Arithmetic — Number System, Percentages, and Simplification

Why This Chapter Matters

Arithmetic is the subject where preparation effort converts most directly and most reliably into exam marks. Unlike current affairs, where the underlying material keeps changing, or reasoning, where questions can be phrased in unfamiliar ways, arithmetic rests on a fixed, learnable set of rules and formulas that behave the same way every single time you apply them. A candidate who genuinely masters number system properties, percentage conversions, and simplification techniques will solve those questions correctly on any exam, in any year, because the underlying mathematics does not change. This makes arithmetic one of the highest-leverage subjects in your entire preparation — every hour spent building genuine fluency here pays off with near-certain returns on exam day, which is a promise very few other subjects on this syllabus can make.

Beyond the exam, basic numerical fluency is also directly useful in the day-to-day work of a Village Revenue Officer or Village Revenue Assistant, who regularly deals with land measurements, revenue calculations, percentage-based assessments, and record-keeping that demands quick, accurate arithmetic. This chapter builds that fluency from the ground up, starting with the number system — the foundation on which every other arithmetic topic rests — moving through percentages, one of the most frequently tested and most practically useful topics on the entire syllabus, and finishing with simplification and the order-of-operations rules that let you solve complex-looking expressions quickly and without error. Throughout, the emphasis is on worked examples, because arithmetic is a subject you learn by doing, not by reading formulas passively.

The Number System: Foundations

Classification of Numbers

Every arithmetic topic ultimately rests on a clear understanding of how numbers are classified, so it is worth being precise about this before moving further. Natural numbers are the counting numbers: 1, 2, 3, 4, and so on, extending infinitely. Whole numbers include all natural numbers plus zero: 0, 1, 2, 3, and so on. Integers extend whole numbers to include negative numbers as well: ..., -3, -2, -1, 0, 1, 2, 3, .... Rational numbers are any number that can be expressed as a fraction of two integers (with a non-zero denominator), which includes all integers, all terminating decimals, and all recurring decimals — for example, 3, -5, 0.75, and 0.333... (which equals 1/3) are all rational. Irrational numbers cannot be expressed as a simple fraction; their decimal expansion is non-terminating and non-recurring, and common examples include the square root of 2 and the mathematical constant pi. Together, rational and irrational numbers make up the real numbers, which is the full set of numbers you will work with throughout your arithmetic preparation.

A further useful classification is prime and composite numbers. A prime number is a natural number greater than 1 that has exactly two distinct factors — 1 and itself. The number 2 is the only even prime number; every other even number greater than 2 is composite because it is divisible by 2 in addition to 1 and itself. The first several primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29 — memorising primes up to 100 is a genuinely useful investment, since many exam questions rely on quick prime identification. A composite number, by contrast, has more than two factors — for example, 12 is composite because it is divisible by 1, 2, 3, 4, 6, and 12.

Divisibility Rules

Divisibility rules let you determine, by inspection, whether one number divides another without performing long division — an essential speed technique for arithmetic exams. A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8). It is divisible by 3 if the sum of its digits is divisible by 3 — for example, message 471: 4+7+1 = 12, which is divisible by 3, so 471 is divisible by 3. It is divisible by 4 if its last two digits form a number divisible by 4 — for 3528, the last two digits are 28, which is divisible by 4, so 3528 is divisible by 4. It is divisible by 5 if its last digit is 0 or 5. It is divisible by 6 if it is divisible by both 2 and 3. It is divisible by 8 if its last three digits form a number divisible by 8. It is divisible by 9 if the sum of its digits is divisible by 9 — for 4527: 4+5+2+7 = 18, divisible by 9, so 4527 is divisible by 9. It is divisible by 10 if its last digit is 0. It is divisible by 11 by an alternating-sum rule: sum the digits in odd positions and the digits in even positions separately, and if the difference between these two sums is 0 or a multiple of 11, the number is divisible by 11 — for example, for 2728: digits from the right are 8, 2, 7, 2; odd-position sum (1st, 3rd from right) = 8+7 = 15, even-position sum (2nd, 4th from right) = 2+2 = 4, difference = 11, which is a multiple of 11, so 2728 is divisible by 11.

HCF and LCM

The Highest Common Factor (HCF, also called the Greatest Common Divisor) of two or more numbers is the largest number that divides all of them exactly. The Lowest Common Multiple (LCM) is the smallest number that is a multiple of all of them. These two concepts are tested constantly, both directly and embedded within word problems about things like bells ringing together, or people completing laps of a track together.

The most reliable method for finding the HCF of two numbers is prime factorisation: break each number down into its prime factors, and the HCF is the product of the common prime factors raised to the lowest power in which they appear across the numbers. For example, to find the HCF of 36 and 60: 36 = 2² × 3², and 60 = 2² × 3 × 5. The common prime factors are 2 and 3; taking the lowest powers, 2² and 3¹, gives HCF = 4 × 3 = 12.

The LCM, using the same prime factorisation, is the product of all prime factors involved, each raised to the highest power in which it appears across the numbers. Using the same example, the primes involved are 2, 3, and 5; the highest powers are 2², 3², and 5¹, so LCM = 4 × 9 × 5 = 180.

A useful shortcut connects HCF and LCM directly: for any two numbers, the product of the HCF and the LCM equals the product of the two numbers themselves. Checking our example: HCF × LCM = 12 × 180 = 2160, and 36 × 60 = 2160 — they match, confirming our calculation. This relationship is extremely useful for quickly finding one value when the other is known, without repeating the full factorisation.

Word problems on HCF typically involve finding the largest measure that divides given quantities exactly — for example, "find the largest length of tape that can exactly measure 48 metres, 60 metres, and 72 metres" is an HCF problem, because you need the largest value that divides all three lengths exactly. Word problems on LCM typically involve finding when repeating events coincide — for example, "three bells ring at intervals of 12, 18, and 24 minutes respectively; if they ring together at 9:00 AM, when will they next ring together?" is an LCM problem, because you need the smallest time interval that is a multiple of all three ringing intervals. For this example, LCM(12, 18, 24): 12 = 2²×3, 18 = 2×3², 24 = 2³×3; highest powers give 2³ × 3² = 8 × 9 = 72 minutes, so the bells will next ring together at 9:00 AM + 72 minutes = 10:12 AM.

Percentages: The Most Practical Topic on the Syllabus

The Core Concept

A percentage is simply a fraction with a denominator of 100 — "percent" literally means "per hundred." Understanding this single fact unlocks the entire topic: 25% means 25/100, which simplifies to 1/4; 50% means 50/100, which simplifies to 1/2; 75% means 75/100, which simplifies to 3/4. Building instant familiarity with the fraction equivalents of common percentages is one of the highest-value things you can do for speed, because it lets you convert percentage calculations into simple fraction multiplication, which is almost always faster than working with decimals under exam time pressure.

A useful reference table to memorise: 100% = 1, 50% = 1/2, 33⅓% = 1/3, 66⅔% = 2/3, 25% = 1/4, 75% = 3/4, 20% = 1/5, 40% = 2/5, 60% = 3/5, 80% = 4/5, 16⅔% = 1/6, 12½% = 1/8, 10% = 1/10, and 5% = 1/20. When you see a question asking for 33⅓% of 900, recognising instantly that 33⅓% = 1/3 lets you compute 900/3 = 300 in seconds, far faster than multiplying by 0.3333 and rounding.

Converting Between Fractions, Decimals, and Percentages

To convert a fraction to a percentage, multiply it by 100: 3/8 as a percentage is (3/8) × 100 = 37.5%. To convert a decimal to a percentage, multiply by 100 (which in practice just means moving the decimal point two places right): 0.42 becomes 42%. To convert a percentage to a decimal, divide by 100 (move the decimal point two places left): 68% becomes 0.68. These conversions should become entirely automatic with practice, since nearly every percentage question requires moving fluently between these three forms.

Finding a Percentage of a Number

To find P% of a number N, calculate (P/100) × N. For example, to find 15% of 640: (15/100) × 640 = 15 × 6.4 = 96. Alternatively, using the building-block method, which is often faster mentally: 10% of 640 is 64, and 5% of 640 is half of that, 32; so 15% = 10% + 5% = 64 + 32 = 96. This building-block approach — computing 10%, 1%, and 5% quickly and combining them — is one of the most useful mental-math techniques for percentage questions and is worth practising until it becomes second nature.

Percentage Increase and Decrease

If a quantity increases from an original value to a new value, the percentage increase is calculated as: (Increase / Original Value) × 100. For example, if a price rises from ₹250 to ₹300, the increase is ₹50, and the percentage increase is (50/250) × 100 = 20%. Similarly, percentage decrease is (Decrease / Original Value) × 100. If a population falls from 5,000 to 4,600, the decrease is 400, and the percentage decrease is (400/5000) × 100 = 8%.

A critical point that trips up many candidates: percentage change is always calculated with respect to the original (initial) value, not the new value. If a quantity increases by 25% and then decreases by 25%, the net result is not zero — because the second percentage is calculated on a different (larger) base than the first. Working through this concretely: start with 100. A 25% increase brings it to 125. A subsequent 25% decrease is calculated on 125, not on the original 100: 25% of 125 = 31.25, so the final value is 125 - 31.25 = 93.75. The net change is a 6.25% decrease, not zero — a classic exam trap that rewards careful, methodical calculation over hasty assumption.

Successive Percentage Change Formula

For two successive percentage changes of a% and b%, the net percentage change is given by the formula: a + b + (ab/100), where increases are taken as positive and decreases as negative. Applying this to the example above, with a = 25 and b = -25: net change = 25 + (-25) + (25 × -25)/100 = 0 + (-625/100) = -6.25%, confirming our step-by-step calculation of a 6.25% net decrease. This formula is extremely valuable for speed on exam day once you trust it, since it collapses a two-step calculation into a single formula application. It generalises naturally to population growth, successive discounts, and successive price increases — any scenario involving two sequential percentage changes applied to a moving base.

Percentage Point Word Problems

A recurring and important category of percentage problems involves finding an original value when a percentage of it is known, often phrased through population, voting, or exam-pass scenarios. For example: "In an election, a candidate got 45% of the total votes and won by 4,000 votes over the only other candidate. Find the total number of votes." Since there are only two candidates, the winner got 45% and the loser got 55%; the difference of 10 percentage points corresponds to 4,000 votes, so 10% of the total = 4,000, meaning 1% = 400, and the total votes = 400 × 100 = 40,000.

Simplification and BODMAS

The Order of Operations

Simplification questions test your ability to correctly evaluate expressions containing multiple operations in the right order, and the entire topic rests on one rule set, commonly remembered by the acronym BODMAS: Brackets, Of, Division, Multiplication, Addition, Subtraction, applied in that priority order. Brackets are resolved first, working from the innermost bracket outward, and within brackets, typically in the order (), then {}, then []. "Of" refers to expressions like "half of 20," which is treated with the same priority as multiplication and division, evaluated left to right along with them. Division and multiplication are then performed, left to right as they appear in the expression — they do not have separate priority levels over each other; whichever appears first, from left to right, is calculated first. Finally, addition and subtraction are performed, again left to right.

A worked example: simplify 24 ÷ 4 × 2 + 6 - 3 × 2. Following BODMAS, there are no brackets, so we move to division and multiplication, working strictly left to right: 24 ÷ 4 = 6, giving 6 × 2 + 6 - 3 × 2 = 12 + 6 - 3 × 2; continuing left to right through the remaining multiplication, 3 × 2 = 6, giving 12 + 6 - 6. Now addition and subtraction, left to right: 12 + 6 = 18, then 18 - 6 = 12. The final answer is 12.

A second worked example involving brackets: simplify [50 - {2 × (8 + 4) - 6}]. Work from the innermost bracket outward. First, (8+4) = 12. Substituting: [50 - {2 × 12 - 6}]. Within the curly brackets, apply BODMAS internally: 2 × 12 = 24, then 24 - 6 = 18. Substituting: [50 - 18] = 32. The final answer is 32.

Common Simplification Traps

Many errors in simplification arise not from misunderstanding BODMAS but from careless sign handling, particularly when a bracket is preceded by a negative sign. When you remove a bracket preceded by a minus sign, every term inside the bracket has its sign flipped: -(a + b - c) becomes -a - b + c. This single rule is responsible for a disproportionate share of simplification errors under exam time pressure, so it is worth deliberately slowing down whenever you encounter a bracket preceded by a subtraction sign, rather than rushing through it.

Another common trap involves the "of" operator being mistakenly given lower priority than it deserves. In an expression like "3/4 of 60 - 10," candidates sometimes mistakenly compute the subtraction first. Correctly, "of" is evaluated at the same priority level as multiplication and division, so it is resolved before the subtraction: 3/4 of 60 = 45, and then 45 - 10 = 35.

Simplification of Fractions and Decimals

Simplification questions frequently combine fractions and decimals in the same expression, and fluency in converting between the two, along with confident fraction arithmetic (finding common denominators, cross-multiplying), is essential. A useful habit is to convert all terms in an expression to a single form — either all fractions or all decimals — before proceeding, rather than juggling mixed forms mid-calculation, since mixed-form calculation is where most careless errors creep in.

Practising a large volume of simplification problems under timed conditions is genuinely the best way to build the speed and accuracy this topic rewards, precisely because the rules themselves are simple and few — mastery here is almost entirely a matter of repetition until the correct order of operations becomes automatic and error-free, even under the time pressure of an actual exam hall.

Bringing It Together

Number system fundamentals, percentages, and simplification are not three isolated topics — they are deeply interconnected, and questions on the actual exam frequently blend them. A percentage word problem might require you to first simplify a complex fractional expression to find a value, then convert it to a percentage, then apply an HCF-based reasoning step to find a final answer involving whole units. Building genuine fluency in each of the three areas covered in this chapter, rather than memorising isolated formulas, is what will let you move smoothly across these blended question types on exam day. The next chapter builds directly on this foundation, extending into ratio, averages, time-speed-distance, and profit-loss — all of which lean heavily on the percentage and simplification skills you have just built here.