Bulgaria NVO Grade 7 Mathematics common mistakes
Bulgaria NVO Grade 7 Mathematics · Understand the traps and how to avoid them.
When adding or subtracting fractions such as 3/4 − 1/6, the candidate adds numerators and denominators directly and writes 2/10, or adds only the numerators and keeps one denominator unchanged.
The candidate over-generalises the whole-number rule that operations act component-wise and has not consolidated the idea that a fraction is a single number, so equal parts of different size are treated as if they were the same unit.
What to do instead
Have the candidate draw the two fractions as bars split into 4 and 6 equal parts, then re-split both bars into twelfths so the parts match; insist on writing the common denominator explicitly before any numerator arithmetic, and give mixed drills that alternate same-denominator and different-denominator cases so the rule is chosen, not applied blindly.
When computing with negative numbers, the candidate evaluates −3² as 9, or writes 5 − (−2) = 3, treating the minus sign in front of a bracket as absorbed rather than as multiplication by −1.
The candidate reads the minus sign as a label of negativity attached to a number rather than as an operator, and has not settled the convention that exponentiation binds tighter than the unary minus.
What to do instead
Separate the two roles explicitly: rewrite every subtraction of a negative as addition of the opposite, 5 − (−2) = 5 + 2, and read −3² aloud as 'the opposite of three squared'. Give paired contrasts (−3)² versus −3² and 7 − 4 versus 7 − (−4) until the candidate states the binding rule before computing.
When simplifying an expression like 3(2x − 5) − 4(x − 1), the candidate distributes into the first bracket only and writes 6x − 15 − 4x − 1, or drops the sign on the second bracket entirely.
The candidate treats the bracket as decoration around the nearest term and does not yet see the factor in front as multiplying every term inside, especially when that factor is negative.
What to do instead
Require an intermediate line in which every product is written out before collecting: 3·2x, 3·(−5), (−4)·x, (−4)·(−1). Use a two-column table (term inside bracket, product with outside factor) for the first few exercises, then fade the table but keep the intermediate line mandatory until the candidate produces it unprompted.
When factorising x² − 9 or x² + 5x, the candidate writes (x − 3)² instead of (x − 3)(x + 3), or factors only the first term and leaves x² + 5x as x(x) + 5x.
The candidate matches the visual pattern of a square to the squared-binomial identity without checking that the middle term vanishes, and lacks a habit of verifying a factorisation by expanding it back.
What to do instead
Make expansion the compulsory last step of every factorisation: multiply the proposed factors out and compare term by term with the original. Contrast (x − 3)² = x² − 6x + 9 with (x − 3)(x + 3) = x² − 9 side by side so the differing middle term becomes the discriminating feature.
When solving 5 − 2x = 11, the candidate moves the 5 across as +5 and obtains −2x = 16, or divides only one side of the equation by a coefficient.
The candidate has memorised 'change the sign when moving across' as a ritual without the underlying idea of applying the same operation to both sides, so the rule misfires when the term is subtracted or the coefficient is negative.
What to do instead
Ban the phrase 'move across' and require each step to be written as an operation on both sides: subtract 5 from both sides, then divide both sides by −2. Have the candidate check by substitution every time, and include equations with negative and fractional coefficients so the balance idea, not the shortcut, carries the work.
When solving 3 − 2x < 7, the candidate divides by −2 and keeps the inequality sign as <, giving x < −2 instead of x > −2.
The candidate treats an inequality as an equation with a decorative symbol and has not connected reversing the sign to the number line picture where multiplying by a negative reflects the order.
What to do instead
Have the candidate test a concrete value on each side of the proposed boundary and check it in the original inequality; also require a number-line sketch with the solution region shaded. Drill pairs such as −2x < 6 and 2x < 6 together so the reversal is noticed as the only difference.
When reading a linear graph, the candidate identifies the slope as the y-intercept or computes slope as change in x over change in y, and misreads a descending line as positive gradient.
The candidate confuses the two parameters of y = kx + b because both are numbers in the same formula, and has not practised extracting information from a picture rather than from an equation.
What to do instead
Have the candidate annotate every graph before answering: mark the intercept on the y-axis, draw a right triangle between two lattice points and label vertical change over horizontal change. Give mixed sets where some questions ask for the intercept and some for the slope so the candidate must decide which feature is being requested.
When finding the third angle of a triangle or an angle in a quadrilateral, the candidate applies the wrong total (180° to a quadrilateral, 360° to a triangle) or ignores that the triangle is isosceles and treats the base angles as unrelated.
The two facts are learned as similar-sounding numbers without an anchor, and the candidate does not habitually mark equal sides and equal angles on the figure before computing.
What to do instead
Anchor the totals by derivation: tear the corners of a paper triangle to form a straight line for 180°, and split a quadrilateral by a diagonal into two triangles for 360°. Require the candidate to mark equal angles on every figure and to write the angle sum equation before substituting numbers.
When computing the area of a composite figure, the candidate adds the areas of overlapping parts twice, or uses the diameter as the radius in A = πr² and in the circumference formula.
The candidate decomposes the figure by eye without a written plan, and the words radius and diameter are not firmly linked to the picture of centre to edge versus edge to edge.
What to do instead
Require a labelled sketch with the decomposition drawn in and each sub-region numbered, then a written sum of the sub-areas with no region counted twice. For circles, insist that every problem begins by writing d = …, r = d/2 = … before any formula is used.
When a chart or table is given, the candidate reads the tallest bar as the largest total regardless of the axis scale, or reports the mean of a data set without dividing by the count, and answers a probability question with a count rather than a ratio.
The candidate reads graphs as pictures rather than as scaled representations, and treats the mean as a synonym for the sum; probability is answered from intuition about which outcome 'should' happen rather than from favourable over total outcomes.
What to do instead
Train axis reading explicitly: state the value of one grid step and the unit before answering, and cover the numbers to predict heights. For the mean, require the written form sum ÷ count with both numbers shown. For probability, require the fraction favourable outcomes over all equally likely outcomes, then simplify, and check that the answer lies between 0 and 1.