Health Education Systems Inc (HESI) A2 Exam Quiz: Dimensional Analysis
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Dimensional AnalysisQuestion 1 of 20

A prescription calls for 0.125 mg of medication to be given twice daily. The medication is available as a 1:4000 solution (w/v). Which dimensional analysis setup correctly determines the volume needed for each dose?

0.125 mg×1000 mg1 g×1 g4000 mL0.125 \text{ mg} \times \frac{1000 \text{ mg}}{1 \text{ g}} \times \frac{1 \text{ g}}{4000 \text{ mL}}
0.125 mg×1 g1000 mg×4000 mL1 g0.125 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}} \times \frac{4000 \text{ mL}}{1 \text{ g}}
0.125 mg×1 g1000 mg×1 g4000 mL0.125 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}} \times \frac{1 \text{ g}}{4000 \text{ mL}}
0.125 mg×1000 mg1 g×4000 mL1 g0.125 \text{ mg} \times \frac{1000 \text{ mg}}{1 \text{ g}} \times \frac{4000 \text{ mL}}{1 \text{ g}}
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Health Education Systems Inc (HESI) A2 Exam Quiz

Health Education Systems Inc (HESI) A2 Exam Quiz: Dimensional Analysis

Practice Dimensional Analysis in Health Education Systems Inc (HESI) A2 Exam 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 Dimensional Analysis, giving you a quick way to practice the rules, question types, and explanations that matter most for Health Education Systems Inc (HESI) A2 Exam.

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 prescription calls for 0.125 mg of medication to be given twice daily. The medication is available as a 1:4000 solution (w/v). Which dimensional analysis setup correctly determines the volume needed for each dose?

  1. 0.125 mg×1000 mg1 g×1 g4000 mL0.125 \text{ mg} \times \frac{1000 \text{ mg}}{1 \text{ g}} \times \frac{1 \text{ g}}{4000 \text{ mL}}
  2. 0.125 mg×1 g1000 mg×4000 mL1 g0.125 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}} \times \frac{4000 \text{ mL}}{1 \text{ g}} (correct answer)
  3. 0.125 mg×1 g1000 mg×1 g4000 mL0.125 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}} \times \frac{1 \text{ g}}{4000 \text{ mL}}
  4. 0.125 mg×1000 mg1 g×4000 mL1 g0.125 \text{ mg} \times \frac{1000 \text{ mg}}{1 \text{ g}} \times \frac{4000 \text{ mL}}{1 \text{ g}}
Explanation: When you encounter medication dosage calculations involving ratio solutions, you need to systematically convert units and apply the concentration correctly using dimensional analysis. A 1:4000 (w/v) solution means 1 gram of medication per 4000 mL of solution. To find the volume needed for 0.125 mg, you must first convert milligrams to grams, then apply the solution concentration. Starting with 0.125 mg, you need 1 g1000 mg\frac{1 \text{ g}}{1000 \text{ mg}} to convert to grams (since 1000 mg = 1 g). This gives you the amount in grams. Then, since the solution contains 1 g per 4000 mL, you multiply by 4000 mL1 g\frac{4000 \text{ mL}}{1 \text{ g}} to find the required volume. Choice A incorrectly converts mg to g by multiplying instead of dividing by 1000, then uses the reciprocal of the concentration ratio. Choice C correctly converts mg to g but then incorrectly inverts the concentration ratio, treating it as if 4000 mL contains 1 g instead of 1 g being in 4000 mL. Choice D makes both errors from A and inverts the concentration improperly. Choice B correctly sets up: 0.125 mg×1 g1000 mg×4000 mL1 g0.125 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}} \times \frac{4000 \text{ mL}}{1 \text{ g}}, which yields 0.5 mL per dose. Remember: Always ensure your conversion factors are set up so unwanted units cancel out, and ratio concentrations like 1:4000 mean 1 part drug in 4000 parts total volume. Check that your dimensional analysis flows logically from the given units to the desired units.

Question 2

To prepare a cleaning solution, a nurse must dilute a concentrate. The instructions state to use 3 fluid ounces (fl oz) of concentrate for every 1 gallon (gal) of water. Which expression correctly sets up the calculation for how many milliliters (mL) of concentrate are needed to prepare a 5-gallon solution? (Use 1 fl oz ≈ 29.57 mL)

  1. 5 gal1×1 gal3 fl oz×29.57 mL1 fl oz\frac{5 \text{ gal}}{1} \times \frac{1 \text{ gal}}{3 \text{ fl oz}} \times \frac{29.57 \text{ mL}}{1 \text{ fl oz}}
  2. 5 gal1×3 fl oz1 gal×29.57 mL1 fl oz\frac{5 \text{ gal}}{1} \times \frac{3 \text{ fl oz}}{1 \text{ gal}} \times \frac{29.57 \text{ mL}}{1 \text{ fl oz}} (correct answer)
  3. 5 gal1×3 fl oz1 gal×1 fl oz29.57 mL\frac{5 \text{ gal}}{1} \times \frac{3 \text{ fl oz}}{1 \text{ gal}} \times \frac{1 \text{ fl oz}}{29.57 \text{ mL}}
  4. 15 gal×3 fl oz1 gal×29.57 mL1 fl oz\frac{1}{5 \text{ gal}} \times \frac{3 \text{ fl oz}}{1 \text{ gal}} \times \frac{29.57 \text{ mL}}{1 \text{ fl oz}}
Explanation: When you encounter dosage calculation problems involving unit conversions, you need to set up a dimensional analysis chain that systematically cancels unwanted units while preserving the desired final unit. The correct approach is option B: 5 gal1×3 fl oz1 gal×29.57 mL1 fl oz\frac{5 \text{ gal}}{1} \times \frac{3 \text{ fl oz}}{1 \text{ gal}} \times \frac{29.57 \text{ mL}}{1 \text{ fl oz}}. This expression starts with 5 gallons of total solution, then applies the concentration ratio (3 fl oz concentrate per 1 gallon), and finally converts fluid ounces to milliliters. Notice how the units cancel perfectly: gal cancels with gal, and fl oz cancels with fl oz, leaving only mL. Option A inverts the concentration ratio incorrectly, showing 1 gal3 fl oz\frac{1 \text{ gal}}{3 \text{ fl oz}} instead of 3 fl oz1 gal\frac{3 \text{ fl oz}}{1 \text{ gal}}. This would give you gallons per fluid ounce rather than the needed fluid ounces of concentrate per gallon of solution. Option C uses the correct concentration ratio but inverts the conversion factor, showing 1 fl oz29.57 mL\frac{1 \text{ fl oz}}{29.57 \text{ mL}} instead of 29.57 mL1 fl oz\frac{29.57 \text{ mL}}{1 \text{ fl oz}}. This would convert milliliters to fluid ounces rather than fluid ounces to milliliters. Option D places 5 gallons in the denominator (15 gal\frac{1}{5 \text{ gal}}), which would divide by the volume instead of multiplying by it, giving an incorrect result. For HESI dosage calculations, always write out your dimensional analysis completely and check that unwanted units cancel. Set up each conversion factor so the unit you want to eliminate appears in both a numerator and denominator.

Question 3

A patient is prescribed a medication with a dosage of 25 mg/kg per day, divided into three equal doses. The patient weighs 198 lbs. Which dimensional analysis setup correctly calculates the number of milligrams per single dose? (Use the conversion factor 1 kg = 2.2 lbs)

  1. 198 lbs1×1 kg2.2 lbs×25 mg1 kg×1 day3 doses\frac{198 \text{ lbs}}{1} \times \frac{1 \text{ kg}}{2.2 \text{ lbs}} \times \frac{25 \text{ mg}}{1 \text{ kg}} \times \frac{1 \text{ day}}{3 \text{ doses}} (correct answer)
  2. 198 lbs1×2.2 lbs1 kg×25 mg1 kg×3 doses1 day\frac{198 \text{ lbs}}{1} \times \frac{2.2 \text{ lbs}}{1 \text{ kg}} \times \frac{25 \text{ mg}}{1 \text{ kg}} \times \frac{3 \text{ doses}}{1 \text{ day}}
  3. 198 lbs1×1 kg2.2 lbs×1 kg25 mg×1 day3 doses\frac{198 \text{ lbs}}{1} \times \frac{1 \text{ kg}}{2.2 \text{ lbs}} \times \frac{1 \text{ kg}}{25 \text{ mg}} \times \frac{1 \text{ day}}{3 \text{ doses}}
  4. 198 lbs1×1 kg2.2 lbs×25 mg1 kg×3 doses1 day\frac{198 \text{ lbs}}{1} \times \frac{1 \text{ kg}}{2.2 \text{ lbs}} \times \frac{25 \text{ mg}}{1 \text{ kg}} \times \frac{3 \text{ doses}}{1 \text{ day}}
Explanation: Dimensional analysis questions test your ability to set up conversion factors correctly so that units cancel properly to give you the desired final unit. When you see a multi-step dosage calculation, work systematically through each conversion needed. To find the milligrams per single dose, you need four steps: convert pounds to kilograms, calculate total daily milligrams, then divide by the number of doses per day. The key is ensuring units cancel correctly at each step. Starting with 198 lbs, you first convert to kg using 1 kg2.2 lbs\frac{1 \text{ kg}}{2.2 \text{ lbs}} (this cancels "lbs" and gives "kg"). Next, multiply by the dosage rate 25 mg1 kg\frac{25 \text{ mg}}{1 \text{ kg}} (this cancels "kg" and gives total daily "mg"). Finally, divide by 3 doses using 1 day3 doses\frac{1 \text{ day}}{3 \text{ doses}} to get mg per single dose. This is exactly what option A shows. Option B incorrectly uses 2.2 lbs1 kg\frac{2.2 \text{ lbs}}{1 \text{ kg}}, which would convert kg to lbs instead of lbs to kg, and multiplies by 3 doses instead of dividing. Option C flips the dosage fraction to 1 kg25 mg\frac{1 \text{ kg}}{25 \text{ mg}}, which would give an incorrect unit. Option D correctly sets up the first three conversions but multiplies by 3 doses rather than dividing, giving you the total daily dose times 3 instead of the single dose. Remember: in dimensional analysis, fractions must be arranged so unwanted units cancel and your target unit remains. Always check that your setup will mathematically produce the unit you want in the final answer.

Question 4

A physician orders heparin to be infused at 1,100 units/hour. The pharmacy provides a bag of 500 mL solution containing 25,000 units of heparin. Which dimensional analysis setup correctly calculates the required IV pump rate in mL/hour?

  1. 1100 units1 hr×25000 units500 mL\frac{1100 \text{ units}}{1 \text{ hr}} \times \frac{25000 \text{ units}}{500 \text{ mL}}
  2. 1100 units1 hr×500 mL25000 units\frac{1100 \text{ units}}{1 \text{ hr}} \times \frac{500 \text{ mL}}{25000 \text{ units}} (correct answer)
  3. 1 hr1100 units×500 mL25000 units\frac{1 \text{ hr}}{1100 \text{ units}} \times \frac{500 \text{ mL}}{25000 \text{ units}}
  4. 25000 units500 mL×1 hr1100 units\frac{25000 \text{ units}}{500 \text{ mL}} \times \frac{1 \text{ hr}}{1100 \text{ units}}
Explanation: When you encounter IV drip calculations, you're working with dimensional analysis to convert between different units while ensuring unwanted units cancel out. The key is setting up your conversion factors so that units cancel properly and you end up with the desired final unit. You need to convert from the ordered rate (1,100 units/hour) to the pump setting (mL/hour). Start with what you know: the ordered dose of 1,100 units per hour. Then you need a conversion factor that relates the concentration of your IV bag - you have 25,000 units dissolved in 500 mL. Answer B correctly sets up this conversion: 1100 units1 hr×500 mL25000 units\frac{1100 \text{ units}}{1 \text{ hr}} \times \frac{500 \text{ mL}}{25000 \text{ units}}. Notice how "units" appears in both the numerator of the first fraction and denominator of the second fraction, so they cancel out, leaving you with mL/hr. Answer A uses the concentration ratio upside down (25000 units500 mL\frac{25000 \text{ units}}{500 \text{ mL}}), which would give you units²/(hr × mL) - the wrong final unit entirely. Answer C flips the dose rate to 1 hr1100 units\frac{1 \text{ hr}}{1100 \text{ units}}, resulting in hr²/units after cancellation - completely wrong units. Answer D combines both errors from A and C, again producing incorrect units that don't make sense for a pump rate. Study tip: Always check your dimensional analysis by writing out the units and confirming they cancel to give you the unit you want. If your setup doesn't produce mL/hr for pump rates, rearrange your conversion factors.

Question 5

A medication is available in grains (gr). The order is for gr iss. The available tablets are 60 mg. Which setup correctly finds the number of tablets to administer? (Use 1 gr = 60 mg; 'iss' is the Roman numeral for 1.5)

  1. 1.5 gr1×1 gr60 mg×1 tablet60 mg\frac{1.5 \text{ gr}}{1} \times \frac{1 \text{ gr}}{60 \text{ mg}} \times \frac{1 \text{ tablet}}{60 \text{ mg}}
  2. 1.5 gr1×60 mg1 gr×60 mg1 tablet\frac{1.5 \text{ gr}}{1} \times \frac{60 \text{ mg}}{1 \text{ gr}} \times \frac{60 \text{ mg}}{1 \text{ tablet}}
  3. 1.5 gr1×1 tablet60 mg\frac{1.5 \text{ gr}}{1} \times \frac{1 \text{ tablet}}{60 \text{ mg}}
  4. 1.5 gr1×60 mg1 gr×1 tablet60 mg\frac{1.5 \text{ gr}}{1} \times \frac{60 \text{ mg}}{1 \text{ gr}} \times \frac{1 \text{ tablet}}{60 \text{ mg}} (correct answer)
Explanation: When you encounter dosage calculations involving unit conversions, you need to set up a dimensional analysis chain that systematically converts from the ordered dose to the number of tablets. Each conversion factor must be arranged so unwanted units cancel out, leaving only the desired unit. The order is for gr iss (1.5 grains), and you have 60 mg tablets available. You need to convert grains → milligrams → tablets using the conversion factors properly oriented. Option D correctly sets up this conversion: 1.5 gr1×60 mg1 gr×1 tablet60 mg\frac{1.5 \text{ gr}}{1} \times \frac{60 \text{ mg}}{1 \text{ gr}} \times \frac{1 \text{ tablet}}{60 \text{ mg}}. Starting with 1.5 gr, the first conversion factor converts grains to milligrams (gr cancels, leaving mg), then the second converts milligrams to tablets (mg cancels, leaving tablets). This gives you: 1.5 × 60 ÷ 60 = 1.5 tablets. Option A incorrectly inverts the grain-to-milligram conversion factor (1 gr60 mg\frac{1 \text{ gr}}{60 \text{ mg}}), which would give you an impossible unit combination and wrong numerical result. Option B inverts the final conversion factor (60 mg1 tablet\frac{60 \text{ mg}}{1 \text{ tablet}}), so milligrams don't cancel out properly, leaving you with mg²/tablet instead of just tablets. Option C skips the essential grain-to-milligram conversion entirely, attempting to go directly from grains to tablets without the intermediate step. Study tip: Always check that your dimensional analysis setup allows units to cancel properly - each unwanted unit should appear once in a numerator and once in a denominator. Write out the units in your calculation to verify the setup before computing the numbers.

Question 6

A pediatric patient requires an antibiotic at a dose of 30 mg/kg/day. The child weighs 33 lbs and the oral suspension is available at a concentration of 250 mg/5 mL. The physician wants to know the total volume in mL the child will receive over a 24-hour period. Which setup is correct? (1 kg = 2.2 lbs)

  1. 33 lbs1×2.2 lbs1 kg×30 mg1 kg×5 mL250 mg\frac{33 \text{ lbs}}{1} \times \frac{2.2 \text{ lbs}}{1 \text{ kg}} \times \frac{30 \text{ mg}}{1 \text{ kg}} \times \frac{5 \text{ mL}}{250 \text{ mg}}
  2. 33 lbs1×1 kg2.2 lbs×30 mg1 kg×5 mL250 mg\frac{33 \text{ lbs}}{1} \times \frac{1 \text{ kg}}{2.2 \text{ lbs}} \times \frac{30 \text{ mg}}{1 \text{ kg}} \times \frac{5 \text{ mL}}{250 \text{ mg}} (correct answer)
  3. 33 lbs1×1 kg2.2 lbs×1 kg30 mg×5 mL250 mg\frac{33 \text{ lbs}}{1} \times \frac{1 \text{ kg}}{2.2 \text{ lbs}} \times \frac{1 \text{ kg}}{30 \text{ mg}} \times \frac{5 \text{ mL}}{250 \text{ mg}}
  4. 33 lbs1×1 kg2.2 lbs×30 mg1 kg×250 mg5 mL\frac{33 \text{ lbs}}{1} \times \frac{1 \text{ kg}}{2.2 \text{ lbs}} \times \frac{30 \text{ mg}}{1 \text{ kg}} \times \frac{250 \text{ mg}}{5 \text{ mL}}
Explanation: Pediatric dosage calculations require systematic dimensional analysis to convert between units and ensure accurate medication administration. When you encounter these problems, identify what you're given, what you need to find, and set up conversion factors so units cancel properly. You need to find the total volume in mL for a 24-hour period. Starting with the child's weight of 33 lbs, you must: convert pounds to kilograms, calculate the total daily dose in mg, then convert mg to mL using the suspension concentration. The correct setup in option B follows this logical sequence: 33 lbs1×1 kg2.2 lbs×30 mg1 kg×5 mL250 mg\frac{33 \text{ lbs}}{1} \times \frac{1 \text{ kg}}{2.2 \text{ lbs}} \times \frac{30 \text{ mg}}{1 \text{ kg}} \times \frac{5 \text{ mL}}{250 \text{ mg}}. Notice how units cancel: lbs cancels with lbs, kg cancels with kg, and mg cancels with mg, leaving only mL as your final answer. Option A incorrectly uses 2.2 lbs1 kg\frac{2.2 \text{ lbs}}{1 \text{ kg}} instead of 1 kg2.2 lbs\frac{1 \text{ kg}}{2.2 \text{ lbs}}, which would multiply the weight rather than convert it properly to kilograms. Option C has 1 kg30 mg\frac{1 \text{ kg}}{30 \text{ mg}} instead of 30 mg1 kg\frac{30 \text{ mg}}{1 \text{ kg}}, placing the dose conversion upside down. Option D uses 250 mg5 mL\frac{250 \text{ mg}}{5 \text{ mL}} instead of 5 mL250 mg\frac{5 \text{ mL}}{250 \text{ mg}}, inverting the final concentration conversion. Remember: in dimensional analysis, always check that your conversion factors are oriented so unwanted units cancel out, leaving only the units you want in your final answer.

Question 7

A patient is receiving a dopamine infusion at 7 mcg/kg/min. The patient weighs 165 lbs. The solution concentration is 400 mg in 250 mL D5W. Which expression correctly sets up the calculation for the infusion pump rate in mL/hr? (1 kg ≈ 2.2 lbs; 1 mg = 1000 mcg)

  1. 7 mcg1 kg⋅min×165 lbs1×1 kg2.2 lbs×1 mg1000 mcg×250 mL400 mg×60 min1 hr\frac{7 \text{ mcg}}{1 \text{ kg} \cdot \text{min}} \times \frac{165 \text{ lbs}}{1} \times \frac{1 \text{ kg}}{2.2 \text{ lbs}} \times \frac{1 \text{ mg}}{1000 \text{ mcg}} \times \frac{250 \text{ mL}}{400 \text{ mg}} \times \frac{60 \text{ min}}{1 \text{ hr}} (correct answer)
  2. 7 mcg1 kg⋅min×165 lbs1×1 kg2.2 lbs×1000 mcg1 mg×250 mL400 mg×1 hr60 min\frac{7 \text{ mcg}}{1 \text{ kg} \cdot \text{min}} \times \frac{165 \text{ lbs}}{1} \times \frac{1 \text{ kg}}{2.2 \text{ lbs}} \times \frac{1000 \text{ mcg}}{1 \text{ mg}} \times \frac{250 \text{ mL}}{400 \text{ mg}} \times \frac{1 \text{ hr}}{60 \text{ min}}
  3. 7 mcg1 kg⋅min×165 lbs1×2.2 lbs1 kg×1 mg1000 mcg×400 mg250 mL×60 min1 hr\frac{7 \text{ mcg}}{1 \text{ kg} \cdot \text{min}} \times \frac{165 \text{ lbs}}{1} \times \frac{2.2 \text{ lbs}}{1 \text{ kg}} \times \frac{1 \text{ mg}}{1000 \text{ mcg}} \times \frac{400 \text{ mg}}{250 \text{ mL}} \times \frac{60 \text{ min}}{1 \text{ hr}}
  4. 7 mcg1 kg⋅min×1165 lbs×1 kg2.2 lbs×1 mg1000 mcg×250 mL400 mg×60 min1 hr\frac{7 \text{ mcg}}{1 \text{ kg} \cdot \text{min}} \times \frac{1}{165 \text{ lbs}} \times \frac{1 \text{ kg}}{2.2 \text{ lbs}} \times \frac{1 \text{ mg}}{1000 \text{ mcg}} \times \frac{250 \text{ mL}}{400 \text{ mg}} \times \frac{60 \text{ min}}{1 \text{ hr}}
Explanation: IV medication dosage calculations require systematic dimensional analysis to convert between units while maintaining proper direction of conversion factors. When setting up these problems, you must carefully track units to ensure they cancel correctly, leading you from the starting dose to the final pump rate. The correct setup in option A follows logical steps: Start with the prescribed dose (7 mcg/kg/min), multiply by the patient's weight (165 lbs), convert pounds to kilograms using 1 kg2.2 lbs\frac{1 \text{ kg}}{2.2 \text{ lbs}}, convert micrograms to milligrams using 1 mg1000 mcg\frac{1 \text{ mg}}{1000 \text{ mcg}}, convert milligrams to volume using the solution concentration 250 mL400 mg\frac{250 \text{ mL}}{400 \text{ mg}}, and finally convert minutes to hours using 60 min1 hr\frac{60 \text{ min}}{1 \text{ hr}}. Option B contains two critical errors: it uses the inverted conversion 1000 mcg1 mg\frac{1000 \text{ mcg}}{1 \text{ mg}} (which converts mg to mcg instead of mcg to mg) and 1 hr60 min\frac{1 \text{ hr}}{60 \text{ min}} (converting hours to minutes instead of minutes to hours). Option C has multiple errors: 2.2 lbs1 kg\frac{2.2 \text{ lbs}}{1 \text{ kg}} converts kg to lbs (backward), and 400 mg250 mL\frac{400 \text{ mg}}{250 \text{ mL}} converts mL to mg instead of mg to mL. Option D incorrectly places the patient's weight in the denominator as 1165 lbs\frac{1}{165 \text{ lbs}}, which would divide by the weight instead of multiplying. Study tip: Always write out the units in dimensional analysis problems and verify they cancel to give your desired final units. Each conversion factor should move you one step closer to mL/hr.

Question 8

A medical device's pressure gauge reads 120 kilopascals (kPa). A manual specifies the operating pressure in pounds per square inch (psi). Which dimensional analysis setup correctly converts the pressure from kPa to psi? (Use 1 kPa = 0.145 psi)

  1. 120 kPa1×1 psi0.145 kPa\frac{120 \text{ kPa}}{1} \times \frac{1 \text{ psi}}{0.145 \text{ kPa}}
  2. 120 kPa1×0.145 psi1 kPa\frac{120 \text{ kPa}}{1} \times \frac{0.145 \text{ psi}}{1 \text{ kPa}} (correct answer)
  3. 1120 kPa×0.145 psi1 kPa\frac{1}{120 \text{ kPa}} \times \frac{0.145 \text{ psi}}{1 \text{ kPa}}
  4. 120 kPa+0.145 psi120 \text{ kPa} + 0.145 \text{ psi}
Explanation: Dimensional analysis questions test your ability to set up unit conversions correctly by canceling unwanted units and keeping desired ones. When converting between units, you must arrange conversion factors so that units cancel algebraically, leaving only the target unit. To convert 120 kPa to psi, you need to eliminate kPa and retain psi. Starting with your given value (120 kPa), you multiply by a conversion factor where kPa appears in the denominator (to cancel with the kPa in your starting value) and psi appears in the numerator (your desired final unit). Since 1 kPa = 0.145 psi, the correct setup is 120 kPa1×0.145 psi1 kPa\frac{120 \text{ kPa}}{1} \times \frac{0.145 \text{ psi}}{1 \text{ kPa}}. The kPa units cancel, leaving 120 × 0.145 = 17.4 psi. Choice A sets up the conversion factor backwards (1 psi0.145 kPa\frac{1 \text{ psi}}{0.145 \text{ kPa}}), which would give an incorrect numerical relationship and wrong final units. Choice C incorrectly places 120 kPa in the denominator of the first fraction, turning your starting value into a reciprocal rather than preserving it as the quantity being converted. Choice D simply adds the values together, which is meaningless since you cannot add different units—this completely ignores the conversion process. For dimensional analysis success, always write out units in your setup and verify they cancel to give your target unit. The numbers should follow the unit relationships, not the other way around.

Question 9

A protocol requires a saline flush of 2.5 mL to be administered over 30 seconds. Which expression correctly sets up the calculation to find the equivalent flow rate in liters per hour (L/hr)?

  1. 2.5 mL30 sec×1 L1000 mL×3600 sec1 hr\frac{2.5 \text{ mL}}{30 \text{ sec}} \times \frac{1 \text{ L}}{1000 \text{ mL}} \times \frac{3600 \text{ sec}}{1 \text{ hr}} (correct answer)
  2. 2.5 mL30 sec×1000 mL1 L×1 hr3600 sec\frac{2.5 \text{ mL}}{30 \text{ sec}} \times \frac{1000 \text{ mL}}{1 \text{ L}} \times \frac{1 \text{ hr}}{3600 \text{ sec}}
  3. 30 sec2.5 mL×1 L1000 mL×3600 sec1 hr\frac{30 \text{ sec}}{2.5 \text{ mL}} \times \frac{1 \text{ L}}{1000 \text{ mL}} \times \frac{3600 \text{ sec}}{1 \text{ hr}}
  4. 2.5 mL30 sec×1 L1000 mL×60 sec1 min\frac{2.5 \text{ mL}}{30 \text{ sec}} \times \frac{1 \text{ L}}{1000 \text{ mL}} \times \frac{60 \text{ sec}}{1 \text{ min}}
Explanation: Unit conversion problems in healthcare require careful attention to setting up dimensional analysis correctly. When converting flow rates, you need to systematically convert each unit while maintaining the proper mathematical relationships. Starting with the given rate of 2.5 mL over 30 seconds, you first express this as 2.5 mL30 sec\frac{2.5 \text{ mL}}{30 \text{ sec}}. To convert to L/hr, you need two additional conversion factors: milliliters to liters, and seconds to hours. Since 1 L = 1000 mL, you multiply by 1 L1000 mL\frac{1 \text{ L}}{1000 \text{ mL}} to convert the numerator. Since 1 hour = 3600 seconds, you multiply by 3600 sec1 hr\frac{3600 \text{ sec}}{1 \text{ hr}} to convert the denominator. This gives you option A, which correctly sets up the calculation. Option B uses the wrong conversion factors - it has 1000 mL1 L\frac{1000 \text{ mL}}{1 \text{ L}} (which would convert liters to milliliters, not the reverse) and 1 hr3600 sec\frac{1 \text{ hr}}{3600 \text{ sec}} (which would convert hours to seconds). Option C flips the initial rate to 30 sec2.5 mL\frac{30 \text{ sec}}{2.5 \text{ mL}}, which would give you the reciprocal of flow rate. Option D stops the time conversion at minutes rather than completing the conversion to hours, using only 60 sec1 min\frac{60 \text{ sec}}{1 \text{ min}}. Remember: in dimensional analysis, units must cancel properly. Write out all units in your conversion factors and verify that unwanted units cancel, leaving only your desired final units.

Question 10

A hospital needs to order 2 cubic yards (yd³) of mulch for landscaping. The supplier sells mulch by the liter (L). Which expression correctly sets up the conversion from cubic yards to liters? (Use 1 yd = 3 ft, 1 ft = 12 in, 1 in = 2.54 cm, 1 L = 1000 cm³)

  1. 2 yd31×3 ft1 yd×12 in1 ft×2.54 cm1 in×1 L1000 cm3\frac{2 \text{ yd}^3}{1} \times \frac{3 \text{ ft}}{1 \text{ yd}} \times \frac{12 \text{ in}}{1 \text{ ft}} \times \frac{2.54 \text{ cm}}{1 \text{ in}} \times \frac{1 \text{ L}}{1000 \text{ cm}^3}
  2. 2 yd31×(3 ft1 yd)3×(12 in1 ft)3×(2.54 cm1 in)3×1 L1000 cm3\frac{2 \text{ yd}^3}{1} \times (\frac{3 \text{ ft}}{1 \text{ yd}})^3 \times (\frac{12 \text{ in}}{1 \text{ ft}})^3 \times (\frac{2.54 \text{ cm}}{1 \text{ in}})^3 \times \frac{1 \text{ L}}{1000 \text{ cm}^3} (correct answer)
  3. 2 yd31×1 yd3(3 ft)3×1 ft3(12 in)3×1 in3(2.54 cm)3×1000 cm31 L\frac{2 \text{ yd}^3}{1} \times \frac{1 \text{ yd}^3}{(3 \text{ ft})^3} \times \frac{1 \text{ ft}^3}{(12 \text{ in})^3} \times \frac{1 \text{ in}^3}{(2.54 \text{ cm})^3} \times \frac{1000 \text{ cm}^3}{1 \text{ L}}
  4. 2 yd31×3 ft31 yd3×12 in31 ft3×2.54 cm31 in3×1 L1000 cm3\frac{2 \text{ yd}^3}{1} \times \frac{3 \text{ ft}^3}{1 \text{ yd}^3} \times \frac{12 \text{ in}^3}{1 \text{ ft}^3} \times \frac{2.54 \text{ cm}^3}{1 \text{ in}^3} \times \frac{1 \text{ L}}{1000 \text{ cm}^3}
Explanation: When converting between cubic units, you must account for the three-dimensional nature of volume. Each linear conversion factor must be cubed because volume involves length × width × height. To convert 2 yd³ to liters, you need to systematically convert through the given relationships. Since you're working with cubic measurements, each conversion factor must be raised to the third power. Starting with 2 yd³, you multiply by (3 ft1 yd)3(\frac{3 \text{ ft}}{1 \text{ yd}})^3 because there are 3 feet per yard in each dimension. This gives you cubic feet. Then multiply by (12 in1 ft)3(\frac{12 \text{ in}}{1 \text{ ft}})^3 to get cubic inches, followed by (2.54 cm1 in)3(\frac{2.54 \text{ cm}}{1 \text{ in}})^3 to get cubic centimeters. Finally, convert to liters using 1 L1000 cm3\frac{1 \text{ L}}{1000 \text{ cm}^3}. Option A incorrectly treats this as a linear conversion, using each factor only once instead of cubing it. This would give you an answer in mixed units, not liters. Option C uses the reciprocals of the correct conversion factors, which would convert from liters back to cubic yards instead. Option D appears similar to B but uses incorrect notation - it shows 3 ft31 yd3\frac{3 \text{ ft}^3}{1 \text{ yd}^3} instead of the cubed conversion factor, which represents a different mathematical relationship. Remember: whenever converting cubic units, cube each linear conversion factor. The exponent applies to the entire fraction, affecting both the numerical value and the units.

Question 11

A patient's height is measured as 5 feet, 8 inches. To calculate BMI, the height must be in meters. Which expression correctly sets up the conversion of the total height in inches to meters? (Use 1 inch = 2.54 cm and 1 m = 100 cm)

  1. (5 ft×12 in1 ft+8 in)×1 m2.54 cm(5 \text{ ft} \times \frac{12 \text{ in}}{1 \text{ ft}} + 8 \text{ in}) \times \frac{1 \text{ m}}{2.54 \text{ cm}}
  2. (5 ft+8 in)×12 in1 ft×2.54 cm1 in×1 m100 cm(5 \text{ ft} + 8 \text{ in}) \times \frac{12 \text{ in}}{1 \text{ ft}} \times \frac{2.54 \text{ cm}}{1 \text{ in}} \times \frac{1 \text{ m}}{100 \text{ cm}}
  3. (5 ft×12 in1 ft+8 in)×2.54 cm1 in×100 cm1 m(5 \text{ ft} \times \frac{12 \text{ in}}{1 \text{ ft}} + 8 \text{ in}) \times \frac{2.54 \text{ cm}}{1 \text{ in}} \times \frac{100 \text{ cm}}{1 \text{ m}}
  4. (5 ft×12 in1 ft+8 in)×2.54 cm1 in×1 m100 cm(5 \text{ ft} \times \frac{12 \text{ in}}{1 \text{ ft}} + 8 \text{ in}) \times \frac{2.54 \text{ cm}}{1 \text{ in}} \times \frac{1 \text{ m}}{100 \text{ cm}} (correct answer)
Explanation: Unit conversion problems in healthcare require systematic dimensional analysis to ensure accurate calculations for clinical assessments like BMI. The key is setting up conversion factors so that unwanted units cancel out, leaving only the desired unit. To convert 5 feet, 8 inches to meters, you need a three-step process. First, convert the mixed units (feet and inches) to total inches: 5 ft×12 in1 ft+8 in=68 inches5 \text{ ft} \times \frac{12 \text{ in}}{1 \text{ ft}} + 8 \text{ in} = 68 \text{ inches}. Then convert inches to centimeters using the given conversion factor: 2.54 cm1 in\frac{2.54 \text{ cm}}{1 \text{ in}}. Finally, convert centimeters to meters: 1 m100 cm\frac{1 \text{ m}}{100 \text{ cm}}. Choice D correctly combines all these steps: (5 ft×12 in1 ft+8 in)×2.54 cm1 in×1 m100 cm(5 \text{ ft} \times \frac{12 \text{ in}}{1 \text{ ft}} + 8 \text{ in}) \times \frac{2.54 \text{ cm}}{1 \text{ in}} \times \frac{1 \text{ m}}{100 \text{ cm}}. Notice how the units cancel properly: inches cancel between the first and second terms, centimeters cancel between the second and third terms, leaving meters. Choice A skips the centimeter step and uses an incorrect conversion factor. Choice B incorrectly adds feet and inches before converting feet to inches, which is mathematically invalid since you can't add different units directly. Choice C has the conversion factor for the final step inverted (100 cm1 m\frac{100 \text{ cm}}{1 \text{ m}}), which would multiply by 100 instead of dividing, giving an answer 10,000 times too large. For HESI dimensional analysis problems, always write out each conversion step separately and verify that units cancel properly before calculating the final answer.

Question 12

A patient's lab result for creatinine is 106 micromol/L (µmol/L). A different hospital system records this value in mg/dL. Which setup correctly converts the value? (Molecular weight of creatinine ≈ 113 g/mol; 1 mol = 1,000,000 µmol; 1 g = 1000 mg; 1 L = 10 dL)

  1. 106 µmol1 L×1 mol106 µmol×113 g1 mol×1000 mg1 g×1 L10 dL\frac{106 \text{ µmol}}{1 \text{ L}} \times \frac{1 \text{ mol}}{10^6 \text{ µmol}} \times \frac{113 \text{ g}}{1 \text{ mol}} \times \frac{1000 \text{ mg}}{1 \text{ g}} \times \frac{1 \text{ L}}{10 \text{ dL}} (correct answer)
  2. 106 µmol1 L×106 µmol1 mol×113 g1 mol×1 g1000 mg×10 dL1 L\frac{106 \text{ µmol}}{1 \text{ L}} \times \frac{10^6 \text{ µmol}}{1 \text{ mol}} \times \frac{113 \text{ g}}{1 \text{ mol}} \times \frac{1 \text{ g}}{1000 \text{ mg}} \times \frac{10 \text{ dL}}{1 \text{ L}}
  3. 106 µmol1 L×1 mol106 µmol×1 mol113 g×1000 mg1 g×1 L10 dL\frac{106 \text{ µmol}}{1 \text{ L}} \times \frac{1 \text{ mol}}{10^6 \text{ µmol}} \times \frac{1 \text{ mol}}{113 \text{ g}} \times \frac{1000 \text{ mg}}{1 \text{ g}} \times \frac{1 \text{ L}}{10 \text{ dL}}
  4. 106 µmol1 L×1 mol106 µmol×113 g1 mol×1 g1000 mg×1 L10 dL\frac{106 \text{ µmol}}{1 \text{ L}} \times \frac{1 \text{ mol}}{10^6 \text{ µmol}} \times \frac{113 \text{ g}}{1 \text{ mol}} \times \frac{1 \text{ g}}{1000 \text{ mg}} \times \frac{1 \text{ L}}{10 \text{ dL}}
Explanation: Unit conversion problems in healthcare require systematic dimensional analysis to ensure patient safety. When converting between different measurement systems, you must carefully track each conversion factor to cancel units properly and arrive at the correct final units. To convert 106 µmol/L to mg/dL, you need to transform both the numerator (µmol to mg) and denominator (L to dL). Starting with 106 µmol1 L\frac{106 \text{ µmol}}{1 \text{ L}}, the correct sequence is: convert µmol to mol, then mol to grams using molecular weight, then grams to milligrams, and finally liters to deciliters. Choice A correctly executes this conversion: 1 mol106 µmol\frac{1 \text{ mol}}{10^6 \text{ µmol}} converts micromoles to moles, 113 g1 mol\frac{113 \text{ g}}{1 \text{ mol}} uses the molecular weight properly, 1000 mg1 g\frac{1000 \text{ mg}}{1 \text{ g}} converts to milligrams, and 1 L10 dL\frac{1 \text{ L}}{10 \text{ dL}} converts the volume units correctly. Choice B inverts the µmol to mol conversion (106 µmol1 mol\frac{10^6 \text{ µmol}}{1 \text{ mol}}) and the mg conversion (1 g1000 mg\frac{1 \text{ g}}{1000 \text{ mg}}), which would multiply instead of converting properly. Choice C incorrectly inverts the molecular weight factor (1 mol113 g\frac{1 \text{ mol}}{113 \text{ g}}), treating it as if molecular weight converts grams to moles rather than moles to grams. Choice D inverts the milligram conversion, which would convert mg to grams instead of grams to mg. Remember: in dimensional analysis, write conversion factors so unwanted units cancel out. Always double-check that your setup will yield the target units before calculating.

Question 13

A powdered medication must be reconstituted. The vial contains 1 gram of powder. Instructions state to add 9.6 mL of sterile water to yield a solution with a concentration of 100 mg/mL. A dose of 250 mg is ordered. Which expression correctly sets up the calculation for the volume to administer in mL?

  1. 250 mg1×1 mL100 mg\frac{250 \text{ mg}}{1} \times \frac{1 \text{ mL}}{100 \text{ mg}} (correct answer)
  2. 250 mg1×100 mg1 mL\frac{250 \text{ mg}}{1} \times \frac{100 \text{ mg}}{1 \text{ mL}}
  3. 1 g1×9.6 mL1×1 dose250 mg\frac{1 \text{ g}}{1} \times \frac{9.6 \text{ mL}}{1} \times \frac{1 \text{ dose}}{250 \text{ mg}}
  4. 250 mg1×9.6 mL1000 mg\frac{250 \text{ mg}}{1} \times \frac{9.6 \text{ mL}}{1000 \text{ mg}}
Explanation: When you encounter medication dosage calculations involving reconstituted powders, focus on the final concentration after mixing, not the original powder amount or the diluent volume used during reconstitution. The key information here is that after reconstitution, you have a solution with a concentration of 100 mg/mL, and you need to deliver 250 mg. This is a straightforward dose-to-volume conversion using dimensional analysis. Answer A correctly sets up the calculation: 250 mg1×1 mL100 mg\frac{250 \text{ mg}}{1} \times \frac{1 \text{ mL}}{100 \text{ mg}}. The milligrams cancel out, leaving you with the volume in mL. This gives you 2.5 mL, which is the correct volume to administer. Answer B has the concentration fraction inverted: 100 mg1 mL\frac{100 \text{ mg}}{1 \text{ mL}} instead of 1 mL100 mg\frac{1 \text{ mL}}{100 \text{ mg}}. This setup would give you mg²/mL instead of mL, making it dimensionally incorrect. Answer C unnecessarily incorporates the original powder mass (1 g) and reconstitution volume (9.6 mL), which are irrelevant once you know the final concentration. This complex setup doesn't lead to the correct volume calculation. Answer D attempts to use the total reconstitution volume (9.6 mL) with the total drug amount (1000 mg), but this approach ignores that you only need a portion of the total solution to deliver the ordered dose. Study tip: For reconstituted medications, always use the final concentration (mg/mL) given in the problem. Ignore the reconstitution details once you have the final concentration—they're distractors designed to complicate a simple dose calculation.

Question 14

A patient's temperature increases by 2.5° Celsius. A nurse wants to document this change in degrees Fahrenheit. Which setup correctly calculates the magnitude of the temperature change in Fahrenheit? (The conversion factor for temperature change is 9°F / 5°C).

  1. (2.5×95)+32(2.5 \times \frac{9}{5}) + 32
  2. (2.5+32)×95(2.5 + 32) \times \frac{9}{5}
  3. 2.5×592.5 \times \frac{5}{9}
  4. 2.5×952.5 \times \frac{9}{5} (correct answer)
Explanation: When you encounter temperature conversion problems in nursing, it's crucial to distinguish between converting absolute temperatures versus temperature changes. For absolute temperature conversions, you use the full formula with the +32 offset, but for temperature changes (increases or decreases), you only need the conversion factor. The correct approach is answer D: 2.5×952.5 \times \frac{9}{5}. Since you're converting a temperature change of 2.5°C to Fahrenheit, you simply multiply by the conversion factor 9°F/5°C. This gives you 2.5×95=4.5°F2.5 \times \frac{9}{5} = 4.5°F. The +32 adjustment is not needed because you're measuring the magnitude of change, not converting between absolute temperature scales. Answer A incorrectly adds 32 to the calculation: (2.5×95)+32(2.5 \times \frac{9}{5}) + 32. This would give 36.5°F, which represents an absolute temperature, not a temperature change. Answer B makes the same error but applies the operations in the wrong order: (2.5+32)×95(2.5 + 32) \times \frac{9}{5}. This completely distorts the conversion by adding 32 to the Celsius change first. Answer C uses the wrong conversion factor: 2.5×592.5 \times \frac{5}{9}. This factor (5°C/9°F) would convert Fahrenheit changes to Celsius, not the reverse. Remember: when converting temperature changes, ignore the +32 offset entirely. Use only the 9/5 factor for Celsius to Fahrenheit changes, or 5/9 for Fahrenheit to Celsius changes. The +32 is only for converting absolute temperatures between scales.

Question 15

A medication's half-life is 8 hours. A patient is given a dose of 500 mg. Which setup correctly calculates how many days it will take for the amount of medication in the body to be reduced to 62.5 mg? (Note: 62.5 mg is 1/8th of 500 mg, which represents 3 half-lives).

  1. 3 half-lives1×1 half-life8 hours×24 hours1 day\frac{3 \text{ half-lives}}{1} \times \frac{1 \text{ half-life}}{8 \text{ hours}} \times \frac{24 \text{ hours}}{1 \text{ day}}
  2. 500 mg62.5 mg×8 hours1 half-life×1 day24 hours\frac{500 \text{ mg}}{62.5 \text{ mg}} \times \frac{8 \text{ hours}}{1 \text{ half-life}} \times \frac{1 \text{ day}}{24 \text{ hours}}
  3. 3 half-lives1×8 hours1 half-life×1 day24 hours\frac{3 \text{ half-lives}}{1} \times \frac{8 \text{ hours}}{1 \text{ half-life}} \times \frac{1 \text{ day}}{24 \text{ hours}} (correct answer)
  4. 62.5 mg×8 hours1 half-life×1 day24 hours62.5 \text{ mg} \times \frac{8 \text{ hours}}{1 \text{ half-life}} \times \frac{1 \text{ day}}{24 \text{ hours}}
Explanation: When you encounter medication half-life problems, you're working with pharmacokinetics—how drugs are processed and eliminated from the body. The key is setting up dimensional analysis to convert between different units while tracking the biological process. Half-life represents the time needed for a drug concentration to decrease by 50%. To go from 500 mg to 62.5 mg, you need to determine how many half-lives this represents: 500 → 250 → 125 → 62.5 mg. That's 3 half-lives, as the question confirms. The correct setup is C: 3 half-lives1×8 hours1 half-life×1 day24 hours\frac{3 \text{ half-lives}}{1} \times \frac{8 \text{ hours}}{1 \text{ half-life}} \times \frac{1 \text{ day}}{24 \text{ hours}}. This multiplies the number of half-lives needed (3) by the duration of each half-life (8 hours), then converts to days. The units cancel properly: half-lives cancel, hours cancel, leaving days. Choice A incorrectly places "8 hours" in the denominator of the second fraction, which would divide by 8 instead of multiply, giving an impossibly small result. Choice B starts with the mg ratio (500/62.5 = 8), but this represents the fold-reduction, not the number of half-lives—it's the wrong starting point. Choice D begins with 62.5 mg, which has no logical connection to calculating time duration. For HESI pharmacology questions, always identify what you're solving for, then work systematically through dimensional analysis. Make sure your units cancel correctly—if you're calculating time, your final units should be time-related, not concentration-related.

Question 16

During a fluid balance check, a nurse notes a patient's intake as one 8-oz cup of coffee and 0.5 liters of water. The patient's urine output is 650 mL. To calculate the net fluid balance in mL, which setup is correct? (1 oz ≈ 30 mL; 1 L = 1000 mL)

  1. (8 oz×30 mL1 oz)+(0.5 L×1 L1000 mL)−650 mL(8 \text{ oz} \times \frac{30 \text{ mL}}{1 \text{ oz}}) + (0.5 \text{ L} \times \frac{1 \text{ L}}{1000 \text{ mL}}) - 650 \text{ mL}
  2. (8 oz×30 mL1 oz)+(0.5 L×1000 mL1 L)−650 mL(8 \text{ oz} \times \frac{30 \text{ mL}}{1 \text{ oz}}) + (0.5 \text{ L} \times \frac{1000 \text{ mL}}{1 \text{ L}}) - 650 \text{ mL} (correct answer)
  3. (8 oz×1 oz30 mL)+(0.5 L×1000 mL1 L)+650 mL(8 \text{ oz} \times \frac{1 \text{ oz}}{30 \text{ mL}}) + (0.5 \text{ L} \times \frac{1000 \text{ mL}}{1 \text{ L}}) + 650 \text{ mL}
  4. (8 oz×30 mL1 oz)−(0.5 L×1000 mL1 L)−650 mL(8 \text{ oz} \times \frac{30 \text{ mL}}{1 \text{ oz}}) - (0.5 \text{ L} \times \frac{1000 \text{ mL}}{1 \text{ L}}) - 650 \text{ mL}
Explanation: Fluid balance calculations are fundamental in nursing practice, requiring you to convert all measurements to the same unit (typically mL) before calculating net balance using the formula: Total Intake - Total Output. Let's work through this step-by-step. First, convert the coffee: 8 oz needs to be multiplied by the conversion factor 30 mL1 oz\frac{30 \text{ mL}}{1 \text{ oz}} to get 240 mL. Next, convert the water: 0.5 L needs to be multiplied by 1000 mL1 L\frac{1000 \text{ mL}}{1 \text{ L}} to get 500 mL. Total intake is 240 + 500 = 740 mL. Finally, subtract the output: 740 mL - 650 mL = 90 mL net positive fluid balance. Option B correctly shows this setup. Option A incorrectly uses 1 L1000 mL\frac{1 \text{ L}}{1000 \text{ mL}} for the liter conversion, which would give you 0.0005 instead of 500 mL—the conversion factor is upside down. Option C makes two errors: it uses the wrong conversion factor for ounces (1 oz30 mL\frac{1 \text{ oz}}{30 \text{ mL}}) and adds the output instead of subtracting it, completely misunderstanding the fluid balance concept. Option D subtracts the water intake from the coffee intake, then subtracts output—this doesn't make physiological sense since all intake should be added together. Remember: always check that your conversion factors will cancel units properly (oz cancels with oz, L cancels with L), and fluid balance always follows the pattern of total intake minus total output. Watch for upside-down conversion factors—a common HESI trap.

Question 17

A patient consumes a meal containing 40 grams of carbohydrates. The nurse needs to calculate the energy provided in kilojoules (kJ). Which setup is correct? (Use 1 gram of carbohydrate ≈ 4 kcal; 1 kcal ≈ 4.184 kJ)

  1. 40 g1×1 g4 kcal×1 kcal4.184 kJ\frac{40 \text{ g}}{1} \times \frac{1 \text{ g}}{4 \text{ kcal}} \times \frac{1 \text{ kcal}}{4.184 \text{ kJ}}
  2. 40 g1×4 kcal1 g×1 kcal4.184 kJ\frac{40 \text{ g}}{1} \times \frac{4 \text{ kcal}}{1 \text{ g}} \times \frac{1 \text{ kcal}}{4.184 \text{ kJ}}
  3. 40 g1×4 kcal1 g×4.184 kJ1 kcal\frac{40 \text{ g}}{1} \times \frac{4 \text{ kcal}}{1 \text{ g}} \times \frac{4.184 \text{ kJ}}{1 \text{ kcal}} (correct answer)
  4. 40 g1×1 g4 kcal×4.184 kJ1 kcal\frac{40 \text{ g}}{1} \times \frac{1 \text{ g}}{4 \text{ kcal}} \times \frac{4.184 \text{ kJ}}{1 \text{ kcal}}
Explanation: When you encounter energy conversion problems in healthcare, you need to carefully set up dimensional analysis to convert step-by-step from your starting unit to your target unit. The key is ensuring each conversion factor is positioned so unwanted units cancel out. You're converting 40 grams of carbohydrates to kilojoules using two given conversion factors: 1 gram carbohydrate = 4 kcal, and 1 kcal = 4.184 kJ. Start with 40 grams, then multiply by conversion factors that will eliminate grams and kcal, leaving only kJ. The correct setup is 40 g1×4 kcal1 g×4.184 kJ1 kcal\frac{40 \text{ g}}{1} \times \frac{4 \text{ kcal}}{1 \text{ g}} \times \frac{4.184 \text{ kJ}}{1 \text{ kcal}}. This works because grams cancel (40 g × g⁻¹), then kcal cancel (kcal × kcal⁻¹), leaving only kJ. The calculation gives: 40 × 4 × 4.184 = 669.44 kJ. Option A incorrectly places grams in the denominator of the first conversion (1 g/4 kcal), which would give you kcal/g instead of eliminating grams. Option B has the same gram problem, plus it inverts the final conversion factor (1 kcal/4.184 kJ), which would cancel out kcal and leave you with an inverted kJ unit. Option D correctly handles the final conversion but makes the same initial error as option A with the gram placement. Remember: in dimensional analysis, the unit you want to eliminate must appear in both numerator and denominator positions so they cancel. Always trace through your setup to verify that unwanted units disappear and only your target unit remains.

Question 18

A wound has a surface area of 15 square centimeters (cm²). Which dimensional analysis setup correctly converts this area to square inches (in²)? (Use 1 inch = 2.54 cm)

  1. 15 cm21×1 in2.54 cm\frac{15 \text{ cm}^2}{1} \times \frac{1 \text{ in}}{2.54 \text{ cm}}
  2. 15 cm21×2.54 cm1 in×2.54 cm1 in\frac{15 \text{ cm}^2}{1} \times \frac{2.54 \text{ cm}}{1 \text{ in}} \times \frac{2.54 \text{ cm}}{1 \text{ in}}
  3. 15 cm21×1 in2.54 cm×1 in2.54 cm\frac{15 \text{ cm}^2}{1} \times \frac{1 \text{ in}}{2.54 \text{ cm}} \times \frac{1 \text{ in}}{2.54 \text{ cm}} (correct answer)
  4. 15 cm21×(1)2 in2(2.54) cm2\frac{15 \text{ cm}^2}{1} \times \frac{(1)^2 \text{ in}^2}{(2.54) \text{ cm}^2}
Explanation: When converting area measurements between units, you're working with squared units, which requires special attention to dimensional analysis. The key is ensuring that both dimensions of the area get converted properly. To convert 15 cm² to square inches, you need to convert each linear dimension from centimeters to inches. Since area involves two dimensions (length × width), you must apply the conversion factor twice - once for each dimension. Choice C is correct: 15 cm21×1 in2.54 cm×1 in2.54 cm\frac{15 \text{ cm}^2}{1} \times \frac{1 \text{ in}}{2.54 \text{ cm}} \times \frac{1 \text{ in}}{2.54 \text{ cm}}. This setup properly converts both dimensions by using the conversion factor (1 in = 2.54 cm) twice. The cm² units cancel out completely, leaving you with in². Choice A fails because it only converts one dimension - you'd end up with mixed units (cm·in) rather than in². Choice B uses the conversion factor upside down, which would convert inches to centimeters instead of centimeters to inches, giving you an incorrect result. Choice D appears mathematically equivalent to choice C, but it's written incorrectly - the exponent notation (1)2(1)^2 and the placement suggest a misunderstanding of how to handle squared conversions. Remember: when converting area units, always square your conversion factor or apply it twice (once for each dimension). For volume conversions, you'd apply it three times. This pattern appears frequently on healthcare exams when dealing with wound measurements, medication calculations, and dosing based on body surface area.

Question 19

A car is traveling at 65 miles per hour. A nurse calculates that this is the same speed at which an automated delivery cart travels down a long hospital corridor, but needs the speed in feet per second to check the cart's calibration. Which expression correctly sets up the conversion? (1 mile = 5280 feet; 1 hour = 3600 seconds)

  1. 65 miles1 hr×5280 ft1 mile×3600 sec1 hr\frac{65 \text{ miles}}{1 \text{ hr}} \times \frac{5280 \text{ ft}}{1 \text{ mile}} \times \frac{3600 \text{ sec}}{1 \text{ hr}}
  2. 65 miles1 hr×1 mile5280 ft×1 hr3600 sec\frac{65 \text{ miles}}{1 \text{ hr}} \times \frac{1 \text{ mile}}{5280 \text{ ft}} \times \frac{1 \text{ hr}}{3600 \text{ sec}}
  3. 65 miles1 hr×5280 ft1 mile×1 hr3600 sec\frac{65 \text{ miles}}{1 \text{ hr}} \times \frac{5280 \text{ ft}}{1 \text{ mile}} \times \frac{1 \text{ hr}}{3600 \text{ sec}} (correct answer)
  4. 65 miles1 hr×5280 ft1 mile×1 min60 sec\frac{65 \text{ miles}}{1 \text{ hr}} \times \frac{5280 \text{ ft}}{1 \text{ mile}} \times \frac{1 \text{ min}}{60 \text{ sec}}
Explanation: Unit conversion problems require careful attention to how fractions are arranged so that unwanted units cancel out, leaving only the desired units. When converting from miles per hour to feet per second, you need to convert miles to feet (multiply by a larger number) and convert hours to seconds (divide by a larger number since there are many seconds in one hour). Starting with 65 miles1 hr\frac{65 \text{ miles}}{1 \text{ hr}}, you need conversion factors arranged so units cancel properly. To convert miles to feet, multiply by 5280 ft1 mile\frac{5280 \text{ ft}}{1 \text{ mile}} - this allows "miles" to cancel. To convert from hours to seconds, multiply by 1 hr3600 sec\frac{1 \text{ hr}}{3600 \text{ sec}} - this allows "hr" to cancel, leaving feet per second. Answer C correctly sets up: 65 miles1 hr×5280 ft1 mile×1 hr3600 sec\frac{65 \text{ miles}}{1 \text{ hr}} \times \frac{5280 \text{ ft}}{1 \text{ mile}} \times \frac{1 \text{ hr}}{3600 \text{ sec}}. When you multiply through, miles and hours cancel, leaving 65×5280 ft3600 sec\frac{65 \times 5280 \text{ ft}}{3600 \text{ sec}}. Answer A incorrectly uses 3600 sec1 hr\frac{3600 \text{ sec}}{1 \text{ hr}}, which would give units of feet-seconds per hour instead of feet per second. Answer B flips both conversion factors wrong - 1 mile5280 ft\frac{1 \text{ mile}}{5280 \text{ ft}} would convert feet to miles (backward), and 1 hr3600 sec\frac{1 \text{ hr}}{3600 \text{ sec}} would convert seconds to hours (also backward). Answer D uses minutes instead of hours in the final conversion factor, which won't cancel with the hours in the original rate. Remember: in unit conversion, arrange fractions so unwanted units appear in both numerator and denominator, allowing them to cancel completely.

Question 20

A medication has a concentration of 250 mg per 5 mL and is to be administered at 12 mg/kg every 8 hours. For a 45 kg patient, which dimensional analysis expression correctly determines the volume needed per dose?

  1. 45 kg×12 mg1 kg⋅dose×250 mg5 mL45 \text{ kg} \times \frac{12 \text{ mg}}{1 \text{ kg} \cdot \text{dose}} \times \frac{250 \text{ mg}}{5 \text{ mL}}
  2. 45 kg×12 mg1 kg⋅dose×1 dose8 hr×5 mL250 mg45 \text{ kg} \times \frac{12 \text{ mg}}{1 \text{ kg} \cdot \text{dose}} \times \frac{1 \text{ dose}}{8 \text{ hr}} \times \frac{5 \text{ mL}}{250 \text{ mg}}
  3. 45 kg×12 mg1 kg⋅dose×5 mL250 mg45 \text{ kg} \times \frac{12 \text{ mg}}{1 \text{ kg} \cdot \text{dose}} \times \frac{5 \text{ mL}}{250 \text{ mg}} (correct answer)
  4. 45 kg×12 mg1 kg×8 hr1 dose×5 mL250 mg45 \text{ kg} \times \frac{12 \text{ mg}}{1 \text{ kg}} \times \frac{8 \text{ hr}}{1 \text{ dose}} \times \frac{5 \text{ mL}}{250 \text{ mg}}
Explanation: Dimensional analysis questions test your ability to set up conversion factors so that unwanted units cancel out, leaving only the desired unit. When you see dosage calculations, focus on what the question is asking for and work systematically through the conversions. The question asks for the volume per dose, so you need to convert from the patient's weight to the final volume in mL. Start with 45 kg, then convert to mg needed per dose using the dosing rate of 12 mg/kg/dose. Finally, convert from mg to mL using the concentration. Choice C correctly sets up this pathway: 45 kg×12 mg1 kg⋅dose×5 mL250 mg45 \text{ kg} \times \frac{12 \text{ mg}}{1 \text{ kg} \cdot \text{dose}} \times \frac{5 \text{ mL}}{250 \text{ mg}}. The units cancel properly (kg cancels with kg, mg cancels with mg), leaving mL/dose as the final unit. Choice A uses the concentration fraction upside down (250 mg5 mL\frac{250 \text{ mg}}{5 \text{ mL}}), which would give you an answer in mg²/(kg·mL·dose) instead of mL/dose. Choice B unnecessarily includes the 8-hour frequency, but since the question asks for volume per dose (not per hour), this extra conversion factor makes the setup incorrect. Choice D also incorrectly includes the 8-hour timing and puts it in the wrong position, which would multiply rather than divide by the frequency. For HESI dosage calculations, always identify your starting point, your desired endpoint, and set up conversion factors so units cancel step by step. Ignore information that doesn't relate to what the question is specifically asking for.