# Original derivations for five paired mathematics guides

Author: article_author. Written on Japan calendar date 2026-10-05, while the verified UTC date is 2026-10-04. The actual work interval was bounded by clock observations 2026-10-04T15:17:43Z and 2026-10-04T15:32:06Z. The filename is a Japan-date batch identifier; checkedAt is 2026-10-04. This is author evidence, not approval.

All thirty Japanese/English example presentations use fifteen existing mathematics IDs. Original bank text and current independent approval hashes were checked before writing. No new bank question, news item, real price or empirical dataset is created. This document contains the entire original mathematical reasoning used by these guide bodies. Suggestions to sketch, label and check are practice instructions, without claims about measured learning effects.

## math-units

ma-0321: the conventional length-unit relation is 1 m = 100 cm. Thus 2.5 m = 2.5×100 cm = 250 cm. More smaller units count the same length, which establishes the multiplication direction rather than a change in length.

ma-0322: the conventional volume-unit relation is 1 L = 1000 mL. Thus 750 mL = (750/1000) L = 0.75 L. Fewer larger units describe the same volume. The relation is a unit definition applied to the given hypothetical volume, not a measurement of a liquid sample.

ma-0323: a 1 m by 1 m square is 100 cm by 100 cm. Its area is 100×100 = 10000 cm², using two length conversion factors. Area is a product of two perpendicular length dimensions; multiplying by only 100 would fail to convert both factors. The article does not claim a length factor is itself an area factor.

## math-expressions

ma-0081: if u+18=45, subtract 18 from both sides to obtain u=27. Substitution gives 27+18=45, satisfying the initial equality.

ma-0083: x pencils at the fictional price 80 yen contribute 80x yen. One eraser contributes 100 yen once. The total is 80x+100 yen. The 100 is an additive fixed amount; it is not another factor applied per pencil. Only 80x scales with x, so the expression does not have direct-proportion form kx for the total.

ma-0086: the question explicitly supplies y=kx, with nonzero input values 1,2,3 and corresponding outputs 4,8,12. Quotients are 4/1=8/2=12/3=4. Thus k=4 is the supplied rule’s multiplier for all stated pairs. Three values alone would not prove a proportional continuation for an unspecified general relationship; the supplied form is essential.

## math-graphs

ma-0093: let t be the numerical number of elapsed minutes, and let y be the numerical volume measured in litres, so y=5t+20. Equivalently the actual volume is (5t+20) L. At t=0 the volume is 20 L. Increasing t by one increases y by five, interpreting the coefficient as 5 L/min; 5t is the added volume after t minutes, not a time or a volume-unit applied only to the constant 20. The article retains this unit interpretation explicitly.

ma-0094: horizontal time axis and vertical cumulative distance-travelled axis imply that a horizontal segment has constant recorded cumulative distance throughout the segment. Under the exercise’s ideal graph interpretation there is no travel during that interval, hence a stop. A horizontal velocity graph could instead represent constant nonzero velocity; the guide does not transfer the stop interpretation to different vertical quantities. The coordinate-zero time is determined by the statement, not by the visible page edge.

ma-0095: the ordered pair (3,−2) gives the horizontal coordinate first, vertical second in the example’s usual axes. From (0,0), a +3 horizontal displacement moves right; a −2 vertical displacement moves down. The sign affects the direction within its coordinate, not the ordering of coordinates.

## math-counting

ma-0331: n cars supply 8n places; to seat 41 people require 8n≥41. For n=5, 8n=40<41. For n=6, 8n=48≥41. Thus 6 is the least integer satisfying capacity. Division by the per-car capacity counts groups, not the capacity itself. A partial last group is needed because all people must be seated; this differs from counting only full groups.

ma-0332: a straight 100 m path at 20 m intervals has 100/20=5 gaps. Positions including both distinct endpoints are 20j m for j=0,1,2,3,4,5, hence six positions. The extra position follows from the open interval arrangement, not an unconditional add-one rule.

ma-0333: a circular circumference 120 m with 20 m spacing has six gaps. Tree positions can be represented by arc lengths 0,20,40,60,80,100 m modulo 120 m. At 120 m the point is the original 0 m point, not a seventh tree. Thus the closed loop has six distinct positions and six gaps. Each scenario counts different objects after the quotient is interpreted.

## math-chance-models

ma-0265: label the red objects R1,R2 and the blue object B. Uniform sequential sampling without replacement has six ordered outcomes: R1R2,R2R1,R1B,BR1,R2B,BR2. Each is probability (1/3)(1/2)=1/6. Four are different-colour outcomes, hence 4/6=2/3. Each of the three unordered pairs merges exactly two ordered outcomes, so each has probability 1/3 and two are favourable. R1R1 and similar self-pairs are impossible without replacement. Either representation gives 2/3 if numerator and denominator use it consistently.

ma-0266: two distinguishable independent fair six-sided dice yield ordered pairs (i,j), i,j in {1,…,6}; each has probability (1/6)(1/6)=1/36. There are 36 pairs. For sum 7, j=7−i lies in {1,…,6} for each i=1,…,6, giving six favourable pairs and probability 6/36=1/6. Sum categories are not equally likely: for example sum 2 has only (1,1), whereas sum 7 has six pairs.

ma-0270: let A be five preceding heads, B the next toss being tails. The stipulated independent fair-coin model gives P(B|A)=P(B)=1/2 since the past does not change the trial rule. This is a conditional consequence of the supplied model, not an empirical claim about unknown coins. Without-replacement sampling changes the remaining collection, so that different trial rule cannot inherit this conclusion.