Linear Equations & Inequalities - Finite Mathematics

1. A company manufactures tablets and laptops. Each tablet requires 2 hours of assembly, and each laptop requires 3 hours. The company has at most 600 assembly hours available per day. The packaging department can handle at most 250 devices (tablets and laptops combined) per day. If the company plans to manufacture exactly 80 laptops tomorrow, what is the maximum number of tablets, $t$, that can be manufactured?
   Answer: 170 Let $t$ be the number of tablets and $l$ be the number of laptops. The constraints are given by two inequalities. Assembly hours: $2t + 3l \le 600$. Packaging capacity: $t + l \le 250$. We are given that $l=80$. We must substitute this value into both inequalities to find the constraints on $t$. From the assembly constraint: $2t + 3(80) \le 600 \implies 2t + 240 \le 600 \implies 2t \le 360 \implies t \le 180$. From the packaging constraint: $t + 80 \le 250 \implies t \le 170$. For both constraints to be satisfied, $t$ must be less than or equal to both 180 and 170. The more restrictive condition is $t \le 170$. Therefore, the maximum number of tablets that can be manufactured is 170.
2. In a certain market, the demand for a product is given by the equation $p = -2q + 300$ and the supply is given by $p = 3q + 150$, where $p$ is the price per unit and $q$ is the number of units. The government decides to impose a tax of $10 per unit on the suppliers. What is the total tax revenue collected by the government at the new market equilibrium?
   Answer: $280 The tax of $10 per unit on suppliers shifts the supply curve vertically upwards by $10. The new supply equation is $p_{new} = (3q + 150) + 10 = 3q + 160$. To find the new equilibrium, set the new supply equal to the demand: $3q + 160 = -2q + 300$. Solving for $q$ gives $5q = 140$, so the new equilibrium quantity is $q_{new} = 28$. The government's tax revenue is the tax per unit multiplied by the number of units sold at the new equilibrium. Tax Revenue = $10 \times $q_{new}$ = 10 \times 28 = $280.
3. A tourist is renting a car for a 5-day trip. Rental Company A charges $40 per day and $0.15 per mile. Rental Company B charges $55 per day and $0.10 per mile. What is the maximum number of full miles the tourist can drive for Company A to be the less expensive option?
   Answer: 1499 Let $m$ be the number of miles driven. The total cost for the 5-day trip for Company A is $C_A = 5(40) + 0.15m = 200 + 0.15m$. The total cost for Company B is $C_B = 5(55) + 0.10m = 275 + 0.10m$. We want to find the range of miles where Company A is less expensive, so we set up the inequality $C_A < C_B$. This gives $200 + 0.15m < 275 + 0.10m$. Subtracting $200$ and $0.10m$ from both sides gives $0.05m < 75$. Dividing by 0.05 gives $m < 1500$. Since the question asks for the maximum number of full miles, we need the largest integer that satisfies this inequality, which is 1499.
4. A chemist needs to create 20 liters of a saline solution with a salt concentration between 20% and 25%, inclusive. To do this, she will mix a 15% salt solution with a 40% salt solution. What is the allowable range for the volume of the 15% solution that she can use?
   Answer: Between 12 and 16 liters. Let $x$ be the volume in liters of the 15% solution. Then the volume of the 40% solution is $20-x$. The total amount of salt in the mixture is $0.15x + 0.40(20-x)$. The total volume is 20 liters. The final concentration is the total salt divided by the total volume, which must be between 0.20 and 0.25. This gives the inequality: $0.20 \le \frac{0.15x + 0.40(20-x)}{20} \le 0.25$. Multiply all parts by 20: $4 \le 0.15x + 8 - 0.40x \le 5$. Simplify the middle expression: $4 \le 8 - 0.25x \le 5$. Subtract 8 from all parts: $-4 \le -0.25x \le -3$. Divide by -0.25 and reverse the inequalities: $\frac{-3}{-0.25} \le x \le \frac{-4}{-0.25}$, which simplifies to $12 \le x \le 16$. So, she must use between 12 and 16 liters of the 15% solution.
5. A manufacturing company produces widgets at a cost of $8 per widget plus a fixed daily cost of $200. If the company's daily budget cannot exceed $1,400, and they must produce at least 50 widgets per day to meet demand, what is the maximum number of widgets they can produce in one day?
   Answer: 150 widgets This is a linear inequality problem involving cost constraints and production requirements. When you see questions about maximum production under budget limits, you need to set up an inequality that represents the total cost staying within the budget. The total daily cost equals the fixed cost plus the variable cost per unit. Here, that's $200 + 8x$, where $x$ is the number of widgets produced. Since the budget cannot exceed $1,400, you have: $200 + 8x \leq 1400$ Solving this inequality: $8x \leq 1200$, so $x \leq 150$. Since they must produce at least 50 widgets (which is satisfied), the maximum number of widgets they can produce is 150. Let's verify: $200 + 8(150) = 200 + 1200 = 1400$, which exactly meets the budget limit. Looking at the wrong answers: Choice A (125 widgets) would cost only $1,200 total, staying well under budget but not maximizing production. This represents the common error of being too conservative. Choice B (175 widgets) would cost $1,600, which exceeds the $1,400 budget by $200. Choice D (200 widgets) would cost $1,800, wildly exceeding the budget by $400. Both B and D represent the trap of ignoring the budget constraint entirely. Strategy tip: In optimization problems with constraints, always check that your answer satisfies all given conditions. Set up the inequality carefully, solve it completely, and verify your final answer by substituting back into the original cost equation.
6. Two machines, A and B, were purchased at the same time. Machine A had an initial value of $40,000 and depreciates linearly at a rate of $2,500 per year. Machine B had an initial value of $32,000 and depreciates linearly at a rate of $1,500 per year. For what values of $t$, the number of years since purchase, is the value of machine B greater than or equal to the value of machine A?
   Answer: $t \ge 8$ Let $V_A(t)$ and $V_B(t)$ be the values of machine A and B after $t$ years. The value functions are $V_A(t) = 40000 - 2500t$ and $V_B(t) = 32000 - 1500t$. We want to find when $V_B(t) \ge V_A(t)$. The inequality is $32000 - 1500t \ge 40000 - 2500t$. To solve for $t$, we can add $2500t$ to both sides and subtract $32000$ from both sides. This gives $2500t - 1500t \ge 40000 - 32000$, which simplifies to $1000t \ge 8000$. Dividing by 1000, we find $t \ge 8$.
7. Two competing phone plans have the following monthly costs: Plan A costs $30 plus $0.05 per minute, and Plan B costs $45 plus $0.02 per minute. For what number of minutes will Plan A cost exactly $3 more than Plan B?
   Answer: 600 minutes Let $m$ be the number of minutes. Plan A costs $30 + 0.05m$ and Plan B costs $45 + 0.02m$. We need Plan A to cost exactly $3 more than Plan B: $30 + 0.05m = (45 + 0.02m) + 3$. Simplifying: $30 + 0.05m = 48 + 0.02m$. Rearranging: $0.05m - 0.02m = 48 - 30$, so $0.03m = 18$, giving $m = 600$. Choice B results from the error $0.06m = 30$ (incorrect coefficient). Choice C comes from solving $0.05m - 0.02m = 45 - 30 - 3$ (subtracting 3 instead of adding). Choice D results from the equation $0.05m = 15$ (multiple algebraic errors).
8. A mobile data plan costs $30 per month, which includes 5 gigabytes (GB) of data. For any data usage exceeding 5 GB, the provider charges $8 per additional GB. If a customer wants to keep their monthly bill to be at most $60, what is the maximum number of gigabytes of data they can use?
   Answer: 8.75 GB This is a piecewise cost function problem where you need to find the maximum data usage within a budget constraint. The cost structure has two parts: a base fee plus overage charges. Let's set up the problem systematically. The base plan costs $30 for up to 5 GB. Any usage beyond 5 GB costs an additional $8 per GB. If the customer uses $x$ GB total where $x > 5$, their bill will be: $30 + 8(x - 5)$. Since they want to spend at most $60, we solve: $30 + 8(x - 5) \leq 60$ Simplifying: $8(x - 5) \leq 30$, so $x - 5 \leq 3.75$, which gives us $x \leq 8.75$ GB. Therefore, the maximum usage is 8.75 GB, making B correct. Now for the wrong answers: A) 3.75 GB represents a common error where students find the maximum additional data (beyond the 5 GB base) but forget to add it to the base amount. C) 7.50 GB might result from setting up the inequality as $30 + 8(x - 5) = 60$ but making an arithmetic error, or from misunderstanding the overage calculation. D) 12.5 GB likely comes from incorrectly thinking you get $60 worth of additional data at $8 per GB ($$60 ÷ 8 = 7.5$$, plus the base 5 GB). Study tip: In piecewise pricing problems, always identify the base amount included in the fixed fee, then carefully set up your inequality for the variable portion. Double-check by substituting your answer back into the original cost structure.
9. A water tank is being filled at a rate of 12 gallons per minute. The tank starts with 50 gallons and has a maximum capacity of 350 gallons. During which time interval will the tank contain between 200 and 300 gallons?
   Answer: Between 12.5 and 20.83 minutes When you encounter rate problems involving tanks or containers, you're working with linear equations where the amount changes at a constant rate over time. The key is setting up an equation that relates the initial amount, rate of change, and time. Here, the tank starts with 50 gallons and fills at 12 gallons per minute, so the amount in the tank after $t$ minutes is: $\text{Amount} = 50 + 12t$ To find when the tank contains between 200 and 300 gallons, you need to solve two inequalities. For 200 gallons: $50 + 12t = 200$, so $12t = 150$ and $t = 12.5$ minutes. For 300 gallons: $50 + 12t = 300$, so $12t = 250$ and $t = 20.83$ minutes. Therefore, the tank contains between 200 and 300 gallons from 12.5 to 20.83 minutes. Choice A (12.5 to 25 minutes) correctly identifies the start time but extends too far. At 25 minutes, the tank would contain $50 + 12(25) = 350$ gallons, which exceeds our upper limit of 300 gallons. Choice C (16.67 to 25 minutes) misses the beginning of the interval and again goes too far. At 16.67 minutes, the tank already contains about 250 gallons, so it missed time when it was between 200-250 gallons. Choice D (10 to 20 minutes) starts too early—at 10 minutes, there are only 170 gallons in the tank. Always solve for both boundary conditions in interval problems. Set up your linear equation, solve for each endpoint separately, then double-check by substituting back into your original equation.
10. A company manufactures a product with a variable cost of $15 per unit and fixed monthly costs of $12,000. The product sells for $35 per unit. For what range of units sold, $x$, will the company achieve a monthly profit that is between $4,000 and $10,000, inclusive?
    Answer: Between 800 and 1,100 units. The profit function is given by $P(x) = \text{Revenue} - \text{Total Cost}$. Revenue is $35x$ and Total Cost is $15x + 12000$. So, $P(x) = 35x - (15x + 12000) = 20x - 12000$. The problem asks for the profit to be between $4,000 and $10,000, inclusive. This can be written as the compound inequality: $4000 \le 20x - 12000 \le 10000$. To solve for $x$, first add 12,000 to all parts: $16000 \le 20x \le 22000$. Then, divide all parts by 20: $800 \le x \le 1100$. Thus, the company must sell between 800 and 1,100 units.
11. An investor is choosing between two funds. Fund A requires an initial investment of $5,000 and has an account maintenance fee of $100 per year. Fund B requires an initial investment of $8,000 and has a maintenance fee of $50 per year. Assuming no other gains or losses, what is the minimum number of full years the funds must be held for the total cost of Fund B to be less than the total cost of Fund A?
    Answer: 61 Let $t$ be the number of years. The total cost for Fund A is $C_A(t) = 5000 + 100t$. The total cost for Fund B is $C_B(t) = 8000 + 50t$. We want to find the minimum integer $t$ for which $C_B(t) < C_A(t)$. Setting up the inequality: $8000 + 50t < 5000 + 100t$. Subtracting $5000$ and $50t$ from both sides gives $3000 < 50t$. Dividing by 50 gives $60 < t$. Since $t$ must be a number of full years, the smallest integer value for $t$ that satisfies this condition is 61.
12. A chemist needs to create a 500 mL solution that is exactly 26% acid. She has two stock solutions available: one is 20% acid and the other is 40% acid. To obtain the desired solution, she mixes a certain amount of the 20% solution, a certain amount of the 40% solution, and a quantity of pure water (0% acid). If she uses 150 mL of the 20% solution, what volume of pure water must she add?
    Answer: 100 mL Let $x$, $y$, and $w$ be the volumes in mL of the 20% solution, 40% solution, and pure water, respectively. The total volume is $x+y+w=500$. The total amount of acid is $0.20x + 0.40y + 0w = 0.26(500)$. We are given $x=150$. The acid equation becomes $0.20(150) + 0.40y = 130$, which simplifies to $30 + 0.40y = 130$. Solving for $y$, we get $0.40y = 100$, so $y = 250$ mL. Now, using the total volume equation, $150 + 250 + w = 500$. This gives $400 + w = 500$, so $w = 100$ mL.
13. A workshop manufactures two products: tables and chairs. Each table requires 2 hours of carpentry and 1 hour of finishing. Each chair requires 3 hours of carpentry and 0.5 hours of finishing. The workshop has a maximum of 120 carpentry hours and 40 finishing hours available per week. If the workshop decides to produce exactly 15 tables, what is the maximum number of chairs it can produce in a week?
    Answer: 30 Let $t$ be the number of tables and $c$ be the number of chairs. The constraints are given by two inequalities: $2t + 3c \le 120$ (carpentry) and $t + 0.5c \le 40$ (finishing). Given $t=15$, we substitute this value into both inequalities. For carpentry: $2(15) + 3c \le 120 \implies 30 + 3c \le 120 \implies 3c \le 90 \implies c \le 30$. For finishing: $15 + 0.5c \le 40 \implies 0.5c \le 25 \implies c \le 50$. To satisfy both constraints, the number of chairs must be less than or equal to both 30 and 50. The more restrictive condition is $c \le 30$, so the maximum number of chairs is 30.
14. A machine is purchased for $65,000 and is expected to have a scrap value of $5,000 after 12 years. Assuming linear depreciation, during which time interval, in years $t$ from the date of purchase, is the book value of the machine between $20,000 and $35,000, inclusive?
    Answer: Between years 6 and 9. First, find the linear depreciation model. The total depreciation over 12 years is $65,000 - 5,000 = $60,000. The annual depreciation is $60,000 / 12 = $5,000 per year. The value of the machine, $V(t)$, after $t$ years is $V(t) = 65,000 - 5,000t$. We want to find the time interval for which $20,000 \le V(t) \le 35,000$. Substitute the expression for $V(t)$: $20,000 \le 65,000 - 5,000t \le 35,000$. Subtract 65,000 from all parts: $-45,000 \le -5,000t \le -30,000$. Divide all parts by -5,000 and reverse the direction of the inequalities: $\frac{-30,000}{-5,000} \le t \le \frac{-45,000}{-5,000}$. This simplifies to $6 \le t \le 9$. The book value is between $20,000 and $35,000 from year 6 to year 9, inclusive.
15. A rectangular garden has a length that is 8 feet more than twice its width. If the perimeter must be at least 76 feet but no more than 100 feet, what are the possible values for the width of the garden?
    Answer: $10 \leq w \leq 14$ When you encounter word problems involving constraints and inequalities, start by translating the verbal descriptions into mathematical expressions, then set up the constraint inequalities. Let's define the width as $w$ feet. Since the length is "8 feet more than twice the width," we write $l = 2w + 8$. The perimeter of a rectangle is $P = 2l + 2w$, so substituting our expression for length gives us $P = 2(2w + 8) + 2w = 4w + 16 + 2w = 6w + 16$. The constraint "at least 76 feet but no more than 100 feet" translates to $76 \leq 6w + 16 \leq 100$. Solving the left inequality: $76 \leq 6w + 16$ gives us $60 \leq 6w$, so $w \geq 10$. Solving the right inequality: $6w + 16 \leq 100$ gives us $6w \leq 84$, so $w \leq 14$. Therefore, $10 \leq w \leq 14$, which is choice B. Choice A ($12 \leq w \leq 16$) sets the minimum width too high and maximum too high. Choice C ($8 \leq w \leq 12$) sets both bounds too low, allowing widths that would make the perimeter less than 76 feet. Choice D ($14 \leq w \leq 18$) sets both bounds too high, allowing widths that would exceed the 100-foot maximum perimeter. Remember to always check your inequality work by substituting boundary values back into the original constraints. This catches sign errors and ensures your solution makes sense in the problem context.
16. A car rental company charges a flat fee of $25 plus $0.18 per mile driven. If Maria's total bill was between $67 and $85, inclusive, what is the range of miles she could have driven?
    Answer: 233 to 333 miles Let $x$ be the miles driven. The total cost is $25 + 0.18x$. We need $67 \leq 25 + 0.18x \leq 85$. Subtracting 25: $42 \leq 0.18x \leq 60$. Dividing by 0.18: $\frac{42}{0.18} \leq x \leq \frac{60}{0.18}$, which gives $233.33... \leq x \leq 333.33...$. Since miles must be whole numbers, the range is 233 to 333 miles. Choice B uses incorrect rounding (rounding up the lower bound). Choice C uses the intermediate values before dividing by 0.18. Choice D incorrectly adds 25 instead of subtracting it.
17. A gym membership has an enrollment fee of $75 and monthly dues of $45. A competing gym has no enrollment fee but charges $55 per month. After how many complete months will the total cost at the first gym be less than the total cost at the second gym?
    Answer: 8 months This is a classic break-even analysis problem where you need to find when one linear cost function becomes smaller than another. Set up equations for each gym's total cost and find where they intersect. For the first gym, total cost = $75 + 45m$ where $m$ is the number of months. For the second gym, total cost = $55m$. You want to find when the first gym costs less, so solve the inequality: $75 + 45m < 55m$. Subtracting $45m$ from both sides gives $75 < 10m$, so $m > 7.5$. Since the question asks for complete months, you need at least 8 months for the first gym to be cheaper. Let's verify: After 7 months, gym 1 costs $75 + 45(7) = 390$ and gym 2 costs $55(7) = 385$. Gym 1 is still more expensive. After 8 months, gym 1 costs $75 + 45(8) = 435$ and gym 2 costs $55(8) = 440$. Now gym 1 is cheaper. Option A) 9 months is incorrect because the first gym becomes cheaper after just 8 months, not 9. Option B) 7 months is wrong because at 7 months, the first gym still costs more ($390 vs $385). Option D) 6 months is incorrect because at 6 months, the first gym costs $345 while the second costs only $330. When solving break-even problems, always check your boundary values by substituting back into the original equations. The inequality often gives you a decimal answer, so pay attention to whether the question asks for "complete" periods.
18. A student's final grade in a course is determined by a weighted average of three exams. The first exam is worth 20%, the second is worth 30%, and the final exam is worth 50% of the grade. The student scored 75 on the first exam and 85 on the second. To receive a final grade of 'B', the student's average must be in the interval $[80, 90)$. What is the range of scores the student must achieve on the final exam to earn a 'B'?
    Answer: $[79, 99)$ Let $x_3$ be the score on the final exam. The final grade $G$ is calculated as $G = 0.20(75) + 0.30(85) + 0.50x_3$. This simplifies to $G = 15 + 25.5 + 0.5x_3 = 40.5 + 0.5x_3$. The condition for a 'B' grade is $80 \le G < 90$. Substituting the expression for $G$, we get the compound inequality $80 \le 40.5 + 0.5x_3 < 90$. Subtracting 40.5 from all parts gives $39.5 \le 0.5x_3 < 49.5$. Multiplying all parts by 2 gives $79 \le x_3 < 99$. Thus, the required range of scores is $[79, 99)$.
19. A sales representative earns a monthly income composed of a $2,000 base salary plus a commission on her total monthly sales, $S$. The commission is 5% on sales up to $40,000. For sales exceeding $40,000, she earns an 8% commission on the portion of sales above $40,000. Which inequality represents the total sales $S$ required for her to earn a monthly income of at least $5,000?
    Answer: $2000 + 0.05(40000) + 0.08(S - 40000) \ge 5000$ First, determine the income at the $40,000 sales threshold: $2000 + 0.05(40000) = $2000 + $2000 = $4000$. Since the target income of $5,000 is greater than $4,000, sales must exceed $40,000. The income $I(S)$ for $S > 40000$ consists of three parts: the base salary ($2000), the commission on the first $40,000 of sales ($0.05 \times 40000$), and the commission on sales above $40,000 ($0.08 \times (S - 40000)$). The total income must be at least $5,000, so the correct inequality is $2000 + 0.05(40000) + 0.08(S - 40000) \ge 5000$.
20. A small business produces widgets. The fixed monthly cost is $12,000, and the variable cost per widget is $15. Each widget is sold for $40. The business owner sets a goal for the monthly profit to be at least 20% of the total monthly revenue. What is the minimum number of widgets that must be produced and sold to meet this goal?
    Answer: 706 Let $x$ be the number of widgets. The revenue function is $R(x) = 40x$ and the cost function is $C(x) = 12000 + 15x$. The profit function is $P(x) = R(x) - C(x) = 40x - (12000 + 15x) = 25x - 12000$. The goal is for the profit to be at least 20% of revenue, which can be written as the inequality $P(x) \ge 0.20 \cdot R(x)$. Substituting the functions gives $25x - 12000 \ge 0.20(40x)$, which simplifies to $25x - 12000 \ge 8x$. Solving for $x$ yields $17x \ge 12000$, so $x \ge \frac{12000}{17} \approx 705.88$. Since the number of widgets must be an integer, the minimum number is 706.

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