IBEW Aptitude Test common mistakes
IBEW Aptitude Test · Understand the traps and how to avoid them.
When a linear equation has fractions such as (x/3) + (x/4) = 7, the candidate multiplies only the first term by the common denominator, or adds the fractions without first clearing denominators, and lands on x = 12 instead of x = 12 checked against the original.
Fractional coefficients get treated as decoration rather than as numbers that must be eliminated consistently. Under time pressure the candidate clears the denominator on the term that looks hardest and forgets the others, so the equation silently stops being equivalent.
What to do instead
Adopt a fixed two-line ritual: write the LCD for every denominator in the equation, multiply every single term on both sides by it (draw a tick over each term as you do), then solve the now-integer equation. Finish by substituting the answer back into the original fractional form, not the cleared form. Practise 10 equations with three different denominators until the tick-every-term step is automatic.
The candidate distributes a negative or a coefficient across a binomial incorrectly, writing −(x − 5) as −x − 5 or 3(x + 2) as 3x + 2, then solves the resulting wrong equation confidently.
The minus sign is processed as an operator between terms rather than as a multiplier of the whole group. This is a habit carried from arithmetic, where a leading minus rarely has to reach inside parentheses, so it fails the first time algebra demands it.
What to do instead
Force the invisible 1: rewrite −(x − 5) as −1(x − 5) before expanding, and say the distribution aloud ('negative x, plus five'). After expanding any expression with a leading sign, do a one-second sign audit of each resulting term. Build a set of 15 expansion drills mixing plus and minus signs outside parentheses.
When solving a system of two equations, the candidate adds or subtracts the equations without first scaling one so a variable's coefficients match, or scales only one side of an equation, producing a plausible but wrong pair of values.
The elimination method is remembered as 'combine the equations' rather than 'make coefficients opposites, then combine'. Scaling only one side is the same one-sided-operation error as in single equations, and it is easy to miss because the arithmetic still looks tidy.
What to do instead
Use a written checklist for every system: (1) pick the variable to kill, (2) find the multiplier for each equation, (3) multiply both sides of each equation, (4) confirm the coefficients are exact opposites, (5) add. Always verify by substituting both values into the equation you did not use for elimination. Drill five systems where the multipliers are different for each equation.
The candidate solves a quadratic such as x² − 5x + 6 = 0 by writing the factors as (x − 6)(x + 1), choosing numbers that multiply to the constant but ignoring the middle-term sum, and reports roots 6 and −1.
The candidate remembers 'two numbers that multiply to the last term' and drops the second condition about the sum. The check that would catch it - expanding the factors back - is skipped because the answer already looks finished.
What to do instead
State both conditions explicitly before factoring: product equals the constant term, sum equals the coefficient of x (with sign). Then always expand the proposed factors mentally or on paper to confirm the middle term. For x² − 5x + 6 the pair must multiply to +6 and add to −5, giving −2 and −3. Practise 12 quadratics with positive and negative constants.
Given a graph, the candidate reads the slope by counting grid squares between two labelled points but inverts the ratio, or reports the y-intercept as the x-intercept, then writes an equation that fits the picture at one point only.
Slope is a ratio and the candidate treats it as a single count; without writing rise over run as a fraction, the two counts blur. Intercept confusion comes from reading the axis labels too quickly and from not distinguishing 'where the line crosses' from 'where the value is zero'.
What to do instead
Always write slope as an explicit fraction rise/run with signed counts before simplifying, and mark the crossing point on each axis with its coordinate pair (0, b) and (a, 0). Then test the finished equation at a third point on the line. Practise converting ten graphs into slope-intercept form and back, including lines with negative and fractional slopes.
The candidate evaluates a function at an expression rather than a number, writing f(x + 2) for f(x) = x² + 1 as x² + 1 + 2 = x² + 3 instead of substituting the whole input into every occurrence of x.
Function notation is read as a label attached to the formula rather than as an instruction to substitute. Because f(3) works fine with casual reading, the error only surfaces with expression inputs, and the candidate has no reason to suspect the method.
What to do instead
Treat the parentheses in f( ) as empty slots: rewrite the formula with visible blanks, f( ) = ( )² + 1, then fill every blank with the same input, using parentheses around it. So f(x + 2) = (x + 2)² + 1. Do ten evaluations alternating numeric and expression inputs, and expand the result each time.
On a passage about a technical procedure, the candidate picks an answer choice that is true in general knowledge or repeats a phrase from the passage, rather than the choice that the passage actually supports, and moves on without locating the evidence.
Familiarity with the topic creates a strong pull toward what the candidate already believes, and matching surface words feels like confirmation. Under timed conditions, returning to the text to verify feels slower than trusting recognition.
What to do instead
For every answer, name the line or sentence in the passage that supports it before selecting; if you cannot point to it, the choice is wrong for this question. Treat answer choices that reuse passage wording as suspects until verified, since they are often traps. Practise ten questions where you write the supporting phrase next to the correct option.
The candidate answers a question asking what the author implies or what can be inferred by choosing a restatement of a sentence that is stated outright, or by choosing a conclusion the passage never commits to.
Inference questions look identical to detail questions on the surface, and the candidate does not separate 'the passage says' from 'the passage allows you to conclude'. The safest-feeling option is the one already written in the text, so it wins by default.
What to do instead
Before answering, classify the question as stated-detail or inference. For inference, eliminate any choice that is directly stated (it may be true but is not an inference) and any choice that goes beyond the passage's scope. The correct choice should be a small, forced step from the text. Drill ten paired questions on one passage, one detail and one inference, to feel the difference.
When a pulley or gear question shows a rope supporting a load through multiple rope segments, the candidate assumes the effort equals the load regardless of the number of supporting strands, or counts every strand including the one being pulled.
The candidate reasons from everyday lifting, where effort feels roughly equal to weight, and has no rule for translating a diagram into a force count. Counting strands is done visually without deciding which strands actually bear on the movable pulley.
What to do instead
Use the rule that the load is shared equally among the rope segments supporting the movable pulley, so effort equals load divided by that count (ignoring friction and rope weight). Count only segments that pull upward on the movable block, and exclude the free end you pull. Work through diagrams with 1, 2, 3, and 4 supporting strands until the count is unambiguous.
On a lever question, the candidate compares the distances from the fulcrum to the load and to the effort but measures both from the same end, or ignores the perpendicular component, and concludes the wrong side is heavier.
Levers are judged by appearance - the longer-looking arm seems stronger - and the candidate does not convert the picture into a torque balance. Distance is measured casually rather than from the fulcrum to each force along the perpendicular.
What to do instead
Write the balance equation before judging: effort × effort arm = load × load arm, with both arms measured from the fulcrum. Compute the products as numbers and compare them. If the arms are at an angle, use the perpendicular distance from the fulcrum to the line of the force. Practise ten lever problems where the answer is not obvious from the drawing, including cases where the effort arm is shorter.