Secondary 4 Mathematics-Science Option Ministerial Examination - Québec common mistakes
Secondary 4 Mathematics-Science Option Ministerial Examination - Québec · Understand the traps and how to avoid them.
When solving an inequality such as -2(x-3) ≤ 4x+8, the candidate divides by the negative coefficient and keeps the inequality sign the same, writing x ≤ 1/3 instead of reversing it to x ≥ 1/3.
The candidate treats inequalities as equations and applies the same inverse-operation routine, because the sign-reversal rule is memorized as a sentence rather than tied to the number line (multiplying by a negative reflects every point, so order flips). Under time pressure the routine wins over the rule.
What to do instead
Before dividing or multiplying by any negative number, force a written checkpoint: draw the number line, mark two test values (e.g. 0 and 4), and verify which side satisfies the original inequality. Practise 10 inequalities where the coefficient of x is negative, and after each, substitute one endpoint and one interior value back into the original statement.
When expanding or factoring, the candidate distributes only to the first term: writes 3(2x-5) = 6x-5, or expands (x+4)² as x²+16, dropping the middle term 8x.
The candidate reads the bracket as a label attached to the first term rather than as a product applied to each term, and the square notation is confused with 'square each thing inside'. These are procedural gaps that survive because the errors are invisible in simple cases.
What to do instead
Use an area model (rectangle split into regions) for every expansion until it is automatic, and always check by substituting x = 2 into both the original and expanded forms. For squares, memorize the pattern (a+b)² = a²+2ab+b² and drill five perfect-square expansions per session, writing the middle term explicitly.
Given a word problem, the candidate writes an equation but solves for the wrong quantity, or answers with the value of x when the question asks for a derived quantity (e.g. the son's age when asked for the father's age, or the width when asked for the area).
The candidate focuses entirely on producing an equation and treats solving as the finish line, so the final read-back step is skipped. Variables are also defined loosely ('let x be the number') without recording what x represents.
What to do instead
Adopt a three-line protocol on every problem: (1) write 'Let x = ___' in words, (2) after solving, write 'The question asks for ___', (3) restate the answer in a full sentence with units. Underline the interrogative phrase in the question before starting.
When working with a step (greatest-integer) function, the candidate plots it as a straight line or a series of disconnected dots, and evaluates f(3) for a rule like f(x) = 2⌊x⌋ + 1 by rounding to the nearest integer instead of down.
The candidate has only seen linear graphs and defaults to a line, and the floor notation is confused with ordinary rounding, which is the rounding they use daily. The open/closed endpoint convention is not connected to the meaning of the intervals.
What to do instead
Build a table of x values including 2.9, 3.0, 3.1 and compute each output by hand, then plot with closed circles on the left end of each step and open circles on the right. State in words what the function does ('the cost jumps by $2 at each full hour').
When finding the rule of an exponential function through two points, the candidate fits a linear rule instead, or computes the base as the difference of the y-values rather than their ratio, and ignores the asymptote when stating the range.
Linear thinking dominates because most prior practice is linear, and 'rate of change' is fused in the candidate's mind with subtraction. The multiplicative structure of exponential growth is not reinforced by everyday experience.
What to do instead
Always test the data first: compute successive differences and successive ratios; if ratios are constant, it is exponential. Practise writing y = a·b^x by dividing the two y-values to get b, then substituting one point to get a. State the range as y > k whenever the horizontal asymptote is y = k.
When simplifying a rational expression such as (x²-9)/(x²+x-6), the candidate cancels the x² terms or the x terms across the fraction, writing (x²-9)/(x²+x-6) = -9/(x-6), and forgets to state the restriction x ≠ -3.
Cancellation is learned as 'cross out matching things', and the candidate does not distinguish factors (which may be cancelled) from terms joined by addition (which may not). Domain restrictions feel like optional decoration.
What to do instead
Require full factoring of numerator and denominator before any cancellation, and physically draw a box around each factor. Rule: only factors may be cancelled, never terms. After simplifying, list excluded values by setting each original denominator factor to zero, and check them against the simplified form.
In a similarity problem, the candidate sets up the proportion by pairing sides that do not correspond (e.g. matching the shortest side of one triangle with the longest of the other), or applies a length ratio directly to an area without squaring it.
The candidate matches numbers by position in the diagram rather than by the angles that define corresponding parts, and the scaling laws for area and volume (k² and k³) are memorized as formulas disconnected from the geometry.
What to do instead
Before writing any proportion, mark the equal angles on both figures and label corresponding vertices in the same order (A↔D, B↔E, C↔F). Write the ratio as k, then explicitly write Area ratio = k² and Volume ratio = k³. Verify with a simple numeric example (double the sides, area quadruples).
When computing the volume or surface area of a composite solid, the candidate adds the volumes of overlapping parts or uses the slant height of a cone as the height in V = (1/3)πr²h, and mixes units (cm with m) within one calculation.
The candidate selects a formula by shape name without checking which dimension the formula requires, and unit conversion is treated as a separate topic rather than part of the measurement. Diagrams with a slant height drawn prominently invite the substitution.
What to do instead
Annotate the figure first: label r, h and slant height a separately, and circle the one the formula needs. For composites, decompose into non-overlapping pieces and state 'subtract the shared part' when relevant. Convert all lengths to one unit before substituting, and carry units through the calculation.
When interpreting a box-and-whisker plot or comparing two distributions, the candidate reads the whisker ends as the minimum and maximum of the data set rather than as the extremes within the outlier fence, and concludes that a wider box means more data rather than more spread in the middle 50%.
The five-number summary is memorized as a list of positions on a drawing without the underlying percentiles, so each mark is interpreted as 'an end of the data'. Percentile language is not connected to the visual.
What to do instead
For every box plot, label each of the five marks with its percentile (0, 25, 50, 75, 100 within the fence) and state in words what fraction of data lies in each region. When comparing, write two separate sentences: one about centre (median) and one about spread (interquartile range).
When computing a probability for a two-stage experiment, the candidate adds probabilities that should be multiplied (e.g. P(two heads) = 1/2 + 1/2 = 1), or uses the without-replacement denominator after a first draw when the problem specifies replacement.
The candidate has no stable decision rule for 'and' versus 'or', and replacement is read as a technical detail rather than something that changes the sample space. Tree diagrams are seen as extra work rather than as the tool that resolves the ambiguity.
What to do instead
Draw a tree diagram for every multi-stage experiment, writing the denominator on each branch and multiplying along paths, adding across paths. Before starting, underline 'with replacement' or 'without replacement' and adjust the second-branch denominator accordingly. Check that all final probabilities sum to 1.