Finite Mathematics Quiz: Linear Equations And Inequalities
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Linear Equations And InequalitiesQuestion 1 of 20

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?

600 minutes
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Finite Mathematics Quiz

Finite Mathematics Quiz: Linear Equations And Inequalities

Practice Linear Equations And Inequalities in Finite Mathematics 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 Finite Mathematics.

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

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?

  1. 600 minutes (correct answer)
  2. 500 minutes
  3. 400 minutes
  4. 300 minutes
Explanation: Let mm be the number of minutes. Plan A costs 30+0.05m30 + 0.05m and Plan B costs 45+0.02m45 + 0.02m. We need Plan A to cost exactly $3 more than Plan B: $30+0.05m=(45+0.02m)+330 + 0.05m = (45 + 0.02m) + 3 .Simplifying:. Simplifying: 30+0.05m=48+0.02m30 + 0.05m = 48 + 0.02m .Rearranging:. Rearranging: 0.05m−0.02m=48−300.05m - 0.02m = 48 - 30 ,so, so 0.03m=180.03m = 18 ,giving, giving m=600m = 600 .ChoiceBresultsfromtheerror. Choice B results from the error 0.06m=300.06m = 30 (incorrectcoefficient).ChoiceCcomesfromsolving(incorrect coefficient). Choice C comes from solving 0.05m−0.02m=45−30−30.05m - 0.02m = 45 - 30 - 3 (subtracting3insteadofadding).ChoiceDresultsfromtheequation(subtracting 3 instead of adding). Choice D results from the equation 0.05m=150.05m = 15 $ (multiple algebraic errors).

Question 2

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?

  1. 125 widgets
  2. 175 widgets
  3. 150 widgets (correct answer)
  4. 200 widgets
Explanation: 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+8x200 + 8x, where xx is the number of widgets produced. Since the budget cannot exceed $1,400, you have: $200+8x≤1400200 + 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.

Question 3

A taxi charges $3.50 for the first mile and $1.25 for each additional mile. Sarah paid between $18 and $22 for her ride. If she traveled a whole number of miles, which of the following could NOT be the distance she traveled?

  1. 13 miles
  2. 14 miles
  3. 15 miles
  4. 16 miles (correct answer)
Explanation: For nn miles, the cost is 3.50+1.25(n−1)3.50 + 1.25(n-1). We need 18≤3.50+1.25(n−1)≤2218 \leq 3.50 + 1.25(n-1) \leq 22. Simplifying: 14.50≤1.25(n−1)≤18.5014.50 \leq 1.25(n-1) \leq 18.50, so 11.6≤n−1≤14.811.6 \leq n-1 \leq 14.8, giving 12.6≤n≤15.812.6 \leq n \leq 15.8. For whole miles: n∈{13,14,15}n \in \{13, 14, 15\}. Let's verify: 13 miles costs 3.50+1.25(12)=18.503.50 + 1.25(12) = 18.50 ✓; 14 miles costs 3.50+1.25(13)=19.753.50 + 1.25(13) = 19.75 ✓; 15 miles costs 3.50+1.25(14)=21.003.50 + 1.25(14) = 21.00 ✓; 16 miles costs 3.50+1.25(15)=22.253.50 + 1.25(15) = 22.25 > 22 ✗. Therefore, 16 miles is NOT possible.

Question 4

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?

  1. Between 800 and 1,100 units. (correct answer)
  2. Between 200 and 500 units.
  3. Between 115 and 286 units.
  4. Any number of units over 600.
Explanation: The profit function is given by P(x)=Revenue−Total CostP(x) = \text{Revenue} - \text{Total Cost}. Revenue is 35x35x and Total Cost is 15x+1200015x + 12000. So, P(x)=35x−(15x+12000)=20x−12000P(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≤20x−12000≤100004000 \le 20x - 12000 \le 10000. To solve for xx, first add 12,000 to all parts: 16000≤20x≤2200016000 \le 20x \le 22000. Then, divide all parts by 20: 800≤x≤1100800 \le x \le 1100. Thus, the company must sell between 800 and 1,100 units.

Question 5

In a competitive market, the price-demand and price-supply equations for a product are p=−2q+100p = -2q + 100 and p=3q+20p = 3q + 20 respectively, where pp is the price and qq is the quantity. If the government imposes a per-unit tax tt on the supplier, what is the range of tax values for which the new equilibrium quantity will be at least 10 units?

  1. 0≤t≤300 \le t \le 30 (correct answer)
  2. t≥30t \ge 30
  3. 0≤t≤100 \le t \le 10
  4. t≥0t \ge 0
Explanation: The tax tt increases the supplier's cost per unit, shifting the supply curve up. The new supply equation is p=3q+20+tp = 3q + 20 + t. To find the new equilibrium, set the new supply equal to the demand: 3q+20+t=−2q+1003q + 20 + t = -2q + 100. Now, solve for qq in terms of tt: 5q=80−t5q = 80 - t, so q=16−t5q = 16 - \frac{t}{5}. The problem requires the new equilibrium quantity to be at least 10 units, so we set up the inequality q≥10q \ge 10. This gives 16−t5≥1016 - \frac{t}{5} \ge 10. Subtract 16 from both sides: −t5≥−6-\frac{t}{5} \ge -6. Multiply by -5 and reverse the inequality sign: t≤30t \le 30. Since a tax cannot be negative, we have t≥0t \ge 0. Combining these gives the range 0≤t≤300 \le t \le 30.

Question 6

A scientific experiment requires that the temperature TCT_C in degrees Celsius be kept in the interval defined by ∣TC−50∣≤5|T_C - 50| \le 5. A monitoring system displays the temperature TFT_F in degrees Fahrenheit, where TF=95TC+32T_F = \frac{9}{5}T_C + 32. What is the corresponding interval for the temperature in degrees Fahrenheit?

  1. [113,122][113, 122]
  2. [122,131][122, 131]
  3. [7.2,12.8][7.2, 12.8]
  4. [113,131][113, 131] (correct answer)
Explanation: When you encounter absolute value inequalities involving unit conversions, you need to solve the inequality first, then apply the conversion formula to find the corresponding interval in the new units. Start by solving ∣TC−50∣≤5|T_C - 50| \le 5. This absolute value inequality means the distance from 50 is at most 5 units, so: −5≤TC−50≤5-5 \le T_C - 50 \le 5. Adding 50 to all parts gives you 45≤TC≤5045 \le T_C \le 50. So the Celsius temperature must be between 45°C and 55°C. Now convert these boundary values to Fahrenheit using TF=95TC+32T_F = \frac{9}{5}T_C + 32:
  • Lower bound: TF=95(45)+32=81+32=113°FT_F = \frac{9}{5}(45) + 32 = 81 + 32 = 113°F
  • Upper bound: TF=95(55)+32=99+32=131°FT_F = \frac{9}{5}(55) + 32 = 99 + 32 = 131°F
Therefore, the Fahrenheit interval is [113, 131], which is answer D. Answer A gives [113, 122], which correctly finds the lower bound but incorrectly calculates the upper bound—likely from using 50°C instead of 55°C for the maximum temperature. Answer B gives [122, 131], which has the correct upper bound but uses an incorrect lower bound, possibly from a calculation error. Answer C gives [7.2, 12.8], which appears to result from forgetting the "+32" in the conversion formula entirely. Remember: when converting intervals between units, always convert the boundary values separately using the complete conversion formula, and double-check that you've solved the original inequality correctly before converting.

Question 7

A salesperson can choose between two salary plans. Plan A offers a fixed salary of $3,000 per month. Plan B offers a base salary of $1,200 per month plus a 5% commission on total monthly sales. For what range of monthly sales, $S$, will the salary from Plan B be at least $600 more than the salary from Plan A?

  1. S \ge \24,000
  2. S \ge \36,000
  3. S \ge \48,000 (correct answer)
  4. S \ge \60,000
Explanation: The salary for Plan A is SA=3000S_A = 3000. The salary for Plan B is SB=1200+0.05SS_B = 1200 + 0.05S. The condition is that Plan B's salary is at least $600 more than Plan A's salary. This can be written as the inequality: $S_B \ge S_A + 600.Substitutingtheexpressionsforthesalariesgives:. Substituting the expressions for the salaries gives: 1200 + 0.05S \ge 3000 + 600.Simplifytherightside:. Simplify the right side: 1200 + 0.05S \ge 3600.Subtract1200frombothsides:. Subtract 1200 from both sides: 0.05S \ge 2400.Tosolvefor. To solve for S,dividebothsidesby0.05:, divide both sides by 0.05: S \ge \frac{2400}{0.05}.Thisgives. This gives S \ge 48000$. So, the monthly sales must be at least $48,000.

Question 8

A tourist rents a car for a 5-day trip. Agency X charges $30 per day and $0.20 per mile. Agency Y charges $50 per day and $0.10 per mile. For which number of miles driven, $m$, will the total cost from Agency X be strictly less than the total cost from Agency Y?

  1. Fewer than 200 miles
  2. Fewer than 500 miles
  3. Fewer than 1,000 miles (correct answer)
  4. More than 1,000 miles
Explanation: First, calculate the total fixed cost for the 5-day trip for each agency. For Agency X, the cost is C_X = (5 \text{ days} \times \30/\text{day}) + $0.20m = 150 + 0.20m.ForAgencyY,thecostis. For Agency Y, the cost is C_Y = (5 \text{ days} \times $50/\text{day}) + $0.10m = 250 + 0.10m.Wewanttofindtherangeofmiles. We want to find the range of miles mforwhichfor whichC_X < C_Y.Setuptheinequality:. Set up the inequality: 150 + 0.20m < 250 + 0.10m.Subtract$0.10m$frombothsides:. Subtract $0.10m$ from both sides: 150 + 0.10m < 250.Subtract150frombothsides:. Subtract 150 from both sides: 0.10m < 100.Divideby0.10:. Divide by 0.10: m < 1000$. Therefore, the total cost from Agency X is less than Agency Y when the number of miles driven is fewer than 1,000.

Question 9

The population of a town was 12,500 at the beginning of 2010 and grew to 14,000 at the beginning of 2015. Assuming the population growth is linear, in what year will the population first exceed 18,000?

  1. 2027
  2. 2028 (correct answer)
  3. 2029
  4. 2030
Explanation: Let tt be the number of years after the beginning of 2010. So, t=0t=0 corresponds to 2010. We have two points: (0,12500)(0, 12500) and (5,14000)(5, 14000). The rate of growth (slope) is r=14000−125005−0=15005=300r = \frac{14000 - 12500}{5 - 0} = \frac{1500}{5} = 300 people per year. The linear model for the population is P(t)=12500+300tP(t) = 12500 + 300t. We want to find the smallest time tt for which P(t)>18000P(t) > 18000. Set up the inequality: 12500+300t>1800012500 + 300t > 18000. Subtract 12500 from both sides: 300t>5500300t > 5500. Divide by 300: t>5500300=553≈18.33t > \frac{5500}{300} = \frac{55}{3} \approx 18.33. This means the population exceeds 18,000 after 18.33 years from the start of 2010. The year is 2010+t2010 + t. At t=18t=18 (start of 2028), the population is 12500+300(18)=1790012500 + 300(18) = 17900. At t=19t=19 (start of 2029), the population is 12500+300(19)=1820012500 + 300(19) = 18200. Since the population exceeds 18,000 when t>18.33t > 18.33, this occurs during the year 2028.

Question 10

The length of a rectangular garden is 5 meters more than its width. If the perimeter of the garden must be greater than 50 meters but no more than 78 meters, what is the valid range for the garden's width, ww?

  1. 10<w≤2210 < w \le 22
  2. 10<w≤1710 < w \le 17 (correct answer)
  3. 12.5<w≤19.512.5 < w \le 19.5
  4. 20<w≤3420 < w \le 34
Explanation: When you encounter word problems involving constraints and inequalities, start by defining your variable and translating the given relationships into mathematical expressions. Let ww represent the width. Since the length is 5 meters more than the width, the length is w+5w + 5. The perimeter of a rectangle is P=2(length+width)=2(w+5+w)=2(2w+5)=4w+10P = 2(\text{length} + \text{width}) = 2(w + 5 + w) = 2(2w + 5) = 4w + 10. The constraint "greater than 50 meters but no more than 78 meters" translates to: 50<4w+10≤7850 < 4w + 10 \leq 78. Solving the left inequality: 50<4w+1050 < 4w + 10, so 40<4w40 < 4w, which gives w>10w > 10. Solving the right inequality: 4w+10≤784w + 10 \leq 78, so 4w≤684w \leq 68, which gives w≤17w \leq 17. Therefore, 10<w≤1710 < w \leq 17, which is answer choice B. Let's examine why the other options are incorrect. Choice A (10<w≤2210 < w \leq 22) has the correct lower bound but an upper bound that's too high—if w=22w = 22, the perimeter would be 4(22)+10=984(22) + 10 = 98 meters, exceeding the 78-meter limit. Choice C (12.5<w≤19.512.5 < w \leq 19.5) likely results from incorrectly setting up the perimeter formula or making algebraic errors. Choice D (20<w≤3420 < w \leq 34) produces perimeters well above the maximum constraint. Remember: always check your inequality symbols carefully. "Greater than" becomes a strict inequality (>>), while "no more than" becomes "less than or equal to" (≤\leq). Also verify your answer by testing boundary values in the original constraint.

Question 11

The temperature TT (in degrees Celsius) of a chemical process is determined by the dial setting SS according to the formula T(S)=2S+10T(S) = 2S + 10. The process is successful only if the temperature is maintained at 80∘C80^\circ C with an absolute error of no more than 5∘C5^\circ C. Which range of dial settings SS will ensure a successful process?

  1. 35≤S≤4535 \le S \le 45
  2. 32.5≤S≤37.532.5 \le S \le 37.5 (correct answer)
  3. 65≤S≤7565 \le S \le 75
  4. 75≤S≤8575 \le S \le 85
Explanation: When you encounter problems involving absolute error or tolerance limits, you're working with absolute value inequalities that need to be translated into practical constraints. The process succeeds when the temperature is within 5°C of 80°C, meaning: ∣T−80∣≤5|T - 80| \leq 5. This absolute value inequality translates to: 75≤T≤8575 \leq T \leq 85. Since temperature relates to dial setting through T(S)=2S+10T(S) = 2S + 10, you need to find which dial settings produce temperatures in this acceptable range. Substituting the temperature limits: For the minimum: 75=2S+1075 = 2S + 10, so 2S=652S = 65, giving S=32.5S = 32.5 For the maximum: 85=2S+1085 = 2S + 10, so 2S=752S = 75, giving S=37.5S = 37.5 Therefore, the dial setting range is 32.5≤S≤37.532.5 \leq S \leq 37.5, which is answer choice B. Let's check why the other options fail: A) 35≤S≤4535 \leq S \leq 45 would produce temperatures from T(35)=80°CT(35) = 80°C to T(45)=100°CT(45) = 100°C. The upper end exceeds the 85°C limit. C) 65≤S≤7565 \leq S \leq 75 would give temperatures from T(65)=140°CT(65) = 140°C to T(75)=160°CT(75) = 160°C, far above the acceptable range. D) 75≤S≤8575 \leq S \leq 85 would produce temperatures from T(75)=160°CT(75) = 160°C to T(85)=180°CT(85) = 180°C, also way too high. Study tip: When solving tolerance problems, always convert the absolute value constraint to a compound inequality first, then work backward through the given function to find the input range. Double-check by plugging your boundary values back into the original function.

Question 12

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?

  1. 9 months
  2. 7 months
  3. 8 months (correct answer)
  4. 6 months
Explanation: 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+45m75 + 45m where mm is the number of months. For the second gym, total cost = 55m55m. You want to find when the first gym costs less, so solve the inequality: 75+45m<55m75 + 45m < 55m. Subtracting 45m45m from both sides gives 75<10m75 < 10m, so m>7.5m > 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)=39075 + 45(7) = 390 and gym 2 costs 55(7)=38555(7) = 385. Gym 1 is still more expensive. After 8 months, gym 1 costs 75+45(8)=43575 + 45(8) = 435 and gym 2 costs 55(8)=44055(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.

Question 13

A company's profit PP (in thousands of dollars) is modeled by P=3x−15P = 3x - 15, where xx is the number of units sold (in hundreds). If the company needs a profit of at least $12,000 to cover operating expenses, what is the minimum number of units they must sell?

  1. 900 units (correct answer)
  2. 2700 units
  3. 270 units
  4. 90 units
Explanation: We need P≥12P \geq 12 thousand dollars. So 3x−15≥123x - 15 \geq 12, which gives 3x≥273x \geq 27, so x≥9x \geq 9. Since xx represents hundreds of units, we need at least 9×100=9009 \times 100 = 900 units. Choice B multiplies by 3 incorrectly (9×3×1009 \times 3 \times 100). Choice C forgets to convert from hundreds (uses 270 instead of 2700, but that's still wrong as the answer is 900). Choice D uses only the value of xx without converting to actual units.

Question 14

A theater has 180 seats divided into two sections. The orchestra section has 30 more seats than twice the number of seats in the balcony section. If xx represents the number of seats in the balcony section, which equation correctly represents this situation?

  1. 2x+(x−30)=1802x + (x - 30) = 180
  2. x+(2x−30)=180x + (2x - 30) = 180
  3. 2x+(x+30)=1802x + (x + 30) = 180
  4. x+(2x+30)=180x + (2x + 30) = 180 (correct answer)
Explanation: When you encounter word problems involving two unknown quantities that must sum to a total, the key is carefully translating each piece of information into mathematical expressions. Let's define our variable: xx = number of seats in the balcony section. Now we need to express the orchestra section in terms of xx. The problem states "The orchestra section has 30 more seats than twice the number of seats in the balcony section." Breaking this down: "twice the balcony" means 2x2x, and "30 more than" that amount means 2x+302x + 30. Since the total seats equal 180, our equation becomes: x+(2x+30)=180x + (2x + 30) = 180. Looking at the wrong answers: Choice A incorrectly represents the orchestra as x−30x - 30, which would mean the orchestra has 30 fewer seats than the balcony—the opposite of what's stated. Choice B makes the orchestra 2x−302x - 30 (30 less than twice the balcony), again reversing the relationship. Choice C sets up the total as 2x+(x+30)=1802x + (x + 30) = 180, which incorrectly assumes the balcony has the extra 30 seats rather than the orchestra. The correct answer is D: x+(2x+30)=180x + (2x + 30) = 180, which properly represents balcony seats (xx) plus orchestra seats (30 more than twice the balcony). Study tip: In word problems, always identify what your variable represents first, then carefully parse phrases like "more than" or "less than" to ensure you're adding or subtracting from the correct quantity. Write out the relationships in plain English before converting to algebra.

Question 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?

  1. 12≤w≤1612 \leq w \leq 16
  2. 10≤w≤1410 \leq w \leq 14 (correct answer)
  3. 8≤w≤128 \leq w \leq 12
  4. 14≤w≤1814 \leq w \leq 18
Explanation: 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 ww feet. Since the length is "8 feet more than twice the width," we write l=2w+8l = 2w + 8. The perimeter of a rectangle is P=2l+2wP = 2l + 2w, so substituting our expression for length gives us P=2(2w+8)+2w=4w+16+2w=6w+16P = 2(2w + 8) + 2w = 4w + 16 + 2w = 6w + 16. The constraint "at least 76 feet but no more than 100 feet" translates to 76≤6w+16≤10076 \leq 6w + 16 \leq 100. Solving the left inequality: 76≤6w+1676 \leq 6w + 16 gives us 60≤6w60 \leq 6w, so w≥10w \geq 10. Solving the right inequality: 6w+16≤1006w + 16 \leq 100 gives us 6w≤846w \leq 84, so w≤14w \leq 14. Therefore, 10≤w≤1410 \leq w \leq 14, which is choice B. Choice A (12≤w≤1612 \leq w \leq 16) sets the minimum width too high and maximum too high. Choice C (8≤w≤128 \leq w \leq 12) sets both bounds too low, allowing widths that would make the perimeter less than 76 feet. Choice D (14≤w≤1814 \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.

Question 16

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?

  1. Between years 3 and 6.
  2. Between years 5 and 8.
  3. Between years 6 and 9. (correct answer)
  4. Between years 9 and 12.
Explanation: 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, $V(t),after, after tyearsisyears isV(t) = 65,000 - 5,000t.Wewanttofindthetimeintervalforwhich. We want to find the time interval for which 20,000 \le V(t) \le 35,000.Substitutetheexpressionfor. Substitute the expression for V(t):: 20,000 \le 65,000 - 5,000t \le 35,000.Subtract65,000fromallparts:. Subtract 65,000 from all parts: -45,000 \le -5,000t \le -30,000.Divideallpartsby−5,000andreversethedirectionoftheinequalities:. Divide all parts by -5,000 and reverse the direction of the inequalities: \frac{-30,000}{-5,000} \le t \le \frac{-45,000}{-5,000}.Thissimplifiesto. This simplifies to 6 \le t \le 9$. The book value is between $20,000 and $35,000 from year 6 to year 9, inclusive.

Question 17

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?

  1. Between 4 and 8 liters.
  2. Between 8 and 10 liters.
  3. Between 10 and 12 liters.
  4. Between 12 and 16 liters. (correct answer)
Explanation: Let xx be the volume in liters of the 15% solution. Then the volume of the 40% solution is 20−x20-x. The total amount of salt in the mixture is 0.15x+0.40(20−x)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≤0.15x+0.40(20−x)20≤0.250.20 \le \frac{0.15x + 0.40(20-x)}{20} \le 0.25. Multiply all parts by 20: 4≤0.15x+8−0.40x≤54 \le 0.15x + 8 - 0.40x \le 5. Simplify the middle expression: 4≤8−0.25x≤54 \le 8 - 0.25x \le 5. Subtract 8 from all parts: −4≤−0.25x≤−3-4 \le -0.25x \le -3. Divide by -0.25 and reverse the inequalities: −3−0.25≤x≤−4−0.25\frac{-3}{-0.25} \le x \le \frac{-4}{-0.25}, which simplifies to 12≤x≤1612 \le x \le 16. So, she must use between 12 and 16 liters of the 15% solution.

Question 18

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, tt, that can be manufactured?

  1. 170 (correct answer)
  2. 180
  3. 250
  4. 360
Explanation: Let tt be the number of tablets and ll be the number of laptops. The constraints are given by two inequalities. Assembly hours: 2t+3l≤6002t + 3l \le 600. Packaging capacity: t+l≤250t + l \le 250. We are given that l=80l=80. We must substitute this value into both inequalities to find the constraints on tt. From the assembly constraint: 2t+3(80)≤600  ⟹  2t+240≤600  ⟹  2t≤360  ⟹  t≤1802t + 3(80) \le 600 \implies 2t + 240 \le 600 \implies 2t \le 360 \implies t \le 180. From the packaging constraint: t+80≤250  ⟹  t≤170t + 80 \le 250 \implies t \le 170. For both constraints to be satisfied, tt must be less than or equal to both 180 and 170. The more restrictive condition is t≤170t \le 170. Therefore, the maximum number of tablets that can be manufactured is 170.

Question 19

A small business is choosing a phone plan. Plan A has a $500 setup fee and costs $45 per month. Plan B has a $200 setup fee and costs $60 per month. After how many full months of service does the total accumulated cost of Plan A become strictly less than the total accumulated cost of Plan B?

  1. After 19 months.
  2. After 20 months.
  3. After 21 months. (correct answer)
  4. After 7 months.
Explanation: Let mm be the number of months. The total cost for Plan A is CA=500+45mC_A = 500 + 45m. The total cost for Plan B is CB=200+60mC_B = 200 + 60m. We want to find when CA<CBC_A < C_B. Set up the inequality: 500+45m<200+60m500 + 45m < 200 + 60m. Subtract $45m$ from both sides: 500<200+15m500 < 200 + 15m. Subtract 200 from both sides: 300<15m300 < 15m. Divide by 15: 20<m20 < m. The number of months mm must be strictly greater than 20. Since the question asks for the number of full months, the first integer value of mm that satisfies this condition is 21.

Question 20

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?

  1. 100 mL (correct answer)
  2. 150 mL
  3. 250 mL
  4. 350 mL
Explanation: Let xx, yy, and ww be the volumes in mL of the 20% solution, 40% solution, and pure water, respectively. The total volume is x+y+w=500x+y+w=500. The total amount of acid is 0.20x+0.40y+0w=0.26(500)0.20x + 0.40y + 0w = 0.26(500). We are given x=150x=150. The acid equation becomes 0.20(150)+0.40y=1300.20(150) + 0.40y = 130, which simplifies to 30+0.40y=13030 + 0.40y = 130. Solving for yy, we get 0.40y=1000.40y = 100, so y=250y = 250 mL. Now, using the total volume equation, 150+250+w=500150 + 250 + w = 500. This gives 400+w=500400 + w = 500, so w=100w = 100 mL.