Health Education Systems Inc (HESI) A2 Exam Quiz: One Variable Equations
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One Variable EquationsQuestion 1 of 20

Solve for k: 8−2(k−3)=k+58 - 2(k - 3) = k + 5

-3
1
3
5
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Health Education Systems Inc (HESI) A2 Exam Quiz

Health Education Systems Inc (HESI) A2 Exam Quiz: One Variable Equations

Practice One Variable Equations 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 One Variable Equations, 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

Solve for k: 8−2(k−3)=k+58 - 2(k - 3) = k + 5

  1. -3
  2. 1
  3. 3 (correct answer)
  4. 5
Explanation: When you encounter linear equations with parentheses and variables on both sides, your goal is to isolate the variable through systematic algebraic steps. Start by distributing the -2 to everything inside the parentheses: 8−2(k−3)=k+58 - 2(k - 3) = k + 5 becomes 8−2k+6=k+58 - 2k + 6 = k + 5. Combine like terms on the left side: 14−2k=k+514 - 2k = k + 5. Next, collect all variable terms on one side by adding 2k to both sides: 14=3k+514 = 3k + 5. Subtract 5 from both sides: 9=3k9 = 3k. Finally, divide by 3: k=3k = 3. You can verify this by substituting back into the original equation: 8−2(3−3)=3+58 - 2(3 - 3) = 3 + 5, which simplifies to 8=88 = 8 ✓ Looking at the wrong answers: (A) -3 likely comes from sign errors when distributing the negative or combining terms. (B) 1 might result from incorrectly handling the coefficient when isolating k, perhaps dividing 9 by something other than 3. (D) 5 could come from forgetting to subtract 5 before dividing, leaving you with the constant term instead of the variable's value. The key strategy for these problems is to work methodically: distribute first, combine like terms, collect variables on one side, then isolate. Always check your answer by substituting back into the original equation—this catches arithmetic mistakes that are easy to make under test pressure.

Question 2

Solve for x: x+53=2x−10\frac{x+5}{3} = 2x - 10

  1. -7
  2. 3
  3. 7 (correct answer)
  4. 9
Explanation: When you encounter an equation with a fraction equal to an algebraic expression, your goal is to isolate the variable by eliminating the fraction and combining like terms systematically. To solve x+53=2x−10\frac{x+5}{3} = 2x - 10, start by eliminating the fraction. Multiply both sides by 3: x+5=3(2x−10)x + 5 = 3(2x - 10). Distribute the 3 on the right side: x+5=6x−30x + 5 = 6x - 30. Now collect all terms with x on one side and constants on the other. Subtract x from both sides: 5=5x−305 = 5x - 30. Add 30 to both sides: 35=5x35 = 5x. Finally, divide by 5: x=7x = 7. You can verify this by substituting back: 7+53=123=4\frac{7+5}{3} = \frac{12}{3} = 4, and 2(7)−10=14−10=42(7) - 10 = 14 - 10 = 4. Both sides equal 4, confirming our answer. Choice A (-7) results from sign errors when moving terms across the equation. Choice B (3) likely comes from incorrectly distributing or combining terms early in the process. Choice D (9) could result from calculation mistakes when isolating x, particularly in the final division step. The key strategy for linear equations with fractions is to eliminate fractions first by multiplying both sides by the denominator, then systematically move all variable terms to one side and constants to the other. Always verify your answer by substituting back into the original equation—this catches arithmetic errors that are common on timed exams like the HESI.

Question 3

A pharmacy's inventory equation is 3(V−12)+2V=4V+63(V - 12) + 2V = 4V + 6, where V represents the number of vials ordered last month. How many vials were ordered?

  1. 40 vials
  2. 48 vials
  3. 45 vials
  4. 42 vials (correct answer)
Explanation: This question tests your ability to solve linear equations, a fundamental algebra skill that appears regularly on the HESI exam. When you encounter an equation with variables, your goal is to isolate the variable by using inverse operations systematically. Starting with 3(V−12)+2V=4V+63(V - 12) + 2V = 4V + 6, first distribute the 3: 3V−36+2V=4V+63V - 36 + 2V = 4V + 6. Combine like terms on the left side: 5V−36=4V+65V - 36 = 4V + 6. Now isolate the variable by subtracting 4V4V from both sides: V−36=6V - 36 = 6. Finally, add 36 to both sides: V=42V = 42. Let's verify: 3(42−12)+2(42)=3(30)+84=90+84=1743(42 - 12) + 2(42) = 3(30) + 84 = 90 + 84 = 174, and 4(42)+6=168+6=1744(42) + 6 = 168 + 6 = 174. Both sides equal 174, confirming our answer. Choice (A) 40 vials results from incorrectly combining terms or making arithmetic errors during distribution. Choice (B) 48 vials likely comes from sign errors when moving terms across the equal sign—a common mistake when students forget to change signs. Choice (C) 45 vials represents errors in the final addition step, possibly adding incorrectly when isolating the variable. When solving linear equations on the HESI, always work systematically: distribute first, combine like terms, then isolate the variable. Most importantly, substitute your answer back into the original equation to verify—this catches calculation errors and builds confidence in your solution.

Question 4

The formula to convert Celsius to Fahrenheit is F=95C+32F = \frac{9}{5}C + 32. A patient has a temperature of 102.2° F. The doctor orders the temperature to be reduced to a target of 38° C. How many degrees Fahrenheit does the patient's temperature need to decrease to meet the target?

  1. 1.0° F
  2. 1.8° F (correct answer)
  3. 2.2° F
  4. 4.2° F
Explanation: Temperature conversion problems on the HESI require careful attention to what's being asked. You're not just converting between scales—you're finding the difference between two temperature values. Start by converting the target temperature from Celsius to Fahrenheit using the given formula: F=95C+32F = \frac{9}{5}C + 32. With the target of 38°C: F=95(38)+32=3425+32=68.4+32=100.4°FF = \frac{9}{5}(38) + 32 = \frac{342}{5} + 32 = 68.4 + 32 = 100.4°F Now find the difference between the current temperature and the target: 102.2°F−100.4°F=1.8°F102.2°F - 100.4°F = 1.8°F The patient's temperature needs to decrease by 1.8°F to reach the target, making (B) 1.8°F correct. Let's examine the wrong answers: (A) 1.0°F likely comes from rounding errors or miscalculating the Celsius conversion. (C) 2.2°F represents a common trap—this is simply the decimal portion of the original temperature (102.2), which students might mistakenly think relates to the answer. (D) 4.2°F could result from calculation errors in the conversion process or confusing the relationship between the numbers given. Study tip: Always convert to the same temperature scale before finding differences. Write out each step clearly: convert the target temperature, then subtract to find the difference. Don't try to shortcut these multi-step problems—the wrong answer choices are specifically designed to catch common computational errors.

Question 5

The perimeter of a rectangular therapy room is 110 feet. The length of the room is 10 feet longer than twice the width. What is the width, w, of the room in feet?

  1. 15 feet (correct answer)
  2. 25 feet
  3. 33 feet
  4. 40 feet
Explanation: When you encounter perimeter problems involving rectangles with algebraic relationships, you need to set up equations using the given constraints and solve systematically. Start by defining your variables and writing the perimeter formula. For a rectangle, P=2l+2wP = 2l + 2w, where ll is length and ww is width. You know the perimeter is 110 feet, and the length is 10 feet longer than twice the width, so l=2w+10l = 2w + 10. Substitute the length expression into the perimeter formula: 110=2(2w+10)+2w110 = 2(2w + 10) + 2w. Simplifying: 110=4w+20+2w=6w+20110 = 4w + 20 + 2w = 6w + 20. Subtract 20 from both sides: 90=6w90 = 6w. Therefore, w=15w = 15 feet. Let's verify: if w=15w = 15, then l=2(15)+10=40l = 2(15) + 10 = 40. The perimeter is 2(40)+2(15)=80+30=1102(40) + 2(15) = 80 + 30 = 110 ✓ Answer choice A (15 feet) is correct. Answer B (25 feet) would give a length of 60 feet and perimeter of 170 feet—too large. Answer C (33 feet) would result in a length of 76 feet and perimeter of 218 feet—way too large. Answer D (40 feet) would make the length 90 feet with a perimeter of 260 feet—massively oversized. For HESI geometry problems, always define your variables clearly, translate word relationships into algebraic expressions, and check your answer by substituting back into the original constraints. This verification step catches calculation errors and confirms your solution makes sense.

Question 6

The concentration of a drug in a patient's bloodstream, C, in mg/L, is modeled by the equation C=15−1.25tC = 15 - 1.25t, where t is the number of hours after administration. How many hours, t, will it take for the concentration to reach 8.75 mg/L?

  1. 3
  2. 5 (correct answer)
  3. 7
  4. 19
Explanation: This question tests your ability to solve linear equations in a healthcare context, where you need to find when a drug concentration reaches a specific level. To find when the concentration reaches 8.75 mg/L, you substitute this value for C in the equation C=15−1.25tC = 15 - 1.25t and solve for t: 8.75=15−1.25t8.75 = 15 - 1.25t Subtract 15 from both sides: 8.75−15=−1.25t8.75 - 15 = -1.25t This gives you: −6.25=−1.25t-6.25 = -1.25t Divide both sides by -1.25: t=−6.25−1.25=5t = \frac{-6.25}{-1.25} = 5 So the answer is B) 5 hours. Let's examine why the other options are incorrect: A) 3 hours would give C=15−1.25(3)=15−3.75=11.25C = 15 - 1.25(3) = 15 - 3.75 = 11.25 mg/L, which is too high. C) 7 hours would give C=15−1.25(7)=15−8.75=6.25C = 15 - 1.25(7) = 15 - 8.75 = 6.25 mg/L, which is too low. D) 19 hours would give C=15−1.25(19)=15−23.75=−8.75C = 15 - 1.25(19) = 15 - 23.75 = -8.75 mg/L. This negative concentration is impossible in real life and shows this answer is far too large. Study tip: When working with linear drug concentration models on the HESI, always check that your answer makes biological sense. Drug concentrations decrease over time and cannot be negative. Also, practice isolating variables by performing the same operation on both sides of the equation—this skill appears frequently in dosage calculations.

Question 7

A patient on a diet is supposed to limit their daily sodium intake. They consumed a breakfast with 350 mg of sodium. For the rest of the day, they can eat x snacks, each containing 120 mg of sodium. If their total sodium intake for the day must be exactly 950 mg, how many snacks can they have?

  1. 4
  2. 5 (correct answer)
  3. 6
  4. 8
Explanation: When you encounter word problems involving dietary restrictions or medication dosages on the HESI exam, you're being tested on your ability to set up and solve linear equations — a critical skill for calculating proper dosages and nutritional limits in healthcare. Let's work through this systematically. You know the patient consumed 350 mg of sodium at breakfast and needs to stay at exactly 950 mg total for the day. This means they have 950−350=600950 - 350 = 600 mg of sodium remaining for snacks. Since each snack contains 120 mg of sodium, you can set up the equation: 120x=600120x = 600, where x is the number of snacks. Solving: x=600÷120=5x = 600 ÷ 120 = 5 snacks. Let's verify: 350+(5×120)=350+600=950350 + (5 × 120) = 350 + 600 = 950 mg ✓ Now for the wrong answers: Choice A (4 snacks) gives you 350+(4×120)=830350 + (4 × 120) = 830 mg total — this falls short of the required 950 mg. Choice C (6 snacks) results in 350+(6×120)=1070350 + (6 × 120) = 1070 mg, exceeding the limit by 120 mg. Choice D (8 snacks) yields 350+(8×120)=1310350 + (8 × 120) = 1310 mg, dangerously overshooting the sodium restriction. Study tip: For HESI calculation problems, always double-check your work by substituting your answer back into the original scenario. In healthcare settings, calculation errors with medications or dietary restrictions can have serious consequences, so developing this verification habit is essential for both exam success and clinical practice.

Question 8

Three-fifths of the patients in a clinic have an appointment with a doctor, and the remaining 16 patients have an appointment with a nurse practitioner. What is the total number of patients, P, in the clinic?

  1. 24
  2. 32
  3. 40 (correct answer)
  4. 48
Explanation: When you encounter word problems involving fractions and remainders, you need to set up an equation that accounts for all parts of the whole. Here, you know that three-fifths of patients see doctors, and the remaining patients (16) see nurse practitioners. Let P represent the total number of patients. If three-fifths see doctors, then the remaining fraction is 1−35=251 - \frac{3}{5} = \frac{2}{5}. This means 25\frac{2}{5} of all patients see nurse practitioners. Since 16 patients see nurse practitioners, you can write: 25×P=16\frac{2}{5} \times P = 16 Solving for P: P=16×52=16×2.5=40P = 16 \times \frac{5}{2} = 16 \times 2.5 = 40 You can verify: If there are 40 total patients, then 35×40=24\frac{3}{5} \times 40 = 24 see doctors, and 40−24=1640 - 24 = 16 see nurse practitioners. ✓ Choice A (24) represents only the number of patients seeing doctors, not the total. This is a common trap where students solve for part of the problem rather than what's actually being asked. Choice B (32) might result from incorrectly assuming that 16 patients represent 12\frac{1}{2} of the total instead of 25\frac{2}{5}. Choice D (48) could come from misreading the fraction as 23\frac{2}{3} instead of 35\frac{3}{5}, leading to 16×3=4816 \times 3 = 48. Always double-check fraction problems by verifying that your answer makes both parts of the problem work. Set up your equation carefully and make sure you're solving for what the question asks—the total, not just one component.

Question 9

A patient's weight is 170 pounds. The doctor wants the patient to lose weight at a rate of 1.5 pounds per week. Which equation represents the patient's weight, W, after x weeks? Using this, after how many weeks will the patient reach a target weight of 149 pounds?

  1. 10
  2. 12
  3. 14 (correct answer)
  4. 16
Explanation: When you encounter word problems involving linear change over time, you need to identify the starting value, the rate of change, and the direction of change to build your equation. The patient starts at 170 pounds and loses 1.5 pounds per week. Since weight is decreasing, this creates the equation: W=170−1.5xW = 170 - 1.5x, where W is the weight after x weeks. To find when the patient reaches 149 pounds, set W = 149 and solve: 149=170−1.5x149 = 170 - 1.5x. Subtracting 170 from both sides gives −21=−1.5x-21 = -1.5x. Dividing by -1.5 yields x=14x = 14 weeks. Let's check why other answers are incorrect. Answer A (10 weeks) would result in a weight of 170−1.5(10)=155170 - 1.5(10) = 155 pounds, which is too high. Answer B (12 weeks) gives 170−1.5(12)=152170 - 1.5(12) = 152 pounds, still above the target. Answer D (16 weeks) produces 170−1.5(16)=146170 - 1.5(16) = 146 pounds, which overshoots the goal by continuing past the target weight. Answer C (14 weeks) is correct because 170−1.5(14)=149170 - 1.5(14) = 149 pounds exactly. For linear word problems on the HESI, always write your equation first using the pattern: starting value ± (rate × time). Pay attention to whether the quantity increases (+) or decreases (-). Then substitute your target value and solve algebraically. Double-check by plugging your answer back into the original equation.

Question 10

A medical supply company offers two billing plans. Plan A has a $50 monthly fee plus $15 per box of supplies. Plan B has a $100 monthly fee plus $10 per box of supplies. For how many boxes of supplies, x, would the total monthly cost be the same for both plans?

  1. 5
  2. 8
  3. 10 (correct answer)
  4. 15
Explanation: This is a classic linear equation problem where you need to find the point where two different cost structures intersect. When you see questions comparing two pricing plans, set up equations for each plan and solve for where they're equal. To solve this, write an equation for each plan's total monthly cost. Plan A costs 50+15x50 + 15x dollars, where xx is the number of boxes. Plan B costs 100+10x100 + 10x dollars. Set these equal to find where the costs are the same: 50+15x=100+10x50 + 15x = 100 + 10x Subtract 10x10x from both sides: 50+5x=10050 + 5x = 100 Subtract 50 from both sides: 5x=505x = 50 Divide by 5: x=10x = 10 You can verify this by substituting back: Plan A at 10 boxes costs 50+15(10)=20050 + 15(10) = 200 dollars. Plan B at 10 boxes costs 100+10(10)=200100 + 10(10) = 200 dollars. Both equal $200, confirming our answer. Choice A (5 boxes) gives you Plan A = $125 and Plan B = $150, so they're not equal. Choice B (8 boxes) gives you Plan A = $170 and Plan B = $180, still not equal. Choice D (15 boxes) gives you Plan A = $275 and Plan B = $250, with Plan B actually cheaper. Remember that for cost comparison problems, always set up your equations carefully and double-check by substituting your answer back into both original equations. The intersection point tells you exactly where the two options become equivalent.

Question 11

A patient's heart rate is measured three times. The second measurement is 5 beats per minute (bpm) higher than the first, and the third measurement is 2 bpm lower than the second. If the sum of the three measurements is 233 bpm, what was the first measurement, x?

  1. 75 bpm (correct answer)
  2. 77 bpm
  3. 78 bpm
  4. 80 bpm
Explanation: When you encounter word problems involving sequential measurements or changes, the key is to define your variables clearly and set up equations that represent the relationships described in the problem. Let's call the first measurement xx. According to the problem, the second measurement is 5 bpm higher than the first, so it equals x+5x + 5. The third measurement is 2 bpm lower than the second, making it (x+5)−2=x+3(x + 5) - 2 = x + 3. Since the sum of all three measurements is 233 bpm, you can write: x+(x+5)+(x+3)=233x + (x + 5) + (x + 3) = 233 Simplifying: 3x+8=2333x + 8 = 233 Solving for x: 3x=2253x = 225, so x=75x = 75 Let's verify: First measurement = 75, second = 80, third = 78. Sum: 75 + 80 + 78 = 233 ✓ Now examining the wrong answers: Choice B (77 bpm) would give you measurements of 77, 82, and 80, totaling 239 bpm—too high. Choice C (78 bpm) results in 78, 83, and 81, summing to 242 bpm. Choice D (80 bpm) produces 80, 85, and 83, totaling 248 bpm. Each of these exceeds the required sum of 233. For HESI math problems involving sequential relationships, always define your starting variable first, then express each subsequent value in terms of that variable. Set up your equation based on the given constraint (like a total sum), and always verify your answer by substituting back into the original conditions.

Question 12

A total of 180 milliliters (mL) of a saline solution needs to be infused over 3 hours. For the first hour, the solution is infused at a rate of 50 mL/hr. To complete the infusion on time, what must be the infusion rate, r, in mL/hr for the remaining 2 hours?

  1. 60
  2. 65 (correct answer)
  3. 70
  4. 130
Explanation: When you encounter infusion rate problems, you're working with the fundamental relationship between volume, rate, and time. These calculations are essential for safe medication administration in clinical practice. Let's break down what happens during this 3-hour infusion. In the first hour, 50 mL is infused at 50 mL/hr. This leaves 180 - 50 = 130 mL remaining to be infused over the next 2 hours. To find the required rate for the remaining time, divide the remaining volume by the remaining time: r=130 mL2 hours=65 mL/hrr = \frac{130 \text{ mL}}{2 \text{ hours}} = 65 \text{ mL/hr} Looking at the wrong answers: Choice A (60 mL/hr) would only deliver 120 mL over 2 hours, leaving 10 mL uninfused and failing to complete the order on time. Choice C (70 mL/hr) would deliver 140 mL over 2 hours, resulting in 10 mL too much being infused, which exceeds the prescribed amount. Choice D (130 mL/hr) represents a common error where students use the remaining volume as the rate, confusing units and potentially creating a dangerous overdose situation. The correct answer is B (65 mL/hr), which ensures exactly 130 mL is delivered over the remaining 2 hours. For HESI infusion problems, always identify what's already been given, calculate what remains, then apply the rate formula. Double-check that your final rate will deliver the exact remaining volume in the exact remaining time—no more, no less.

Question 13

A supply clerk is inventorying boxes of sterile gloves. He has 4 large cases, each containing the same number of boxes, b. He removes 15 boxes from each large case to be sent to another department. The total number of boxes remaining in the 4 cases is 340. How many boxes, b, were originally in each case?

  1. 70
  2. 85
  3. 100 (correct answer)
  4. 115
Explanation: This is a linear equation problem that tests your ability to translate a word problem into mathematical expressions and solve for an unknown variable. Let's set up the equation step by step. Each of the 4 cases originally contained bb boxes. After removing 15 boxes from each case, each case has (b−15)(b - 15) boxes remaining. Since there are 4 cases total, the expression for all remaining boxes is 4(b−15)=3404(b - 15) = 340. Now solve for bb: 4(b−15)=3404(b - 15) = 340 4b−60=3404b - 60 = 340 4b=4004b = 400 b=100b = 100 Let's verify: If each case originally had 100 boxes, removing 15 from each leaves 85 boxes per case. With 4 cases: 4×85=3404 × 85 = 340 boxes remaining. ✓ Looking at the wrong answers: Choice A (70) would leave only 4(70−15)=2204(70-15) = 220 boxes remaining, far short of 340. Choice B (85) would result in 4(85−15)=2804(85-15) = 280 boxes remaining, still too few. Choice D (115) would give 4(115−15)=4004(115-15) = 400 boxes remaining, which exceeds the given total of 340. When solving word problems involving inventory or quantities, always define your variable clearly, set up the equation based on the final condition described, and verify your answer by substituting back into the original scenario. Pay close attention to whether the problem asks for an original amount, final amount, or the change between them.

Question 14

A solution is being heated in a lab. Its initial temperature is 12°C. The temperature increases by 2.5°C per minute. Another solution starts at 30°C and cools at a rate of 1.5°C per minute. After how many minutes, m, will the two solutions have the same temperature?

  1. 3.0
  2. 4.5 (correct answer)
  3. 7.2
  4. 10.5
Explanation: When you encounter problems involving two quantities changing at different rates over time, you're working with linear equations. The key is setting up expressions for each quantity's value at time m, then finding when they're equal. For the heating solution: it starts at 12°C and increases 2.5°C per minute, so its temperature after m minutes is 12+2.5m12 + 2.5m. For the cooling solution: it starts at 30°C and decreases 1.5°C per minute, so its temperature is 30−1.5m30 - 1.5m. To find when temperatures are equal, set the expressions equal: 12+2.5m=30−1.5m12 + 2.5m = 30 - 1.5m. Adding 1.5m to both sides: 12+4m=3012 + 4m = 30. Subtracting 12 from both sides: 4m=184m = 18. Therefore: m=4.5m = 4.5 minutes. Let's check why the other answers are wrong. Choice A (3.0) gives temperatures of 19.5°C and 25.5°C respectively—not equal. Choice C (7.2) results in 30°C and 19.2°C—the heating solution would actually be hotter than the cooling one. Choice D (10.5) gives 38.25°C and 14.25°C—an even larger temperature difference. For rate problems like this, always write an expression for each changing quantity, set them equal at the target condition, and solve algebraically. Double-check your answer by substituting back into both original expressions—they should give the same result. This type of problem appears frequently on standardized tests, so mastering the setup is crucial for success.

Question 15

A hospital charges a flat fee of $2,200 for a surgery. In addition, there is a charge of $900 for each day the patient is admitted for recovery. If a patient's total bill was $5,800, for how many days, d, was the patient admitted?

  1. 3
  2. 4 (correct answer)
  3. 6
  4. 9
Explanation: When you encounter word problems involving total costs with both fixed and variable components, you need to set up an equation that separates these different charges. Here, the hospital has a fixed surgery fee of $2,200 plus a variable daily charge of $900 per day. With a total bill of $5,800, you can write: $2,200+900d=5,8002,200 + 900d = 5,800 $ To solve for the number of days (d), first subtract the fixed surgery cost: 900d = 5,800 - 2,200 = 3,600 Then divide by the daily rate: d = 3,600 ÷ 900 = 4 \text{ days} You can verify: $2,200 + (4 × $900) = $2,200 + $3,600 = $5,800 ✓ Looking at the wrong answers: Choice A (3 days) would give you $2,200 + $2,700 = $4,900, which is $900 short of the actual bill. Choice C (6 days) would result in $2,200 + $5,400 = $7,600, exceeding the bill by $1,800. Choice D (9 days) would cost $2,200 + $8,100 = $10,300, more than double the actual charges. For HESI math problems involving costs, always identify what's fixed versus what varies with time or quantity. Set up your equation systematically, separating these components, then solve step by step. Double-check by substituting your answer back into the original scenario to ensure it produces the given total.

Question 16

Solve for the variable n in the following equation: 25n−6=2\frac{2}{5}n - 6 = 2

  1. 3.2
  2. 10
  3. 20 (correct answer)
  4. 35
Explanation: When you encounter linear equations with fractions, your goal is to isolate the variable through systematic algebraic operations. Start by eliminating constants, then deal with coefficients. To solve 25n−6=2\frac{2}{5}n - 6 = 2, first add 6 to both sides to isolate the term containing n: 25n=2+6=8\frac{2}{5}n = 2 + 6 = 8. Now you need to solve 25n=8\frac{2}{5}n = 8. To eliminate the fraction coefficient, multiply both sides by the reciprocal 52\frac{5}{2}: n=8×52=402=20n = 8 \times \frac{5}{2} = \frac{40}{2} = 20. You can verify this: 25(20)−6=8−6=2\frac{2}{5}(20) - 6 = 8 - 6 = 2 ✓ Answer A (3.2) likely results from incorrectly dividing 8 by 25\frac{2}{5} instead of multiplying by its reciprocal, or from calculation errors with decimal conversion. Answer B (10) suggests you may have forgotten to add 6 in the first step, solving 25n=2\frac{2}{5}n = 2 instead of the correct 25n=8\frac{2}{5}n = 8. Answer D (35) probably comes from multiplying 8 by both parts of the fraction separately (8 × 2 = 16, then 16 + something), showing confusion about fraction operations. Remember that dividing by a fraction is the same as multiplying by its reciprocal. When you see abn=c\frac{a}{b}n = c, multiply both sides by ba\frac{b}{a} to get n=c×ban = c \times \frac{b}{a}. Always check your answer by substituting back into the original equation.

Question 17

The relationship between a patient's heart rate before and after exercise is given by H+204=H−102\frac{H + 20}{4} = \frac{H - 10}{2}, where H is the resting heart rate in beats per minute. What is the resting heart rate?

  1. 70 bpm
  2. 80 bpm
  3. 90 bpm
  4. 100 bpm (correct answer)
Explanation: Cross multiply: 2(H + 20) = 4(H - 10). Expand: 2H + 40 = 4H - 40. Subtract 2H: 40 = 2H - 40. Add 40: 80 = 2H. Divide by 2: H = 100 bpm. Choice A results from sign errors. Choice B comes from arithmetic errors in cross multiplication. Choice C results from incorrectly moving terms between sides.

Question 18

A lab technician's daily sample count follows 5S−2(S+4)=225S - 2(S + 4) = 22, where S represents the number of samples processed in the morning shift. How many samples were processed in the morning?

  1. 8 samples
  2. 10 samples (correct answer)
  3. 12 samples
  4. 14 samples
Explanation: Expand: 5S - 2S - 8 = 22. Combine like terms: 3S - 8 = 22. Add 8: 3S = 30. Divide by 3: S = 10 samples. Choice A results from distribution errors with the negative sign. Choice C comes from adding 8 instead of subtracting. Choice D results from incorrectly combining the S terms.

Question 19

A healthcare facility charges a base fee of $85 plus $12 per hour for equipment rental. If a patient's total bill was $157, how many hours was the equipment rented?

  1. 6 hours (correct answer)
  2. 7 hours
  3. 8 hours
  4. 9 hours
Explanation: Set up the equation: 85 + 12h = 157. Subtract 85 from both sides: 12h = 72. Divide by 12: h = 6. Choice B results from adding instead of subtracting the base fee. Choice C comes from using the wrong base fee ($73 instead of $85). Choice D results from calculation errors in the division step.

Question 20

A medication dosage formula is given by D=2W+153D = \frac{2W + 15}{3}, where D is the dosage in mg and W is the patient's weight in kg. If a patient requires a dosage of 25 mg, what is their weight?

  1. 22.5 kg
  2. 30 kg (correct answer)
  3. 35 kg
  4. 37.5 kg
Explanation: Substitute D = 25: 25 = (2W + 15)/3. Multiply both sides by 3: 75 = 2W + 15. Subtract 15: 60 = 2W. Divide by 2: W = 30 kg. Choice A results from forgetting to subtract 15. Choice C comes from multiplying by 2 instead of dividing. Choice D results from adding 15 instead of subtracting.