Linear Equations & Inequalities - IB Mathematics Applications and Interpretation

1. A farmer has 1200 square meters of land to plant two crops, corn and beans. Each corn plant requires 1.5 square meters of land, and each bean plant requires 0.8 square meters. The farmer plants 400 corn plants. What is the maximum number of bean plants they can grow?
   Answer: 750 Let $c$ be the number of corn plants and $b$ be the number of bean plants. The total area used is given by $1.5c + 0.8b$. This area must be less than or equal to the total available land, 1200 square meters. So, $1.5c + 0.8b \le 1200$. The farmer plants 400 corn plants, so $c = 400$. Substitute this value into the inequality: $1.5(400) + 0.8b \le 1200$. This becomes $600 + 0.8b \le 1200$. Subtract 600 from both sides: $0.8b \le 600$. Divide by 0.8 to solve for $b$: $b \le \frac{600}{0.8} = 750$. The maximum number of bean plants the farmer can grow is 750.
2. 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?
   Answer: 9.38 GB 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) plus additional data charges ($8 per GB over 5 GB). If a customer uses $x$ GB total, and $x > 5$, their bill equals $40 + 8(x - 5)$. Since they want their bill to be at most $75, you set up the inequality: $40 + 8(x - 5) \leq 75$. 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.
3. 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?
   Answer: 9 Let $n$ be the number of downloads. This problem has two constraints. First, the cost of Plan B must be cheaper than Plan A: $7 + 1.25n < 19$. Solving this gives $1.25n < 12$, so $n < 9.6$. Second, the cost of Plan B must be no more than $25: $7 + 1.25n \le 25$. Solving this gives $1.25n \le 18$, so $n \le 14.4$. Both conditions must be met. We need $n < 9.6$ AND $n \le 14.4$. The more restrictive condition is $n < 9.6$. Since the number of downloads must be an integer, the maximum number of downloads is 9.
4. 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?
   Answer: 150 Let $D$ be the total number of deliveries made in the 5-day week. The total base pay for the week is $5 \times \$50 = \$250$. The total earnings from deliveries is $3D$. The total gross income for the week is $250 + 3D$. The total expenses for the week are $5 \times \$20 = \$100$. The net income is the gross income minus the total expenses: $Net\,Income = (250 + 3D) - 100 = 150 + 3D$. The driver wants the net income to be at least $600. So, we set up the inequality $150 + 3D \ge 600$. Subtracting 150 from both sides gives $3D \ge 450$. Dividing by 3 gives $D \ge 150$. The minimum number of deliveries required is 150.
5. 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?
   Answer: 80 Let $F$ be the score on the Final Exam. The weighted average for the course grade is given by $0.20 \times 90 + 0.35 \times 75 + 0.45 \times F$. The student wants this grade to be at least 80. We set up the inequality: $0.20(90) + 0.35(75) + 0.45F \ge 80$. Calculating the known parts: $18 + 26.25 + 0.45F \ge 80$, which simplifies to $44.25 + 0.45F \ge 80$. Subtracting 44.25 from both sides gives $0.45F \ge 35.75$. Dividing by 0.45 gives $F \ge \frac{35.75}{0.45} \approx 79.44$. Since the score must be an integer, the minimum score required is 80.
6. 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?
   Answer: 13 Let $w$ be the number of weeks. Anna's savings can be modeled as $A(w) = 200 + 15w$. Ben's savings can be modeled as $B(w) = 80 + 25w$. We want to find the smallest integer $w$ for which $B(w) > A(w)$. This gives the inequality $80 + 25w > 200 + 15w$. To solve for $w$, we rearrange the terms: $25w - 15w > 200 - 80$, which simplifies to $10w > 120$. Dividing by 10 gives $w > 12$. Since $w$ must be an integer representing full weeks, the smallest integer value for $w$ that satisfies this condition is 13.
7. A swimming pool with a capacity of 50,000 liters is initially 15% full. Water is added at a constant rate, $R$, in liters per minute. The pool must be filled to at least 90% capacity in no more than 4 hours. Which inequality correctly models the minimum required rate $R$?
   Answer: $R \ge 156.25$ First, calculate the initial volume of water: $0.15 \times 50000 = 7500$ liters. The target volume is at least 90% of capacity: $0.90 \times 50000 = 45000$ liters. The volume of water that needs to be added is $45000 - 7500 = 37500$ liters. The time available is 4 hours, which is $4 \times 60 = 240$ minutes. The rate $R$ is in liters per minute. The total volume added is $R \times time$. We need to add at least 37500 liters, so $R \times 240 \ge 37500$. Solving for $R$, we get $R \ge \frac{37500}{240} = 156.25$. The minimum required rate is 156.25 liters per minute.
8. 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?
   Answer: 120 Let $s$ be the number of standard tickets sold and $p$ be the number of premium tickets sold. We can set up a system of two linear equations. From the total number of tickets sold: $s + p = 420$. From the total revenue: $40s + 75p = 21000$. From the first equation, we can express $s$ as $s = 420 - p$. Substitute this into the second equation: $40(420 - p) + 75p = 21000$. Expand the equation: $16800 - 40p + 75p = 21000$. Simplify: $16800 + 35p = 21000$. Subtract 16800 from both sides: $35p = 4200$. Solve for $p$: $p = \frac{4200}{35} = 120$. Therefore, 120 premium tickets were sold.
9. 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?
   Answer: 22.5 m Let $w$ be the width of the garden in meters. The length, $l$, is given by $l = w + 5$. The perimeter of a rectangle is $P = 2l + 2w$. Substituting the expression for $l$, we get $P = 2(w + 5) + 2w = 2w + 10 + 2w = 4w + 10$. The problem states that the perimeter must be no more than 100 meters, so $P \le 100$. This gives the inequality $4w + 10 \le 100$. Subtracting 10 from both sides gives $4w \le 90$. Dividing by 4 gives $w \le 22.5$. The maximum possible width is 22.5 meters.
10. 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?
    Answer: 17.50 km Let $d$ be the distance traveled in kilometers. The total cost of the taxi trip is $C = 3.50 + 1.80d$. The passenger starts with $40 and wants to have at least $5 left, which means the maximum they can spend on the taxi is $40 - 5 = 35$ dollars. So, the cost must be less than or equal to $35: $3.50 + 1.80d \le 35$. Subtracting 3.50 from both sides gives $1.80d \le 31.50$. Dividing by 1.80 gives $d \le \frac{31.50}{1.80} = 17.5$. The maximum distance is 17.50 km.
11. 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?
    Answer: 80 ml Let $x$ be the volume (in ml) of the 20% solution and $y$ be the volume of the 40% solution. The total volume is 200 ml, so $x + y = 200$. The total amount of acid is 28% of 200 ml, which is $0.28 \times 200 = 56$ ml. The amount of acid from each stock solution is $0.20x$ and $0.40y$. So, the second equation is $0.20x + 0.40y = 56$. From the first equation, $x = 200 - y$. Substitute this into the second equation: $0.20(200 - y) + 0.40y = 56$. This gives $40 - 0.20y + 0.40y = 56$. Simplifying, we get $40 + 0.20y = 56$. Then $0.20y = 16$, and $y = \frac{16}{0.20} = 80$. So, 80 ml of the 40% acid solution must be used.
12. 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?
    Answer: 300 Let $x$ be the number of units sold. The revenue is $R = 25x$. The total cost is $C = 3000 + 10x$. The profit is $P = R - C = 25x - (3000 + 10x) = 15x - 3000$. The business wants the profit to be at least $1500, so we set up the inequality $P \ge 1500$. This gives $15x - 3000 \ge 1500$. Adding 3000 to both sides gives $15x \ge 4500$. Dividing by 15 gives $x \ge 300$. The minimum number of units that must be sold is 300.
13. A company is selling a new product. The total revenue, $R$, is given by $R(x) = 45x$ and the total cost, $C$, is given by $C(x) = 2000 + 20x$, where $x$ 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?
    Answer: 125 Let $x$ be the number of units sold. Profit, $P$, is Revenue minus Cost: $P(x) = R(x) - C(x) = 45x - (2000 + 20x) = 25x - 2000$. The condition is that profit must be at least 20% of revenue: $P(x) \ge 0.20 \times R(x)$. Substituting the expressions for $P(x)$ and $R(x)$ gives $25x - 2000 \ge 0.20(45x)$. This simplifies to $25x - 2000 \ge 9x$. Rearranging the inequality: $25x - 9x \ge 2000$, which is $16x \ge 2000$. Solving for $x$ gives $x \ge \frac{2000}{16} = 125$. The minimum number of units that must be sold is 125.
14. 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?
    Answer: 200 Let $n$ be the number of items printed. The cost for PrintCorp is $C_P = 100 + 1.50n$. The cost for QualityPrint is $C_Q = 2.00n$. To find when the costs are the same, we set $C_P = C_Q$: $100 + 1.50n = 2.00n$. To solve for $n$, subtract $1.50n$ from both sides: $100 = 2.00n - 1.50n$, which simplifies to $100 = 0.50n$. Dividing both sides by 0.50 gives $n = \frac{100}{0.50} = 200$. The cost is the same for 200 items.
15. Two car rental companies offer different pricing schemes. Company A charges a flat fee of $50 per day plus $0.10 per kilometer. Company B charges $30 per day plus $0.30 per kilometer. For a single-day rental, for what range of kilometers driven is Company A cheaper than Company B?
    Answer: More than 100 km When comparing linear pricing models, you need to find the break-even point where costs are equal, then determine which company is cheaper on either side of that point. Let's set up cost equations for both companies. Company A costs $C_A = 50 + 0.10k$ and Company B costs $C_B = 30 + 0.30k$, where $k$ is kilometers driven. To find when Company A is cheaper, solve the inequality $50 + 0.10k < 30 + 0.30k$. Rearranging: $50 - 30 < 0.30k - 0.10k$, which gives us $20 < 0.20k$, so $k > 100$. Let's verify with examples: At 50 km, Company A costs $50 + 0.10(50) = \$55$ while Company B costs $30 + 0.30(50) = \$45$. Company B is cheaper. At 150 km, Company A costs $50 + 0.10(150) = \$65$ while Company B costs $30 + 0.30(150) = \$75$. Company A is cheaper. Option A is incorrect because for fewer than 100 km, Company B is actually cheaper due to its lower daily fee. Option B is wrong because at exactly 100 km, both companies cost the same ($60), so neither is cheaper. Option C introduces 400 km, which has no relevance to our break-even calculation of 100 km. Option D correctly identifies that Company A becomes cheaper when you drive more than 100 kilometers, as the lower per-kilometer rate eventually overcomes the higher daily fee. Strategy tip: In linear cost comparison problems, always find the break-even point first, then test values on both sides to confirm which option is cheaper in each range.
16. 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 $S$ GB must be kept between 0.5 seconds and 0.8 seconds. Which compound inequality represents the possible file sizes $S$ in GB?
    Answer: $150 \le S \le 250$ The total processing time, $T$, in milliseconds (ms) is given by the linear model $T(S) = 50 + 3S$, where $S$ is the file size in GB. The required time range is between 0.5 and 0.8 seconds. We must convert this to milliseconds: $0.5 \, s = 500 \, ms$ and $0.8 \, s = 800 \, ms$. So, the compound inequality for the time is $500 \le T(S) \le 800$. Substituting the expression for $T(S)$ gives $500 \le 50 + 3S \le 800$. To solve for $S$, first subtract 50 from all parts: $500 - 50 \le 3S \le 800 - 50$, which simplifies to $450 \le 3S \le 750$. Finally, divide all parts by 3: $\frac{450}{3} \le S \le \frac{750}{3}$, which gives $150 \le S \le 250$.
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?
    Answer: 13 Let $t$ be the number of years. The population of the first town, $P_1$, can be modeled as $P_1(t) = 25000 + 400t$. The population of the second town, $P_2$, can be modeled as $P_2(t) = 32000 - 150t$. We want to find when $P_1(t) > P_2(t)$. This gives the inequality $25000 + 400t > 32000 - 150t$. Rearranging the terms to solve for $t$: $400t + 150t > 32000 - 25000$, which simplifies to $550t > 7000$. Dividing by 550 gives $t > \frac{7000}{550} \approx 12.727$. 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.
18. The temperature $C$ in degrees Celsius is related to the temperature $F$ in degrees Fahrenheit by the formula $C = \frac{5}{9}(F - 32)$. A chemical process requires the temperature to be kept in the range $20^{\circ}C \le C \le 30^{\circ}C$. What is the corresponding temperature range in degrees Fahrenheit?
    Answer: $68^{\circ}F \le F \le 86^{\circ}F$ We need to solve the compound inequality $20 \le \frac{5}{9}(F - 32) \le 30$. First, multiply all parts of the inequality by $\frac{9}{5}$: $20 \times \frac{9}{5} \le F - 32 \le 30 \times \frac{9}{5}$. This simplifies to $36 \le F - 32 \le 54$. Now, add 32 to all parts of the inequality: $36 + 32 \le F \le 54 + 32$. This gives the final range $68 \le F \le 86$. So the corresponding temperature range is from $68^{\circ}F$ to $86^{\circ}F$.

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