1.1.1 Place Value and Prime Structure

Every positive integer has a decimal expansion in powers of $10$. A three-digit string $abc$ denotes $a \times 100 + b \times 10 + c$, and the general principle extends to any length. This expansion is the source of all divisibility reasoning, because divisibility by a divisor $d$ depends only on the remainders of powers of $10$ modulo $d$.

The Fundamental Theorem of Arithmetic states that every integer greater than $1$ factorises uniquely into primes, up to ordering. All HCF and LCM reasoning, all fraction reduction, and much of counting theory rest on this uniqueness. A number of the form $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$ carries its full divisor structure in its exponents, and the count of divisors follows as $(e_1+1)(e_2+1)\cdots(e_k+1)$.

1.1.2 Why the Divisibility Tests Work

The tests for $3$ and $9$ work because $10 \equiv 1 \pmod 9$. Hence $100 \equiv 1 \pmod 9$, $1000 \equiv 1 \pmod 9$, and so on for every power of $10$. Each digit therefore contributes exactly its face value to the remainder modulo $9$. A number and its digit sum leave the same remainder on division by $9$, and since $3$ divides $9$, the same holds modulo $3$. A digit sum of $18$ thus guarantees divisibility by both $3$ and $9$, while a digit sum of $14$ leaves remainder $5$ modulo $9$ and remainder $2$ modulo $3$.

The test for $11$ works because $10 \equiv -1 \pmod{11}$. Powers of $10$ therefore alternate between $+1$ and $-1$ modulo $11$, so digits in even and odd positions enter with opposite signs. The alternating sum of the digits leaves the same remainder as the number itself modulo $11$. An alternating difference of $0$ or $\pm 11$ or $\pm 22$ signals divisibility by $11$.

Tests based on terminal digits work for a different reason. Since $100$ is divisible by $4$ and by $25$, only the last two digits affect divisibility by $4$ or $25$. Since $1000$ is divisible by $8$, only the last three digits affect divisibility by $8$. Since $10$ is divisible by $2$ and $5$, only the last digit affects divisibility by $2$, $5$, and $10$.

Divisor	Test	Reason
$2$	Last digit even	$10$ is a multiple of $2$
$3$	Digit sum divisible by $3$	$10 \equiv 1 \pmod 9$
$4$	Last two digits divisible by $4$	$100$ is a multiple of $4$
$5$	Last digit $0$ or $5$	$10$ is a multiple of $5$
$8$	Last three digits divisible by $8$	$1000$ is a multiple of $8$
$9$	Digit sum divisible by $9$	$10 \equiv 1 \pmod 9$
$10$	Last digit $0$	$10$ is a multiple of $10$
$11$	Alternating digit-sum difference is $0$ or a multiple of $11$	$10 \equiv -1 \pmod{11}$
$25$	Last two digits $00$, $25$, $50$, or $75$	$100$ is a multiple of $25$

Composite tests combine coprime factors. Divisibility by $6$ is divisibility by $2$ and $3$ together. Divisibility by $12$ is divisibility by $3$ and $4$ together. Divisibility by $15$ is divisibility by $3$ and $5$ together. The combination is valid exactly when the two component divisors share no common factor.

1.2.1 HCF and LCM from Prime Exponents

The highest common factor of a set is the greatest integer dividing every member. The lowest common multiple is the smallest positive integer that is a multiple of every member. In prime-exponent form, the HCF takes the minimum exponent of each prime across the set, while the LCM takes the maximum exponent. A value such as $\text{HCF}(12, 18) = 6$ arises because $12 = 2^2 \times 3$ and $18 = 2 \times 3^2$ share $2^1 \times 3^1$ at minimum exponents, while their LCM $2^2 \times 3^2 = 36$ takes maximum exponents.

For exactly two positive integers $a$ and $b$, the identity $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$ holds, because minimum plus maximum exponents reconstruct the total exponent of each prime across the two numbers. The identity does not extend to three or more numbers, where pairwise products and triple products interact in more complex ways.

1.2.2 HCF and LCM of Fractions

Fractions are handled by separating numerators and denominators. The HCF of a set of fractions in lowest terms equals the HCF of the numerators divided by the LCM of the denominators, because a common divisor of fractions must divide each numerator while any feasible denominator must be a multiple of each given denominator. The LCM of fractions reverses the construction, taking the LCM of the numerators over the HCF of the denominators. Reduction to lowest terms before applying these rules is essential, since unreduced forms misstate the numerator and denominator structure.

1.2.3 Fraction Arithmetic and Ordering

Addition and subtraction of fractions require a common denominator, with the LCM of the denominators as the natural choice because it keeps numerators smallest. Multiplication multiplies numerators and denominators directly. Division multiplies by the reciprocal. Ordering of fractions is preserved under cross-multiplication for positive denominators: $a/b < c/d$ exactly when $ad < bc$. A decimal expansion terminates exactly when the reduced denominator contains no prime other than $2$ and $5$, because only those primes divide powers of $10$.

1.3.1 The Meaning of Percent

A statement of $x\%$ means $x/100$ of a specified base. The base is the quantity that $100\%$ denotes, and every percentage computation is a statement about a ratio to that base. An increase of $20\%$ on a base of $500$ adds $100$ to reach $600$, while the same $20\%$ on a base of $600$ adds $120$ to reach $720$. The percentage is identical and the absolute increment differs, because the base differs.

1.3.2 Successive-Change Derivation

Successive percentage changes multiply rather than add, because each change redefines the base for the next. A change of $a\%$ multiplies the current value by the factor $(1 + a/100)$, with a fall represented by a negative $a$. Two successive changes $a\%$ and $b\%$ therefore multiply by $(1+a/100)(1+b/100)$, which expands to $1 + (a+b)/100 + ab/10000$. The net percentage change is $a + b + ab/100$. A rise of $20\%$ followed by a fall of $20\%$ gives $20 - 20 + (20)(-20)/100 = -4$, so a starting value of $100$ passes through $120$ and settles at $96$.

Quick example — Successive change: Start from Rs 400 with a rise of 25% followed by a fall of 20%.

Given factors are $1 + 25/100 = 1.25$ and $1 - 20/100 = 0.80$, with net change $25 - 20 + 25 \times (-20)/100 = 0$.

Running value moves as $400 \times 1.25 = 500$ and then $500 \times 0.80 = 400$, so the final value is Rs 400.

The trap is adding the rates to get plus 5 percent, while the cross term $25 \times (-20)/100 = -5$ cancels it exactly.

1.3.3 Base-Trap Analysis

The base trap is the error of treating percentages with different bases as directly addible. A rise and an equal fall never cancel, because the fall acts on the enlarged base. A salary moving from $100$ units to $120$ units and back by the same absolute $20$ units rises $20\%$ and then falls $16.67\%$, since the second base is $120$. Any computation that adds or subtracts percentages must first confirm a common base. Conversion between fraction and percentage forms supports base discipline by making the underlying ratio visible.

Fraction	Percentage	Fraction	Percentage
$1/2$	$50\%$	$1/8$	$12.5\%$
$1/3$	$33.33\%$	$1/9$	$11.11\%$
$1/4$	$25\%$	$1/10$	$10\%$
$1/5$	$20\%$	$1/11$	$9.09\%$
$1/6$	$16.67\%$	$1/12$	$8.33\%$
$1/7$	$14.29\%$	$1/16$	$6.25\%$

A change expressed as a fraction converts directly: a rise by $1/4$ is a $25\%$ rise, and a fall by $1/5$ is a $20\%$ fall. Reading divisions such as $96/600 = 16/100$ as $16\%$ follows the same logic.

1.4.1 Ratio and Proportion

A ratio $a:b$ states that two quantities stand in $a$ parts to $b$ parts of a common unit. Splitting a total of $56$ in the ratio $5:2$ divides it into $7$ equal parts of $8$ each, giving $40$ and $16$. A proportion $a:b = c:d$ asserts equality of two ratios, equivalently $ad = bc$ by cross-multiplication. Direct proportion keeps the ratio constant as quantities grow together; inverse proportion keeps the product constant as one quantity grows while the other shrinks.

1.4.2 Partnership Theory

Partnership profit divides in proportion to effective capital contribution, measured as capital multiplied by time. A contribution of $6000$ held for $4$ months carries the same weight as $4000$ held for $6$ months, because both products equal $24000$ unit-months. Sleeping and working partner distinctions modify the division only through prior agreement on salary or commission; the residual profit still follows the capital-time ratio.

1.4.3 Alligation and Proof of the Cross Rule

Alligation solves mixture problems where two ingredients of unit values $p$ and $q$, with $p < m < q$, are blended to attain mean value $m$. Let the quantities be $x$ of the cheaper and $y$ of the dearer. Conservation of total value gives $px + qy = m(x+y)$. Rearranging yields $x(m - p) = y(q - m)$, so $x:y = (q-m):(m-p)$. Each quantity is proportional to the diagonally opposite gap between the target mean and the other ingredient. A blend of values $60$ and $90$ targeting $72$ therefore mixes in the ratio $(90-72):(72-60) = 18:12 = 3:2$, and the weighted mean $(3 \times 60 + 2 \times 90)/5$ restores $72$.

The cross diagram encodes this proof visually, with the two ingredient values at the ends, the target mean in the centre, and the crossed differences as the mixing ratio.

The diagram reads as follows. The upper diagonal joins the cheaper value to the lower gap, and the lower diagonal joins the dearer value to the upper gap. The crossing is the rearranged conservation equation, not a separate mnemonic.

Quick example — Alligation mix: Blend stock at Rs 30 with stock at Rs 50 to hit a mean of Rs 44.

Opposite gaps are $50 - 44 = 6$ and $44 - 30 = 14$, so the mixing ratio is $6:14 = 3:7$.

Check by weighted mean gives $(3 \times 30 + 7 \times 50)/10 = 440/10 = 44$, so the ratio holds.

The insight is that the larger share always belongs to the ingredient nearer the target mean.

1.5.1 Profit and Loss on Cost

Profit percent and loss percent are defined on cost price unless a different base is stated explicitly. With cost $CP$ and selling price $SP$, profit percent is $(SP-CP)/CP \times 100$ and loss percent is $(CP-SP)/CP \times 100$. A cost of $200$ sold at $240$ earns $20\%$, because the surplus $40$ is measured against the cost $200$. The selling price corresponding to a profit of $r\%$ is $CP \times (1 + r/100)$, and for a loss of $r\%$ it is $CP \times (1 - r/100)$.

1.5.2 Discount on Marked Price

Discount percent is defined on marked price $MP$. A discount of $d\%$ gives $SP = MP \times (1 - d/100)$. Successive discounts $d_1\%$ and $d_2\%$ multiply to $SP = MP \times (1-d_1/100)(1-d_2/100)$, which is the successive-change formula with negative rates. Successive discounts of $20\%$ and $10\%$ on a marked price of $2000$ give $2000 \times 0.80 \times 0.90 = 1440$, equivalent to a single discount of $28\%$. Profit and discount combine through $MP \times (1-d/100) = CP \times (1+r/100)$, linking the two bases in one equation.

1.5.3 Faulty Weights and Hidden Bases

A dishonest weighing instrument shifts the quantity base while the price appears unchanged. Selling $800$ grams as $1000$ grams at cost price earns $(1000-800)/800 \times 100 = 25\%$, because the true cost covers only $800$ grams. The reasoning is identical to profit theory with the quantity delivered playing the role of cost.

1.6.1 Simple Interest as Linear Growth

Simple interest accrues only on the original principal $P$ at rate $R\%$ per annum for time $T$ years: $SI = P \times R \times T/100$. The amount grows linearly, adding the fixed increment $PR/100$ each year. A principal of $12000$ at $10\%$ for $2$ years earns $2400$ in simple interest.

1.6.2 Compound Interest as Geometric Growth

Compound interest accrues on principal plus accumulated interest. The amount after $n$ years at annual rate $R\%$ is $A = P(1+R/100)^n$, and compound interest is $A - P$. Annual compounding multiplies by the growth factor each year; half-yearly compounding halves the rate and doubles the number of periods; quarterly compounding quarters the rate and quadruples the periods.

1.6.3 The Two-Year Gap Proof

Over two years, simple interest is $2PR/100$ while the compound amount is $P(1 + 2R/100 + (R/100)^2)$ by binomial expansion. Subtracting gives $CI - SI = P(R/100)^2$ exactly. The gap is one year's interest earned on the first year's interest. A principal of $12000$ at $10\%$ gives a gap of $12000 \times 0.01 = 120$, matching the difference between $2520$ of compound interest and $2400$ of simple interest. Over three years the gap generalises to $P(R/100)^2(3 + R/100)$.

Quick example — CI-SI gap: A principal of Rs 10,000 runs 2 years at 15%.

Simple interest is $10000 \times 15 \times 2/100 = 3000$ and the gap formula gives $10000 \times (15/100)^2 = 10000 \times 0.0225 = 225$.

Compound interest is then $3000 + 225 = 3225$, with amount $10000 + 3225 = 13225$.

The insight is that the entire gap is one year of interest on the first year interest, here $1500 \times 15/100 = 225$.

1.6.4 Rule of 72 Justification

Doubling under compounding solves $(1+R/100)^T = 2$, so $T = \ln 2 / \ln(1+R/100)$. For small rates, $\ln(1+R/100)$ is close to $R/100$, giving $T \approx 69.3/R$. The constant $72$ replaces $69.3$ because $72$ is divisible by $2, 3, 4, 6, 8, 9,$ and $12$, making mental division exact for the most common rates, while the approximation error stays small across single-digit and low double-digit rates. A rate of $8\%$ gives $72/8 = 9$ years against an exact value near $9.0$ years.

1.7.1 Rate Theory

Work theory treats output per unit time as a rate. A worker completing a job in $n$ days performs $1/n$ of the job per day. Combined workers add rates, so times combine harmonically rather than additively.

1.7.2 Why the LCM Method Works

Fractions such as $1/6 + 1/4$ are correct but slow. Choosing total work as $\text{LCM}(6,4) = 12$ units converts each rate to an integer: $2$ units per day and $3$ units per day, jointly $5$ units per day, so the time together is $12/5$ days. The LCM is the smallest total divisible by every individual time, which guarantees integer daily rates. The method is a change of unit, from fractions of the job to absolute work units, and the final division restores the time.

Quick example — LCM work combo: Worker A finishes in 8 days and worker B finishes in 12 days.

Take total work as $\text{LCM}(8, 12) = 24$ units, so daily rates are $24/8 = 3$ and $24/12 = 2$ units.

Joint rate is $3 + 2 = 5$ units per day, so time together is $24/5 = 4.8$ days.

The point of the LCM is integer daily rates, with no fractions until the final division.

1.7.3 Efficiency Ratios and Negative Work

An efficiency statement such as $A$ being twice as good as $B$ is a rate ratio $2:1$, so work units split in that ratio over any common period. Pipes follow the same theory with signs: filling pipes contribute positive rates and emptying pipes contribute negative rates. A tank with a $6$-hour filler and a $4$-hour filler empties through a leak expressed as a negative rate, and the net rate determines the filling time. Alternating-day schedules and leaving-early arrangements are handled by accumulating integer work units day by day against the LCM total.

1.8.1 The Fundamental Relation

Speed is distance divided by time, $v = d/t$, with consistent units throughout. Conversion between kilometres per hour and metres per second multiplies by $5/18$, since $1$ km is $1000$ m and $1$ hour is $3600$ s. A speed of $72$ km/h equals $20$ m/s.

1.8.2 Equal-Distance Average Speed and Harmonic-Mean Proof

When equal distances are covered at speeds $x$ and $y$, total distance is $2d$ and total time is $d/x + d/y$. Average speed is $2d/(d/x + d/y) = 2xy/(x+y)$, the harmonic mean. Speeds of $40$ and $60$ over equal distances give $2 \times 40 \times 60/100 = 48$, below the arithmetic mean $50$. The harmonic mean always lies below the arithmetic mean for distinct positive speeds, because equal time is spent at neither speed: more time is spent at the slower speed, dragging the average down. The number line below shows the harmonic average sitting strictly between the two speeds but closer to the slower one.

For three equal segments at speeds $x$, $y$, $z$, the average generalises to $3/(1/x + 1/y + 1/z)$.

Quick example — Harmonic average: Equal stretches are covered at 60 km/h and 30 km/h.

The mean formula gives $2 \times 60 \times 30/(60 + 30) = 3600/90 = 40$ km/h.

With 60 km each way, times are $60/60 = 1$ h and $60/30 = 2$ h, so average is $120/3 = 40$ km/h.

The average sits below the arithmetic 45 km/h because more time is spent at the slower speed.

1.8.3 Relative Motion, Trains, and Boats

Relative speed governs meetings and chases. Objects moving toward each other close at the sum of speeds; one chasing another closes at the difference. A train crossing a bridge or platform covers its own length plus the length of the structure, because the crossing is complete only when the last carriage clears the far end. A $240$ m train at $20$ m/s crossing a $360$ m bridge covers $600$ m in $30$ s. A train crossing a pole or a person covers only its own length.

Boat-and-stream theory is relative motion in a moving medium. With still-water speed $b$ and stream speed $s$, downstream speed is $b+s$ and upstream speed is $b-s$. Adding the two observed speeds recovers $2b$; subtracting recovers $2s$.

1.9.1 Plane and Solid Formulae with Derivations

Area measures two-dimensional extent and volume measures three-dimensional capacity. The rectangle $l \times b$ and the cuboid $l \times b \times h$ are the primitive forms. The triangle area $bh/2$ is half the enclosing rectangle. The circle area $\pi r^2$ follows from dissection into narrowing sectors that reassemble toward a rectangle of sides $\pi r$ and $r$. The cylinder volume $\pi r^2 h$ stacks circular layers. The cone volume $(1/3)\pi r^2 h$ is one third of the enclosing cylinder, established by comparison of cross-sections at every height. The sphere surface $4\pi r^2$ and volume $(4/3)\pi r^3$ follow the same cross-sectional comparison against cylinder and cone. The hemisphere volume $(2/3)\pi r^3$ is half the sphere.

The space diagonal of a cuboid, $\sqrt{l^2+b^2+h^2}$, applies the Pythagoras relation twice: once across the base and once vertically against that base diagonal.

1.9.2 Scale Laws with Proofs

When every length of a figure scales by a factor $k$, each product of two lengths scales by $k^2$ and each product of three lengths scales by $k^3$. Area therefore scales as $k^2$ and volume as $k^3$, regardless of shape. Doubling lengths quadruples area and multiplies volume by $8$. A sphere of radius $r$ growing to $2r$ moves from surface $4\pi r^2$ to $16\pi r^2$ and from volume $(4/3)\pi r^3$ to $(32/3)\pi r^3$.

Scaling	Length	Area	Volume
Factor $k$	$k$	$k^2$	$k^3$
Doubling ($k=2$)	$2$	$4$	$8$
Tripling ($k=3$)	$3$	$9$	$27$
Halving ($k=1/2$)	$1/2$	$1/4$	$1/8$

1.10.1 Arithmetic Progression and the Pairing Proof

An arithmetic progression adds a fixed difference $d$ at each step: $a, a+d, a+2d, \ldots$ with $n$th term $a+(n-1)d$. The sum pairs the first term with the last, the second with the second-last, and so on, each pair totalling $2a+(n-1)d$. With $n/2$ such pairs, the sum is $S_n = n/2 \times (2a+(n-1)d) = n/2 \times (\text{first}+\text{last})$. The integers $1$ through $100$ sum to $100/2 \times 101 = 5050$ by this pairing.

1.10.2 Geometric Progression and the Telescoping Idea

A geometric progression multiplies by a fixed ratio $r$: $a, ar, ar^2, \ldots$ with $n$th term $ar^{n-1}$. Subtracting $rS_n$ from $S_n$ cancels every interior term, leaving $S_n(1-r) = a(1-r^n)$, so $S_n = a(r^n-1)/(r-1)$ for $r \ne 1$. The cancellation is the telescoping idea: shifted copies of the same sum annihilate each other except at the boundary. When the absolute value of $r$ is below $1$, $r^n$ vanishes in the limit and the infinite sum is $a/(1-r)$. A first term of $1$ with ratio $1/2$ sums to $2$.

Result	AP ($d$)	GP ($r$)
$n$th term	$a+(n-1)d$	$ar^{n-1}$
Sum of $n$ terms	$n/2 \times (2a+(n-1)d)$	$a(r^n-1)/(r-1)$
Sum via endpoints	$n/2 \times (\text{first}+\text{last})$	$a/(1-r)$ for infinite sum with $\lvert r \rvert < 1$
Proof idea	Symmetric pairing	Shift and subtract, interior terms cancel

1.10.3 Quadratic Relations and Identities

For $ax^2+bx+c = 0$ with $a \ne 0$, the sum of roots is $-b/a$ and the product is $c/a$. These Vieta relations recover the equation from its roots and factorisation from its coefficients. The identities $(a+b)^2 = a^2+2ab+b^2$, $(a-b)^2 = a^2-2ab+b^2$, and $a^2-b^2 = (a+b)(a-b)$ factor the great majority of algebraic expressions in timed tests. Averages theory connects here as well: the mean of $n$ values is their sum divided by $n$, so a corrected total changes the mean by the correction divided by $n$. Correcting a mis-entered $64$ to $46$ across $20$ readings lowers the total by $18$ and the mean by $18/20 = 0.9$, moving an average of $45$ to $44.1$.

1.11.1 The Multiplication and Addition Principles

Counting rests on two principles. The multiplication principle states that a procedure with $m$ outcomes for its first stage and $n$ outcomes for its second stage, chosen independently, has $mn$ combined outcomes. The addition principle states that disjoint alternatives add: $m$ outcomes of one kind and $n$ of another give $m+n$ outcomes when exactly one kind is selected. Permutations count ordered selections: $P(n,r) = n!/(n-r)!$. Combinations count unordered selections: $C(n,r) = n!/(r!(n-r)!)$. Arranging $5$ distinct objects gives $5! = 120$ orders, while choosing $3$ from $9$ gives $84$ subsets.

1.11.2 Complement Logic and Case Splitting

Complement logic computes a difficult count as a total minus its negation. Committees with at least one member of a given kind are counted as all committees minus committees with none of that kind: $C(9,3) - C(5,3) = 84 - 10 = 74$. The method applies whenever the forbidden class is simpler than the permitted class. Case splitting partitions the desired count into disjoint subcases that add. Restrictions such as inclusion or exclusion of named individuals split naturally into cases with the individual present and cases with the individual absent.

1.11.3 Probability as Favourable over Total

Probability on a finite sample space with equally likely outcomes is the ratio of favourable outcomes to total outcomes. Two fair dice generate $6 \times 6 = 36$ equally likely ordered pairs by the multiplication principle, and $6$ of those pairs sum to $7$. A standard deck provides $52$ equally likely single-card outcomes, with $13$ per suit, $4$ per rank, $12$ face cards, and $26$ cards of each colour. Sums of $9$ arise from $4$ of the $36$ dice pairs, giving $4/36 = 1/9$.

1.11.4 Independence, Mutual Exclusivity, and Conditional Structure

Independent events multiply: the probability of both $A$ and $B$ occurring is $P(A)P(B)$ when neither affects the other. Mutually exclusive events add: the probability of $A$ or $B$ is $P(A)+P(B)$ when both cannot occur together. Draws without replacement are conditionally structured, because each draw alters the composition for the next. The general addition law $P(A \cup B) = P(A)+P(B)-P(A \cap B)$ corrects for double-counted overlap.

1.12.1 The Reading Order

Data interpretation theory prescribes reading the stem before the data. The stem names the required relationship, such as growth, share, ratio, or average, and the data supply the two numbers that instantiate it. Units, years, and scale factors are fixed before any arithmetic, because a table headed in thousand tonnes reports $165$ as $165000$ tonnes. Approximation replaces exact long division wherever options are spaced, since timed tests reward the first two significant figures rather than full precision.

1.12.2 Growth, Share, Ratio, and Average Relationships

Growth is $( \text{later} - \text{earlier})/\text{earlier} \times 100$, with the earlier period as base. A product moving from $80$ to $104$ grows $30\%$. Share is $\text{part}/\text{total} \times 100$ within a single period. A product contributing $96$ to a total of $600$ holds $16\%$. Ratio compares two parts within a period without reference to the total: values $150$ and $165$ stand in the ratio $10:11$. Average divides a total across its components. Column totals computed once feed all four relationships: the total $725$ for the latest year in a sequence $550$, $600$, $650$, $725$ simultaneously identifies the highest year, anchors every share, and supplies the denominator for every average.

1.12.3 Base Discipline and Reuse

Every DI relationship carries its own base. Growth uses the earlier year. Share uses the same-year total. Ratio uses neither, comparing parts directly. Confusing these bases is the DI analogue of the percentage base trap. Computational reuse follows from this structure: one pass over the table establishes row and column totals, and every subsequent relationship reuses those totals rather than recomputing them.

Number theory supplies divisibility reasoning from place-value remainders, with HCF and LCM as minimum and maximum prime-exponent selections. Percentage theory reduces every change to a multiplicative factor on an explicit base, with successive changes multiplying and the net following $a+b+ab/100$. Ratio theory converts parts to equations, partnership theory weights capital by time, and alligation theory derives the cross rule from value conservation. Profit theory measures surplus on cost while discount theory measures reduction on marked price, linked by a single equation. Interest theory contrasts linear simple growth with geometric compound growth, with the two-year gap $P(R/100)^2$ and the Rule of $72$ as consequences. Work theory converts times to integer rates through the LCM total, speed theory averages equal distances harmonically through $2xy/(x+y)$, and mensuration theory scales areas by $k^2$ and volumes by $k^3$. Progression theory sums by pairing and by telescoping cancellation, counting theory builds on multiplication and complement logic, probability theory divides favourable by total outcomes, and DI theory organises growth, share, ratio, and average around strict base discipline.

Later chapters build directly on this foundation. Profit, interest, and growth reuse the base discipline developed for percentages. Work and speed reuse rate reasoning and harmonic structure. Mensuration scale laws reuse exponent reasoning from number theory, and data interpretation reuses every relationship in this chapter under time pressure.