Kagoshima Entrance Mathematics common mistakes
Kagoshima Entrance Mathematics · Understand the traps and how to avoid them.
When simplifying an expression such as 3(2x−4)−2(x−5), the candidate distributes only the first factor and writes 6x−12−2x−5, dropping the sign change produced by the minus sign in front of the second bracket.
The candidate treats the subtraction sign as belonging only to the 2 and not to the whole bracket, because the bracket is processed left to right as a visual unit rather than as a multiplier of every term inside it. This is especially likely when the bracket contains a negative constant, since the two negatives that should multiply into a plus are easy to lose.
What to do instead
Rewrite every subtraction of a bracket as addition of the negative before simplifying: 3(2x−4)+(−2)(x−5). Then draw an arrow from the outside multiplier to each interior term, and check by substituting x=1 into both the original and the simplified expression; both must give the same number.
The candidate solves a linear equation by dividing one side by a coefficient but not the other, for example turning 4x−6=2x+8 into x−6=8 after dividing only the left-hand terms containing x.
The candidate applies an operation to the terms that visibly contain the unknown and treats the remaining numbers as untouched, which is a partial application of the equality principle rather than a misunderstanding of arithmetic. Working quickly under time pressure reinforces the habit because the intermediate step looks tidy.
What to do instead
Adopt a fixed two-column layout: write the operation performed on both sides on a separate line before carrying it out, for example 'subtract 2x from both sides' then 'add 6 to both sides'. Finish by substituting the answer back into the original equation, which catches an asymmetric division immediately.
When asked for the equation of a line parallel or perpendicular to a given line, the candidate reuses the given slope for a perpendicular line, or changes the slope but forgets to substitute the known point to find the intercept.
The candidate remembers that parallel lines share a slope but has not consolidated the reciprocal-and-negate rule for perpendicular slopes, and often stops as soon as a slope is found because the slope feels like the answer. The intercept is then guessed from the diagram or left as the original line's intercept.
What to do instead
Write the rule explicitly before computing: parallel slope m, and perpendicular slope −1/m when m is finite and nonzero, with slope product −1 as a check. A horizontal line is perpendicular to a vertical line, whose slope is undefined. Always finish with the point-substitution step y−y₁=m(x−x₁) and verify that the resulting line passes through the stated point and has the required slope product.
The candidate reads a gradient or intercept from a graph or a word problem with the axes swapped, reporting the y-intercept as the rate of change, or interpreting a negative gradient as a negative quantity rather than a decreasing one.
The candidate matches numbers to the most visually prominent feature of the graph instead of to the meaning of the variables, and word problems in Japanese frequently describe change per unit in a clause placed after the quantity, so the pairing is misread. The distinction between a value being negative and a relationship being decreasing is also conceptually thin.
What to do instead
Before calculating, write down what each axis represents and what one unit on each axis means in the context, then state in words what the gradient and intercept represent. Check plausibility: the sign and units of a rate must fit the conditions actually given, and an intercept describing a starting amount should match the initial condition in the text.
When finding the coordinates of the intersection of a line and a parabola, the candidate solves the resulting quadratic and reports only the positive root, discarding the negative one as impossible.
The candidate carries an assumption from earlier geometry work that lengths and coordinates are positive, and a negative solution feels like an error rather than a second intersection point. If the quadratic is solved by taking a square root, the ± is also frequently omitted.
What to do instead
Treat every intersection as a system: substitute, arrange into standard form, factor or use the formula, and explicitly write both roots with a ± where a square root is taken. Then plot or sketch both graphs and confirm that the number of solutions matches the number of crossing points before selecting an answer.
The candidate answers a maximum or minimum question about a quadratic by using the vertex value as the maximum when the coefficient of x² is positive, or by reporting the x-coordinate of the vertex when the question asks for the maximum value.
The vertex is memorised as 'the top of the parabola', so the sign of the leading coefficient is not checked, and the distinction between where the extreme occurs and what the extreme value is collapses when answers are written quickly. Completing the square is also done mechanically without reading off the vertex.
What to do instead
After completing the square to the form a(x−p)²+q, state the sign of a in words first: positive means a minimum at q, negative means a maximum at q. Then reread the question and underline whether it asks for the value of the function or the value of x at which it occurs, and answer exactly that.
In a circle or parallel-line angle problem, the candidate uses the inscribed angle theorem with the wrong arc, or applies the alternate segment theorem to an angle that is not between a tangent and a chord.
Angle theorems are recalled as names and rough pictures rather than as precise conditions about which points and lines must be involved, so a visually similar configuration is treated as satisfying the theorem. Diagrams in these problems are often crowded, making it hard to identify which arc an angle subtends.
What to do instead
For each angle used, write a short label naming the two points that define the arc it intercepts and the vertex, then apply the theorem only if the vertex lies on the circle or at the centre as required. Redraw the relevant portion of the figure enlarged and separately, so the arc-angle relationship is unambiguous before any calculation.
When proving two triangles congruent or similar, the candidate lists three equal parts that do not correspond to a valid condition, for example two sides and a non-included angle, or three angles for congruence.
The candidate counts matching marks on the figure without checking their arrangement, because the congruence conditions are remembered as counts of equal elements rather than as named configurations. Similarity and congruence conditions also get mixed, since both involve matching angles.
What to do instead
Before writing the proof, name the condition being used explicitly (two sides and the included angle, two angles and the included side, three sides, or for similarity two equal angles) and check that the equal parts sit in exactly that arrangement. Write the correspondence in the vertex order, such as △ABC ≡ △DEF, so corresponding parts cannot drift.
In a similarity problem involving lengths, the candidate writes the ratio of areas or volumes as the ratio of corresponding lengths, or applies a length ratio to an area without squaring it.
The scale factor is the number the candidate has just calculated, so it is applied directly to whatever quantity is asked, and the exponent relationship between length, area, and volume is not held as a principle. Problems that give an area ratio and ask for a length are especially prone to this.
What to do instead
Write the chain length ratio a:b, area ratio a²:b², volume ratio a³:b³ at the start of every similarity problem, then identify which level the given quantity belongs to and which level the answer requires, converting in one explicit step with the exponent shown.
When asked for a probability from grouped data or a two-way table, the candidate divides by the total number of entries instead of by the relevant subgroup, or treats non-equally-likely outcomes as equally likely.
The total is the most available denominator in the table, and conditional-style questions phrased in Japanese often place the restricting condition late in the sentence, so the candidate has already selected the denominator before finishing reading. For dice and coin problems, outcomes of different structure are counted as single cases.
What to do instead
Underline the population being conditioned on before touching the numbers, and write the fraction as 'favourable cases within that population over the size of that population'. For dice and coins, enumerate outcomes in a systematic grid so that equally likely elementary outcomes, not composite events, are counted.