NCEA L1 Maths common mistakes
NCEA L1 Maths · Understand the traps and how to avoid them.
When solving a linear equation such as 5x − 3 = 2x + 9, the candidate moves a term across the equals sign but keeps its sign unchanged, writing 5x − 2x = 9 − 3 and then 3x = 6 giving x = 2 instead of x = 4.
The candidate has memorised 'move it to the other side' as a physical shuffle rather than as adding or subtracting the same quantity from both sides. Because the operation is not seen as balancing, the sign change feels optional, and the error is invisible when checking because the candidate re-does the same shuffle.
What to do instead
Force the balancing language every time: write the operation under both sides, e.g. '+3' beneath each side, then '−2x' beneath each side. Practise five equations where the candidate must write the intermediate line before simplifying. Check by substituting the answer back into the original equation, not the rearranged one.
Expanding (x + 3)² as x² + 9, or expanding 2(x + 4) as 2x + 4, by multiplying only the first term inside the bracket.
The candidate reads the bracket as a label attached to the first term rather than as a factor applied to the whole group. With squares, the candidate pattern-matches to the 'square each thing' rule learned for (2x)² and over-applies it.
What to do instead
Draw arrows from the outside factor to each term inside the bracket before writing anything. For squares, rewrite (x + 3)² as (x + 3)(x + 3) and use the four-arrow grid every time until it is automatic. Substitute x = 1 into both the original and the expanded expression as a numerical check: they must give the same value.
On a question asking for the gradient of the line through (2, 5) and (6, 13), the candidate computes (6 − 2)/(13 − 5) = 0.5, inverting the rise-over-run order.
The formula is remembered as 'difference in x over difference in y' or as a vague 'subtract the coordinates' without a fixed order. The candidate also fails to notice the answer is wrong because the line visibly rises steeply and 0.5 is shallow.
What to do instead
Anchor the formula to the phrase 'change in y on top' and write Δy/Δx above the working before substituting. Always subtract in the same direction: (y₂ − y₁)/(x₂ − x₁). Sketch the two points roughly and sanity-check whether the gradient should be steep, shallow, positive, or negative before finalising.
When asked to solve 2x + 1 > 7, the candidate divides by a negative number later in a multi-step problem and forgets to flip the inequality, or treats the inequality as an equation and gives a single value such as x = 3 instead of a range.
Inequalities are practised far less than equations, so the candidate defaults to equation habits. The flip rule is memorised as an arbitrary rule rather than as a consequence of multiplying by a negative, so it is dropped under time pressure.
What to do instead
Test the flip rule numerically each time: if −2x > 6, try x = −4 and x = 4 in the original to see which side works, confirming x < −3. Always present the answer as an inequality or interval and mark it on a number line so the range is explicit.
Finding the vertex of y = x² − 6x + 5 by reading off (0, 5) as the turning point, or completing the square incorrectly as (x − 3)² + 5 without adjusting the constant.
The candidate confuses the y-intercept with the vertex because both are single 'special' points on the parabola. Completing the square is a multi-step procedure where the candidate forgets that adding 9 inside the bracket must be compensated outside.
What to do instead
Use the vertex formula x = −b/(2a) as a reliable route, then substitute to get y. Alternatively, complete the square with an explicit compensation line: x² − 6x + 5 = (x² − 6x + 9) − 9 + 5 = (x − 3)² − 4, so the vertex is (3, −4). Sketch the parabola and check the vertex sits between the x-intercepts.
In a right triangle with the opposite side 7 and hypotenuse 12, the candidate computes 12/7 for sin θ, or uses cos instead of sin because the side labels were assigned from the wrong angle.
SOHCAHTOA is recited but not applied to a labelled diagram. The candidate identifies 'opposite' relative to the wrong angle, especially when the triangle is rotated or the angle is at the top.
What to do instead
Before choosing a ratio, mark the given angle, then physically trace the side opposite it and the side adjacent to it, labelling O, A, H on the diagram. Write the ratio as a fraction of named sides first (opposite/hypotenuse), then substitute numbers. Check the answer is less than 1 for sin and cos.
Calculating the mean of grouped data by averaging the class midpoints without weighting by frequency, or reporting the midpoint of the modal class as the median.
The candidate treats grouped data like a simple list and does not connect frequency to weighting. The distinction between mean, median, and mode collapses because all three are answered by 'find the middle of the table'.
What to do instead
Write the frequency table with an extra column for midpoint × frequency and sum that column before dividing by total frequency. For the median, compute the cumulative frequency and locate the position (n+1)/2, then state the class containing it. Label the final answer with which average it is.
When asked for the probability of drawing two red counters without replacement from a bag, the candidate multiplies 4/10 × 4/10 instead of 4/10 × 3/9.
The candidate applies the independent-events multiplication rule to a dependent situation because the two cases look identical on the page. The phrase 'without replacement' is read but not translated into a changed denominator.
What to do instead
Before multiplying, redraw the bag after the first draw and write the new counts explicitly. Use a tree diagram and label each branch with the updated fraction. State in words 'the total has dropped by one' as a check before multiplying.
Reading a box-and-whisker plot, the candidate reports the range as the length of the box (the interquartile range) or claims 50% of the data lies between the minimum and the lower quartile.
The five-number summary is memorised as a list without linking each feature to a proportion of the data. The box is visually dominant, so it is assumed to represent the whole spread.
What to do instead
Annotate every plot with the five values and the percentage of data in each section: 25% in each whisker and 25% in each half of the box. Answer range and interquartile range as separate labelled calculations, and check the IQR is smaller than the range.
Describing a distribution as 'roughly symmetric' when it is clearly skewed, or reporting an association as causation, e.g. 'increased study time caused higher marks' from a scatter plot.
The candidate uses everyday language loosely and has not been taught to justify shape or association with evidence. Scatter plots invite causal storytelling because the trend looks purposeful.
What to do instead
Require a two-part answer: a claim plus evidence, e.g. 'the distribution is skewed to the right because the tail of the data extends towards higher values and the mean exceeds the median'. For scatter plots, state the direction and strength of the association and explicitly note that association does not establish causation without a controlled comparison.