Checking Units & Reasonableness - Finite Mathematics

A company manufactures solid steel ball bearings. It plans to introduce a new, larger model with double the weight (and thus double the volume). Assuming the cost of the protective coating is directly proportional to the surface area, and the original coating costs $0.32 per bearing, what is the most reasonable cost to coat the new, larger bearing?
Answer: $0.51 This problem tests your understanding of how geometric scaling affects different measurements - a key concept in finite mathematics applications. When a sphere doubles in volume, you need to find how its surface area changes. Since volume scales with the cube of linear dimensions and surface area scales with the square, there's a specific relationship. If volume doubles, then the linear scale factor is $\sqrt[3]{2} \approx 1.26$ . The surface area then increases by this factor squared: $(\sqrt[3]{2})^2 = \sqrt[3]{4} \approx 1.587$ . Therefore, the new coating cost is $\$0.32 \times 1.587 = \$0.508$ , which rounds to $0.51. Let's examine why the other answers represent common mistakes: A) $0.64 assumes surface area doubles when volume doubles, but this incorrectly applies linear scaling to area measurements. B) $0.40 might result from using an incorrect scaling factor of 1.25 ($ $\$0.32 \times 1.25$ $), possibly from approximating the cube root relationship poorly. C) $1.28 applies the volume scaling directly to cost ($ $\$0.32 \times 4 = \$1.28$ $), incorrectly assuming that doubling volume means quadrupling surface area. The key insight is recognizing that when volume changes by a factor, surface area changes by that factor raised to the $\frac{2}{3}$ power. For finite mathematics problems involving geometric scaling, always identify what measurement you're given (volume, area, or length) and what you need to find, then apply the appropriate scaling relationship rather than assuming direct proportionality.
A doctor prescribes a drug with a dosage of 15 mg per kg of body weight. The drug is supplied as a liquid with a concentration of 5.0 grams per liter. For a patient weighing 176 pounds, which is the most reasonable volume of the liquid drug to administer? (Use the conversion 1 kg ≈ 2.2 lbs).
Answer: 240 mL This is a multi-step unit conversion problem that tests your ability to work systematically through dosage calculations. When you encounter medical dosage problems, always identify what units you're given, what units you need, and plan your conversion pathway before calculating. First, convert the patient's weight from pounds to kilograms: $176 \text{ lbs} \times \frac{1 \text{ kg}}{2.2 \text{ lbs}} = 80 \text{ kg}$ Next, calculate the total drug dose needed: $80 \text{ kg} \times 15 \frac{\text{mg}}{\text{kg}} = 1200 \text{ mg}$ Now convert this to grams since the concentration is given in grams per liter: $1200 \text{ mg} = 1.2 \text{ g}$ Finally, use the concentration to find the volume needed: $\frac{1.2 \text{ g}}{5.0 \text{ g/L}} = 0.24 \text{ L} = 240 \text{ mL}$ Answer C (240 mL) is correct. Answer A (48 mL) likely results from an error in the weight conversion or forgetting to convert mg to grams. Answer B (1.2 mL) represents the mass in grams rather than the volume, showing confusion between mass and volume units. Answer D (240 L) makes the same calculation error as C but fails to convert liters to milliliters—this volume would be enormous and clearly unreasonable for a single dose. Strategy tip: In dosage problems, always check if your final answer passes the "reasonableness test." A few hundred milliliters is typical for liquid medications, while liters would be excessive.
The weekly profit $P$ , in thousands of dollars, for a company is modeled by the function $P(x) = -2x^2 + 80x - 300$ , where $x$ is the number of units produced, in hundreds. A manager correctly calculates the production level that maximizes profit. Which of the following values is the most plausible maximum weekly profit?
Answer: $500,000 The maximum profit occurs at the vertex of the parabola. The x-coordinate of the vertex is $x = -b/(2a) = -80/(2(-2)) = 20$ . Since $x$ is in hundreds of units, this corresponds to 2,000 units. The maximum profit is $P(20) = -2(20)^2 + 80(20) - 300 = -800 + 1600 - 300 = 500$ . Since $P$ is in thousands of dollars, the maximum profit is $500 \times $1,000 = $500,000. This is a monetary value, not a number of units, and is the result of the full calculation.
Pipe A can fill a swimming pool in 12 hours, and Pipe B can fill the same pool in 8 hours. Pipe C can drain the pool completely in 4 hours. If the pool is empty and all three pipes are opened simultaneously, a student calculates the time it will take to fill the pool. Which of the following outcomes is the most reasonable?
Answer: The pool will never be completely filled. To determine the outcome, we must compare the combined filling rate to the drainage rate. The rates are: Pipe A fills at $1/12$ pool/hour, Pipe B fills at $1/8$ pool/hour, and Pipe C drains at $1/4$ pool/hour. The combined filling rate is $1/12 + 1/8 = 2/24 + 3/24 = 5/24$ pool/hour. The drainage rate is $1/4 = 6/24$ pool/hour. Since the drainage rate ( $6/24$ ) is greater than the combined filling rate ( $5/24$ ), water is removed faster than it is added. Therefore, the pool will never be completely filled.
An alloy is created by mixing 100 cm³ of Metal A with 300 cm³ of Metal B. Metal A has a density of 8 g/cm³ and Metal B has a density of 12 g/cm³. A scientist calculates the density of the resulting alloy. Which of the following is the most reasonable density for the alloy? (Assume the final volume is the sum of the initial volumes).
Answer: 11 g/cm³ When you encounter density problems involving mixtures, you need to find the weighted average density based on the masses and total volume of the components. First, calculate the mass of each metal using $\text{mass} = \text{density} \times \text{volume}$ : - Metal A: $8 \text{ g/cm³} \times 100 \text{ cm³} = 800 \text{ g}$ - Metal B: $12 \text{ g/cm³} \times 300 \text{ cm³} = 3600 \text{ g}$ The total mass is $800 + 3600 = 4400 \text{ g}$ , and the total volume is $100 + 300 = 400 \text{ cm³}$ . Therefore, the alloy's density is $\frac{4400 \text{ g}}{400 \text{ cm³}} = 11 \text{ g/cm³}$ , making D correct. Choice A (10 g/cm³) represents a simple arithmetic average of the two densities, ignoring that Metal B contributes three times more volume than Metal A. Choice B (4400 g) gives you the total mass rather than density—this catches students who forget to divide by volume. Choice C (12.5 g/cm³) is higher than either individual metal's density, which is impossible since the alloy density must fall between the component densities. Remember that in mixture problems, the resulting density will always be between the individual densities, weighted toward whichever component contributes more mass. Since Metal B has both higher density and greater volume, the alloy density should be closer to 12 g/cm³ than to 8 g/cm³.
A cargo drone flies from a distribution center to a delivery point 60 miles away and returns along the same path. The drone flies at 90 mph with a tailwind on the outbound trip and returns at 30 mph against a headwind. A student is asked to calculate the average speed for the entire round trip. Which of the following is the most reasonable answer?
Answer: 45 mph The correct answer is the average speed, which must have units of miles per hour. Average speed is total distance divided by total time. The outbound trip takes $60 \text{ miles} / 90 \text{ mph} = 2/3$ hours. The return trip takes $60 \text{ miles} / 30 \text{ mph} = 2$ hours. The total distance is 120 miles and the total time is $2 + 2/3 = 8/3$ hours. The average speed is $120 \text{ miles} / (8/3 \text{ hours}) = 360/8 = 45$ mph. This value is reasonably between the two speeds.
An insurance company offers a one-year policy for a specific type of equipment. The policy premium is $250. Based on historical data, the company estimates the following probabilities for claims during the year: a 1% chance of a total loss claim of $10,000, a 5% chance of a partial damage claim of $2,000, and a 94% chance of no claim. Which of the following is a reasonable estimate for the company's expected profit per policy?
Answer: $50 The expected profit is the premium minus the expected payout. The expected payout is calculated by multiplying each claim amount by its probability and summing the results: $E(\text{Payout}) = (0.01)(\$10,000) + (0.05)(\$2,000) + (0.94)(\$0) = \$100 + \$100 + \$0 = \$200$ . The expected profit is then $E( $\text{Profit}$ ) = \text{Premium} - E( $\text{Payout}$ ) = $250 - $200 = $50. This is a reasonable positive profit for the insurer.
Water is pumped into a vertical cylindrical tank with a radius of 4 meters at a constant rate of $2\pi$ cubic meters per minute. An analyst is calculating the rate at which the water level is rising. Which of the following is a reasonable value for this rate?
Answer: $0.125$ meters per minute The question asks for the rate the water level is rising, which must have units of length per time. The volume $V$ of water in the tank is $V = \pi r^2 h$ . Taking the derivative with respect to time gives $dV/dt = \pi r^2 (dh/dt)$ . We are given $dV/dt = 2\pi$ m³/min and $r=4$ m. Substituting these values gives $2\pi = \pi (4^2) (dh/dt)$ , so $2\pi = 16\pi (dh/dt)$ . Solving for $dh/dt$ gives $dh/dt = 2\pi / (16\pi) = 1/8 = 0.125$ m/min.
A lab technician attempts to create 100 mL of an 80% acid solution using two available stock solutions: a 20% acid solution and a 70% acid solution. Which statement correctly identifies a fundamental flaw in the plan based on a reasonableness check, even before any calculations are performed?
Answer: The task is impossible because the target concentration of 80% is higher than the concentration of either stock solution. When mixing two solutions of different concentrations, the resulting concentration must lie strictly between the concentrations of the two initial solutions. In this case, any mixture of a 20% solution and a 70% solution must have a final concentration between 20% and 70%. The target concentration of 80% is outside this range, so the plan is fundamentally impossible. This reasonableness check can be done without setting up or solving equations.
The weights of five puppies in a litter are recorded as 410 g, 450 g, 390 g, 430 g, and 4200 g. A supervisor notes that the last measurement is an obvious error and that the correct weight for that puppy was 470 g. Which of the following is the most reasonable value for the corrected mean weight of the litter?
Answer: 430 g The original data contains a significant outlier (4200 g) that unreasonably skews the mean. To find the corrected mean, replace the erroneous value with the correct one and recalculate. The corrected set of weights is 410, 450, 390, 430, and 470 g. The sum of these weights is $410 + 450 + 390 + 430 + 470 = 2150$ g. The corrected mean is the sum divided by the number of puppies: $2150 / 5 = 430$ g. This value is centered within the range of the individual puppy weights and is therefore a reasonable mean.

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