Solve Algebra Word Problems - International Brotherhood of Electrical Workers Exam IBEW

What is the output voltage if a transformer has $N_p:N_s=5:1$ and $V_p=120\,V$ ?
Answer: $24\,V$ This question tests the ability to solve algebraic word problems in electrical applications. Algebra involves understanding how to form and solve equations based on given data, crucial in electrical contexts for calculating loads, voltages, and power. In this scenario, the apprentice is asked to find the output voltage of a transformer, using provided data such as turns ratio $N_p$ : $N_s$ =5:1 and primary voltage $V_p$ =120 V. The correct choice, B, is accurate because it applies the transformer voltage ratio formula $V_s$ = $V_p$ * ( $N_s$ / $N_p$ ) = 120 (1/5) = 24 V, considering all data points. A common error, as in choice A, occurs when inverting the ratio, leading to 120 5 = 600 V. Teaching strategies include reinforcing the understanding of key formulas through practical exercises and encouraging checking calculations through reverse-engineering problems. Visual aids like circuit diagrams can help in visualizing data correctly.
An electrician buys 3 light switches and 5 outlets for a total of $40. If each outlet costs $5, what is the cost of one light switch?
Answer: $5 This is a basic algebra word problem that tests your ability to set up and solve equations with multiple unknowns - a skill you'll use frequently when calculating material costs and electrical loads on the job. Since you know each outlet costs $5 and there are 5 outlets, the outlets cost $5 \times \$5 = \$25$ total. With a total purchase of $40, the remaining amount spent on switches is $\$40 - \$25 = \$15$ . Since this $15 covers 3 light switches, each switch costs $\$15 \div 3 = \$5$ . Looking at the wrong answers: Choice A ($3) would make the total cost $3 \times \$3 + 5 \times \$5 = \$9 + \$25 = \$34$ , which is $6 short of the given total. Choice C ($8) would result in $3 \times \$8 + 5 \times \$5 = \$24 + \$25 = \$49$ , exceeding the total by $9. Choice D ($15) represents the total cost of all three switches, not the cost per switch - this would make the grand total $3 \times \$15 + \$25 = \$70$ , far too high. The correct answer is B ($5). For similar problems on the IBEW exam, always identify what you know first, then work systematically toward what you need to find. Material cost calculations are common in electrical work, so practice breaking down totals into individual component costs - this skill applies whether you're pricing switches or calculating wire footage needs.
An electrician needs to install a cable that is 450 centimeters long. The cable is sold by the foot. If 1 inch is approximately 2.5 centimeters, and there are 12 inches in a foot, approximately how many feet of cable are needed?
Answer: 15 feet Unit conversion problems are common on the IBEW exam, and they test your ability to set up conversion factors systematically. When converting between metric and imperial units, always work step-by-step through each conversion rather than trying to do it all at once. Start with what you know: 450 centimeters needs to be converted to feet. You have two conversion factors: 1 inch = 2.5 centimeters, and 12 inches = 1 foot. Set up your conversions to cancel units properly: $450 \text{ cm} \times \frac{1 \text{ inch}}{2.5 \text{ cm}} \times \frac{1 \text{ foot}}{12 \text{ inches}}$ First, convert centimeters to inches: $450 ÷ 2.5 = 180 \text{ inches}$ Then convert inches to feet: $180 ÷ 12 = 15 \text{ feet}$ This confirms answer B is correct. Looking at the wrong answers: A) 12 feet would result from incorrectly using 37.5 cm per foot (450 ÷ 12), mixing up the conversion process. C) 18 feet likely comes from using 25 cm per foot instead of 30 cm per foot (450 ÷ 25). D) 37.5 feet results from only converting centimeters to inches (450 ÷ 2.5 = 180) but forgetting the final step to convert inches to feet, then somehow getting 37.5. For unit conversion success on the IBEW exam, always write out your conversion factors as fractions, ensure units cancel properly, and double-check by working backwards from your answer. Practice converting between metric and imperial measurements since these appear frequently in electrical work.
The sum of three consecutive odd integers is 111. What is the largest of the three integers?
Answer: 39 When you encounter consecutive integer problems, you're dealing with algebra word problems that require setting up equations based on number patterns. Consecutive odd integers follow the pattern: if the first is $n$ , then the next two are $n+2$ and $n+4$ . Let's call the first odd integer $x$ . The three consecutive odd integers are $x$ , $x+2$ , and $x+4$ . Since their sum is 111: $x + (x+2) + (x+4) = 111$ Simplifying: $3x + 6 = 111$ Subtracting 6: $3x = 105$ Dividing by 3: $x = 35$ So the three integers are 35, 37, and 39. The largest is 39, making C correct. Let's examine why the other answers are wrong. Choice A (35) gives you the smallest of the three consecutive odd integers, not the largest—this is a common trap where students solve correctly but answer the wrong part of the question. Choice B (37) is the middle integer of our sequence, another distractor for students who might get confused about which value the question asks for. Choice D (41) would be the largest if you incorrectly assumed the three integers were 37, 39, and 41, which sum to 117—this happens when students make arithmetic errors in their setup or calculations. Always double-check consecutive integer problems by verifying your three numbers actually sum to the given total, and make sure you're identifying the specific value the question asks for (smallest, largest, or middle).
An electrician has a box of fasteners containing a mix of screws and nails. The number of screws is 8 less than twice the number of nails. If there are a total of 112 fasteners, how many screws are there?
Answer: 72 This is a classic algebraic word problem that tests your ability to translate written relationships into mathematical equations. When you encounter problems involving "relationships between quantities," set up variables and use the given constraints to create equations. Let's define our variables: let $n$ = number of nails and $s$ = number of screws. From the problem, we know two things: the screws equal 8 less than twice the nails ( $s = 2n - 8$ ), and the total fasteners equal 112 ( $s + n = 112$ ). Now substitute the first equation into the second: $(2n - 8) + n = 112$ . Simplifying: $3n - 8 = 112$ , so $3n = 120$ , which gives us $n = 40$ nails. Therefore, $s = 2(40) - 8 = 72$ screws. Looking at the wrong answers: Answer (A) 40 represents the number of nails, not screws—this is a common trap where students solve for the wrong variable. Answer (B) 52 likely comes from incorrectly setting up the relationship as $s = 2n + 8$ instead of $s = 2n - 8$ , which would give 28 nails and 84 screws—but 52 doesn't fit this scenario cleanly either. Answer (D) 80 might result from solving $2n - 8 = 112$ directly without considering the total constraint, giving $n = 60$ and $s = 112$ , but this violates the total of 112 fasteners. Always define your variables clearly and double-check which quantity the question asks for. Many IBEW word problems deliberately ask for the less obvious variable to test your careful reading.
An electrical contractor charges a $75 service fee plus $60 per hour for labor. If a job costs a total of $315, how many hours of labor were required?
Answer: 4 hours This is a linear equation problem that tests your ability to set up and solve cost scenarios you'll encounter as an electrical worker. When you see a problem with a fixed fee plus an hourly rate, you're dealing with the equation: Total Cost = Fixed Fee + (Hourly Rate × Hours). Let's set up the equation with the given information. The contractor charges $75 as a service fee plus $60 per hour for labor, and the total job cost is $315. So: $315 = 75 + 60h$ , where $h$ represents hours of labor. To solve, subtract the service fee from both sides: $315 - 75 = 60h$ , which gives us $240 = 60h$ . Dividing both sides by 60: $h = 240 ÷ 60 = 4$ hours. Let's verify: $75 + (60 × 4) = 75 + 240 = 315$ ✓ Now for the wrong answers: Choice A (3.5 hours) would cost $75 + (60 × 3.5) = 285$ , which is $30 short. Choice C (4.5 hours) would cost $75 + (60 × 4.5) = 345$ , which is $30 over budget. Choice D (5.25 hours) would cost $75 + (60 × 5.25) = 390$ , which is $75 over budget. Study tip: For IBEW cost problems, always identify the fixed costs versus variable costs first, then set up your equation systematically. Double-check by plugging your answer back into the original scenario—this catches calculation errors quickly.
An apprentice scores 78, 85, and 91 on the first three tests. What score must the apprentice get on the fourth test to have an average score of exactly 86?
Answer: 90 When you encounter an average problem like this, you're working with the fundamental relationship that average equals the sum of all values divided by the number of values. Here, you need to find what fourth test score will produce an exact average of 86. Start by setting up the equation. For four tests to average 86, the total points must be $86 \times 4 = 344$ points. You already know three scores: 78, 85, and 91, which sum to 254 points. Therefore, the fourth test score must be $344 - 254 = 90$ points. Let's verify: $(78 + 85 + 91 + 90) \div 4 = 344 \div 4 = 86$ ✓ Looking at the wrong answers: Choice A (86) represents a common misconception that the needed score equals the target average. This would actually give you an average of 85, not 86. Choice B (88) might seem reasonable since it's close to the target average, but it results in an average of 85.5. Choice D (92) overshoots the target, giving you an average of 86.5. The key insight is that since the apprentice's current average after three tests is $254 \div 3 = 84.67$ , which is below the target of 86, the fourth score must be higher than 86 to "pull up" the overall average. Study tip: For IBEW average problems, always calculate the required total first, then subtract what you already have. This systematic approach prevents arithmetic errors and helps you work efficiently under test pressure.
The length of a rectangular work area is 5 feet more than twice its width. If the perimeter of the area is 82 feet, what is the width of the area?
Answer: 12 feet When you encounter word problems involving geometric shapes and relationships, the key is translating the written descriptions into mathematical equations you can solve. Let's define the width as $w$ . The problem states the length is "5 feet more than twice the width," so length = $2w + 5$ . Since perimeter equals twice the length plus twice the width, you get: $2w + 2(2w + 5) = 82$ . Solving this equation: $2w + 4w + 10 = 82$ , which simplifies to $6w + 10 = 82$ . Subtracting 10 from both sides gives $6w = 72$ , so $w = 12$ feet. Let's verify: if width = 12 feet, then length = $2(12) + 5 = 29$ feet. The perimeter would be $2(12) + 2(29) = 24 + 58 = 82$ feet ✓ Answer A (12 feet) is correct. Answer B (13 feet) likely comes from making an arithmetic error in the algebraic manipulation, perhaps incorrectly handling the coefficient of $w$ . Answer C (29 feet) is actually the length of the rectangle, not the width—this tests whether you're answering the right question. Answer D (41 feet) might result from confusing the perimeter formula or incorrectly setting up the initial equation. For IBEW geometry problems, always define your variable clearly, translate each phrase methodically into mathematical expressions, and double-check your final answer by substituting back into the original conditions. Many wrong answers on these exams come from solving correctly but answering the wrong question.
An electrician cuts a 38-foot piece of conduit into two pieces. One piece is 6 feet longer than the other. What is the length of the shorter piece?
Answer: 16 feet This is a classic algebra word problem that tests your ability to set up and solve equations with unknowns—a skill you'll use regularly when calculating electrical loads and wire lengths in the field. Let's define our variables systematically. If the shorter piece has length $x$ feet, then the longer piece must be $x + 6$ feet (since it's 6 feet longer). Because both pieces came from the original 38-foot conduit, we can write: $x + (x + 6) = 38$ Solving this equation: $2x + 6 = 38$ , so $2x = 32$ , which gives us $x = 16$ feet for the shorter piece. We can verify: the longer piece is $16 + 6 = 22$ feet, and $16 + 22 = 38$ feet total. ✓ Looking at the wrong answers: Choice A (13 feet) would make the longer piece 19 feet, totaling only 32 feet—you'd be missing 6 feet of conduit. Choice C (19 feet) represents the longer piece, not the shorter one that the question asks for. Choice D (22 feet) would make the total 50 feet if it were the shorter piece, which exceeds your original 38-foot length. For word problems like this, always define your variable as what the question is asking for (here, the shorter piece), then express everything else in terms of that variable. Double-check by substituting your answer back into the original conditions—this catches calculation errors and ensures you answered the right question.
Two service vans leave the main office at the same time, traveling in opposite directions. One van travels at 45 mph, and the other travels at 55 mph. How many hours will it take for them to be 400 miles apart?
Answer: 4 hours When you encounter problems involving objects moving in opposite directions, you're dealing with relative motion where the distances traveled by each object add together. This is a classic "separation rate" problem that appears frequently on technical exams. Since both vans leave at the same time and travel in opposite directions, their combined speed determines how quickly they separate. The first van travels at 45 mph while the second travels at 55 mph, giving you a combined separation rate of $45 + 55 = 100$ mph. To find when they'll be 400 miles apart, use the formula: Time = Distance ÷ Rate. So: $\text{Time} = \frac{400 \text{ miles}}{100 \text{ mph}} = 4 \text{ hours}$ Looking at the wrong answers: Choice A (3.5 hours) would only put them 350 miles apart since $100 \times 3.5 = 350$ . Choice C (4.5 hours) overshoots the target, resulting in 450 miles of separation. Choice D (8 hours) is a common trap that occurs when students mistakenly divide 400 by 50 (the average of the two speeds) instead of using the combined separation rate. The key strategy for these problems is recognizing that when objects move in opposite directions, you always add their speeds to get the separation rate. If they were moving in the same direction, you'd subtract the speeds instead. Remember: opposite directions = add speeds, same direction = subtract speeds.
A supply house offers a 15% discount on all purchases over $500. An electrician's bill for materials comes to $620. How much will the bill be after the discount?
Answer: $527.00 When you encounter discount problems on the IBEW exam, you're working with percentage calculations that directly relate to real-world electrical contracting scenarios where material costs and discounts significantly impact job profitability. To find the discounted price, you need to calculate 15% of the original bill and subtract it from the total. The original bill is $620, which qualifies for the discount since it exceeds $500. First, calculate the discount amount: $\$620 \times 0.15 = \$93$ Then subtract the discount from the original bill: $\$620 - \$93 = \$527$ The final bill after discount is $527.00, making C correct. Looking at the wrong answers: A ($93.00) represents the discount amount itself, not the final bill—this catches students who stop halfway through the calculation. B ($505.00) might result from incorrectly subtracting the full $115 difference between $620 and $505, possibly confusing this with a different type of discount structure. D ($530.00) is close to the correct answer but represents a calculation error, perhaps from rounding mistakes or using the wrong percentage. Remember that discount problems always require two steps: calculate the discount amount, then subtract it from the original price. Don't stop at just finding the discount—the question typically asks for the final price after applying the discount. This type of problem appears regularly on trade exams since material cost management is essential in electrical work.
A spool of wire is $3/4$ full. After an electrician uses 30 feet of wire, the spool is $1/2$ full. How many feet of wire does the spool hold when it is full?
Answer: 120 feet When you encounter word problems involving fractions of containers or spools, set up equations based on the relationship between the changes described. Here, you need to find the total capacity by working with the difference between two fractional states. Let's call the full capacity of the spool $x$ feet. Initially, the spool was $\frac{3}{4}$ full, so it contained $\frac{3}{4}x$ feet of wire. After using 30 feet, it became $\frac{1}{2}$ full, containing $\frac{1}{2}x$ feet of wire. The key insight is that the 30 feet used represents the difference between these two states: $\frac{3}{4}x - 30 = \frac{1}{2}x$ Solving for $x$ : $\frac{3}{4}x - \frac{1}{2}x = 30$ $\frac{3}{4}x - \frac{2}{4}x = 30$ $\frac{1}{4}x = 30$ $x = 120$ So the spool holds 120 feet when full, making C correct. Answer A (60 feet) likely comes from doubling the 30 feet used, incorrectly assuming this represents half the spool's capacity. Answer B (90 feet) might result from adding 30 feet to 60, or from miscalculating the fractional relationship. Answer D (150 feet) could come from incorrectly treating 30 feet as $\frac{1}{5}$ of the total capacity. For IBEW wire calculation problems, always define your variable clearly (total capacity), translate the word problem into mathematical relationships, and use the given change (30 feet) to bridge between the two states described.
A rectangular panel has a perimeter of 56 inches. Its length is 4 inches less than three times its width. What is the length of the panel?
Answer: 20 inches When you encounter perimeter problems with constraints, you're working with a system of equations. Set up variables for the unknown dimensions, then translate the given information into mathematical relationships. Let's call the width $w$ and length $l$ . From the problem, you know two things: the perimeter is 56 inches, and the length is 4 inches less than three times the width. This gives you: - $2l + 2w = 56$ (perimeter formula) - $l = 3w - 4$ (length relationship) Substitute the second equation into the first: $2(3w - 4) + 2w = 56$ . Expanding: $6w - 8 + 2w = 56$ , so $8w = 64$ and $w = 8$ inches. Therefore, $l = 3(8) - 4 = 20$ inches. Choice A (8 inches) is the width, not the length. This is a common trap where test-makers include the other variable as a distractor. Choice C (24 inches) would result if you miscalculated the length as $3w$ instead of $3w - 4$ , forgetting to subtract 4. Choice D (28 inches) is half the perimeter, which might tempt you if you confused perimeter with the sum of length and width rather than twice that sum. Choice B (20 inches) correctly represents the length when you properly set up and solve the system of equations. For IBEW perimeter problems, always define your variables clearly, write both constraint equations, and double-check by verifying your answer satisfies the original conditions. Watch for answer choices that represent other parts of your calculation.
According to Ohm's Law (V = IR), voltage is the product of current and resistance. A circuit has a constant resistance. When the voltage is 24 volts, the current is 2 amperes. If the voltage is increased to 60 volts, what will the new current be?
Answer: 5 amperes When you encounter Ohm's Law problems on the IBEW exam, you're dealing with one of the fundamental relationships in electrical work: $V = IR$ . This equation tells you that voltage, current, and resistance are mathematically linked, and if you know two values, you can always find the third. Since the resistance remains constant in this circuit, you can use the relationship between the two scenarios. First, find the resistance using the initial conditions: $R = \frac{V}{I} = \frac{24V}{2A} = 12Ω$ . Now apply Ohm's Law to the new voltage: $I = \frac{V}{R} = \frac{60V}{12Ω} = 5A$ . Alternatively, you can use proportional reasoning. Since voltage and current are directly proportional when resistance is constant, if voltage increases by a factor of 2.5 (from 24V to 60V), current must also increase by the same factor: $2A × 2.5 = 5A$ . Choice A (3 amperes) might result from incorrectly adding the voltage increase to the current. Choice B (4 amperes) could come from doubling the original current without accounting for the full voltage increase. Choice D (12 amperes) represents a common error of confusing the resistance value (12Ω) with the current. Remember this key pattern for IBEW exam success: in Ohm's Law problems with constant resistance, voltage and current always change proportionally. Set up a ratio or calculate the resistance first—both methods work reliably and help you avoid the arithmetic traps built into the wrong answers.
Calculate the voltage drop across section B when $I=2\,A$ , $R_A=3\,\Omega$ , $R_B=x\,\Omega$ , and $V_T=18\,V$ .
Answer: $12\,V$ This question tests the ability to solve algebraic word problems in electrical applications. Algebra involves understanding how to form and solve equations based on given data, crucial in electrical contexts for calculating loads, voltages, and power. In this scenario, the apprentice is asked to calculate the voltage drop across section B in a series circuit, using provided data such as current I=2 A, resistance $R_A$ =3 Ω, $R_B$ =x Ω, and total voltage $V_T$ =18 V. The correct choice, B, is accurate because it applies Ohm's Law correctly by first finding x from total resistance ( $V_T$ /I = 9 Ω, so x=6 Ω) and then $V_B$ = I * x = 12 V, considering all data points. A common error, as in choice A, occurs when mistakenly calculating only the drop across $R_A$ without solving for x. Teaching strategies include reinforcing the understanding of key formulas through practical exercises and encouraging checking calculations through reverse-engineering problems. Visual aids like circuit diagrams can help in visualizing data correctly.
Two electricians are paid a total of $720 for a job. The first electrician worked 15 hours and the second worked 25 hours. If they are paid proportionally to the hours they worked, how much should the second electrician receive?
Answer: $450 When you encounter a proportional payment problem, you're dealing with ratios and parts of a whole. The key insight is that each person's share should match their contribution relative to the total work done. First, find the total hours worked: $15 + 25 = 40$ hours. The second electrician worked 25 out of these 40 hours, so their share should be $\frac{25}{40} = \frac{5}{8}$ of the total payment. Calculate this: $\frac{5}{8} \times \$720 = \$450$ . You can verify this by checking the first electrician's share: $\frac{15}{40} \times \$720 = \$270$ . Notice that $\$270 + \$450 = \$720$ ✓ Answer A ($270) represents the first electrician's correct payment, not the second electrician's. This is a common trap where students calculate the right amount but assign it to the wrong person. Answer B ( $360) would result from splitting the money equally ($ $\$720 ÷ 2 = \$360$ $), ignoring the proportional requirement entirely. This misses the whole point of the problem. Answer D ($500) doesn't correspond to any logical calculation based on the given ratios. It might tempt students who make arithmetic errors or confuse the setup. For proportional payment problems on the IBEW exam, always follow this pattern: find total units (hours, work completed, etc.), determine each person's fraction of the total, then multiply that fraction by the total payment. Double-check that all shares add up to the original amount.
Determine the power consumption if $V=24\,V$ and current splits: $I_1=3\,A$ and $I_2=2\,A$ .
Answer: $120\,W$ This question tests the ability to solve algebraic word problems in electrical applications. Algebra involves understanding how to form and solve equations based on given data, crucial in electrical contexts for calculating loads, voltages, and power. In this scenario, the apprentice is asked to determine the total power consumption in a circuit with splitting currents, using provided data such as voltage V=24 V, $I_1$ =3 A, and $I_2$ =2 A. The correct choice, B, is accurate because it applies the power formula P = V I_total correctly, where I_total = 5 A, yielding 120 W, considering all data points. A common error, as in choice D, occurs when using only one branch current, like P = V $I_2$ = 24 W. Teaching strategies include reinforcing the understanding of key formulas through practical exercises and encouraging checking calculations through reverse-engineering problems. Visual aids like circuit diagrams can help in visualizing data correctly.
What is the total load on the circuit if three resistors $4\,\Omega$ , $6\,\Omega$ , and $R=2\,\Omega$ are in series?
Answer: $12\,\Omega$ This question tests the ability to solve algebraic word problems in electrical applications. Algebra involves understanding how to form and solve equations based on given data, crucial in electrical contexts for calculating loads, voltages, and power. In this scenario, the apprentice is asked to find the total load (resistance) of a series circuit, using provided data such as resistors of 4 Ω, 6 Ω, and 2 Ω. The correct choice, A, is accurate because it applies the series resistance formula correctly by summing them up to 12 Ω, considering all data points. A common error, as in choice B, occurs when confusing series with parallel and incorrectly taking the reciprocal. Teaching strategies include reinforcing the understanding of key formulas through practical exercises and encouraging checking calculations through reverse-engineering problems. Visual aids like circuit diagrams can help in visualizing data correctly.
An experienced electrician can install 12 outlets in 3 hours. A new apprentice can install 8 outlets in 4 hours. How many more outlets can the experienced electrician install in an 8-hour workday compared to the apprentice?
Answer: 16 outlets When you encounter work rate problems on the IBEW exam, you need to establish each worker's rate per hour, then scale up to the full workday to make comparisons. First, calculate each worker's hourly rate. The experienced electrician installs 12 outlets in 3 hours, so their rate is $\frac{12 \text{ outlets}}{3 \text{ hours}} = 4 \text{ outlets per hour}$ . The apprentice installs 8 outlets in 4 hours, so their rate is $\frac{8 \text{ outlets}}{4 \text{ hours}} = 2 \text{ outlets per hour}$ . Now scale these rates to an 8-hour workday. The experienced electrician would install $4 \times 8 = 32 \text{ outlets}$ , while the apprentice would install $2 \times 8 = 16 \text{ outlets}$ . The difference is $32 - 16 = 16 \text{ outlets}$ . Looking at the wrong answers: Choice A (2 outlets) represents the difference in their hourly rates, not the 8-hour difference. Choice B (8 outlets) might come from incorrectly using half the workday or confusing intermediate calculations. Choice D (24 outlets) could result from finding the apprentice's 8-hour total (16) and adding the experienced electrician's hourly rate (4), then doubling it—a common calculation error. For IBEW work rate problems, always follow this three-step approach: convert given information to hourly rates, multiply by the target time period, then find the difference. This systematic method prevents calculation errors and ensures you're comparing equivalent time periods.
In a training class, the ratio of apprentices to journeymen is 5 to 2. If there are 42 people in the class, how many are apprentices?
Answer: 30 When you encounter ratio problems on the IBEW exam, you're working with proportional relationships that are common in electrical work, from wire sizing to crew compositions. The key is translating the ratio into actual quantities. Here, you have a 5:2 ratio of apprentices to journeymen, meaning for every 5 apprentices, there are 2 journeymen. This creates groups of 7 people total (5 + 2 = 7). To find how many complete groups fit into 42 people, divide: $42 ÷ 7 = 6$ groups. Since each group contains 5 apprentices, multiply: $6 × 5 = 30$ apprentices. You can verify this works: 30 apprentices + 12 journeymen = 42 total people, and 30:12 simplifies to 5:2. Answer choice (A) 6 represents the number of complete ratio groups, not the number of apprentices—a common mix-up when students stop calculating too early. Choice (B) 12 is actually the number of journeymen (6 groups × 2 journeymen per group). Choice (C) 21 might tempt you if you incorrectly think apprentices make up half the class, but that would only work if the ratio were 1:1, not 5:2. The correct answer is (D) 30. Remember this pattern: when given a ratio and total quantity, find the "group size" by adding the ratio parts, determine how many complete groups exist, then multiply by the specific part you need. This systematic approach prevents the calculation errors that often trip up test-takers on ratio problems.

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