 4

1. $x=2$
2. $x=3$
3. $x=5$ (correct answer)
4. $x=6$
5. $x=-3$

Explanation: This question tests finding values that satisfy a system of linear inequalities. The correct approach is to solve each inequality separately and find the intersection of the solution sets. For 2x + 1 > 7, subtract 1 and divide by 2 to get x > 3; combined with x ≤ 5, the range is 3 < x ≤ 5. Among the options, x = 5 satisfies both as 2(5) + 1 = 11 > 7 and 5 ≤ 5. This is justified because it fits the intersection precisely. A common incorrect option like x = 6 fails as it exceeds x ≤ 5. Another error might be selecting x = 3, which does not satisfy the strict inequality >7.

Question 5

1. 120
2. 150
3. 170
4. 180 (correct answer)
5. 200

Explanation: This question tests solving linear equations in a cost model. The correct approach is to isolate the variable for pages after subtracting the fixed cost. Given 50 + 0.2p = 86, subtract 50 to get 0.2p = 36, then p = 180. This is justified as 50 + 0.2(180) = 86. A common incorrect option like 170 might come from dividing 86 by 0.2 without subtracting. Another error could be using 0.2 incorrectly, leading to 150 or 200.

Question 6

1. $2$
2. $3$
3. $4$ (correct answer)
4. $9$
5. $13$

Explanation: This question tests solving systems of linear equations. The correct approach is to use elimination by subtracting equations. From 2x + y = 13 and x + y = 9, subtract to get x = 4. This is justified by substituting back: for x = 4, y = 5, and 2(4) + 5 = 13. A common incorrect option like 3 might come from adding instead of subtracting. Another error could be solving for y first, leading to 9 or 13.

Question 7

1. $-8$
2. $1$
3. $8$ (correct answer)
4. $16$
5. $-1$

Explanation: This question tests solving linear equations in one variable. The correct approach is to isolate the variable x by moving terms involving x to one side and constants to the other. Starting with 5x - 7 = 3x + 9, subtract 3x from both sides to get 2x - 7 = 9, then add 7 to both sides yielding 2x = 16. Dividing both sides by 2 gives x = 8. This solution is justified because substituting x = 8 back into the original equation gives 5(8) - 7 = 33 and 3(8) + 9 = 33, which are equal. A common incorrect option like 16 might result from forgetting to divide by 2 after isolating the term. Another error could be mishandling signs, leading to negative values like -8 or -1.

Question 8

1. $x\ge 5$ (correct answer)
2. $y\ge 5$
3. $x\le 5$
4. $y\le 0$
5. $x=y$

Explanation: This question tests linear equations combined with inequalities. Given x + y = 10 and x ≥ y, we need to determine what must be true. Since x ≥ y and x + y = 10, we can substitute y = 10 - x into the inequality: x ≥ 10 - x. Adding x to both sides gives 2x ≥ 10, so x ≥ 5. This means x must be at least 5, making choice A correct. Choice B (y ≥ 5) is false because if x = 6 and y = 4, the conditions are satisfied but y < 5.

Question 9

1. $x \le 4$
2. $x \le -4$
3. $x \ge 4$
4. $x \ge -4$ (correct answer)
5. $x=-4$

Explanation: This question tests solving linear inequalities with attention to inequality direction. Starting with 3 - 2x ≤ 11, we first subtract 3 from both sides to get -2x ≤ 8. When dividing both sides by -2, we must reverse the inequality sign, yielding x ≥ -4. This means x can be any value greater than or equal to -4. A common error is forgetting to flip the inequality sign when dividing by a negative number, which would incorrectly give x ≤ -4.

Question 10

1. $t\le -6$
2. $t\ge -6$ (correct answer)
3. $t\le 6$
4. $t\ge 6$
5. $t=-6$

Explanation: This question tests solving linear inequalities with variables on both sides. Starting with 2(t - 3) ≤ 4t + 6, we first expand the left side to get 2t - 6 ≤ 4t + 6. Subtracting 2t from both sides gives -6 ≤ 2t + 6. Subtracting 6 from both sides yields -12 ≤ 2t. Dividing by 2 gives -6 ≤ t, which is equivalent to t ≥ -6. A common error would be to incorrectly move terms, potentially reversing the inequality sign when it shouldn't be reversed.

Question 11

1. 20
2. 25
3. 30 (correct answer)
4. 35
5. 45

Explanation: This question tests solving linear equations with decimals. The correct approach is to isolate x by moving terms. For 0.4x + 6 = 0.1x + 15, subtract 0.1x to get 0.3x + 6 = 15, then subtract 6 yielding 0.3x = 9, so x = 30. This is justified as 0.4(30) + 6 = 18 equals 0.1(30) + 15 = 18. A common incorrect option like 25 might arise from miscalculating the difference. Another error could be multiplying decimals wrong, leading to 35 or 45.

Question 12

1. $4$ (correct answer)
2. $6$
3. $8$
4. $12$
5. $-4$

Explanation: This question tests solving systems of linear equations with two variables. The correct approach is to use substitution or elimination to find the value of y. From the system x + y = 14 and x - 2y = 2, subtract the second from the first to get 3y = 12, so y = 4. This is justified by substituting back: x = 10, and checking x - 2y = 10 - 8 = 2. A common incorrect option like 6 might arise from adding instead of subtracting the equations. Another error could be solving for x instead, leading to larger values like 12.

Question 13

1. 3
2. 4
3. 5 (correct answer)
4. 6
5. 7

Explanation: This question tests solving linear equations in a cost model. The correct approach is to set up the equation with fixed and variable costs, then solve for months. Given 20 + 15m = 95, subtract 20 to get 15m = 75, then m = 5. This is justified as the total is 20 + 15(5) = 95. A common incorrect option like 6 might come from dividing 95 by 15 without subtracting. Another error could be adding fees incorrectly, leading to 4 or 7.

Question 14

1. 6
2. 8
3. 10
4. 12 (correct answer)
5. 16

Explanation: This question tests systems of linear equations for coin problems. The correct approach is to set up equations for number and value, then solve for quarters. Let q be quarters and d dimes: q + d = 22 and 25q + 10d = 400; simplify to 5q + 2d = 80, substitute d = 22 - q to get 5q + 2(22 - q) = 80, so 3q + 44 = 80, 3q = 36, q = 12. This is justified as 12 quarters and 10 dimes total 22 coins and $4.00. A common incorrect option like 10 might come from ignoring the value difference. Another error could be dividing totals incorrectly, leading to 8 or 16.

Question 15

1. $0$
2. $2$
3. $3$
4. $4$ (correct answer)
5. $6$

Explanation: This question tests solving linear equations with parentheses. The correct approach is to distribute and isolate the variable. For 7 - 3(x - 2) = 1, distribute to get 7 - 3x + 6 = 1, then 13 - 3x = 1. Subtract 13 to yield -3x = -12, so x = 4. This is justified as 7 - 3(4 - 2) = 7 - 6 = 1. A common incorrect option like 3 might arise from forgetting to distribute the negative. Another error could be mishandling constants, leading to 2 or 6.

Question 16

1. 10
2. 12 (correct answer)
3. 14
4. 17
5. 7

Explanation: This question tests solving linear equations modeled from a cost scenario. The correct approach is to set up the equation with the flat fee and per-mile charge, then isolate the variable for miles. Given 5 + 2m = 29, subtract 5 to get 2m = 24, then divide by 2 yielding m = 12. This is justified as the total fare is 5 + 2(12) = 29. A common incorrect option like 14 might come from dividing 29 by 2 without subtracting the fee. Another error could be subtracting incorrectly, leading to 10 or 17.

Question 17

1. $x<4$ (correct answer)
2. $x\le 4$
3. $x>4$
4. $x\ge 4$
5. $x=4$

Explanation: This question tests solving linear inequalities with fractions. The correct approach is to multiply both sides by the denominator and isolate x. For (2x - 5)/3 < 1, multiply by 3 to get 2x - 5 < 3, then 2x < 8, so x < 4. This is justified because testing x = 3 gives (6 - 5)/3 = 1/3 < 1, while x = 4 gives (8 - 5)/3 = 1 not less. A common incorrect option like x ≤ 4 fails for the strict inequality at x = 4. Another error might be not reversing signs unnecessarily, leading to x > 4.

Question 18

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

Explanation: This question tests solving systems of linear equations. The correct approach is to add or substitute to find y. From x + y = 5 and 2x - y = 7, add to get 3x = 12, x = 4, then y = 1. This is justified as 4 + 1 = 5 and 8 - 1 = 7. A common incorrect option like 2 might come from subtracting equations wrongly. Another error could be sign mishandling, leading to -1 or 3.

Question 19

1. $4$ (correct answer)
2. $\frac{23}{7}$
3. $\frac{28}{7}$
4. $\frac{18}{7}$
5. $5$

Explanation: This question tests solving a linear equation involving the cost of notebooks. To find the price per notebook, we need to set up an equation: 7 notebooks at $n each, minus a $5 coupon, equals $23 paid. This gives us 7n - 5 = 23. Adding 5 to both sides yields 7n = 28, and dividing by 7 gives n = 4. Therefore, each notebook costs $4. A common error would be to add the coupon instead of subtracting it, which would give 7n + 5 = 23, leading to n = 18/7.