IBEW: Electrical Training Alliance Aptitude Test Quiz: Solve Linear Equations

20 questions · exam conditions

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Solve Linear EquationsQuestion 1 of 20

Solve for $x$ : $3(x + 2) = 3x - 1$

x = 0

x = -7/6

No solution

All real numbers

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IBEW: Electrical Training Alliance Aptitude Test Quiz

IBEW: Electrical Training Alliance Aptitude Test Quiz: Solve Linear Equations

Practice Solve Linear Equations 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 Linear Equations, 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

Solve for $x$ : $3(x + 2) = 3x - 1$

$x = 0$
$x = -7/6$
No solution (correct answer)
All real numbers

Explanation: When you encounter an equation like this on the IBEW exam, you're dealing with a linear equation that may have one solution, no solution, or infinitely many solutions. The key is to carefully distribute and simplify to see what emerges. Let's solve

3(x + 2) = 3x - 1

step by step. First, distribute the 3 on the left side:

3x + 6 = 3x - 1

. Now subtract

3x

from both sides to isolate the variable terms:

6 = -1

. This gives us a mathematical impossibility - 6 can never equal -1. When you arrive at a false statement like

6 = -1

, it means the original equation has no solution. This is answer choice C. Let's examine why the other options are wrong. Choice A suggests

x = 0

, but if you substitute 0 into the original equation, you get

3(0 + 2) = 6

on the left and

3(0) - 1 = -1

on the right, so

6 \neq -1

. Choice B proposes

x = -7/6

, but substituting this value also fails to satisfy the equation. Choice D claims all real numbers work, which would only be true if we had arrived at an identity like

0 = 0

after simplification. Remember this pattern for the IBEW exam: when solving linear equations, if you eliminate all variables and get a false statement, there's no solution. If you get a true statement like

0 = 0

, then all real numbers are solutions. Otherwise, you'll have exactly one solution.

Question 2

Solve for $b$ : $4b + 7 - b = 16$

$b = 3$ (correct answer)
$b = 1.8$
$b = 9$
$b = -3$

Explanation: When you encounter linear equations on the IBEW exam, your goal is to isolate the variable by combining like terms and using inverse operations systematically. Start by simplifying the left side of the equation

4b + 7 - b = 16

. Combine the like terms containing

b

4b - b = 3b

. This gives you

3b + 7 = 16

. Next, isolate the term with the variable by subtracting 7 from both sides:

3b + 7 - 7 = 16 - 7

, which simplifies to

3b = 9

. Finally, solve for

b

by dividing both sides by 3:

b = \frac{9}{3} = 3

. Let's examine why the other answers are incorrect. Answer B (

b = 1.8

) likely results from incorrectly dividing 9 by 5 instead of 3, possibly from miscounting the coefficients when combining like terms. Answer C (

b = 9

) represents stopping too early in the solution process—this is the value of

3b

, not

b

itself. Answer D (

b = -3

) could come from sign errors when combining terms or performing operations. You can verify that A is correct by substituting back into the original equation:

4(3) + 7 - 3 = 12 + 7 - 3 = 16

✓ For IBEW algebra problems, always work methodically: combine like terms first, then isolate the variable using inverse operations. Double-check your work by substituting your answer back into the original equation—this catches calculation errors that could cost you points.

Question 3

A heater draws $12$ amperes on a $120$ volt circuit. Using Ohm's Law with resistance $R$ in ohms, form $120=12R$ and solve for $R$ to check the element resistance before installation.

10 ohms (correct answer)
1,440 ohms
108 ohms
12 ohms

Explanation: This question tests solving single-variable linear equations, a fundamental concept in algebra necessary for electrical calculations. Single-variable linear equations involve finding the value of an unknown variable that satisfies the equation. Techniques include isolating the variable through inverse operations like addition, subtraction, multiplication, and division. In this problem, the equation 120=12R represents checking heater element resistance using Ohm's Law, and solving it involves dividing both sides by 12. Choice A is correct because solving the equation step-by-step gives R=120/12, leading to R=10 ohms. This shows understanding of isolating the variable correctly. Choice B is incorrect because it demonstrates multiplying instead of dividing, resulting in 1440 ohms. This often happens when students overlook the inverse operation for isolation. To help students: Emphasize checking units and logical consistency within the problem. Practice solving similar equations with incremental complexity to build confidence. Encourage double-checking calculations to avoid arithmetic errors.

Question 4

Solve for $x$ : $\frac{1}{2}x + 3 = \frac{1}{4}x + 5$

$x = 8$ (correct answer)
$x = -8$
$x = 32$
$x = 8/3$

Explanation: When you encounter linear equations with fractions on the IBEW exam, your goal is to isolate the variable by eliminating fractions and combining like terms systematically. Start by clearing the fractions to make the algebra cleaner. Multiply every term by 4 (the least common denominator of 2 and 4):

4 \cdot \frac{1}{2}x + 4 \cdot 3 = 4 \cdot \frac{1}{4}x + 4 \cdot 5

This gives you:

2x + 12 = x + 20

Now subtract

x

from both sides:

2x - x + 12 = 20

, which simplifies to

x + 12 = 20

Finally, subtract 12 from both sides:

x = 8

Choice A (

x = 8

) is correct. You can verify by substituting back:

\frac{1}{2}(8) + 3 = 4 + 3 = 7

and

\frac{1}{4}(8) + 5 = 2 + 5 = 7

✓ Choice B (

x = -8

) results from sign errors during the solving process. If you substitute -8, you get

\frac{1}{2}(-8) + 3 = -1

and

\frac{1}{4}(-8) + 5 = 3

, which aren't equal. Choice C (

x = 32

) likely comes from incorrectly multiplying by 8 instead of the LCD of 4, or making arithmetic errors when combining terms. Choice D (

x = \frac{8}{3}

) suggests you may have made errors in fraction manipulation or incorrectly cross-multiplied. For IBEW math problems, always clear fractions first by multiplying by the LCD, then solve systematically. Double-check your answer by substituting back into the original equation—this catches most algebraic mistakes.

Question 5

Find the value of $x$ that solves the equation: $0.5x + 1.2 = 3.7$

$x = 1.25$
$x = 2.5$
$x = 5$ (correct answer)
$x = 9.8$

Explanation: When you encounter linear equations with decimals on the IBEW exam, treat them exactly like any other algebraic equation—the decimal points don't change the fundamental solving process. To solve

0.5x + 1.2 = 3.7

, you need to isolate

x

by using inverse operations. First, subtract 1.2 from both sides:

0.5x + 1.2 - 1.2 = 3.7 - 1.2

, which gives you

0.5x = 2.5

. Then divide both sides by 0.5:

x = \frac{2.5}{0.5} = 5

. You can verify this by substituting back:

0.5(5) + 1.2 = 2.5 + 1.2 = 3.7

✓ Looking at the wrong answers: Choice A (

x = 1.25

) would give you

0.5(1.25) + 1.2 = 0.625 + 1.2 = 1.825

, which is far from 3.7. Choice B (

x = 2.5

) represents a common error where students stop after the first step, mistaking

0.5x = 2.5

as the final answer instead of solving for

x

. Choice D (

x = 9.8

) likely comes from incorrectly adding 1.2 to 3.7 instead of subtracting, then multiplying by 0.5 instead of dividing. For IBEW electrical calculations, you'll frequently work with decimal coefficients in voltage, current, and resistance problems. Practice the division of decimals mentally or remember that dividing by 0.5 is the same as multiplying by 2—this can speed up your calculations during the exam.

Question 6

What is the solution to the equation $5x - 3 = 8$ ?

$x = 1$
$x = 11/5$ (correct answer)
$x = 5/11$
$x = 11$

Explanation: Linear equations like

5x - 3 = 8

are fundamental to electrical work, appearing in circuit analysis, voltage calculations, and power formulas. When you encounter any linear equation, your goal is to isolate the variable by performing the same operation on both sides. Start by adding 3 to both sides to eliminate the constant term on the left:

5x - 3 + 3 = 8 + 3

, which simplifies to

5x = 11

. Next, divide both sides by 5 to isolate x:

\frac{5x}{5} = \frac{11}{5}

, giving you

x = \frac{11}{5}

. Let's examine why the other options are incorrect. Choice A (

x = 1

) comes from a common error where students might subtract incorrectly or forget to divide by the coefficient. If you substitute

x = 1

back into the original equation:

5(1) - 3 = 2

, not 8. Choice C (

x = \frac{5}{11}

) represents flipping the fraction incorrectly—this happens when students rush through the division step. Choice D (

x = 11

) occurs when students forget to divide by the coefficient of x entirely, stopping after finding that

5x = 11

. The correct answer is B:

x = \frac{11}{5}

. Always check your work by substituting your answer back into the original equation. Here:

5(\frac{11}{5}) - 3 = 11 - 3 = 8

✓. This verification step catches arithmetic errors and builds confidence—essential skills when working with electrical formulas where precision matters for safety and functionality.

Question 7

Solve for $x$ in the equation: $4(x - 3) = 20$

$x = 8$ (correct answer)
$x = 5$
$x = 7$
$x = 2$

Explanation: This problem tests your ability to solve linear equations with parentheses, a fundamental algebra skill you'll use constantly in electrical calculations involving circuits, power, and resistance. To solve

4(x - 3) = 20

, you need to isolate

x

using inverse operations. Start by distributing the 4 to everything inside the parentheses:

4 \cdot x - 4 \cdot 3 = 20

, which gives you

4x - 12 = 20

. Next, add 12 to both sides to eliminate the constant term:

4x - 12 + 12 = 20 + 12

, so

4x = 32

. Finally, divide both sides by 4:

x = 8

. You can verify this by substituting back:

4(8 - 3) = 4(5) = 20

✓ Looking at the wrong answers: B)

x = 5

is what you get if you forget to add 12 back after distributing—a common error where students stop at

4x - 12 = 20

and incorrectly think

4x = 20

. C)

x = 7

might result from arithmetic mistakes during the solving process. D)

x = 2

could happen if you incorrectly distributed or made sign errors along the way. For IBEW success, master the systematic approach: distribute first, combine like terms, then use inverse operations to isolate the variable. Always check your answer by substituting back into the original equation—this catches calculation errors and builds confidence in electrical problem-solving where precision is critical.

Question 8

Find the value of $x$ that satisfies the equation: $10 - (2x + 1) = 3$

$x = 4$
$x = -3$
$x = 3$ (correct answer)
$x = 6$

Explanation: When you encounter linear equations with parentheses like this one, your primary task is to isolate the variable through systematic algebraic manipulation. Start by simplifying both sides of the equation step by step. Begin with the original equation:

10 - (2x + 1) = 3

. First, distribute the negative sign through the parentheses:

10 - 2x - 1 = 3

. This simplifies to

9 - 2x = 3

. Next, isolate the term containing

x

by subtracting 9 from both sides:

-2x = 3 - 9

, which gives you

-2x = -6

. Finally, divide both sides by -2 to solve for

x

x = \frac{-6}{-2} = 3

. Therefore, answer C is correct. Let's examine why the other options are wrong. If you chose A (

x = 4

), you likely made an error when distributing the negative sign or combining like terms. Substituting

x = 4

back into the original equation gives

10 - (8 + 1) = 1

, not 3. Option B (

x = -3

) might result from sign errors during the solving process—perhaps forgetting that dividing two negative numbers yields a positive result. Option D (

x = 6

) could occur if you incorrectly handled the parentheses or made arithmetic mistakes during simplification. Always verify your answer by substituting it back into the original equation. For IBEW exam success, practice distributing negative signs carefully and double-check your arithmetic at each step—small calculation errors are common traps in electrical work scenarios.

Question 9

Find the value of $n$ in the following equation: $\frac{2}{3}n - 1 = 5$

$n = 4$
$n = 9$ (correct answer)
$n = 8$
$n = 6$

Explanation: When you encounter linear equations like this one, you're solving for an unknown variable using algebraic manipulation. The key is to isolate the variable by performing the same operations on both sides of the equation. Starting with

\frac{2}{3}n - 1 = 5

, first add 1 to both sides to eliminate the constant term:

\frac{2}{3}n = 6

. Next, multiply both sides by

\frac{3}{2}

(the reciprocal of

\frac{2}{3}

) to isolate

n

n = 6 \times \frac{3}{2} = \frac{18}{2} = 9

. You can verify this by substituting back:

\frac{2}{3}(9) - 1 = 6 - 1 = 5

✓ Let's examine why the other answers fail. Choice A (

n = 4

) gives us

\frac{2}{3}(4) - 1 = \frac{8}{3} - 1 = \frac{5}{3} \approx 1.67

, which doesn't equal 5. Choice C (

n = 8

) yields

\frac{2}{3}(8) - 1 = \frac{16}{3} - 1 = \frac{13}{3} \approx 4.33

, still incorrect. Choice D (

n = 6

) produces

\frac{2}{3}(6) - 1 = 4 - 1 = 3

, not 5. For IBEW exam success, master the two-step process for linear equations: first isolate the variable term by adding or subtracting constants, then isolate the variable itself by multiplying or dividing. Always check your answer by substituting it back into the original equation—this catches calculation errors and builds confidence in your solution.

Question 10

Solve the equation for $x$ : $7 - 2x = 5x - 14$

$x = -3$
$x = 1$
$x = 3$ (correct answer)
$x = -1$

Explanation: Linear equations like this are fundamental to electrical calculations you'll encounter in the field, from calculating voltage drops to determining current flow. When you see an equation with variables on both sides, your goal is to isolate the variable by collecting like terms. Start by moving all terms with

x

to one side and all constants to the other. From

7 - 2x = 5x - 14

, add

2x

to both sides:

7 = 7x - 14

. Next, add

14

to both sides:

21 = 7x

. Finally, divide both sides by

7

x = 3

. You can verify this by substituting back into the original equation:

7 - 2(3) = 1

and

5(3) - 14 = 1

. Both sides equal

1

, confirming our answer. Let's examine why the other options are incorrect. Choice A (

x = -3

) would give you

7 - 2(-3) = 13

on the left side but

5(-3) - 14 = -29

on the right side—these don't match. Choice B (

x = 1

) produces

7 - 2(1) = 5

on the left but

5(1) - 14 = -9

on the right. Choice D (

x = -1

) yields

7 - 2(-1) = 9

on the left and

5(-1) - 14 = -19

on the right. For IBEW exam success, always work systematically: collect like terms, isolate the variable, then verify your answer by substitution. This methodical approach prevents algebraic errors that can cost points on electrical calculations.

Question 11

A certain number is doubled, and then 5 is subtracted from the result. If the final answer is 23, what is the original number?

9
14 (correct answer)
28
51

Explanation: This is a classic "working backwards" algebra problem that tests your ability to set up and solve linear equations. When you see a problem describing a sequence of operations performed on an unknown number, you need to translate the words into mathematical expressions. Let's call the original number

x

. The problem states: "A certain number is doubled" gives us

2x

. Then "5 is subtracted from the result" gives us

2x - 5

. Since "the final answer is 23," we can write the equation:

2x - 5 = 23

. To solve:

2x - 5 = 23

, add 5 to both sides:

2x = 28

, then divide by 2:

x = 14

. Let's verify: double 14 to get 28, subtract 5 to get 23 ✓ Looking at the wrong answers: Choice A) 9 would give you

2(9) - 5 = 18 - 5 = 13

, not 23. Choice C) 28 is a trap answer—this is what you get after doubling the original number but before subtracting 5. If you used 28 as your final answer, you'd get

2(28) - 5 = 56 - 5 = 51

, which is way too large. Choice D) 51 is another trap—this would be the result if you mistakenly added 5 instead of subtracting it from the doubled number. For word problems like this, always define your variable clearly, translate each operation step-by-step into mathematical notation, and check your answer by substituting back into the original problem. This verification step catches most calculation errors.

Question 12

Solve the following equation for $a$ : $5a + 2 = 3a + 10$

$a = 6$
$a = 4$ (correct answer)
$a = 1.5$
$a = -4$

Explanation: When solving linear equations like this one, you're working to isolate the variable on one side of the equation. The goal is to get all terms with

a

on one side and all constants on the other. Starting with

5a + 2 = 3a + 10

, first subtract

3a

from both sides to collect the

a

terms:

5a - 3a + 2 = 3a - 3a + 10

, which simplifies to

2a + 2 = 10

. Next, subtract 2 from both sides to isolate the term with

a

2a + 2 - 2 = 10 - 2

, giving you

2a = 8

. Finally, divide both sides by 2:

a = 4

. You can verify this by substituting back into the original equation:

5(4) + 2 = 20 + 2 = 22

and

3(4) + 10 = 12 + 10 = 22

. Both sides equal 22, confirming our solution. Choice A (

a = 6

) results from incorrectly adding

3a

to the left side instead of subtracting it. Choice C (

a = 1.5

) comes from making an error when dividing, possibly confusing the final step. Choice D (

a = -4

) typically results from sign errors during the algebraic manipulation, such as incorrectly handling the subtraction steps. For IBEW electrical calculations, you'll encounter similar algebraic manipulation when working with Ohm's law variations, power formulas, and circuit analysis. Always double-check your algebra by substituting your answer back into the original equation—this catches computational errors before they cost you points.

Question 13

Find the value of $m$ that solves the equation: $15 - 3m = 6$

$m = -7$
$m = -3$
$m = 7$
$m = 3$ (correct answer)

Explanation: Linear equations are fundamental in electrical work, appearing in calculations for voltage drops, current relationships, and circuit analysis. When you encounter an equation like

15 - 3m = 6

, you need to isolate the variable using inverse operations. Start by getting all terms with the variable on one side and constants on the other. Subtract 15 from both sides:

15 - 3m - 15 = 6 - 15

, which simplifies to

-3m = -9

. Now divide both sides by -3 to isolate

m

m = \frac{-9}{-3} = 3

. Let's verify: substituting

m = 3

back into the original equation gives

15 - 3(3) = 15 - 9 = 6

✓ Looking at the wrong answers: Choice A (

m = -7

) would give you

15 - 3(-7) = 15 + 21 = 36

, not 6. This represents a sign error during solving. Choice B (

m = -3

) yields

15 - 3(-3) = 15 + 9 = 24

, suggesting confusion about which operations to perform. Choice C (

m = 7

) produces

15 - 3(7) = 15 - 21 = -6

, which is the opposite sign of the target value—likely from an error when handling the subtraction of 15. The correct answer is D:

m = 3

. For IBEW exam success, always verify your solution by substituting back into the original equation. This catches arithmetic errors and confirms your answer, especially crucial when working with the mathematical relationships you'll use daily in electrical calculations.

Question 14

Solve for $w$ in the equation: $5(2w - 1) - 3 = 12$

$w = 0.4$
$w = 1.6$
$w = 1.4$
$w = 2$ (correct answer)

Explanation: When you encounter multi-step algebraic equations like this one, your goal is to isolate the variable by systematically undoing the operations in reverse order. This requires careful attention to the order of operations and algebraic manipulation. Let's solve

5(2w - 1) - 3 = 12

step by step. First, add 3 to both sides:

5(2w - 1) = 15

. Next, divide both sides by 5:

2w - 1 = 3

. Then add 1 to both sides:

2w = 4

. Finally, divide by 2:

w = 2

. You can verify this by substituting back:

5(2(2) - 1) - 3 = 5(3) - 3 = 12

✓ Now let's examine why the other answers are incorrect. Choice A (

w = 0.4

) likely results from calculation errors in the division steps—perhaps dividing incorrectly when isolating the variable. Choice B (

w = 1.6

) might come from distributing incorrectly at the beginning or making arithmetic mistakes when combining like terms. Choice C (

w = 1.4

) could result from errors in handling the order of operations, possibly adding or subtracting incorrectly in the middle steps. The correct answer is D:

w = 2

. For IBEW exam success, always work systematically through algebraic equations: distribute first, combine like terms, then isolate the variable using inverse operations. Most importantly, always substitute your answer back into the original equation to verify it works—this catches calculation errors and builds confidence in your solution.

Question 15

Solve for $y$ : $6(y - 2) + 8 = 4(y + 3)$

$y = 8$ (correct answer)
$y = 10$
$y = 4$
$y = -4$

Explanation: When you encounter linear equations with variables on both sides, your goal is to isolate the variable by systematically eliminating terms through inverse operations. Start by distributing on both sides:

6(y - 2) + 8 = 4(y + 3)

becomes

6y - 12 + 8 = 4y + 12

, which simplifies to

6y - 4 = 4y + 12

. Next, collect like terms by subtracting

4y

from both sides:

2y - 4 = 12

. Add 4 to both sides:

2y = 16

. Finally, divide by 2:

y = 8

. Let's verify: substituting

y = 8

into the original equation gives

6(8 - 2) + 8 = 6(6) + 8 = 44

on the left, and

4(8 + 3) = 4(11) = 44

on the right. Both sides equal 44, confirming our answer. Looking at the wrong choices: Choice B (

y = 10

) likely results from incorrectly adding instead of subtracting when moving terms across the equals sign. Choice C (

y = 4

) could come from distribution errors or miscalculating

16 ÷ 2

. Choice D (

y = -4

) might result from sign errors during the solving process or incorrectly handling the negative terms. For IBEW exam success, always verify your algebraic solutions by substituting back into the original equation. This catches arithmetic mistakes and ensures you haven't introduced errors while manipulating the equation. Practice distributing carefully and tracking positive/negative signs—these are the most common sources of errors in linear equation problems.

Question 16

Find the value of $x$ in the equation: $\frac{x + 5}{2} = 9$

$x = 4$
$x = 13$ (correct answer)
$x = 8$
$x = 23$

Explanation: When you encounter linear equations with fractions like this one, your goal is to isolate the variable

x

by performing the same operations on both sides of the equation. Starting with

\frac{x + 5}{2} = 9

, you need to eliminate the fraction first. Multiply both sides by 2 to clear the denominator:

2 \cdot \frac{x + 5}{2} = 2 \cdot 9

. This simplifies to

x + 5 = 18

. Now subtract 5 from both sides:

x + 5 - 5 = 18 - 5

, giving you

x = 13

. You can verify this by substituting back:

\frac{13 + 5}{2} = \frac{18}{2} = 9

✓ Looking at the wrong answers: Choice A (

x = 4

) likely comes from subtracting 5 from 9 instead of properly solving the equation—this ignores the fraction entirely. Choice C (

x = 8

) might result from the error

9 - 5 + 4 = 8

, showing confused arithmetic when trying to work backwards from the answer. Choice D (

x = 23

) could come from multiplying 9 by 2 and then adding 5 (

18 + 5 = 23

), which reverses the correct order of operations. For IBEW exam success with algebraic equations, always work systematically: clear fractions first by multiplying both sides by the denominator, then isolate the variable using inverse operations. Double-check your answer by substituting it back into the original equation—this catches calculation errors and confirms you've solved correctly.

Question 17

During a rough-in, an electrician pulls wire at a steady rate of $18$ feet per minute. If the total pulled is $144$ feet, let $t$ be time in minutes and use $18t=144$ to determine how long the pull takes.

8 minutes (correct answer)
126 minutes
18 minutes
2 minutes

Explanation: This question tests solving single-variable linear equations, a fundamental concept in algebra necessary for electrical calculations. Single-variable linear equations involve finding the value of an unknown variable that satisfies the equation. Techniques include isolating the variable through inverse operations like addition, subtraction, multiplication, and division. In this problem, the equation 18t=144 represents determining time for pulling wire at a given rate, and solving it involves dividing both sides by 18. Choice A is correct because solving the equation step-by-step gives t=144/18, leading to t=8 minutes. This shows understanding of isolating the variable correctly. Choice D is incorrect because it demonstrates dividing incorrectly, such as 18/144=2 instead of the reverse. This often happens when students overlook the variable's position. To help students: Emphasize checking units and logical consistency within the problem. Practice solving similar equations with incremental complexity to build confidence. Encourage double-checking calculations to avoid arithmetic errors.

Question 18

While testing a branch circuit, the meter reads $V=48$ volts and $I=6$ amperes. Using Ohm's Law with resistance $R$ in ohms, set up $48=6R$ and solve for $R$ to verify the conductor load is within spec.

6 ohms
8 ohms (correct answer)
42 ohms
288 ohms

Explanation: This question tests solving single-variable linear equations, a fundamental concept in algebra necessary for electrical calculations. Single-variable linear equations involve finding the value of an unknown variable that satisfies the equation. Techniques include isolating the variable through inverse operations like addition, subtraction, multiplication, and division. In this problem, the equation 48=6R represents verifying conductor load using Ohm's Law, and solving it involves dividing both sides by 6. Choice B is correct because solving the equation step-by-step gives R=48/6, leading to R=8 ohms. This shows understanding of isolating the variable correctly. Choice A is incorrect because it demonstrates using multiplication instead of division, resulting in 6 ohms. This often happens when students overlook the inverse operation needed. To help students: Emphasize checking units and logical consistency within the problem. Practice solving similar equations with incremental complexity to build confidence. Encourage double-checking calculations to avoid arithmetic errors.

Question 19

A crew installs 48 feet of cable tray at 6 feet per hour; determine time $t$ in $6t=48$ .

7 hours
8 hours (correct answer)
288 hours
8 feet

Explanation: This question tests solving single-variable linear equations, a fundamental concept in algebra necessary for electrical calculations. Single-variable linear equations involve finding the value of an unknown variable that satisfies the equation. Techniques include isolating the variable through inverse operations like addition, subtraction, multiplication, and division. In this problem, the equation 6t = 48 represents the time to install 48 feet of cable tray at 6 feet per hour, and solving it involves dividing total length by rate. Choice B is correct because solving the equation step-by-step gives t = 48 / 6 = 8 hours, leading to t = 8 hours. This shows understanding of isolating the variable correctly. Choice A is incorrect because it demonstrates a misdivision; this often happens when students overlook the rate. To help students: Emphasize checking units and logical consistency within the problem. Practice solving similar equations with incremental complexity to build confidence. Encourage double-checking calculations to avoid arithmetic errors.

Question 20

A 12-ohm resistor carries 2.5 amperes; determine voltage $V$ using $V=IR$ .

9.5 volts
30 volts (correct answer)
4.8 volts
30 ohms

Explanation: This question tests solving single-variable linear equations, a fundamental concept in algebra necessary for electrical calculations. Single-variable linear equations involve finding the value of an unknown variable that satisfies the equation. Techniques include isolating the variable through inverse operations like addition, subtraction, multiplication, and division. In this problem, the equation V = 2.5 * 12 represents the voltage across a 12-ohm resistor with 2.5 amperes, and solving it involves multiplying current by resistance. Choice B is correct because solving the equation step-by-step gives V = 2.5 * 12 = 30 volts, leading to V = 30 volts. This shows understanding of isolating the variable correctly. Choice A is incorrect because it demonstrates a multiplication error; this often happens when students overlook decimal points. To help students: Emphasize checking units and logical consistency within the problem. Practice solving similar equations with incremental complexity to build confidence. Encourage double-checking calculations to avoid arithmetic errors.