Linear Equations and Inequalities - Quantitative Reasoning

1. Which inequality is NOT equivalent to $x < -2$?
   Answer: $-x < 2$ Multiplying or dividing an inequality by a negative number reverses the sign, so -x < 2 becomes x > -2, not x < -2; therefore it is the one that is not equivalent. The tempting one is -2x > 4, because the greater-than sign makes it look different, but dividing by -2 flips it to x < -2, so it truly is equivalent.
2. A printing company charges $50 to set up a job plus $0.20 per page printed. If the total cost was $86, how many pages were printed?
   Answer: 180 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.
3. Real number $t$ satisfies $2(t-3)\le 4t+6$. For which value(s) of $t$ is the inequality true?
   Answer: $t\ge -6$ 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.
4. What is the value of $x$ if $0.4x+6=0.1x+15$?
   Answer: 30 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.
5. Two numbers $x$ and $y$ satisfy the system $\begin{cases}x+y=14\\x-2y=2\end{cases}$ What is the value of $y$?
   Answer: $4$ 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.
6. Numbers $x$ and $y$ satisfy the system $\begin{cases}2x+y=13\\x+y=9\end{cases}$ What is the value of $x$?
   Answer: $4$ 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.
7. What is the value of $x$ if $7-3(x-2)=1$?
   Answer: $4$ 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.
8. Real numbers $x$ and $y$ satisfy $x+y=10$ and $x\ge y$. Which of the following must be true?
   Answer: $x\ge 5$ 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.
9. For which value(s) of $x$ is the inequality $\dfrac{x-5}{3} > 2$ true?
   Answer: $x>11$ 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.
10. A store sells notebooks for $n$ dollars each. If a customer buys 7 notebooks and uses a coupon worth $5 off the total, the amount paid is $23. What is the value of $n$?
    Answer: $4$ 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.
11. For what value of $a$ is the solution of $ax - 1 > 2$ given by $x < -3$?
    Answer: $a = -1$ Rewrite as ax > 3. Since the solution is x < -3, a must be negative, and dividing by a flips the inequality: x < 3/a. Set 3/a = -3, so a = -1. The tempting wrong choice -3 confuses the coefficient with the boundary; it would give x < -1.
12. For what value of $k$ does $kx - 4 = 3x + k$ have no solution?
    Answer: $k = 3$ Rearrange to (k - 3)x = k + 4. No linear equation has no solution only when the coefficient of x is 0 and the remaining constant is not 0, so k - 3 = 0 gives k = 3; then 0x = 7 is impossible. The tempting trap is k = -4, which makes the constants match on both sides, but the x terms still leave one solution.
13. For which value(s) of $x$ is the inequality $\frac{2x-5}{3}<1$ true?
    Answer: $x<4$ 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.
14. A car rental company charges a flat fee of $F$ dollars plus $0.25$ dollars per mile driven. If a customer drove $m$ miles and was charged a total of $37.50$ dollars, which of the following equations represents this situation?
    Answer: $F+0.25m=37.50$ 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.
15. For which value(s) of $x$ is the inequality $3-2x \le 11$ true?
    Answer: $x \ge -4$ 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.
16. A jar contains only quarters and dimes, totaling 22 coins. The total value of the coins is $4.00. How many quarters are in the jar?
    Answer: 12 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.
17. In a system of equations, $x$ and $y$ satisfy $\begin{cases}x+y=9\\2x+2y=18\end{cases}$ Which of the following must be true?
    Answer: The system has infinitely many solutions. 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.
18. A gym charges a one-time sign-up fee of $20 and then $15 per month. If the total cost after $m$ months is $95, what is the value of $m$?
    Answer: 5 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.
19. A taxi charges a flat fee of $5 plus $2 per mile. If the total fare for a ride was $29, how many miles was the ride?
    Answer: 12 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.
20. An airport shuttle charges $2.75 plus $1.50 per half-mile. A ride costs $16.25. How many miles is it?
    Answer: 4.5 miles First subtract the flat fee: 16.25 - 2.75 = 13.50. At 1.50 per half-mile, 13.50 / 1.50 = 9 half-miles. Since each half-mile is 0.5 mile, 9 half-miles = 4.5 miles. The tempting trap is stopping at 9 and calling it miles, but that is the number of half-miles, not miles.

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