# Article mathematics: original derivations

Author: article_author. Written 2026-10-04. This is newly authored mathematical reasoning for the five English mathematics guides and the three new Japanese counterparts. It does not approve any article. The conditions are the original fictional exercise conditions, not real survey results or current prices. Pedagogical instructions describe practice, without claims about measured learning effects. This document was written after reading the corresponding per-ID derivations in docs/quiz/math-en-20261004-authoring.md and the actual current approved English bank.

## math-everyday

For ma-0002, 0.7 = 0.700. The candidates have common thousandths units: 700, 650, 90 and 705 thousandths. Thus 705 thousandths is greatest. Zeroes appended to the right of a decimal fractional part add zero smaller-place units, preserving its value.

For ma-0005, four groups of six contain 6+6+6+6 = 6×4 = 24 items. Six is a group size, four is a group count. The sum 6+4 is not the item count of those groups.

For ma-0006, 29 = 6×4+5, with 0≤5<6. Thus four full boxes and five loose items describe the given collection. Five loose items cannot fill a six-item box. This establishes full-box count; it does not establish a minimum box count for packing all items, which is a different question.

## math-geometry

For ma-0404, use the Euclidean circle definitions: a radius joins its centre to a point of its circumference; a diameter is a segment with both endpoints on the circumference that passes through the centre. AB satisfies both diameter conditions. OA, OB and OC each have the centre as an endpoint, hence are radii. Rotation changes coordinates without changing incidence or endpoints.

For ma-0167, through one vertex of a Euclidean triangle draw a line parallel to the opposite side. Alternate interior angles place the two other interior angles adjacent to the angle at that vertex along a straight line. Their sum is therefore 180°. The missing angle is 180−50−60 = 70°.

For ma-0170, cut the triangular overhang of a parallelogram along a perpendicular to the selected base and translate it to the opposite end. This yields a rectangle with the same base and perpendicular height; cutting and translation preserve area. Area is 6×4 = 24 cm². The oblique side 5 cm is not the 4 cm perpendicular height. Length units multiply into square units.

## math-fractions-ratios

For ma-0010, a whole partitioned into eight equal pieces has pieces each representing 1/8. Three represent 3×1/8 = 3/8. Merely counting three unequal pieces would not specify 3/8 of the whole.

For ma-0011, multiplying numerator and denominator of a fraction a/b by the same nonzero c preserves its quotient: (ac)/(bc) = a/b, for b≠0. In particular 8/12 = (2×4)/(3×4) = 2/3. Changing only the numerator or only the denominator generally changes this quotient.

For ma-0017, let the two parts contain 2u and 3u items. Their sum is 5u = 30, so u = 6; the two parts are 12 and 18, summing to 30. Part-to-part is 12/18 = 2/3; first-part-to-whole is 12/30 = 2/5. These are different comparisons, not conflicting descriptions.

## math-data

The arithmetic mean of n observations is their sum divided by n. For ma-0246, 2+4+9 = 15 and n=3, giving 5; 3×5 = 15 reconstructs the total. A mean need not occur among the observations, as this list demonstrates.

For ma-0247, the sorted five values 1,3,8,10,20 have two entries on either side of the third entry 8. Thus 8 is the median by the middle-position definition. Summing the values would instead be a step toward their arithmetic mean.

For ma-0250, range is explicitly defined as maximum minus minimum; max=15 and min=4 give 11. Middle observations do not enter that endpoint subtraction. The three summaries retain different information by their respective definitions; no claim about real-world statistical estimates is made.

## math-probability

For ma-0261, the fair six-sided die has six equally likely atomic outcomes 1,2,3,4,5,6. The even event is {2,4,6}, giving 3/6 = 1/2.

For ma-0262, distinguishable coins with independent equally likely heads/tails have the product outcome set {HH,HT,TH,TT}. Each ordered outcome has probability (1/2)×(1/2) = 1/4. HT differs from TH because the coin identities are fixed.

For ma-0263, two independent fair tosses have the same four equally likely ordered outcomes. At least one head is {HH,HT,TH}, giving 3/4. Its complement is {TT}, giving 1−1/4 = 3/4. Exactly one head is {HT,TH}, a different event. The categories no heads, one head, two heads have probabilities 1/4,1/2,1/4, hence counting category names as equal outcomes would be invalid.