Finite Mathematics Quiz: Translating To Mathematical Expressions
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Translating To Mathematical ExpressionsQuestion 1 of 20

A investment portfolio consists of stocks, bonds, and cash. The value of the stock portion changes by r%r\% annually, bonds change by s%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 VV is the initial portfolio value, which expression represents the dollar amount moved from stocks to achieve this rebalancing?

0.6V(1+r100)0.6V0.6V\left(1 + \frac{r}{100}\right) - 0.6V
[0.6V(1+r100)+0.3V(1+s100)+0.1V]0.6V\left[0.6V\left(1 + \frac{r}{100}\right) + 0.3V\left(1 + \frac{s}{100}\right) + 0.1V\right] - 0.6V
0.6V(r100)0.6[0.3V(s100)+0.1V]0.6V\left(\frac{r}{100}\right) - 0.6\left[0.3V\left(\frac{s}{100}\right) + 0.1V\right]
0.6V(1+r100)0.6[0.6V(1+r100)+0.3V(1+s100)+0.1V]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]
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Finite Mathematics Quiz

Finite Mathematics Quiz: Translating To Mathematical Expressions

Practice Translating To Mathematical Expressions 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 Translating To Mathematical Expressions, 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

A investment portfolio consists of stocks, bonds, and cash. The value of the stock portion changes by r%r\% annually, bonds change by s%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 VV is the initial portfolio value, which expression represents the dollar amount moved from stocks to achieve this rebalancing?

  1. 0.6V(1+r100)0.6V0.6V\left(1 + \frac{r}{100}\right) - 0.6V
  2. [0.6V(1+r100)+0.3V(1+s100)+0.1V]0.6V\left[0.6V\left(1 + \frac{r}{100}\right) + 0.3V\left(1 + \frac{s}{100}\right) + 0.1V\right] - 0.6V
  3. 0.6V(r100)0.6[0.3V(s100)+0.1V]0.6V\left(\frac{r}{100}\right) - 0.6\left[0.3V\left(\frac{s}{100}\right) + 0.1V\right]
  4. 0.6V(1+r100)0.6[0.6V(1+r100)+0.3V(1+s100)+0.1V]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] (correct answer)
Explanation: 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(1+r100)0.6V\left(1 + \frac{r}{100}\right)
  • Bonds: 0.3V(1+s100)0.3V\left(1 + \frac{s}{100}\right)
  • Cash: 0.1V0.1V
The total portfolio value becomes 0.6V(1+r100)+0.3V(1+s100)+0.1V0.6V\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[0.6V(1+r100)+0.3V(1+s100)+0.1V]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.

Question 2

A subscription service charges a monthly fee plus an additional cost per hour of usage. In January, Sarah used the service for 12 hours and paid $47. In February, she used it for 8 hours and paid $35. If $xx representsthemonthlyfeeandrepresents the monthly fee and yy $ represents the cost per hour, which system of equations correctly models this situation?

  1. x+12y=47x + 12y = 47 and x+8y=35x + 8y = 35 (correct answer)
  2. 12x+y=4712x + y = 47 and 8x+y=358x + y = 35
  3. x+12y=35x + 12y = 35 and x+8y=47x + 8y = 47
  4. 12(x+y)=4712(x + y) = 47 and 8(x+y)=358(x + y) = 35
Explanation: The monthly fee xx is charged regardless of usage, and the hourly cost yy is multiplied by the number of hours used. For January: monthly fee + (12 hours × hourly rate) = $47, giving $x+12y=47x + 12y = 47 .ForFebruary:monthlyfee+(8hours×hourlyrate)=. For February: monthly fee + (8 hours × hourly rate) = 35, giving x+8y=35x + 8y = 35. Choice B incorrectly multiplies the monthly fee by hours. Choice C reverses the total costs. Choice D incorrectly assumes the monthly fee and hourly rate are added together before multiplying by hours.

Question 3

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)C(h) forparkingfor parking hh $ hours?

  1. (correct answer)
Explanation: For the first hour (0<h10 < h \leq 1), the cost is 3.Foradditionalhoursbeyondthefirst(3. For additional hours beyond the first ( 1<h51 < h \leq 5 ),thecostis$3forthefirsthourplus$2foreachadditionalhour,giving$), 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.

Question 4

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 $dd GBofdata,whichexpressioncorrectlyrepresentstheirbillwhenGB of data, which expression correctly represents their bill when d>5d > 5 $?

  1. 25+5d+8(d5)25 + 5d + 8(d - 5)
  2. 25+5(3)+8(d5)25 + 5(3) + 8(d - 5) (correct answer)
  3. 25+5(d2)+8(d5)25 + 5(d - 2) + 8(d - 5)
  4. 25+8d25 + 8d
Explanation: When d>5d > 5, the customer pays: base fee (25)+chargesforthe3GBinthemiddletier(25) + charges for the 3 GB in the middle tier ( 5×3=155 \times 3 = 15 )+chargesfordatabeyond5GB() + charges for data beyond 5 GB ( 8(d5)8(d - 5) ).Thefirst2GBarefree,sotheydontcontributetothecost.Thisgives). The first 2 GB are free, so they don't contribute to the cost. This gives 25+15+8(d5)=25+5(3)+8(d5)25 + 15 + 8(d - 5) = 25 + 5(3) + 8(d - 5) .ChoiceAincorrectlycharges$5foralldataused.ChoiceCcharges$5fordatafrom2GBto$. 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.

Question 5

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 P0P_0 and tt represents years, which expression best models the population after tt years (where tt is a multiple of 5)?

  1. P0(1.02)t+1000t5P_0(1.02)^t + 1000\left\lfloor\frac{t}{5}\right\rfloor (correct answer)
  2. (P0+1000t5)(1.02)t\left(P_0 + 1000\left\lfloor\frac{t}{5}\right\rfloor\right)(1.02)^t
  3. (P0+1000)(1.02)t(P_0 + 1000)(1.02)^t
  4. P0(1.02)t+1000tP_0(1.02)^t + 1000t
Explanation: The population grows exponentially at 2% annually, giving the base growth of P0(1.02)tP_0(1.02)^t. Additionally, every 5 years, 1000 people are added. Since tt is a multiple of 5, there will be t5\frac{t}{5} such additions, contributing 1000×t5=1000t51000 \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.

Question 6

A delivery service uses the following pricing structure:

• Base delivery fee: $8 • Distance charge: $0.75 per mile for the first 10 miles • Distance charge: $0.50 per mile for miles 11-20 • Distance charge: $0.25 per mile for any miles beyond 20 • Rush delivery surcharge: additional 25% of total delivery cost • Weekend surcharge: additional $3 (applied before rush surcharge if both apply)

Based on the pricing structure above, which expression correctly represents the total cost for a rush weekend delivery that travels 25 miles?

  1. 1.25[(8+0.75×10+0.50×10+0.25×5)+3]1.25[(8 + 0.75 \times 10 + 0.50 \times 10 + 0.25 \times 5) + 3] (correct answer)
  2. 1.25[8+0.75×25+3]1.25[8 + 0.75 \times 25 + 3]
  3. (8+0.75×10+0.50×10+0.25×5)+3+0.25(8 + 0.75 \times 10 + 0.50 \times 10 + 0.25 \times 5) + 3 + 0.25
  4. 1.25×8+0.75×10+0.50×10+0.25×5+31.25 \times 8 + 0.75 \times 10 + 0.50 \times 10 + 0.25 \times 5 + 3
Explanation: For a 25-mile delivery: base fee (8)+first10miles(8) + first 10 miles ( 0.75×10=7.500.75 \times 10 = 7.50 )+next10miles() + next 10 miles ( 0.50×10=5.000.50 \times 10 = 5.00 )+final5miles() + final 5 miles ( 0.25×5=1.250.25 \times 5 = 1.25 )+weekendsurcharge($3).Thissubtotalis$(8+7.50+5.00+1.25+3)=$24.75.Therushsurchargeof25) + 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.

Question 7

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 xx liters of Solution A with yy liters of Solution B. Then, the chemist adds zz liters of pure water (0% acid) to this new mixture. Which expression represents the final concentration of acid in the mixture, as a percentage?

  1. 100(0.15x+0.40y)x+y+z\frac{100(0.15x + 0.40y)}{x+y+z} (correct answer)
  2. 100(0.15x+0.40y)x+y\frac{100(0.15x + 0.40y)}{x+y}
  3. 15x+40yx+y+z\frac{15x + 40y}{x+y+z}
  4. 100(0.15x+0.40y+z)x+y+z\frac{100(0.15x + 0.40y + z)}{x+y+z}
Explanation: 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 0.15x0.15x liters. The amount of acid from Solution B is 0.40y0.40y liters. The amount of acid from the pure water is 00. The total amount of acid is 0.15x+0.40y0.15x + 0.40y. The total volume of the final mixture is the sum of the volumes of all components: x+y+zx + y + z. The concentration is Total AcidTotal Volume×100=0.15x+0.40yx+y+z×100\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, zz, 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, zz, to the amount of acid in the numerator.

Question 8

A company manufactures two types of bicycles: standard and deluxe. Let ss be the number of standard models and dd 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?

  1. 2s+5d4002s + 5d \le 400
  2. 2s+3d1902s + 3d \ge 190
  3. d2sd \le 2s
  4. s3ds \ge 3d (correct answer)
Explanation: Let's translate each condition into a mathematical inequality:
  1. Assembly time: 2s+5d4002s + 5d \le 400. This matches option A.
  2. Finishing time: 1s+1.5d951s + 1.5d \ge 95. Multiplying this inequality by 2 to clear the decimal gives 2s+3d1902s + 3d \ge 190. This matches option B.
  3. Demand constraint 1: 'standard models... at least half the number of deluxe models' translates to s0.5ds \ge 0.5d. Multiplying by 2 gives 2sd2s \ge d, or d2sd \le 2s. This matches option C.
  4. Demand constraint 2: 'standard models... no more than three times the number of deluxe models' translates to s3ds \le 3d. Option D states s3ds \ge 3d, which contradicts the 'no more than' condition. Therefore, it is not a valid constraint.

Question 9

An employee contributes PP dollars at the end of each year for 10 years to a retirement account earning an annual interest rate of rr, compounded annually. After 10 years, the employee increases the annual contribution to 1.5P1.5P dollars for the next 15 years. Let i=ri=r be the interest rate per period. Which expression represents the total value of the account after the full 25-year period?

  1. P[(1+i)101i](1+i)15+1.5P[(1+i)151i]P \left[ \frac{(1+i)^{10} - 1}{i} \right] (1+i)^{15} + 1.5P \left[ \frac{(1+i)^{15} - 1}{i} \right] (correct answer)
  2. P[(1+i)101i]+1.5P[(1+i)151i]P \left[ \frac{(1+i)^{10} - 1}{i} \right] + 1.5P \left[ \frac{(1+i)^{15} - 1}{i} \right]
  3. P[(1+i)251i]+0.5P[(1+i)151i]P \left[ \frac{(1+i)^{25} - 1}{i} \right] + 0.5P \left[ \frac{(1+i)^{15} - 1}{i} \right]
  4. 1.3P[(1+i)251i]1.3P \left[ \frac{(1+i)^{25} - 1}{i} \right]
Explanation: This problem involves two separate annuities. The future value of an ordinary annuity is given by FV=PMT(1+i)n1iFV = PMT \frac{(1+i)^n - 1}{i}.
  1. The first annuity consists of payments of PP for 10 years. Its value at the 10-year mark is P(1+i)101iP \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(1+i)101i)(1+i)15(P \frac{(1+i)^{10} - 1}{i})(1+i)^{15}.
  2. The second annuity consists of payments of 1.5P1.5P for 15 years (from year 11 to 25). Its future value at the end of the 25-year period is 1.5P(1+i)151i1.5P \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 PP plus a separate 15-year annuity for the 'extra' 0.5P0.5P. 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.

Question 10

An organization consists of MM managers and EE engineers. Of these, MfM_f are female managers and EfE_f are female engineers. A committee of NN people is to be randomly selected from all employees. Which expression represents the probability that the selected committee consists entirely of female engineers, assuming NEfN \le E_f?

  1. (EfN)\binom{E_f}{N}
  2. (EfN)(Mf+EfN)\frac{\binom{E_f}{N}}{\binom{M_f+E_f}{N}}
  3. (EfM+E)N\left(\frac{E_f}{M+E}\right)^N
  4. (EfN)(M+EN)\frac{\binom{E_f}{N}}{\binom{M+E}{N}} (correct answer)
Explanation: 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 EfE_f female engineers: (EfN)\binom{E_f}{N}. The total possible outcomes are the ways to choose N people from all M+EM + E employees: (M+EN)\binom{M+E}{N}. Therefore, the probability is (EfN)(M+EN)\frac{\binom{E_f}{N}}{\binom{M+E}{N}}, which is answer D. Looking at the wrong answers: A gives (EfN)\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 (Mf+EfN)\binom{M_f+E_f}{N} in the denominator, incorrectly assuming we're only selecting from female employees rather than all employees. C applies (EfM+E)N\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.

Question 11

A manufacturer produces an item at a cost of CC. 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 CC?

  1. C+0.20C+0.25C+0.40CC + 0.20C + 0.25C + 0.40C
  2. (1+0.20+0.25+0.40)C(1 + 0.20 + 0.25 + 0.40)C
  3. (1.20)(1.25)(1.40)C(1.20)(1.25)(1.40)C (correct answer)
  4. (1.20+1.25+1.40)C(1.20 + 1.25 + 1.40)C
Explanation: 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 CC. When selling to the wholesaler with a 20% markup, the price becomes C+0.20C=1.20CC + 0.20C = 1.20C. The wholesaler then applies a 25% markup to their cost of 1.20C1.20C, making the retailer's price (1.20C)×1.25=1.20×1.25×C(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 (1.20×1.25×C)×1.40=1.20×1.25×1.40×C(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(1.20)(1.25)(1.40)C. Answer A incorrectly adds the markup amounts (0.20C+0.25C+0.40C0.20C + 0.25C + 0.40C) to the original cost, but markups after the first one aren't based on the original cost CC. Answer B makes the same error by factoring out CC 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.

Question 12

A university is organizing its course schedule. Let mm be the number of mathematics courses and pp 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?

  1. {40m+p50m15m13p\begin{cases} 40 \le m+p \le 50 \\ m \ge 15 \\ m \ge \frac{1}{3}p \end{cases}
  2. {40<m+p<50m>15p13m\begin{cases} 40 < m+p < 50 \\ m > 15 \\ p \ge \frac{1}{3}m \end{cases}
  3. {40m+p50m15p13m\begin{cases} 40 \le m+p \le 50 \\ m \ge 15 \\ p \ge \frac{1}{3}m \end{cases} (correct answer)
  4. {m+p40m+p50m15p13m\begin{cases} m+p \ge 40 \\ m+p \le 50 \\ m \le 15 \\ p \le \frac{1}{3}m \end{cases}
Explanation: 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 40m+p5040 \le m+p \le 50. Next, "mathematics courses must be at least 15" translates to m15m \ge 15. Finally, "physics courses must be no less than one-third of mathematics courses" means p13mp \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 m13pm \ge \frac{1}{3}p instead of p13mp \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 m15m \le 15 instead of m15m \ge 15, and writes p13mp \le \frac{1}{3}m instead of p13mp \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.

Question 13

A bakery sells two types of cakes, chocolate and vanilla. Let CC be the event that a randomly selected customer buys a chocolate cake, and VV be the event that they buy a vanilla cake. The bakery knows that P(C)=0.6P(C) = 0.6 and P(V)=0.5P(V) = 0.5. The probability that a customer buys both types of cake is P(CV)=0.2P(C \cap V) = 0.2. Which expression represents the probability that a randomly selected customer buys a vanilla cake but not a chocolate cake?

  1. P(V)P(C)P(V) - P(C)
  2. P(V)P(CV)P(V) - P(C \cap V) (correct answer)
  3. 1P(CV)1 - P(C \cup V)
  4. 1P(C)1 - P(C)
Explanation: 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 VCcV \cap C^c, which represents customers who buy vanilla cake AND do not buy chocolate cake. To find P(VCc)P(V \cap C^c), think of event VV as being split into two non-overlapping parts: customers who buy vanilla AND chocolate (VCV \cap C), and customers who buy vanilla BUT NOT chocolate (VCcV \cap C^c). Since these parts don't overlap, we have P(V)=P(VC)+P(VCc)P(V) = P(V \cap C) + P(V \cap C^c). Rearranging this equation: P(VCc)=P(V)P(VC)=0.50.2=0.3P(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)=0.50.6=0.1P(V) - P(C) = 0.5 - 0.6 = -0.1, which is impossible since probabilities cannot be negative. Option C calculates 1P(CV)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 1P(C)=0.41 - 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(AB)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.

Question 14

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 $Dbethefixeddailyfeeandbe the fixed daily fee andM$ be the fee per mile. Which system of linear equations correctly models this situation?

  1. {100D+M=75160D+M=93\begin{cases} 100D + M = 75 \\ 160D + M = 93 \end{cases}
  2. {D+100M=75D+160M=93\begin{cases} D + 100M = 75 \\ D + 160M = 93 \end{cases} (correct answer)
  3. {D=75100MD=160M93\begin{cases} D = 75 - 100M \\ D = 160M - 93 \end{cases}
  4. {100(D+M)=75160(D+M)=93\begin{cases} 100(D+M) = 75 \\ 160(D+M) = 93 \end{cases}
Explanation: 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 DD represents the fixed daily fee and MM 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 = 75For160milescosting$93:$D+160M=93 For 160 miles costing $93: $D + 160M = 93 This gives us system B, which correctly captures that the daily fee DD 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=160M93D = 160M - 93 instead of D=93160MD = 93 - 160M. Choice D treats (D+M)(D + M) as if it's a combined rate that gets multiplied by miles, which doesn't match the problem structure where DD 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.

Question 15

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 $nn unitswhereunits where n>100n > 100 $, which expression correctly represents the total cost per unit?

  1. 500n+120.10n100\frac{500}{n} + 12 - 0.10\left\lfloor\frac{n}{100}\right\rfloor
  2. 500+12n0.10nn\frac{500 + 12n - 0.10n}{n}
  3. 500n+120.10n100\frac{500}{n} + 12 - \frac{0.10n}{100}
  4. 500+n(120.10n100)n\frac{500 + n\left(12 - 0.10\left\lfloor\frac{n}{100}\right\rfloor\right)}{n} (correct answer)
Explanation: 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 $nn units,thereareunits, there are n100\left\lfloor\frac{n}{100}\right\rfloor completegroupsof100,sothevariablecostperunitbecomescomplete groups of 100, so the variable cost per unit becomes 120.10n10012 - 0.10\left\lfloor\frac{n}{100}\right\rfloor .Thetotalvariablecostis. The total variable cost is n×(120.10n100)n \times \left(12 - 0.10\left\lfloor\frac{n}{100}\right\rfloor\right) .Therefore,totalcostis. Therefore, total cost is 500+n(120.10n100)500 + n\left(12 - 0.10\left\lfloor\frac{n}{100}\right\rfloor\right) ,andcostperunitisthistotaldividedby, and cost per unit is this total divided by nn $, 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.

Question 16

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 $xx representsthenumberofwidgetsproduced(whererepresents the number of widgets produced (where x>200x > 200 $), which expression represents the monthly profit?

  1. 15x8x20000.50x2005015x - 8x - 2000 - 0.50\left\lfloor\frac{x-200}{50}\right\rfloor
  2. 15x8x20000.50xx2005015x - 8x - 2000 - 0.50x\left\lfloor\frac{x-200}{50}\right\rfloor
  3. 15x(8+0.50x20050)x200015x - \left(8 + 0.50\left\lfloor\frac{x-200}{50}\right\rfloor\right)x - 2000 (correct answer)
  4. 15x8(200)(8.50x20050)(x200)200015x - 8(200) - \left(8.50\left\lfloor\frac{x-200}{50}\right\rfloor\right)(x-200) - 2000
Explanation: Profit = Revenue - Total Cost. Revenue = 15x15x. The cost per widget increases by $0.50 for every complete group of 50 widgets beyond 200, so the cost per widget becomes $8+0.50x200508 + 0.50\left\lfloor\frac{x-200}{50}\right\rfloor .Totalproductioncostisthisperunitcosttimes. Total production cost is this per-unit cost times xx widgets.Totalcost=productioncost+fixedcost=widgets. Total cost = production cost + fixed cost = (8+0.50x20050)x+2000\left(8 + 0.50\left\lfloor\frac{x-200}{50}\right\rfloor\right)x + 2000 .Therefore,profit=. Therefore, profit = 15x(8+0.50x20050)x200015x - \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.

Question 17

A company manufactures smart thermostats. The monthly cost of production includes a fixed cost of $15,000. The production cost per unit is $80 for the first 1,000 units produced in a month. For any additional units produced beyond 1,000, the cost per unit is reduced by 25%. Let $xbethetotalnumberofthermostatsproducedinamonth,wherebe the total number of thermostats produced in a month, wherex > 1000.Whichexpressionrepresentsthetotalmonthlycost,. Which expression represents the total monthly cost, C(x)$?

  1. C(x)=60x+35000C(x) = 60x + 35000 (correct answer)
  2. C(x)=60x+15000C(x) = 60x + 15000
  3. C(x)=80x5000C(x) = 80x - 5000
  4. C(x)=80x+15000C(x) = 80x + 15000
Explanation: The total cost C(x)C(x) is the sum of the fixed cost, the cost for the first 1,000 units, and the cost for the units produced beyond 1,000. The reduced cost for units beyond 1,000 is $80 \times (1 - 0.25) = 80 \times 0.75 = $60 per unit. The number of units in this tier is $(x - 1000).So,thetotalcostiscalculatedas:. So, the total cost is calculated as: C(x) = \text{Fixed Cost} + (Cost per unit for first 1000\text{Cost per unit for first 1000}) \times 1000 + (Reduced cost\text{Reduced cost}) \times (x - 1000) C(x) = 15000 + (80)(1000) + (60)(x - 1000) C(x) = 15000 + 80000 + 60x - 60000 C(x) = 35000 + 60xTherefore,thecorrectexpressionisTherefore, the correct expression isC(x) = 60x + 35000.DistractorBincorrectlyappliesthediscountedrateof$60toall$x. Distractor B incorrectly applies the discounted rate of $60 to all $x units. Distractor C incorrectly swaps the quantities for the two cost tiers. Distractor D ignores the quantity discount entirely.

Question 18

The population of a city is P0P_0. Each year, the population increases by a factor of gg. However, at the end of each year, after the growth has occurred, a fixed number of people, KK, emigrates from the city. Which recurrence relation describes the population PnP_n after nn years?

  1. Pn=Pn1+gKP_n = P_{n-1} + g - K
  2. Pn=g(Pn1K)P_n = g(P_{n-1} - K)
  3. Pn=gPn1KP_n = g P_{n-1} - K (correct answer)
  4. Pn=(gK)Pn1P_n = (g-K)P_{n-1}
Explanation: 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 gg" (meaning it gets multiplied by gg), and then "a fixed number KK emigrates" (meaning KK is subtracted). So if the population at the start of a year is Pn1P_{n-1}, after growth it becomes gPn1g \cdot P_{n-1}, and after emigration it becomes gPn1Kg \cdot P_{n-1} - K. This gives us Pn=gPn1KP_n = g P_{n-1} - K, which is choice C. Choice A treats the growth factor gg as an additive increase rather than multiplication, which misinterprets "increases by a factor of gg." Choice B applies the emigration before the growth, reversing the order described in the problem—it subtracts KK first, then multiplies by gg. Choice D incorrectly combines the growth factor and emigration into a single multiplicative term (gK)(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.

Question 19

A company's revenue R(x)R(x) from selling xx units is given by R(x)=pxR(x) = px, where the price pp is not constant but depends on the number of units sold according to the linear demand equation p=abxp = a - bx, where aa and bb are positive constants. The cost C(x)C(x) of producing xx units consists of a fixed cost FF and a variable cost of vv per unit. Which expression represents the company's profit function, π(x)\pi(x)?

  1. π(x)=axFvx\pi(x) = ax - F - vx
  2. π(x)=(abx)x(F+vx)\pi(x) = (a-bx)x - (F+vx) (correct answer)
  3. π(x)=pxF\pi(x) = px - F
  4. π(x)=(abx)xF\pi(x) = (a-bx)x - F
Explanation: 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)=pxR(x) = px and the demand equation gives us p=abxp = a - bx, you can substitute to get R(x)=(abx)xR(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 FF (incurred regardless of production level) plus variable costs that increase with production: C(x)=F+vxC(x) = F + vx, where vv is the variable cost per unit. The profit function is simply π(x)=R(x)C(x)=(abx)x(F+vx)\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, π(x)=axFvx\pi(x) = ax - F - vx, ignores the demand relationship entirely—it treats price as constant at aa rather than variable at (abx)(a-bx). Choice C, π(x)=pxF\pi(x) = px - F, leaves the price undefined and completely omits variable costs. Choice D, π(x)=(abx)xF\pi(x) = (a-bx)x - F, correctly captures revenue but forgets variable production costs. Study tip: In profit problems, always write out π(x)=R(x)C(x)\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.

Question 20

A survey of students revealed their usage of three social media platforms. Let TT, FF, and CC be the sets of students who use each platform. The number of students in various categories is given by T|T|, F|F|, C|C|, TF|T \cap F|, TC|T \cap C|, FC|F \cap C|, and TFC|T \cap F \cap C|. Which expression correctly calculates the number of students who use exactly one of these three platforms?

  1. TF+TC+FC3TFC|T \cap F| + |T \cap C| + |F \cap C| - 3|T \cap F \cap C|
  2. T+F+C(TF+TC+FC)|T|+|F|+|C| - (|T \cap F| + |T \cap C| + |F \cap C|)
  3. TFC(TF+TC+FC)+TFC|T \cup F \cup C| - (|T \cap F| + |T \cap C| + |F \cap C|) + |T \cap F \cap C|
  4. T+F+C2(TF+TC+FC)+3TFC|T|+|F|+|C| - 2(|T \cap F| + |T \cap C| + |F \cap C|) + 3|T \cap F \cap C| (correct answer)
Explanation: 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|T| + |F| + |C|. However, this counts students in multiple platforms more than once. Students in two-way intersections (like TF|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(TF+TC+FC)2(|T \cap F| + |T \cap C| + |F \cap C|). Students in the three-way intersection TFC|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 3TFC3|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 TFC|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+C2(TF+TC+FC)+3TFC|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.