IB Mathematics: Applications and Interpretation Quiz: Linear Equations And Inequalities
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Linear Equations And InequalitiesQuestion 1 of 20

A student's final grade in a course is determined by a weighted average of three assessments: Homework (20%), Midterm Exam (35%), and Final Exam (45%). The student scored 90 on Homework and 75 on the Midterm Exam. What is the minimum integer score the student must get on the Final Exam to achieve an overall course grade of at least 80?

78
79
80
81
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Linear Equations And Inequalities

Practice Linear Equations And Inequalities in IB Mathematics: Applications and Interpretation 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 IB Mathematics: Applications and Interpretation.

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

A student's final grade in a course is determined by a weighted average of three assessments: Homework (20%), Midterm Exam (35%), and Final Exam (45%). The student scored 90 on Homework and 75 on the Midterm Exam. What is the minimum integer score the student must get on the Final Exam to achieve an overall course grade of at least 80?

  1. 78
  2. 79
  3. 80 (correct answer)
  4. 81
Explanation: Let FF be the score on the Final Exam. The weighted average for the course grade is given by 0.20×90+0.35×75+0.45×F0.20 \times 90 + 0.35 \times 75 + 0.45 \times F. The student wants this grade to be at least 80. We set up the inequality: 0.20(90)+0.35(75)+0.45F≥800.20(90) + 0.35(75) + 0.45F \ge 80. Calculating the known parts: 18+26.25+0.45F≥8018 + 26.25 + 0.45F \ge 80, which simplifies to 44.25+0.45F≥8044.25 + 0.45F \ge 80. Subtracting 44.25 from both sides gives 0.45F≥35.750.45F \ge 35.75. Dividing by 0.45 gives F≥35.750.45≈79.44F \ge \frac{35.75}{0.45} \approx 79.44. Since the score must be an integer, the minimum score required is 80.

Question 2

Two printing companies offer quotes for a job. PrintCorp charges a $100 setup fee plus $1.50 per item. QualityPrint charges no setup fee but costs $2.00 per item. For which number of items is the total cost from both companies exactly the same?

  1. 50
  2. 100
  3. 150
  4. 200 (correct answer)
Explanation: Let nn be the number of items printed. The cost for PrintCorp is CP=100+1.50nC_P = 100 + 1.50n. The cost for QualityPrint is CQ=2.00nC_Q = 2.00n. To find when the costs are the same, we set CP=CQC_P = C_Q: 100+1.50n=2.00n100 + 1.50n = 2.00n. To solve for nn, subtract 1.50n1.50n from both sides: 100=2.00n−1.50n100 = 2.00n - 1.50n, which simplifies to 100=0.50n100 = 0.50n. Dividing both sides by 0.50 gives n=1000.50=200n = \frac{100}{0.50} = 200. The cost is the same for 200 items.

Question 3

A taxi service charges an initial fee of $3.50 and $1.80 per kilometer. A passenger has $40 and must have at least $5 remaining after the trip. What is the maximum distance, in kilometers, the passenger can travel?

  1. 17.50 km (correct answer)
  2. 19.44 km
  3. 20.28 km
  4. 22.22 km
Explanation: Let dd be the distance traveled in kilometers. The total cost of the taxi trip is C=3.50+1.80dC = 3.50 + 1.80d. The passenger starts with $40 and wants to have at least $5 left, which means the maximum they can spend on the taxi is 40−5=3540 - 5 = 35 dollars. So, the cost must be less than or equal to $35: 3.50+1.80d≤353.50 + 1.80d \le 35. Subtracting 3.50 from both sides gives 1.80d≤31.501.80d \le 31.50. Dividing by 1.80 gives d≤31.501.80=17.5d \le \frac{31.50}{1.80} = 17.5. The maximum distance is 17.50 km.

Question 4

A company is selling a new product. The total revenue, RR, is given by R(x)=45xR(x) = 45x and the total cost, CC, is given by C(x)=2000+20xC(x) = 2000 + 20x, where xx is the number of units sold. The company wants its profit to be at least 20% of its total revenue. What is the minimum number of units that must be sold?

  1. 80
  2. 125 (correct answer)
  3. 126
  4. 500
Explanation: Let xx be the number of units sold. Profit, PP, is Revenue minus Cost: P(x)=R(x)−C(x)=45x−(2000+20x)=25x−2000P(x) = R(x) - C(x) = 45x - (2000 + 20x) = 25x - 2000. The condition is that profit must be at least 20% of revenue: P(x)≥0.20×R(x)P(x) \ge 0.20 \times R(x). Substituting the expressions for P(x)P(x) and R(x)R(x) gives 25x−2000≥0.20(45x)25x - 2000 \ge 0.20(45x). This simplifies to 25x−2000≥9x25x - 2000 \ge 9x. Rearranging the inequality: 25x−9x≥200025x - 9x \ge 2000, which is 16x≥200016x \ge 2000. Solving for xx gives x≥200016=125x \ge \frac{2000}{16} = 125. The minimum number of units that must be sold is 125.

Question 5

The temperature CC in degrees Celsius is related to the temperature FF in degrees Fahrenheit by the formula C=59(F−32)C = \frac{5}{9}(F - 32). A chemical process requires the temperature to be kept in the range 20∘C≤C≤30∘C20^{\circ}C \le C \le 30^{\circ}C. What is the corresponding temperature range in degrees Fahrenheit?

  1. 4∘F≤F≤22∘F4^{\circ}F \le F \le 22^{\circ}F
  2. 68∘F≤F≤86∘F68^{\circ}F \le F \le 86^{\circ}F (correct answer)
  3. 72∘F≤F≤104∘F72^{\circ}F \le F \le 104^{\circ}F
  4. 92∘F≤F≤116∘F92^{\circ}F \le F \le 116^{\circ}F
Explanation: We need to solve the compound inequality 20≤59(F−32)≤3020 \le \frac{5}{9}(F - 32) \le 30. First, multiply all parts of the inequality by 95\frac{9}{5}: 20×95≤F−32≤30×9520 \times \frac{9}{5} \le F - 32 \le 30 \times \frac{9}{5}. This simplifies to 36≤F−32≤5436 \le F - 32 \le 54. Now, add 32 to all parts of the inequality: 36+32≤F≤54+3236 + 32 \le F \le 54 + 32. This gives the final range 68≤F≤8668 \le F \le 86. So the corresponding temperature range is from 68∘F68^{\circ}F to 86∘F86^{\circ}F.

Question 6

A conference hall has two types of seating: 350 standard seats and 150 premium seats. A standard ticket costs $40 and a premium ticket costs $75. A total of 420 tickets were sold. If the total revenue from ticket sales was $21,000, how many premium tickets were sold?

  1. 80
  2. 120 (correct answer)
  3. 300
  4. 340
Explanation: Let ss be the number of standard tickets sold and pp be the number of premium tickets sold. We can set up a system of two linear equations. From the total number of tickets sold: s+p=420s + p = 420. From the total revenue: 40s+75p=2100040s + 75p = 21000. From the first equation, we can express ss as s=420−ps = 420 - p. Substitute this into the second equation: 40(420−p)+75p=2100040(420 - p) + 75p = 21000. Expand the equation: 16800−40p+75p=2100016800 - 40p + 75p = 21000. Simplify: 16800+35p=2100016800 + 35p = 21000. Subtract 16800 from both sides: 35p=420035p = 4200. Solve for pp: p=420035=120p = \frac{4200}{35} = 120. Therefore, 120 premium tickets were sold.

Question 7

Anna and Ben are saving money. Anna starts with $200 in her account and saves $15 each week. Ben starts with $80 and saves $25 each week. After how many full weeks will Ben's total savings first be greater than Anna's?

  1. 11
  2. 12
  3. 13 (correct answer)
  4. 14
Explanation: Let ww be the number of weeks. Anna's savings can be modeled as A(w)=200+15wA(w) = 200 + 15w. Ben's savings can be modeled as B(w)=80+25wB(w) = 80 + 25w. We want to find the smallest integer ww for which B(w)>A(w)B(w) > A(w). This gives the inequality 80+25w>200+15w80 + 25w > 200 + 15w. To solve for ww, we rearrange the terms: 25w−15w>200−8025w - 15w > 200 - 80, which simplifies to 10w>12010w > 120. Dividing by 10 gives w>12w > 12. Since ww must be an integer representing full weeks, the smallest integer value for ww that satisfies this condition is 13.

Question 8

An algorithm requires 50 milliseconds for setup. After setup, it processes data at a rate of 3 milliseconds per gigabyte (GB). The total time to run the algorithm on a file of size SS GB must be kept between 0.5 seconds and 0.8 seconds. Which compound inequality represents the possible file sizes SS in GB?

  1. 150≤S≤250150 \le S \le 250 (correct answer)
  2. 450≤S≤750450 \le S \le 750
  3. 0.15≤S≤0.250.15 \le S \le 0.25
  4. 0.45≤S≤0.750.45 \le S \le 0.75
Explanation: The total processing time, TT, in milliseconds (ms) is given by the linear model T(S)=50+3ST(S) = 50 + 3S, where SS is the file size in GB. The required time range is between 0.5 and 0.8 seconds. We must convert this to milliseconds: 0.5 s=500 ms0.5 \, s = 500 \, ms and 0.8 s=800 ms0.8 \, s = 800 \, ms. So, the compound inequality for the time is 500≤T(S)≤800500 \le T(S) \le 800. Substituting the expression for T(S)T(S) gives 500≤50+3S≤800500 \le 50 + 3S \le 800. To solve for SS, first subtract 50 from all parts: 500−50≤3S≤800−50500 - 50 \le 3S \le 800 - 50, which simplifies to 450≤3S≤750450 \le 3S \le 750. Finally, divide all parts by 3: 4503≤S≤7503\frac{450}{3} \le S \le \frac{750}{3}, which gives 150≤S≤250150 \le S \le 250.

Question 9

A mobile phone plan costs $40 per month, which includes 5 GB of data. Any additional data used is charged at a rate of $8 per GB. A customer wants their monthly bill to be at most $75. What is the maximum total amount of data, in GB, they can use?

  1. 4.38 GB
  2. 5.00 GB
  3. 14.38 GB
  4. 9.38 GB (correct answer)
Explanation: This is a linear inequality problem involving cost modeling. When you see questions about plans with base costs plus variable charges, you need to set up an inequality to find the maximum or minimum values within given constraints. The monthly cost structure is: base cost (40)plusadditionaldatacharges(40) plus additional data charges (8 per GB over 5 GB). If a customer uses xx GB total, and x>5x > 5, their bill equals 40+8(x−5)40 + 8(x - 5). Since they want their bill to be at most $75, you set up the inequality: $40+8(x−5)≤7540 + 8(x - 5) \leq 75 $. Solving: 40 + 8x - 40 \leq 75 , which simplifies to 8x \leq 75 , so x \leq 9.375 GB. Since data usage is typically measured to reasonable precision, this rounds to 9.38 GB, making D correct. Option A (4.38 GB) falls below the 5 GB included in the base plan, meaning the customer would only pay $40—well under their $75 budget. This represents the error of subtracting from the base data allowance instead of adding to it. Option B (5.00 GB) is exactly the included data amount, again resulting in only the $40 base cost. This reflects confusion about what the question is asking—the maximum possible usage, not the included amount. Option C (14.38 GB) results from adding 9.38 to 5, double-counting the calculation. This represents the common error of adding the excess data to the total instead of recognizing that 9.38 GB is the total. Remember: in cost problems with base amounts plus overages, your final answer represents the total quantity, not just the excess above the base allowance.

Question 10

A small business sells a product for $25 per unit. The variable cost to produce one unit is $10, and the total fixed costs are $3000 per month. The business wants to make a profit of at least $1500 per month. What is the minimum number of units that must be sold?

  1. 100
  2. 200
  3. 300 (correct answer)
  4. 301
Explanation: Let xx be the number of units sold. The revenue is R=25xR = 25x. The total cost is C=3000+10xC = 3000 + 10x. The profit is P=R−C=25x−(3000+10x)=15x−3000P = R - C = 25x - (3000 + 10x) = 15x - 3000. The business wants the profit to be at least $1500, so we set up the inequality P≥1500P \ge 1500. This gives 15x−3000≥150015x - 3000 \ge 1500. Adding 3000 to both sides gives 15x≥450015x \ge 4500. Dividing by 15 gives x≥300x \ge 300. The minimum number of units that must be sold is 300.

Question 11

A company is selling a new product. The total revenue, RR, is given by R(x)=45xR(x) = 45x and the total cost, CC, is given by C(x)=2000+20xC(x) = 2000 + 20x, where xx is the number of units sold. The company wants its profit to be at least 20% of its total revenue. What is the minimum number of units that must be sold?

  1. 80
  2. 125 (correct answer)
  3. 126
  4. 500
Explanation: Let xx be the number of units sold. Profit, PP, is Revenue minus Cost: P(x)=R(x)−C(x)=45x−(2000+20x)=25x−2000P(x) = R(x) - C(x) = 45x - (2000 + 20x) = 25x - 2000. The condition is that profit must be at least 20% of revenue: P(x)≥0.20×R(x)P(x) \ge 0.20 \times R(x). Substituting the expressions for P(x)P(x) and R(x)R(x) gives 25x−2000≥0.20(45x)25x - 2000 \ge 0.20(45x). This simplifies to 25x−2000≥9x25x - 2000 \ge 9x. Rearranging the inequality: 25x−9x≥200025x - 9x \ge 2000, which is 16x≥200016x \ge 2000. Solving for xx gives x≥200016=125x \ge \frac{2000}{16} = 125. The minimum number of units that must be sold is 125.

Question 12

The temperature CC in degrees Celsius is related to the temperature FF in degrees Fahrenheit by the formula C=59(F−32)C = \frac{5}{9}(F - 32). A chemical process requires the temperature to be kept in the range 20∘C≤C≤30∘C20^{\circ}C \le C \le 30^{\circ}C. What is the corresponding temperature range in degrees Fahrenheit?

  1. 4∘F≤F≤22∘F4^{\circ}F \le F \le 22^{\circ}F
  2. 68∘F≤F≤86∘F68^{\circ}F \le F \le 86^{\circ}F (correct answer)
  3. 72∘F≤F≤104∘F72^{\circ}F \le F \le 104^{\circ}F
  4. 92∘F≤F≤116∘F92^{\circ}F \le F \le 116^{\circ}F
Explanation: We need to solve the compound inequality 20≤59(F−32)≤3020 \le \frac{5}{9}(F - 32) \le 30. First, multiply all parts of the inequality by 95\frac{9}{5}: 20×95≤F−32≤30×9520 \times \frac{9}{5} \le F - 32 \le 30 \times \frac{9}{5}. This simplifies to 36≤F−32≤5436 \le F - 32 \le 54. Now, add 32 to all parts of the inequality: 36+32≤F≤54+3236 + 32 \le F \le 54 + 32. This gives the final range 68≤F≤8668 \le F \le 86. So the corresponding temperature range is from 68∘F68^{\circ}F to 86∘F86^{\circ}F.

Question 13

A delivery driver earns a base pay of $50 per day, plus $3 for each delivery made. The driver's daily expenses for fuel are $20. What is the minimum number of deliveries the driver must make in a 5-day week to achieve a net income of at least $600 for the week?

  1. 117
  2. 120
  3. 150 (correct answer)
  4. 200
Explanation: Let DD be the total number of deliveries made in the 5-day week. The total base pay for the week is 5 \times \50 = $250.Thetotalearningsfromdeliveriesis. The total earnings from deliveries is 3D.Thetotalgrossincomefortheweekis. The total gross income for the week is 250 + 3D.Thetotalexpensesfortheweekare. The total expenses for the week are 5 \times $20 = $100.Thenetincomeisthegrossincomeminusthetotalexpenses:. The net income is the gross income minus the total expenses: Net,Income = (250 + 3D) - 100 = 150 + 3D.Thedriverwantsthenetincometobeatleast$600.So,wesetuptheinequality. The driver wants the net income to be at least $600. So, we set up the inequality 150 + 3D \ge 600.Subtracting150frombothsidesgives. Subtracting 150 from both sides gives 3D \ge 450.Dividingby3gives. Dividing by 3 gives D \ge 150$. The minimum number of deliveries required is 150.

Question 14

A rectangular garden has a length that is 5 meters greater than its width. The perimeter of the garden must be no more than 100 meters. What is the maximum possible width of the garden in meters?

  1. 22.5 m (correct answer)
  2. 25.0 m
  3. 27.5 m
  4. 45.0 m
Explanation: Let ww be the width of the garden in meters. The length, ll, is given by l=w+5l = w + 5. The perimeter of a rectangle is P=2l+2wP = 2l + 2w. Substituting the expression for ll, we get P=2(w+5)+2w=2w+10+2w=4w+10P = 2(w + 5) + 2w = 2w + 10 + 2w = 4w + 10. The problem states that the perimeter must be no more than 100 meters, so P≤100P \le 100. This gives the inequality 4w+10≤1004w + 10 \le 100. Subtracting 10 from both sides gives 4w≤904w \le 90. Dividing by 4 gives w≤22.5w \le 22.5. The maximum possible width is 22.5 meters.

Question 15

A mobile phone plan costs $40 per month, which includes 5 GB of data. Any additional data used is charged at a rate of $8 per GB. A customer wants their monthly bill to be at most $75. What is the maximum total amount of data, in GB, they can use?

  1. 4.38 GB
  2. 5.00 GB
  3. 14.38 GB
  4. 9.38 GB (correct answer)
Explanation: This is a linear inequality problem involving cost modeling. When you see questions about plans with base costs plus variable charges, you need to set up an inequality to find the maximum or minimum values within given constraints. The monthly cost structure is: base cost (40)plusadditionaldatacharges(40) plus additional data charges (8 per GB over 5 GB). If a customer uses xx GB total, and x>5x > 5, their bill equals 40+8(x−5)40 + 8(x - 5). Since they want their bill to be at most $75, you set up the inequality: $40+8(x−5)≤7540 + 8(x - 5) \leq 75 $. Solving: 40 + 8x - 40 \leq 75 , which simplifies to 8x \leq 75 , so x \leq 9.375 GB. Since data usage is typically measured to reasonable precision, this rounds to 9.38 GB, making D correct. Option A (4.38 GB) falls below the 5 GB included in the base plan, meaning the customer would only pay $40—well under their $75 budget. This represents the error of subtracting from the base data allowance instead of adding to it. Option B (5.00 GB) is exactly the included data amount, again resulting in only the $40 base cost. This reflects confusion about what the question is asking—the maximum possible usage, not the included amount. Option C (14.38 GB) results from adding 9.38 to 5, double-counting the calculation. This represents the common error of adding the excess data to the total instead of recognizing that 9.38 GB is the total. Remember: in cost problems with base amounts plus overages, your final answer represents the total quantity, not just the excess above the base allowance.

Question 16

A small business sells a product for $25 per unit. The variable cost to produce one unit is $10, and the total fixed costs are $3000 per month. The business wants to make a profit of at least $1500 per month. What is the minimum number of units that must be sold?

  1. 100
  2. 200
  3. 300 (correct answer)
  4. 301
Explanation: Let xx be the number of units sold. The revenue is R=25xR = 25x. The total cost is C=3000+10xC = 3000 + 10x. The profit is P=R−C=25x−(3000+10x)=15x−3000P = R - C = 25x - (3000 + 10x) = 15x - 3000. The business wants the profit to be at least $1500, so we set up the inequality P≥1500P \ge 1500. This gives 15x−3000≥150015x - 3000 \ge 1500. Adding 3000 to both sides gives 15x≥450015x \ge 4500. Dividing by 15 gives x≥300x \ge 300. The minimum number of units that must be sold is 300.

Question 17

A town's population is 25,000 and is growing by 400 people per year. A neighboring town has a population of 32,000 and is decreasing by 150 people per year. In how many full years will the population of the first town first exceed the population of the second town?

  1. 12
  2. 13 (correct answer)
  3. 14
  4. 15
Explanation: Let tt be the number of years. The population of the first town, P1P_1, can be modeled as P1(t)=25000+400tP_1(t) = 25000 + 400t. The population of the second town, P2P_2, can be modeled as P2(t)=32000−150tP_2(t) = 32000 - 150t. We want to find when P1(t)>P2(t)P_1(t) > P_2(t). This gives the inequality 25000+400t>32000−150t25000 + 400t > 32000 - 150t. Rearranging the terms to solve for tt: 400t+150t>32000−25000400t + 150t > 32000 - 25000, which simplifies to 550t>7000550t > 7000. Dividing by 550 gives t>7000550≈12.727t > \frac{7000}{550} \approx 12.727. Since the question asks for the number of full years, the population of the first town will exceed the second during the 13th year. So, after 13 full years, this condition will be met.

Question 18

A subscription service offers two plans. Plan A has a flat fee of $19 per month. Plan B has a monthly fee of $7 plus $1.25 per download. A customer wants to spend no more than $25 per month. What is the maximum number of downloads they can make under Plan B, such that Plan B is still a cheaper option than Plan A?

  1. 9 (correct answer)
  2. 10
  3. 14
  4. 15
Explanation: Let nn be the number of downloads. This problem has two constraints. First, the cost of Plan B must be cheaper than Plan A: 7+1.25n<197 + 1.25n < 19. Solving this gives 1.25n<121.25n < 12, so n<9.6n < 9.6. Second, the cost of Plan B must be no more than $25: 7+1.25n≤257 + 1.25n \le 25. Solving this gives 1.25n≤181.25n \le 18, so n≤14.4n \le 14.4. Both conditions must be met. We need n<9.6n < 9.6 AND n≤14.4n \le 14.4. The more restrictive condition is n<9.6n < 9.6. Since the number of downloads must be an integer, the maximum number of downloads is 9.

Question 19

A delivery driver earns a base pay of $50 per day, plus $3 for each delivery made. The driver's daily expenses for fuel are $20. What is the minimum number of deliveries the driver must make in a 5-day week to achieve a net income of at least $600 for the week?

  1. 117
  2. 120
  3. 150 (correct answer)
  4. 200
Explanation: Let DD be the total number of deliveries made in the 5-day week. The total base pay for the week is 5 \times \50 = $250.Thetotalearningsfromdeliveriesis. The total earnings from deliveries is 3D.Thetotalgrossincomefortheweekis. The total gross income for the week is 250 + 3D.Thetotalexpensesfortheweekare. The total expenses for the week are 5 \times $20 = $100.Thenetincomeisthegrossincomeminusthetotalexpenses:. The net income is the gross income minus the total expenses: Net,Income = (250 + 3D) - 100 = 150 + 3D.Thedriverwantsthenetincometobeatleast$600.So,wesetuptheinequality. The driver wants the net income to be at least $600. So, we set up the inequality 150 + 3D \ge 600.Subtracting150frombothsidesgives. Subtracting 150 from both sides gives 3D \ge 450.Dividingby3gives. Dividing by 3 gives D \ge 150$. The minimum number of deliveries required is 150.

Question 20

A chemist needs to create a 200 ml solution that is 28% acid. She has two stock solutions available: one is 20% acid and the other is 40% acid. How many milliliters of the 40% acid solution must she use?

  1. 40 ml
  2. 80 ml (correct answer)
  3. 120 ml
  4. 160 ml
Explanation: Let xx be the volume (in ml) of the 20% solution and yy be the volume of the 40% solution. The total volume is 200 ml, so x+y=200x + y = 200. The total amount of acid is 28% of 200 ml, which is 0.28×200=560.28 \times 200 = 56 ml. The amount of acid from each stock solution is 0.20x0.20x and 0.40y0.40y. So, the second equation is 0.20x+0.40y=560.20x + 0.40y = 56. From the first equation, x=200−yx = 200 - y. Substitute this into the second equation: 0.20(200−y)+0.40y=560.20(200 - y) + 0.40y = 56. This gives 40−0.20y+0.40y=5640 - 0.20y + 0.40y = 56. Simplifying, we get 40+0.20y=5640 + 0.20y = 56. Then 0.20y=160.20y = 16, and y=160.20=80y = \frac{16}{0.20} = 80. So, 80 ml of the 40% acid solution must be used.