Arithmetic — Number System, Percentages, and Simplification

Why This Chapter Matters

Arithmetic forms the quantitative backbone of the Grama/Ward Sachivalayam examination, and within Arithmetic, the number system, percentages, and simplification techniques covered in this chapter are the foundation on which every other quantitative topic — ratios, averages, profit and loss, time and work — is built. These are also, happily, among the most reliably scorable topics in the entire examination, because unlike current affairs or even some reasoning questions, arithmetic problems have a single, objectively correct answer that can be reached through a clear, learnable method rather than recall of a fact that might be forgotten under pressure. Examiners favour this area because it tests logical thinking and numerical fluency that are genuinely useful in the day-to-day work of a Sachivalayam employee, who must routinely calculate scheme eligibility thresholds, verify percentage-based subsidy entitlements, and cross-check numerical entries in beneficiary records — making fluency in this chapter both an exam necessity and a practical job skill.

The Number System — Classification and Core Properties

A firm grip on the number system is essential because many simplification and reasoning questions hinge on correctly classifying numbers or applying their basic properties. Natural numbers are the counting numbers starting from 1 (1, 2, 3, and so on), while whole numbers include zero along with all natural numbers (0, 1, 2, 3, and so on). Integers extend this further to include negative numbers as well, giving the complete set of positive and negative whole numbers along with zero. Rational numbers are any numbers that can be expressed as a fraction of two integers (with a non-zero denominator), which includes all integers, all terminating decimals, and all repeating decimals; irrational numbers, by contrast, cannot be expressed as such a fraction and have non-terminating, non-repeating decimal expansions, with the classic examples being the square root of 2 and the mathematical constant pi. Real numbers encompass both rational and irrational numbers together, forming the complete number line that most arithmetic questions operate within.

Prime numbers are numbers greater than 1 that have exactly two distinct positive divisors: 1 and themselves. The first several prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29, and candidates should note that 2 is the only even prime number, since every other even number is divisible by 2 and therefore has at least three divisors. Composite numbers are numbers greater than 1 that have more than two divisors — in other words, any positive integer greater than 1 that is not prime. The number 1 itself is neither prime nor composite, a small but frequently tested fact. Divisibility rules are extremely useful shortcuts worth memorising thoroughly: a number is divisible by 2 if its last digit is even; by 3 if the sum of its digits is divisible by 3; by 4 if its last two digits form a number divisible by 4; by 5 if its last digit is 0 or 5; by 6 if it is divisible by both 2 and 3; by 8 if its last three digits form a number divisible by 8; by 9 if the sum of its digits is divisible by 9; by 10 if its last digit is 0; and by 11 through the alternating-sum rule, where the difference between the sum of digits in odd positions and the sum of digits in even positions (counted from the right) is either 0 or a multiple of 11.

LCM (Least Common Multiple) and HCF (Highest Common Factor, also called GCD or Greatest Common Divisor) are two of the most frequently tested number-system concepts. The HCF of two or more numbers is the largest number that divides all of them exactly, while the LCM is the smallest number that is exactly divisible by all of them. A very useful relationship connects the two for any pair of numbers: the product of the LCM and HCF of two numbers equals the product of the two numbers themselves. For instance, if two numbers are 12 and 18, their HCF is 6 (since 6 is the largest number dividing both 12 and 18 exactly) and their LCM is 36 (since 36 is the smallest number that both 12 and 18 divide into exactly); checking the relationship, the product of the two numbers, 12 times 18, equals 216, and the product of their LCM and HCF, 36 times 6, also equals 216, confirming the rule. This relationship is especially useful in problems that give one of the four values (the two numbers, their LCM, or their HCF) and ask for another, since it allows the missing value to be found through simple division rather than a lengthy factorisation process. The standard method for finding HCF is prime factorisation (breaking each number into its prime factors and multiplying together the lowest power of every common prime factor) or the division method (repeatedly dividing the larger number by the smaller and continuing with the remainder until the remainder becomes zero, at which point the last non-zero divisor is the HCF). The standard method for finding LCM by prime factorisation is to multiply together the highest power of every prime factor that appears in any of the numbers.

Fractions and decimals form another core area, and candidates should be fluent in converting between the two, since many percentage problems are essentially fraction problems in disguise. A proper fraction has a numerator smaller than its denominator (such as three-fourths), while an improper fraction has a numerator equal to or greater than its denominator (such as seven-fourths), which can also be expressed as a mixed number (such as one and three-fourths). Simplifying a fraction means dividing both the numerator and denominator by their HCF until no common factor greater than 1 remains. When comparing fractions with different denominators, converting them to a common denominator (typically their LCM) or converting each fraction to its decimal equivalent are both reliable methods, and speed in this conversion is extremely valuable in a timed exam, so candidates should memorise the decimal equivalents of common fractions such as one-half (0.5), one-third (approximately 0.333), one-fourth (0.25), one-fifth (0.2), three-fourths (0.75), and similar simple fractions up to at least eighths and tenths, since this single piece of memorisation dramatically speeds up percentage calculations, as explained in the next section.

Percentages — Concept, Conversions, and Applications

Percentage is simply a way of expressing a number as a fraction of 100, and the word itself derives from the Latin "per centum," meaning "per hundred." Converting a fraction to a percentage involves multiplying it by 100; converting a percentage to a fraction involves dividing it by 100 and simplifying. This bidirectional fluency is the single most important skill in this chapter, because a large share of Sachivalayam exam arithmetic questions — not just those explicitly labelled "percentage" — are more easily and quickly solved when converted into fraction form using the memorised equivalents mentioned above rather than through direct percentage-formula calculation. For example, since 25 percent is equivalent to the fraction one-fourth, finding 25 percent of any number can be done instantly by simply dividing that number by 4, which is faster and less error-prone than multiplying by 0.25 or by 25 and then dividing by 100. Similarly, 50 percent equals one-half, so finding 50 percent of a number just means halving it; 20 percent equals one-fifth, so finding 20 percent of a number means dividing by 5; and 12.5 percent equals one-eighth, so finding 12.5 percent of a number means dividing by 8.

The core percentage formula that underlies almost every question in this area is: to find what percentage one quantity is of another, divide the first quantity by the second and multiply by 100. For instance, if a candidate wants to know what percentage 45 is of 180, dividing 45 by 180 gives 0.25, and multiplying by 100 gives 25 percent. Conversely, to find a given percentage of a quantity, multiply the quantity by the percentage and divide by 100 — for instance, 30 percent of 250 is found by multiplying 250 by 30 and dividing by 100, giving 75.

Percentage increase and decrease problems are extremely common and deserve careful attention because they are also one of the most common sources of careless errors. To increase a number by a certain percentage, the standard approach is to multiply the original number by (100 plus the percentage) divided by 100; to decrease a number by a certain percentage, multiply by (100 minus the percentage) divided by 100. For example, increasing 400 by 15 percent means multiplying 400 by 115 and dividing by 100, which gives 460; decreasing 400 by 15 percent means multiplying 400 by 85 and dividing by 100, which gives 340. A classic and frequently tested trap involves successive percentage changes: if a quantity is first increased by a certain percentage and then decreased by the same percentage (or vice versa), the net result is never a return to the original value, but always a net decrease, and the size of this net decrease can be calculated directly using the rule that if a number is changed by plus-a-percent and then by minus-a-percent, the net percentage change is minus (a squared, divided by 100) percent. Applying this to a concrete case: if a number is increased by 20 percent and then decreased by 20 percent, the net change is a decrease of 20 times 20 divided by 100, which is 4, so the net effect is a 4 percent decrease from the original value, regardless of what the original number actually was. This is a genuinely important shortcut, since working the problem out longhand — increasing a number by 20 percent and then decreasing that new, larger number by 20 percent — takes considerably longer and is more error-prone than applying this direct formula, and examiners are aware that this shortcut exists, which is precisely why successive-percentage-change questions appear so frequently.

Another commonly tested application is the "percentage of a percentage" type question, sometimes phrased as comparing two quantities where one is a certain percentage more or less than the other. If quantity A is 25 percent more than quantity B, this does not mean B is 25 percent less than A — this is a frequent and important error to avoid. If A is 25 percent more than B, then A equals B multiplied by 125 and divided by 100, so B equals A multiplied by 100 and divided by 125, which simplifies to B being 20 percent less than A, not 25 percent less. This asymmetry arises because percentage increase and percentage decrease are always calculated relative to different base values (the increase is relative to the smaller original quantity, while the corresponding decrease, viewed from the other direction, is relative to the larger new quantity), and candidates must train themselves to always identify clearly which quantity is being used as the base (the "of" quantity) in any percentage comparison problem, since misidentifying the base is the single most common source of error in this entire topic.

Population growth and depreciation problems extend the percentage-change concept across multiple time periods and are common in this exam. If a population or value increases by a fixed percentage each year, the value after several years is found by applying the increase formula repeatedly, once for each year, which is mathematically equivalent to raising the growth multiplier to the power of the number of years. For instance, if a town's population is 10,000 and it grows at 10 percent per year, the population after one year is 10,000 multiplied by 1.1, giving 11,000; after two years, that new value is again multiplied by 1.1, giving 12,100; and after three years, multiplying again by 1.1 gives 13,310. Depreciation problems work identically but with a multiplier below 1, reflecting a percentage decrease compounded each year — for example, a machine worth 50,000 rupees depreciating at 10 percent per year would be worth 50,000 multiplied by 0.9, or 45,000, after one year, and 45,000 multiplied by 0.9, or 40,500, after two years.

Simplification — Order of Operations and Efficient Calculation

Simplification questions test a candidate's ability to correctly evaluate an arithmetic expression involving multiple operations, and the foundation of correct simplification is the BODMAS (or PEMDAS in some international textbooks) rule, which specifies the order in which operations must be carried out: Brackets first, then Orders (meaning powers and roots), then Division and Multiplication (carried out left to right, since these two operations are of equal priority), and finally Addition and Subtraction (also carried out left to right, since these two are likewise of equal priority). Within brackets, if an expression has multiple nested brackets, the innermost bracket is always simplified first, working outward. A very common source of error is treating division and multiplication, or addition and subtraction, as having a fixed priority relative to each other rather than correctly applying them in left-to-right order as they appear in the expression — for instance, in the expression 20 divided by 4 multiplied by 2, the correct approach is to work strictly left to right, first dividing 20 by 4 to get 5, and then multiplying by 2 to get 10; treating multiplication as automatically taking priority over division (multiplying 4 by 2 first to get 8, then dividing 20 by 8) would produce the wrong answer of 2.5, a mistake that examiners specifically design questions to catch.

Beyond correctly applying BODMAS, speed in simplification comes from recognising patterns that allow shortcuts rather than working through every operation mechanically. Recognising perfect squares and cubes up to reasonably high values (knowing, for instance, that 13 squared is 169, 17 squared is 289, and that the cubes of small numbers like 2 through 10 are 8, 27, 64, 125, 216, 343, 512, 729, and 1000) saves considerable time in questions involving squares, cubes, or square roots. Algebraic identities, though the operations here are purely numerical, still provide valuable shortcuts: the identity that the square of a sum equals the sum of the squares plus twice the product, and its counterpart for a difference, can be used to quickly compute the squares of numbers close to round figures — for example, 98 squared can be computed as (100 minus 2) squared, which by the identity equals 100 squared minus 2 times 100 times 2, plus 2 squared, giving 10,000 minus 400 plus 4, which equals 9,604, a calculation that is considerably faster than direct long multiplication of 98 by 98. Similarly, the difference-of-squares identity (that the product of a sum and a difference of the same two numbers equals the difference of their squares) is useful for quickly multiplying two numbers that are equally spaced above and below a round number — for instance, 97 multiplied by 103 can be seen as (100 minus 3) multiplied by (100 plus 3), which by the identity equals 100 squared minus 3 squared, giving 10,000 minus 9, which equals 9,991.

Simplification of fractions within a larger expression also benefits from looking for common factors before performing multiplication, rather than multiplying everything out first and simplifying at the end. When an expression involves multiplying several fractions together, it is almost always faster to cancel common factors between any numerator and any denominator across the entire expression before multiplying, since this keeps the numbers smaller and more manageable throughout the calculation and reduces the chance of arithmetic slips. Candidates should also become comfortable working confidently with recurring decimals, recognising, for instance, that 0.333... repeating equals exactly one-third, and that such conversions are often the fastest route through a complex-looking decimal expression that is secretly built from simple fractions.

A Worked Simplification Example

Consider an expression requiring careful application of BODMAS: 15 plus 6 multiplied by the quantity (8 minus 3), all divided by 3, minus 2. Working through this step by step: first, the bracket is resolved, giving 8 minus 3 equals 5; next, multiplication is carried out before division and addition since it appears first in reading order among the same-priority group after brackets, giving 6 multiplied by 5 equals 30; the expression now reads 15 plus 30, divided by 3, minus 2; division is carried out next, but only on the portion it directly applies to — if the original expression intends 30 divided by 3 as a standalone term rather than the entire sum of 15 and 30 being divided by 3, the placement of any further brackets in the original question becomes critical, which is exactly why exam-setters are careful (and candidates must be equally careful) about exactly which terms a division or multiplication symbol applies to. Assuming the intended reading is 15 plus (30 divided by 3) minus 2, this becomes 15 plus 10 minus 2, working left to right for the final addition and subtraction, giving 25 minus 2, which equals 23. This example illustrates why reading an expression's structure carefully — identifying exactly which terms are grouped together by implicit or explicit brackets — is just as important as knowing the BODMAS rule itself, since misreading the grouping is a far more common cause of error among exam candidates than not knowing the rule in the first place.

Common Mistakes and Exam Strategy for This Chapter

The most frequent error across this entire topic area is rushing into calculation before fully reading and understanding what a question is asking, particularly in percentage-comparison problems where correctly identifying the base quantity is essential, as discussed above. A disciplined habit of writing down, even briefly, exactly what is being compared to what before beginning any calculation prevents a large share of otherwise avoidable mistakes. A second common error is inconsistent handling of decimal points during multiplication and division, especially under time pressure — candidates should develop the habit of estimating the approximate size of an answer before calculating precisely, since this makes it easy to catch a decimal-point error (an answer that is ten or a hundred times too large or too small is usually obviously wrong once a rough estimate has been made in advance). Third, candidates frequently forget to simplify their final answer or to convert it back to the form the question is asking for — for instance, computing a percentage correctly but forgetting to append the percent sign or convert a fraction to the percentage the question actually requested. Building speed and accuracy in this chapter comes primarily from consistent daily practice with a genuinely wide variety of problems, memorising the small set of tables, squares, cubes, and fraction-decimal equivalents mentioned throughout this chapter, and developing the discipline of a brief sanity check on every final answer before moving to the next question — habits that, once ingrained through repetition, make this one of the fastest and most reliably scoring sections of the entire examination.