IBEW: Electrical Training Alliance Aptitude Test Quiz: Solve Algebra Word Problems
20 questions · exam conditions
0:00
Solve Algebra Word ProblemsQuestion 1 of 20

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?

13 feet
16 feet
19 feet
22 feet
← Back to quizzes

IBEW: Electrical Training Alliance Aptitude Test Quiz

IBEW: Electrical Training Alliance Aptitude Test Quiz: Solve Algebra Word Problems

Practice Solve Algebra Word Problems in IBEW: Electrical Training Alliance Aptitude Test 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 Solve Algebra Word Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for IBEW: Electrical Training Alliance Aptitude Test.

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

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?

  1. 13 feet
  2. 16 feet (correct answer)
  3. 19 feet
  4. 22 feet
Explanation: 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 xx feet, then the longer piece must be x+6x + 6 feet (since it's 6 feet longer). Because both pieces came from the original 38-foot conduit, we can write: x+(x+6)=38x + (x + 6) = 38 Solving this equation: 2x+6=382x + 6 = 38, so 2x=322x = 32, which gives us x=16x = 16 feet for the shorter piece. We can verify: the longer piece is 16+6=2216 + 6 = 22 feet, and 16+22=3816 + 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.

Question 2

Four years ago, Sarah was twice as old as her brother, Tom. The sum of their current ages is 38. How old is Tom now?

  1. 10 years old
  2. 14 years old (correct answer)
  3. 18 years old
  4. 24 years old
Explanation: Age relationship problems test your ability to set up equations using variables and work with time shifts. When you see "years ago" or "years from now," you need to track how each person's age changes relative to the given conditions. Let's define variables: Let Tom's current age be tt and Sarah's current age be ss. Four years ago, Tom was (t−4)(t-4) and Sarah was (s−4)(s-4). The problem states that four years ago, Sarah was twice as old as Tom, so: s−4=2(t−4)s-4 = 2(t-4), which simplifies to s−4=2t−8s-4 = 2t-8, or s=2t−4s = 2t-4. We also know their current ages sum to 38: s+t=38s + t = 38. Substituting the first equation into the second: (2t−4)+t=38(2t-4) + t = 38, which gives us 3t−4=383t-4 = 38, so 3t=423t = 42, and t=14t = 14. Tom is currently 14 years old. Let's verify: If Tom is 14, then Sarah is 38−14=2438-14 = 24. Four years ago, Tom was 10 and Sarah was 20. Indeed, 20 is twice 10. Choice A (10) represents Tom's age four years ago, not his current age. Choice C (18) would make Sarah 20 currently, but four years ago Sarah would have been 16 and Tom 14 — Sarah wouldn't have been twice Tom's age then. Choice D (24) represents Sarah's current age, not Tom's. Study tip: In age problems, always define what your variables represent clearly (current ages vs. past/future ages) and double-check your answer by substituting back into the original conditions.

Question 3

An electrician has a length of wire that is 60 feet long. He cuts the wire into two pieces such that one piece is one-fourth the length of the other. What is the length of the longer piece?

  1. 12 feet
  2. 15 feet
  3. 45 feet
  4. 48 feet (correct answer)
Explanation: This problem tests your ability to set up and solve linear equations with word problems—a skill you'll use regularly when calculating wire lengths, conduit runs, and material requirements on electrical jobs. Let's define variables: let the longer piece be xx feet and the shorter piece be yy feet. From the problem, we know two things: the total length is 60 feet (x+y=60x + y = 60), and one piece is one-fourth the length of the other. Since we want the longer piece, the shorter piece must be one-fourth of the longer piece: y=14xy = \frac{1}{4}x. Substituting the second equation into the first: x+14x=60x + \frac{1}{4}x = 60. This gives us 54x=60\frac{5}{4}x = 60. Multiplying both sides by 45\frac{4}{5}: x=60×45=48x = 60 \times \frac{4}{5} = 48 feet. Let's check why the other answers are wrong. Answer A (12 feet) is actually one-fourth of the correct answer—this comes from accidentally solving for the shorter piece instead of the longer one. Answer B (15 feet) results from incorrectly setting up the fraction relationship, perhaps thinking one piece is one-third of the other. Answer C (45 feet) is close but comes from arithmetic errors in the algebraic manipulation, possibly from incorrectly handling the fraction. When solving wire-cutting problems on the IBEW exam, always define which piece you're solving for clearly, set up your fraction relationships carefully, and verify your answer makes sense by checking that both pieces add up to the original length.

Question 4

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?

  1. 6
  2. 12
  3. 21
  4. 30 (correct answer)
Explanation: 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=642 ÷ 7 = 6 groups. Since each group contains 5 apprentices, multiply: 6×5=306 × 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.

Question 5

A number is doubled and then increased by 12. The result is 50. What is the original number?

  1. 19 (correct answer)
  2. 25
  3. 31
  4. 38
Explanation: When you encounter word problems that describe operations performed on an unknown number, your best approach is to translate the words into a mathematical equation and solve systematically. Let's call the original number xx. The problem states that this number is "doubled and then increased by 12," giving us 2x+122x + 12. Since "the result is 50," we can write the equation: 2x+12=502x + 12 = 50. To solve: subtract 12 from both sides to get 2x=382x = 38, then divide by 2 to find x=19x = 19. Let's verify: doubling 19 gives 38, and adding 12 gives 50. ✓ Now let's check why the other options don't work. Option B (25): If we double 25 and add 12, we get 2(25)+12=622(25) + 12 = 62, which is too large. Option C (31): Doubling gives us 62, plus 12 equals 74—way too high. Option D (38): This gives us 2(38)+12=882(38) + 12 = 88, which is nearly double what we need. These incorrect answers likely represent common algebraic mistakes. Someone might choose B by incorrectly thinking 2x=502x = 50 and solving x=25x = 25, forgetting about the "+12" entirely. Options C and D are further off and probably come from more fundamental setup errors. For IBEW math problems, always set up your equation carefully by identifying what operations are performed in what order, then work backwards algebraically. Double-check your answer by substituting it back into the original word problem.

Question 6

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?

  1. 3.5 hours
  2. 4 hours (correct answer)
  3. 4.5 hours
  4. 5.25 hours
Explanation: 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+60h315 = 75 + 60h ,where, where hh $ 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.

Question 7

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?

  1. 12 feet (correct answer)
  2. 13 feet
  3. 29 feet
  4. 41 feet
Explanation: 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 ww. The problem states the length is "5 feet more than twice the width," so length = 2w+52w + 5. Since perimeter equals twice the length plus twice the width, you get: 2w+2(2w+5)=822w + 2(2w + 5) = 82. Solving this equation: 2w+4w+10=822w + 4w + 10 = 82, which simplifies to 6w+10=826w + 10 = 82. Subtracting 10 from both sides gives 6w=726w = 72, so w=12w = 12 feet. Let's verify: if width = 12 feet, then length = 2(12)+5=292(12) + 5 = 29 feet. The perimeter would be 2(12)+2(29)=24+58=822(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 ww. 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.

Question 8

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?

  1. 40
  2. 52
  3. 72 (correct answer)
  4. 80
Explanation: 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 nn = number of nails and ss = number of screws. From the problem, we know two things: the screws equal 8 less than twice the nails (s=2n−8s = 2n - 8), and the total fasteners equal 112 (s+n=112s + n = 112). Now substitute the first equation into the second: (2n−8)+n=112(2n - 8) + n = 112. Simplifying: 3n−8=1123n - 8 = 112, so 3n=1203n = 120, which gives us n=40n = 40 nails. Therefore, s=2(40)−8=72s = 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+8s = 2n + 8 instead of s=2n−8s = 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=1122n - 8 = 112 directly without considering the total constraint, giving n=60n = 60 and s=112s = 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.

Question 9

An apprentice spends 1/31/3 of his workday pulling wire. He then spends 1/21/2 of the remaining time installing fixtures. If he has 3 hours left in his workday after these tasks, what is the total length of his workday in hours?

  1. 6 hours
  2. 8 hours
  3. 9 hours (correct answer)
  4. 12 hours
Explanation: When you encounter word problems involving fractions of time and remaining portions, work systematically through each step and track what portion of the original whole remains after each task. Let's call the total workday xx hours. The apprentice spends 13x\frac{1}{3}x hours pulling wire, leaving x−13x=23xx - \frac{1}{3}x = \frac{2}{3}x hours remaining. Next, he spends half of this remaining time installing fixtures: 12×23x=13x\frac{1}{2} \times \frac{2}{3}x = \frac{1}{3}x hours. After both tasks, the time left is: x−13x−13x=13xx - \frac{1}{3}x - \frac{1}{3}x = \frac{1}{3}x hours. Since this remaining time equals 3 hours, we have 13x=3\frac{1}{3}x = 3, so x=9x = 9 hours. Let's verify: In a 9-hour day, he pulls wire for 3 hours, leaving 6 hours. He then installs fixtures for half of that (3 hours), leaving exactly 3 hours remaining. ✓ Answer A (6 hours) would mean he pulls wire for 2 hours, installs fixtures for 2 hours, with 2 hours left—but the problem states 3 hours remain. Answer B (8 hours) gives 2⅔ hours pulling wire, 2⅔ hours on fixtures, and 2⅔ hours remaining—not the required 3 hours. Answer D (12 hours) results in 4 hours for each activity and 4 hours remaining, which is too much. For fraction word problems on the IBEW exam, always define your variable clearly, convert each step into mathematical expressions, and verify your answer by working forward through the original scenario.

Question 10

The formula for power (P) in a circuit is P=I2RP = I^2R, where I is current and R is resistance. If the resistance in a circuit is 8 ohms and it consumes 72 watts of power, what is the current (I) in amperes?

  1. 3 amperes (correct answer)
  2. 4 amperes
  3. 9 amperes
  4. 576 amperes
Explanation: When you encounter power calculations on the IBEW exam, you're working with one of the fundamental relationships in electrical theory. The power formula P=I2RP = I^2R is one of three ways to calculate power, and you'll need to manipulate it algebraically to solve for the unknown variable. Given that power (P) = 72 watts and resistance (R) = 8 ohms, you need to solve for current (I). Start by substituting the known values: 72=I2×872 = I^2 \times 8. Divide both sides by 8: 9=I29 = I^2. Take the square root of both sides: I=3I = 3 amperes. Looking at the wrong answers: Choice B (4 amperes) likely comes from incorrectly using P=IRP = IR instead of P=I2RP = I^2R, which would give 72=I×872 = I \times 8, so I=9I = 9, but then mistakenly thinking this equals 4. Choice C (9 amperes) is the result you'd get if you forgot to take the square root after finding I2=9I^2 = 9. Choice D (576 amperes) appears to come from multiplying 72 × 8 instead of dividing, showing a fundamental algebraic error. The correct answer is A (3 amperes). Remember this key strategy for IBEW power problems: always identify which power formula fits your given variables (P=VIP = VI, P=I2RP = I^2R, or P=V2/RP = V^2/R), then use careful algebra to isolate the unknown. Double-check by substituting your answer back into the original equation—here, 32×8=723^2 \times 8 = 72 watts confirms you're correct.

Question 11

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?

  1. 3 amperes
  2. 4 amperes
  3. 5 amperes (correct answer)
  4. 12 amperes
Explanation: When you encounter Ohm's Law problems on the IBEW exam, you're dealing with one of the fundamental relationships in electrical work: V=IRV = 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=VI=24V2A=12ΩR = \frac{V}{I} = \frac{24V}{2A} = 12Ω. Now apply Ohm's Law to the new voltage: I=VR=60V12Ω=5AI = \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=5A2A × 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.

Question 12

A motor is running at a certain speed. If the speed is increased by 150 RPM (revolutions per minute), the new speed is 40 RPM more than double the original speed. What was the original speed of the motor?

  1. 55 RPM
  2. 95 RPM
  3. 110 RPM (correct answer)
  4. 190 RPM
Explanation: When you encounter word problems involving relationships between speeds or other quantities, your key strategy is to translate the English description into mathematical equations. This type of algebraic reasoning is fundamental for electrical work calculations. Let's define the original speed as xx RPM. The problem states that when the speed increases by 150 RPM, the new speed becomes x+150x + 150. This new speed is also described as "40 RPM more than double the original speed," which translates to 2x+402x + 40. Setting up the equation: x+150=2x+40x + 150 = 2x + 40 Solving for xx: 150−40=2x−x150 - 40 = 2x - x 110=x110 = x So the original speed was 110 RPM. Let's verify: if the original speed is 110 RPM, then increasing by 150 gives us 260 RPM. Double the original speed (220) plus 40 also equals 260 RPM. ✓ Looking at the wrong answers: Choice A (55 RPM) would give you 205 RPM when increased by 150, but double 55 plus 40 equals only 150 RPM. Choice B (95 RPM) results in 245 RPM versus 230 RPM. Choice D (190 RPM) creates 340 RPM versus 420 RPM. For IBEW exam success, always translate word problems into clear algebraic equations before attempting calculations. Write down what you know, define your variable, and set up the relationship described in the problem. This systematic approach prevents arithmetic errors and helps you tackle more complex electrical calculations with confidence.

Question 13

A team of two electricians, a senior and a junior, completed a job. The senior electrician worked 4 hours more than the junior electrician. Together, they worked a total of 22 hours. For how many hours did the senior electrician work?

  1. 9 hours
  2. 11 hours
  3. 13 hours (correct answer)
  4. 18 hours
Explanation: When you encounter word problems involving two people working different amounts of time, you're dealing with a system of equations. The key is translating the verbal descriptions into mathematical relationships. Let's define variables: let jj = hours the junior electrician worked, and ss = hours the senior electrician worked. The problem gives you two pieces of information that become equations:
  1. "The senior electrician worked 4 hours more than the junior": s=j+4s = j + 4
  2. "Together, they worked a total of 22 hours": s+j=22s + j = 22
Now substitute the first equation into the second: (j+4)+j=22(j + 4) + j = 22. Simplifying: 2j+4=222j + 4 = 22, so 2j=182j = 18, which means j=9j = 9. Therefore, the senior worked s=9+4=13s = 9 + 4 = 13 hours. Looking at the wrong answers: Choice A (9 hours) is the junior electrician's time—this happens when you solve for the wrong variable or mix up which person the question asks about. Choice B (11 hours) comes from incorrectly averaging the total hours (22 ÷ 2 = 11), but this ignores the 4-hour difference between workers. Choice D (18 hours) results from subtracting 4 from 22 instead of properly setting up the equations—a common arithmetic mistake under time pressure. The correct answer is C (13 hours). For IBEW word problems, always define your variables clearly, translate each sentence into an equation, and double-check which person or quantity the question is asking about. Many wrong answers exploit mix-ups between related values.

Question 14

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?

  1. 12 feet
  2. 15 feet (correct answer)
  3. 18 feet
  4. 37.5 feet
Explanation: 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 cm×1 inch2.5 cm×1 foot12 inches450 \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 inches450 ÷ 2.5 = 180 \text{ inches} Then convert inches to feet: 180÷12=15 feet180 ÷ 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.

Question 15

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?

  1. 3.5 hours
  2. 4 hours (correct answer)
  3. 4.5 hours
  4. 8 hours
Explanation: 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=10045 + 55 = 100 mph. To find when they'll be 400 miles apart, use the formula: Time = Distance ÷ Rate. So: Time=400 miles100 mph=4 hours\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 100×3.5=350100 \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.

Question 16

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?

  1. 2 outlets
  2. 8 outlets
  3. 16 outlets (correct answer)
  4. 24 outlets
Explanation: 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 12 outlets3 hours=4 outlets per hour\frac{12 \text{ outlets}}{3 \text{ hours}} = 4 \text{ outlets per hour}. The apprentice installs 8 outlets in 4 hours, so their rate is 8 outlets4 hours=2 outlets per hour\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 4×8=32 outlets4 \times 8 = 32 \text{ outlets}, while the apprentice would install 2×8=16 outlets2 \times 8 = 16 \text{ outlets}. The difference is 32−16=16 outlets32 - 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.

Question 17

A crew of 4 electricians can wire a small building in 5 days. To finish the job faster, the contractor adds 1 more electrician to the crew. Assuming all electricians work at the same rate, how many days will it take the 5-person crew to wire the building?

  1. 3 days
  2. 4 days (correct answer)
  3. 5 days
  4. 6.25 days
Explanation: When you encounter work rate problems on the IBEW exam, remember that the key principle is: total work remains constant, but adding workers changes the time needed to complete it. Start by finding the total work units. If 4 electricians complete the job in 5 days, the total work equals 4×5=204 \times 5 = 20 "electrician-days." This means the building requires 20 days of work from one electrician, or equivalent combinations that multiply to 20. Now with 5 electricians working at the same rate, you divide the total work by the new crew size: 20 electrician-days5 electricians=4 days\frac{20 \text{ electrician-days}}{5 \text{ electricians}} = 4 \text{ days}. The 5-person crew will complete the job in 4 days. Looking at the wrong answers: Choice A (3 days) would require 5×3=155 \times 3 = 15 electrician-days, which underestimates the work needed. Choice C (5 days) assumes adding a worker doesn't reduce the time at all—this ignores the inverse relationship between crew size and completion time. Choice D (6.25 days) represents a common error where students incorrectly calculate 54×5=6.25\frac{5}{4} \times 5 = 6.25, but this formula has no logical basis in work rate problems. For IBEW work rate questions, always remember the formula: Workers×Time=Constant Work\text{Workers} \times \text{Time} = \text{Constant Work}. Calculate the total work first, then solve for the unknown variable. This systematic approach prevents calculation errors and helps you recognize when answer choices don't make mathematical sense.

Question 18

A project requires running conduit up a wall and across a ceiling. The length of conduit needed for the ceiling is 10 feet more than the length needed for the wall. If a total of 52 feet of conduit is used, what is the length of the ceiling section?

  1. 21 feet
  2. 26 feet
  3. 31 feet (correct answer)
  4. 42 feet
Explanation: This is a classic word problem that tests your ability to translate written descriptions into algebraic equations - a skill you'll use regularly when calculating conduit runs and material requirements on electrical jobs. Let's define our variables: let the wall length be ww feet and the ceiling length be cc feet. From the problem, we know two key relationships: the ceiling section is 10 feet longer than the wall section (c=w+10c = w + 10), and the total conduit used is 52 feet (w+c=52w + c = 52). Substituting the first equation into the second: w+(w+10)=52w + (w + 10) = 52. This simplifies to 2w+10=522w + 10 = 52, so 2w=422w = 42 and w=21w = 21 feet. Therefore, the ceiling section is c=21+10=31c = 21 + 10 = 31 feet. Looking at the wrong answers: Choice A (21 feet) is the length of the wall section, not the ceiling - this catches students who solve correctly but answer the wrong question. Choice B (26 feet) is the average of the two sections (522=26\frac{52}{2} = 26), which some students mistakenly calculate when they don't properly account for the 10-foot difference. Choice D (42 feet) comes from subtracting 10 from the total instead of using proper algebra (52−10=4252 - 10 = 42). When tackling word problems on the IBEW exam, always define your variables clearly, write out both relationships as equations, and double-check that you're answering what the question actually asks for. Many incorrect answers deliberately target common algebraic mistakes.

Question 19

An electrician's petty cash box contains only $5 bills and $10 bills. There are 25 bills in total, and their combined value is $165. How many $5 bills are in the box?

  1. 8
  2. 12
  3. 15
  4. 17 (correct answer)
Explanation: This is a classic system of equations problem that tests your ability to translate word problems into mathematical relationships. When you see problems involving two types of items with different values and constraints on both quantity and total value, you'll need to set up two equations with two variables. Let's define variables: let xx = number of $5 bills and $yy =numberof= number of10 bills. From the problem, we can write two equations: x+y=25x + y = 25 (total bills) and 5x+10y=1655x + 10y = 165 (total value). From the first equation, y=25−xy = 25 - x. Substituting into the second equation: 5x+10(25−x)=1655x + 10(25 - x) = 165. Simplifying: 5x+250−10x=1655x + 250 - 10x = 165, which gives us −5x=−85-5x = -85, so x=17x = 17. We can verify: 17 five-dollar bills and 8 ten-dollar bills gives us 17(5)+8(10)=85+80=16517(5) + 8(10) = 85 + 80 = 165. Looking at the wrong answers: Choice A (8) likely comes from calculating the number of $10 bills instead of $5 bills. Choice B (12) might result from arithmetic errors in the substitution process. Choice C (15) could come from incorrectly setting up the value equation or making calculation mistakes during solving. For IBEW math problems, always define your variables clearly and double-check your answer by substituting back into both original conditions. These systematic equation problems appear frequently, so practice translating word problems into mathematical relationships and solving them step-by-step.

Question 20

A contractor purchases a total of 50 light fixtures. Some are LED fixtures costing $25 each, and the rest are fluorescent fixtures costing $15 each. If the total cost was $1050, how many LED fixtures were purchased?

  1. 20
  2. 25
  3. 30 (correct answer)
  4. 35
Explanation: This is a classic system of equations problem that frequently appears on electrical trade exams. When you see two types of items with different costs totaling a fixed amount and cost, you're dealing with a two-variable linear system. Let's define variables: let xx = LED fixtures and yy = fluorescent fixtures. You can set up two equations from the given information:
  • Total fixtures: x+y=50x + y = 50
  • Total cost: 25x+15y=105025x + 15y = 1050
From the first equation, y=50−xy = 50 - x. Substituting into the cost equation: 25x+15(50−x)=105025x + 15(50 - x) = 1050 25x+750−15x=105025x + 750 - 15x = 1050 10x=30010x = 300 x=30x = 30 So 30 LED fixtures were purchased, confirming answer C. Looking at the wrong answers: A) 20 LED fixtures would give a total cost of 25(20)+15(30)=500+450=95025(20) + 15(30) = 500 + 450 = 950, which is $100 short. B) 25 LED fixtures would cost $25(25)+15(25)=625+375=100025(25) + 15(25) = 625 + 375 = 1000 ,still$50short.D)35LEDfixtureswouldcost$, still $50 short. D) 35 LED fixtures would cost $25(35) + 15(15) = 875 + 225 = 1100$$, which exceeds the budget by $50. For IBEW exam success, always set up your system methodically: identify what each variable represents, write both constraint equations clearly, and substitute to solve. These material cost problems are common in electrical work, so practicing the algebraic setup will serve you well both on the exam and in real project estimating.