Translating to Mathematical Expressions - Finite Mathematics

A rental car company charges a daily fee plus a fee per mile driven. For a one-day rental, driving 100 miles costs $75, and driving 160 miles costs $93. Let $D$ be the fixed daily fee and $M$ be the fee per mile. Which system of linear equations correctly models this situation?
Answer: $\begin{cases} D + 100M = 75 \\ D + 160M = 93 \end{cases}$ When you encounter word problems involving fixed costs plus variable costs, you need to identify what stays constant versus what changes with usage. Here, the rental company charges a daily fee (fixed) plus a per-mile fee (variable), so your total cost equation follows the pattern: Total Cost = Fixed Fee + (Rate × Usage). Let's set up the equation structure. Since $D$ represents the fixed daily fee and $M$ represents the fee per mile, the total cost for driving any number of miles would be: Daily fee + (Miles driven × Fee per mile) = Total cost. For 100 miles costing $75: $D + 100M = 75$ For 160 miles costing $93: $D + 160M = 93$ This gives us system B, which correctly captures that the daily fee $D$ remains constant while the mileage charge (miles × rate per mile) varies. Choice A incorrectly multiplies the daily fee by the miles driven, suggesting the daily fee somehow depends on mileage. Choice C rearranges one equation correctly but botches the second by writing $D = 160M - 93$ instead of $D = 93 - 160M$ . Choice D treats $(D + M)$ as if it's a combined rate that gets multiplied by miles, which doesn't match the problem structure where $D$ is fixed regardless of miles driven. Study tip: In cost problems, always identify fixed versus variable components first. Fixed costs appear as standalone terms in your equation, while variable costs get multiplied by the quantity that changes.
An organization consists of $M$ managers and $E$ engineers. Of these, $M_f$ are female managers and $E_f$ are female engineers. A committee of $N$ people is to be randomly selected from all employees. Which expression represents the probability that the selected committee consists entirely of female engineers, assuming $N \le E_f$ ?
Answer: $\frac{\binom{E_f}{N}}{\binom{M+E}{N}}$ When you encounter a probability question involving selecting groups from a larger population, you're dealing with combinations and the fundamental principle that probability equals favorable outcomes divided by total possible outcomes. To find the probability that all N committee members are female engineers, you need to count how many ways this can happen and divide by the total ways to select any N people from the organization. The favorable outcomes are the ways to choose N people from the $E_f$ female engineers: $\binom{E_f}{N}$ . The total possible outcomes are the ways to choose N people from all $M + E$ employees: $\binom{M+E}{N}$ . Therefore, the probability is $\frac{\binom{E_f}{N}}{\binom{M+E}{N}}$ , which is answer D. Looking at the wrong answers: A gives $\binom{E_f}{N}$ , which counts favorable outcomes but ignores the denominator entirely—this isn't a probability since it's not a fraction between 0 and 1. B uses $\binom{M_f+E_f}{N}$ in the denominator, incorrectly assuming we're only selecting from female employees rather than all employees. C applies $\left(\frac{E_f}{M+E}\right)^N$ , which would be correct if we were selecting with replacement (like flipping a coin N times), but committee selection is without replacement. Remember: for "without replacement" probability problems, use combinations in both numerator and denominator. The denominator should always reflect the total population you're selecting from, not a subset.
A concert venue has three seating sections with different pricing. Orchestra seats cost $x$ each, mezzanine seats cost $y$ each, and balcony seats cost $z$ each. On Friday, they sold twice as many mezzanine seats as orchestra seats, and three times as many balcony seats as mezzanine seats, for a total revenue of $18,400. If $n$ represents the number of orchestra seats sold, which equation correctly models Friday's revenue?
Answer: $nx + 2ny + 6nz = 18400$ When you encounter word problems involving multiple variables and relationships between quantities, the key is to systematically translate each piece of information into mathematical expressions before combining them into a single equation. Let's work through this step-by-step. You're told that $n$ represents the number of orchestra seats sold. Since they sold "twice as many mezzanine seats as orchestra seats," the number of mezzanine seats is $2n$ . They sold "three times as many balcony seats as mezzanine seats," so the number of balcony seats is $3 \times (2n) = 6n$ . Now you can calculate the revenue from each section: orchestra seats contribute $nx$ dollars, mezzanine seats contribute $(2n) \times y = 2ny$ dollars, and balcony seats contribute $(6n) \times z = 6nz$ dollars. The total revenue equation becomes $nx + 2ny + 6nz = 18400$ . Looking at the wrong answers: Choice A uses $3nz$ instead of $6nz$ , incorrectly interpreting "three times as many balcony seats as mezzanine seats" as simply $3n$ rather than $3 \times (2n)$ . Choice B uses $2y + 3z$ instead of $2ny + 6nz$ , forgetting to multiply the prices by the actual quantities sold. Choice D makes the same error as B while also using the incorrect balcony calculation from A. Study tip: In multi-step relationship problems, always define your quantities in order and check that each subsequent calculation builds on the previous ones. Write out "number of seats × price per seat = revenue" for each section to avoid missing the quantity-price multiplication.
A university is organizing its course schedule. Let $m$ be the number of mathematics courses and $p$ be the number of physics courses offered. The total number of courses must be between 40 and 50, inclusive. The number of mathematics courses must be at least 15. The number of physics courses offered must be no less than one-third of the number of mathematics courses. Which of the following systems of inequalities correctly represents all of these constraints?
Answer: $\begin{cases} 40 \le m+p \le 50 \\ m \ge 15 \\ p \ge \frac{1}{3}m \end{cases}$ When you encounter word problems involving multiple constraints, your goal is to translate each written condition into a precise mathematical inequality, paying careful attention to phrases like "inclusive," "at least," and "no less than." Let's work through each constraint systematically. The problem states the total courses must be "between 40 and 50, inclusive." The word "inclusive" means the endpoints are included, so we need $40 \le m+p \le 50$ . Next, "mathematics courses must be at least 15" translates to $m \ge 15$ . Finally, "physics courses must be no less than one-third of mathematics courses" means $p \ge \frac{1}{3}m$ . Option C correctly captures all three constraints with the proper inequalities and relationships. Option A makes a critical error in the third inequality, writing $m \ge \frac{1}{3}p$ instead of $p \ge \frac{1}{3}m$ . This reverses the relationship and would mean mathematics courses must be at least one-third of physics courses, which contradicts the problem. Option B uses strict inequalities ( $<$ and $>$ ) instead of inclusive ones ( $\le$ and $\ge$ ), ignoring the word "inclusive" for the total courses and "at least" for mathematics courses. Option D contains multiple errors: it separates the compound inequality unnecessarily, uses $m \le 15$ instead of $m \ge 15$ , and writes $p \le \frac{1}{3}m$ instead of $p \ge \frac{1}{3}m$ , creating upper bounds where the problem specifies lower bounds. Always identify key phrases like "inclusive," "at least," and "no less than" – they directly determine whether you need $\le$ or $\ge$ in your inequalities.
Based on the pricing structure above, which expression correctly represents the total cost for a rush weekend delivery that travels 25 miles?
Answer: $1.25[(8 + 0.75 \times 10 + 0.50 \times 10 + 0.25 \times 5) + 3]$ For a 25-mile delivery: base fee ( $8) + first 10 miles ($ $0.75 \times 10 = 7.50$ $) + next 10 miles ($ $0.50 \times 10 = 5.00$ $) + final 5 miles ($ $0.25 \times 5 = 1.25$ $) + weekend surcharge ($3). This subtotal is $(8 + 7.50 + 5.00 + 1.25 + 3) = $24.75. The rush surcharge of 25% is applied to the entire cost including the weekend surcharge, giving $1.25 \times 24.75 = 1.25[(8 + 0.75 \times 10 + 0.50 \times 10 + 0.25 \times 5) + 3]$ . Choice B uses the wrong rate for all miles. Choice C adds 25% as $0.25 instead of multiplying by 1.25. Choice D applies the 25% rush surcharge only to the base fee.
A investment portfolio consists of stocks, bonds, and cash. The value of the stock portion changes by $r\%$ annually, bonds change by $s\%$ annually, and cash remains constant. Initially, stocks represent 60% of the portfolio, bonds represent 30%, and cash represents 10%. After one year, the portfolio is rebalanced so that stocks again represent 60% of the total value. If $V$ is the initial portfolio value, which expression represents the dollar amount moved from stocks to achieve this rebalancing?
Answer: $0.6V\left(1 + \frac{r}{100}\right) - 0.6\left[0.6V\left(1 + \frac{r}{100}\right) + 0.3V\left(1 + \frac{s}{100}\right) + 0.1V\right]$ Portfolio rebalancing problems require you to track how each component changes, then determine what adjustments are needed to restore target allocations. Let's work through this systematically. After one year, the portfolio components are worth: - Stocks: $0.6V\left(1 + \frac{r}{100}\right)$ - Bonds: $0.3V\left(1 + \frac{s}{100}\right)$ - Cash: $0.1V$ The total portfolio value becomes $0.6V\left(1 + \frac{r}{100}\right) + 0.3V\left(1 + \frac{s}{100}\right) + 0.1V$ . To rebalance to 60% stocks, you need the stock portion to equal 60% of this new total value: $0.6\left[0.6V\left(1 + \frac{r}{100}\right) + 0.3V\left(1 + \frac{s}{100}\right) + 0.1V\right]$ . The amount moved from stocks is the current stock value minus the target stock value, which gives us answer D. Choice A only compares current stock value to the original investment, ignoring that the target has changed due to overall portfolio growth. Choice B calculates the total portfolio value, not the amount moved from stocks. Choice C uses incorrect formulations—it applies only the percentage changes (without the base amounts) and uses a nonsensical subtraction. When tackling rebalancing problems, always identify three key values: current allocation, target allocation based on new total value, and the difference between them. The "amount moved" is simply current minus target for that asset class.
A company's profit depends on the number of units sold and follows this pattern: profit increases by $15 for each unit sold up to 100 units, but decreases by $2 for each unit beyond 100 due to production inefficiencies. If $P$ represents profit in dollars and $n$ represents units sold, which expression correctly models the profit when more than 100 units are sold?
Answer: $P = 15(100) + 13(n - 100)$ for $n > 100$ For units beyond 100, the profit is: profit from first 100 units + net profit from additional units. The first 100 units contribute $15 \times 100 = 1500$ dollars. Each additional unit contributes $15 - 2 = 13$ dollars (the $15 increase minus the $2 decrease). So $P = 1500 + 13(n - 100)$ . Choice A doesn't account for the net effect correctly. Choice C applies the decrease to all units, not just those beyond 100. Choice D ignores the continued $15 increase per unit.
A survey of students revealed their usage of three social media platforms. Let $T$ , $F$ , and $C$ be the sets of students who use each platform. The number of students in various categories is given by $|T|$ , $|F|$ , $|C|$ , $|T \cap F|$ , $|T \cap C|$ , $|F \cap C|$ , and $|T \cap F \cap C|$ . Which expression correctly calculates the number of students who use exactly one of these three platforms?
Answer: $|T|+|F|+|C| - 2(|T \cap F| + |T \cap C| + |F \cap C|) + 3|T \cap F \cap C|$ When you encounter set problems involving "exactly one" condition, you need to use the inclusion-exclusion principle while carefully accounting for overlaps. The key insight is that students counted in intersections are using more than one platform, so we must subtract them appropriately. To find students using exactly one platform, start with the total users of each platform: $|T| + |F| + |C|$ . However, this counts students in multiple platforms more than once. Students in two-way intersections (like $|T \cap F|$ ) are counted twice in our sum but should be counted zero times since they use more than one platform. So we subtract $2(|T \cap F| + |T \cap C| + |F \cap C|)$ . Students in the three-way intersection $|T \cap F \cap C|$ are counted three times in our original sum, subtracted three times in our correction, leaving them with zero count. But these students are also counted in each two-way intersection, so our subtraction removes them three additional times. Since they should have zero count (not negative three), we add back $3|T \cap F \cap C|$ . Choice A only addresses the three-way intersection without considering the individual platform counts. Choice B subtracts intersections once instead of twice, failing to properly eliminate students using multiple platforms. Choice C attempts to use the total universe $|T \cup F \cup C|$ , but this approach doesn't directly isolate the "exactly one" condition. Choice D correctly implements the inclusion-exclusion logic: $|T| + |F| + |C| - 2(|T \cap F| + |T \cap C| + |F \cap C|) + 3|T \cap F \cap C|$ . Strategy tip: For "exactly one" problems, remember the pattern: add individual sets, subtract twice the pairwise intersections, then add back three times the triple intersection.
A manufacturer produces widgets at a cost of $8 each. The fixed monthly costs are $2000, and each widget sells for $15. However, for every 50 widgets produced beyond 200 widgets, the production cost per widget increases by $0.50 due to overtime labor. If $x$ represents the number of widgets produced (where $x > 200$ ), which expression represents the monthly profit?
Answer: $15x - \left(8 + 0.50\left\lfloor\frac{x-200}{50}\right\rfloor\right)x - 2000$ Profit = Revenue - Total Cost. Revenue = $15x$ . The cost per widget increases by $0.50 for every complete group of 50 widgets beyond 200, so the cost per widget becomes $8 + 0.50\left\lfloor\frac{x-200}{50}\right\rfloor$ . Total production cost is this per-unit cost times $x$ widgets. Total cost = production cost + fixed cost = $\left(8 + 0.50\left\lfloor\frac{x-200}{50}\right\rfloor\right)x + 2000$ . Therefore, profit = $15x - \left(8 + 0.50\left\lfloor\frac{x-200}{50}\right\rfloor\right)x - 2000$ . Choice A only adds the increase once, not per widget. Choice B multiplies incorrectly. Choice D unnecessarily complicates the calculation.
A company manufactures two types of bicycles: standard and deluxe. Let $s$ be the number of standard models and $d$ be the number of deluxe models produced each week. Each standard model requires 2 hours of assembly time, and each deluxe model requires 5 hours. The company has a maximum of 400 assembly hours available per week. For finishing, each standard model takes 1 hour and each deluxe model takes 1.5 hours; a minimum of 95 finishing hours must be used. Due to market demand, the number of standard models produced must be at least half the number of deluxe models produced but no more than three times the number of deluxe models. Which of the following is NOT a valid constraint for this production plan?
Answer: $s \ge 3d$ Let's translate each condition into a mathematical inequality: 1. Assembly time: $2s + 5d \le 400$ . This matches option A. 2. Finishing time: $1s + 1.5d \ge 95$ . Multiplying this inequality by 2 to clear the decimal gives $2s + 3d \ge 190$ . This matches option B. 3. Demand constraint 1: 'standard models... at least half the number of deluxe models' translates to $s \ge 0.5d$ . Multiplying by 2 gives $2s \ge d$ , or $d \le 2s$ . This matches option C. 4. Demand constraint 2: 'standard models... no more than three times the number of deluxe models' translates to $s \le 3d$ . Option D states $s \ge 3d$ , which contradicts the 'no more than' condition. Therefore, it is not a valid constraint.
A manufacturer produces an item at a cost of $C$ . The manufacturer sells it to a wholesaler at a 20% markup over its cost. The wholesaler then sells it to a retailer at a 25% markup over the price they paid. Finally, the retailer sells the item to a customer at a 40% markup over the price they paid. Which expression represents the final price paid by the customer in terms of $C$ ?
Answer: $(1.20)(1.25)(1.40)C$ When you encounter sequential markup problems, you're dealing with compound percentage increases where each markup is applied to the new, higher price from the previous step. Let's trace through each transaction step by step. The manufacturer's cost is $C$ . When selling to the wholesaler with a 20% markup, the price becomes $C + 0.20C = 1.20C$ . The wholesaler then applies a 25% markup to their cost of $1.20C$ , making the retailer's price $(1.20C) \times 1.25 = 1.20 \times 1.25 \times C$ . Finally, the retailer marks up their cost by 40%, so the customer pays $(1.20 \times 1.25 \times C) \times 1.40 = 1.20 \times 1.25 \times 1.40 \times C$ . Answer C correctly represents this sequential multiplication: $(1.20)(1.25)(1.40)C$ . Answer A incorrectly adds the markup amounts ( $0.20C + 0.25C + 0.40C$ ) to the original cost, but markups after the first one aren't based on the original cost $C$ . Answer B makes the same error by factoring out $C$ from the incorrect addition. Answer D adds the markup factors (1.20 + 1.25 + 1.40) instead of multiplying them, which completely misses how sequential markups compound. Study tip: For any chain of percentage changes (markups, discounts, growth rates), multiply the factors rather than adding them. Each percentage change applies to the result of the previous change, creating a compounding effect that requires multiplication.
The population of a city is $P_0$ . Each year, the population increases by a factor of $g$ . However, at the end of each year, after the growth has occurred, a fixed number of people, $K$ , emigrates from the city. Which recurrence relation describes the population $P_n$ after $n$ years?
Answer: $P_n = g P_{n-1} - K$ When you encounter population growth problems with both multiplicative growth and fixed changes, you need to carefully track the order of operations described in the problem. Let's trace through what happens each year: The population starts at some value, then "increases by a factor of $g$ " (meaning it gets multiplied by $g$ ), and then "a fixed number $K$ emigrates" (meaning $K$ is subtracted). So if the population at the start of a year is $P_{n-1}$ , after growth it becomes $g \cdot P_{n-1}$ , and after emigration it becomes $g \cdot P_{n-1} - K$ . This gives us $P_n = g P_{n-1} - K$ , which is choice C. Choice A treats the growth factor $g$ as an additive increase rather than multiplication, which misinterprets "increases by a factor of $g$ ." Choice B applies the emigration before the growth, reversing the order described in the problem—it subtracts $K$ first, then multiplies by $g$ . Choice D incorrectly combines the growth factor and emigration into a single multiplicative term $(g-K)$ , which doesn't represent the sequential process described. The key strategy here is to translate the word problem step-by-step in the exact order given. "Increases by a factor" means multiplication, and "after the growth has occurred" tells you the sequence matters. Always set up recurrence relations by following the chronological order of events within each time period.
A city's population growth follows this pattern: it increases by 2% annually, but every 5 years, an additional 1000 people move to the city due to a recurring economic development program. If the initial population is $P_0$ and $t$ represents years, which expression best models the population after $t$ years (where $t$ is a multiple of 5)?
Answer: $P_0(1.02)^t + 1000\left\lfloor\frac{t}{5}\right\rfloor$ The population grows exponentially at 2% annually, giving the base growth of $P_0(1.02)^t$ . Additionally, every 5 years, 1000 people are added. Since $t$ is a multiple of 5, there will be $\frac{t}{5}$ such additions, contributing $1000 \times \frac{t}{5} = 1000\left\lfloor\frac{t}{5}\right\rfloor$ people (the floor function ensures we count complete 5-year periods). These are separate additive effects. Choice B incorrectly compounds the 1000-person additions with the 2% growth. Choice C assumes only one addition of 1000. Choice D adds 1000 people every year instead of every 5 years.
A company's revenue $R(x)$ from selling $x$ units is given by $R(x) = px$ , where the price $p$ is not constant but depends on the number of units sold according to the linear demand equation $p = a - bx$ , where $a$ and $b$ are positive constants. The cost $C(x)$ of producing $x$ units consists of a fixed cost $F$ and a variable cost of $v$ per unit. Which expression represents the company's profit function, $\pi(x)$ ?
Answer: $\pi(x) = (a-bx)x - (F+vx)$ Profit maximization problems require understanding how revenue and cost functions combine. When you see a business scenario with variable pricing and costs, focus on building each component systematically. Start with the revenue function. Since $R(x) = px$ and the demand equation gives us $p = a - bx$ , you can substitute to get $R(x) = (a - bx)x$ . This captures how revenue depends on both the price per unit and quantity sold. Next, build the cost function. The total cost includes fixed costs $F$ (incurred regardless of production level) plus variable costs that increase with production: $C(x) = F + vx$ , where $v$ is the variable cost per unit. The profit function is simply $\pi(x) = R(x) - C(x) = (a-bx)x - (F+vx)$ , which matches answer choice B. Let's examine why the other options fall short. Choice A, $\pi(x) = ax - F - vx$ , ignores the demand relationship entirely—it treats price as constant at $a$ rather than variable at $(a-bx)$ . Choice C, $\pi(x) = px - F$ , leaves the price undefined and completely omits variable costs. Choice D, $\pi(x) = (a-bx)x - F$ , correctly captures revenue but forgets variable production costs. Study tip: In profit problems, always write out $\pi(x) = R(x) - C(x)$ explicitly and build each component separately. This systematic approach prevents you from accidentally omitting cost or revenue elements, especially when pricing isn't constant.
A telecommunications company offers a data plan where customers pay a base fee plus charges based on data usage. The plan includes 2 GB free, then $5 per GB for the next 3 GB, and $8 per GB for any usage beyond 5 GB total. If the base fee is $25 and a customer uses $d$ GB of data, which expression correctly represents their bill when $d > 5$ ?
Answer: $25 + 5(3) + 8(d - 5)$ When $d > 5$ , the customer pays: base fee ( $25) + charges for the 3 GB in the middle tier ($ $5 \times 3 = 15$ $) + charges for data beyond 5 GB ($ $8(d - 5)$ $). The first 2 GB are free, so they don't contribute to the cost. This gives $25 + 15 + 8(d - 5) = 25 + 5(3) + 8(d - 5)$ . Choice A incorrectly charges $5 for all data used. Choice C charges $5 for data from 2 GB to $d$ GB, overlapping with the $8 charge. Choice D ignores the tiered structure and only applies the highest rate.
An employee contributes $P$ dollars at the end of each year for 10 years to a retirement account earning an annual interest rate of $r$ , compounded annually. After 10 years, the employee increases the annual contribution to $1.5P$ dollars for the next 15 years. Let $i=r$ be the interest rate per period. Which expression represents the total value of the account after the full 25-year period?
Answer: $P \left[ \frac{(1+i)^{10} - 1}{i} \right] (1+i)^{15} + 1.5P \left[ \frac{(1+i)^{15} - 1}{i} \right]$ This problem involves two separate annuities. The future value of an ordinary annuity is given by $FV = PMT \frac{(1+i)^n - 1}{i}$ . 1. The first annuity consists of payments of $P$ for 10 years. Its value at the 10-year mark is $P \frac{(1+i)^{10} - 1}{i}$ . This amount then sits in the account and accrues interest for the remaining 15 years. So, its value at the end of 25 years is $(P \frac{(1+i)^{10} - 1}{i})(1+i)^{15}$ . 2. The second annuity consists of payments of $1.5P$ for 15 years (from year 11 to 25). Its future value at the end of the 25-year period is $1.5P \frac{(1+i)^{15} - 1}{i}$ . 3. The total value is the sum of these two parts, which matches option A. Distractor B incorrectly fails to calculate the 15 years of interest earned on the value of the first annuity. Distractor C incorrectly models the contributions as a single 25-year annuity of $P$ plus a separate 15-year annuity for the 'extra' $0.5P$ . Distractor D incorrectly uses a weighted average of the payment amounts over the entire 25 years, which is not how time value of money works.
A bakery sells two types of cakes, chocolate and vanilla. Let $C$ be the event that a randomly selected customer buys a chocolate cake, and $V$ be the event that they buy a vanilla cake. The bakery knows that $P(C) = 0.6$ and $P(V) = 0.5$ . The probability that a customer buys both types of cake is $P(C \cap V) = 0.2$ . Which expression represents the probability that a randomly selected customer buys a vanilla cake but not a chocolate cake?
Answer: $P(V) - P(C \cap V)$ When you encounter probability questions about events that can overlap, focus on carefully defining what "but not" means in set notation. The phrase "vanilla but not chocolate" translates to $V \cap C^c$ , which represents customers who buy vanilla cake AND do not buy chocolate cake. To find $P(V \cap C^c)$ , think of event $V$ as being split into two non-overlapping parts: customers who buy vanilla AND chocolate ( $V \cap C$ ), and customers who buy vanilla BUT NOT chocolate ( $V \cap C^c$ ). Since these parts don't overlap, we have $P(V) = P(V \cap C) + P(V \cap C^c)$ . Rearranging this equation: $P(V \cap C^c) = P(V) - P(V \cap C) = 0.5 - 0.2 = 0.3$ . This confirms that answer B is correct. Let's examine why the other options fail. Option A gives $P(V) - P(C) = 0.5 - 0.6 = -0.1$ , which is impossible since probabilities cannot be negative. Option C calculates $1 - P(C \cup V)$ , which represents customers who buy neither type of cake, not those who buy vanilla but not chocolate. Option D gives $1 - P(C) = 0.4$ , representing customers who don't buy chocolate at all, but this includes those who buy neither cake and those who buy only vanilla. Remember this pattern: "A but not B" always equals $P(A) - P(A \cap B)$ . This formula works because you're taking all of event A and subtracting the portion that overlaps with B.
A chemist has two acid solutions. Solution A is 15% acid by volume, and Solution B is 40% acid by volume. The chemist first mixes $x$ liters of Solution A with $y$ liters of Solution B. Then, the chemist adds $z$ liters of pure water (0% acid) to this new mixture. Which expression represents the final concentration of acid in the mixture, as a percentage?
Answer: $\frac{100(0.15x + 0.40y)}{x+y+z}$ The final concentration is the total amount of acid divided by the total volume of the mixture, multiplied by 100 to express it as a percentage. The amount of acid from Solution A is $0.15x$ liters. The amount of acid from Solution B is $0.40y$ liters. The amount of acid from the pure water is $0$ . The total amount of acid is $0.15x + 0.40y$ . The total volume of the final mixture is the sum of the volumes of all components: $x + y + z$ . The concentration is $\frac{\text{Total Acid}}{\text{Total Volume}} \times 100 = \frac{0.15x + 0.40y}{x+y+z} \times 100$ . This matches option A. Distractor B incorrectly omits the volume of water, $z$ , from the denominator. Distractor C uses whole numbers (15, 40) for percentages in the numerator, which would require division by 100 in the expression, or it represents the answer as parts-per-hundred instead of a percentage value. Distractor D incorrectly adds the volume of water, $z$ , to the amount of acid in the numerator.
A manufacturing process has a setup cost of $500 per batch regardless of batch size. The variable cost per unit is $12, but due to economies of scale, this cost decreases by $0.10 for every 100 units produced in a batch (applied to all units in the batch). If a batch contains $n$ units where $n > 100$ , which expression correctly represents the total cost per unit?
Answer: $\frac{500 + n\left(12 - 0.10\left\lfloor\frac{n}{100}\right\rfloor\right)}{n}$ When analyzing cost-per-unit problems with both fixed and variable components, you need to carefully identify how each cost behaves and then divide the total cost by the number of units. Let's build the correct expression step by step. The total cost has three components: a fixed setup cost of $500, plus variable costs that depend on batch size. The base variable cost is $12 per unit, but this decreases by $0.10 for every complete group of 100 units in the batch. Since we have $n$ units, there are $\left\lfloor\frac{n}{100}\right\rfloor$ complete groups of 100, so the variable cost per unit becomes $12 - 0.10\left\lfloor\frac{n}{100}\right\rfloor$ . The total variable cost is $n \times \left(12 - 0.10\left\lfloor\frac{n}{100}\right\rfloor\right)$ . Therefore, total cost is $500 + n\left(12 - 0.10\left\lfloor\frac{n}{100}\right\rfloor\right)$ , and cost per unit is this total divided by $n$ , giving us answer D. Choice A incorrectly treats the setup cost per unit and variable cost reduction as separate additive terms, rather than dividing total cost by units. Choice B replaces the floor function with a simple linear reduction $\frac{0.10n}{100}$ , which doesn't capture that discounts only apply for complete hundreds of units. Choice C makes the same floor function error as B and uses the wrong structure for cost per unit. Remember: for cost-per-unit problems, always build the total cost first, then divide by quantity. Pay attention to whether discounts apply continuously or in discrete steps—that determines whether you need a floor function.
A parking garage charges $3 for the first hour, $2 for each additional hour up to 5 hours total, and a flat rate of $12 for any stay exceeding 5 hours. Which piecewise function correctly represents the cost $C(h)$ for parking $h$ hours?
Answer: $C(h) = \begin{cases} 3 & \text{if } 0 < h \leq 1 \\ 3 + 2(h-1) & \text{if } 1 < h \leq 5 \\ 12 & \text{if } h > 5 \end{cases}$ For the first hour ( $0 < h \leq 1$ ), the cost is $3. For additional hours beyond the first ($ $1 < h \leq 5$ $), the cost is $3 for the first hour plus $2 for each additional hour, giving $3 + 2(h-1)$ . For stays over 5 hours, the cost is the flat rate of $12. Choice A incorrectly charges $3 per hour for the first hour. Choice C uses an incorrect formula for the middle range. Choice D charges $2 for all hours beyond the first, not just additional hours.

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