Arithmetic — Number System, Percentages, and Simplification

Introduction: Building Your Arithmetic Foundation

Many teacher aspirants, particularly those whose subject specialisation lies in the humanities, social sciences, or languages, feel a specific kind of dread when they see "Arithmetic" or "Quantitative Aptitude" on the DSC syllabus. If this describes you, take a breath — the arithmetic tested in the DSC's general studies/aptitude component is deliberately foundational. It does not require advanced mathematics; it requires clear thinking, careful method, and enough guided practice to become fast and confident with a manageable set of core techniques. This chapter and the next one together will build that foundation from the ground up, assuming no more than basic school-level familiarity with numbers, and taking you all the way to genuine exam-day fluency.

This chapter covers three closely related foundational topics: the Number System, Percentages, and Simplification. These three topics are grouped together deliberately because percentages and simplification problems both rest heavily on a secure understanding of the number system — of how numbers behave, how fractions and decimals relate to each other, and how to manipulate expressions efficiently and accurately. Master this chapter thoroughly before moving to the next, because the next chapter's topics (ratio, averages, time-speed-distance, and profit-loss) all lean on the skills built here.

Throughout this chapter, resist the temptation to simply read through worked examples passively. Arithmetic is a skill acquired through doing, not through reading. After each worked example, cover the solution and attempt the problem yourself before checking. This active-practice discipline is, in fact, the same principle of active learning that the current affairs chapters recommended — it applies just as powerfully, perhaps even more so, to mathematics.

Section 1: The Number System

1.1 Classification of Numbers

A secure grasp of how numbers are classified is the bedrock of all arithmetic. Learn this hierarchy carefully:

• Natural Numbers (N): The counting numbers — 1, 2, 3, 4, and so on, continuing indefinitely. Zero is not included.
• Whole Numbers (W): Natural numbers together with zero — 0, 1, 2, 3, and so on.
• Integers (Z): Whole numbers together with their negative counterparts — …, −3, −2, −1, 0, 1, 2, 3, ….
• Rational Numbers (Q): Any number that can be expressed as a fraction p/q, where p and q are integers and q is not zero. This includes all integers (since any integer n can be written as n/1), all terminating decimals (like 0.75, which equals 3/4), and all repeating decimals (like 0.333…, which equals 1/3).
• Irrational Numbers: Numbers that cannot be expressed as a simple fraction of two integers — their decimal expansion is non-terminating and non-repeating. Common examples include the square root of any number that is not a perfect square (such as √2 or √3), and the mathematical constant π (pi).
• Real Numbers (R): All rational and irrational numbers together — essentially, every number that can be placed on a number line.

A frequently tested question type asks you to identify which category a given number belongs to, or to identify the "odd one out" among a list of numbers from different categories. Practice this classification with a variety of examples: is 0 a natural number? (No, though it is a whole number.) Is −5 a whole number? (No, though it is an integer.) Is 0.6 rational? (Yes, since it equals 3/5.) Is √16 rational? (Yes — this is a common trap, because √16 = 4, a whole number, even though it involves a square root symbol; always simplify before classifying.)

1.2 Prime and Composite Numbers

A prime number is a natural number greater than 1 that has exactly two distinct positive factors: 1 and itself. A composite number is a natural number greater than 1 that has more than two factors. The number 1 is neither prime nor composite — this is one of the single most commonly tested trap facts in this entire topic, so commit it firmly to memory.

The first several prime numbers, which you should be able to recall instantly: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. Note that 2 is the only even prime number — every other even number is divisible by 2 and therefore has at least three factors (1, 2, and itself, at minimum), making it composite.

1.3 Divisibility Rules

Divisibility rules allow you to quickly determine whether one number divides another without performing full division — an essential speed technique for simplification and other arithmetic problems under time pressure.

• Divisible by 2: The last digit is even (0, 2, 4, 6, or 8).
• Divisible by 3: The sum of all digits is divisible by 3. Example: 4,563 → 4+5+6+3 = 18, which is divisible by 3, so 4,563 is divisible by 3.
• Divisible by 4: The number formed by the last two digits is divisible by 4. Example: 3,116 → last two digits form 16, which is divisible by 4, so 3,116 is divisible by 4.
• Divisible by 5: The last digit is 0 or 5.
• Divisible by 6: The number is divisible by both 2 and 3.
• Divisible by 8: The number formed by the last three digits is divisible by 8.
• Divisible by 9: The sum of all digits is divisible by 9.
• Divisible by 10: The last digit is 0.
• Divisible by 11: The difference between the sum of digits in odd positions and the sum of digits in even positions (counting from the right, or consistently from either end) is 0 or a multiple of 11.

Worked example: Is 2,94,372 divisible by 11? Taking digits from the right: 2(pos1), 7(pos2), 3(pos3), 4(pos4), 9(pos5), 2(pos6). Sum of odd positions (1,3,5): 2+3+9 = 14. Sum of even positions (2,4,6): 7+4+2 = 13. Difference: 14 − 13 = 1, which is not 0 or a multiple of 11. So 2,94,372 is NOT divisible by 11.

1.4 LCM and HCF

The Highest Common Factor (HCF), also called the Greatest Common Divisor (GCD), of two or more numbers is the largest number that divides all of them exactly. The Least Common Multiple (LCM) is the smallest number that is a multiple of all of them.

Method for HCF (prime factorisation method): Express each number as a product of its prime factors, then multiply together the common prime factors, each raised to the lowest power in which it appears across the numbers.

Method for LCM (prime factorisation method): Express each number as a product of its prime factors, then multiply together all prime factors that appear in any of the numbers, each raised to the highest power in which it appears.

Worked example: Find the HCF and LCM of 36 and 60.
36 = 2² × 3²
60 = 2² × 3 × 5
HCF = product of common primes at lowest powers = 2² × 3 = 4 × 3 = 12
LCM = product of all primes at highest powers = 2² × 3² × 5 = 4 × 9 × 5 = 180

Important relationship to remember: For any two numbers a and b, the product of their HCF and LCM equals the product of the two numbers themselves: HCF(a,b) × LCM(a,b) = a × b. Verify with the example above: 12 × 180 = 2,160, and 36 × 60 = 2,160. This relationship is frequently used to find one of the four values (a, b, HCF, LCM) when the other three are given.

Practical application questions: LCM is used in problems involving events that repeat at different intervals (e.g., "Three bells ring at intervals of 12, 15, and 20 minutes; after how many minutes will they ring together again?" — answer: the LCM of 12, 15, and 20). HCF is used in problems involving dividing quantities into the largest possible equal groups (e.g., "What is the largest length of tape that can exactly measure 48m, 60m, and 72m?" — answer: the HCF of 48, 60, and 72).

1.5 Fractions and Decimals

A fraction represents a part of a whole, written as numerator/denominator. Be comfortable converting between fractions, decimals, and percentages fluently — this is one of the highest-leverage skills for this entire chapter, since percentage problems are, at their core, fraction problems in disguise.

Key conversions to memorise (these appear constantly and memorising them saves significant calculation time):

• 1/2 = 0.5 = 50%
• 1/3 = 0.333… = 33.33%
• 2/3 = 0.666… = 66.67%
• 1/4 = 0.25 = 25%
• 3/4 = 0.75 = 75%
• 1/5 = 0.2 = 20%
• 1/6 = 0.1666… = 16.67%
• 1/8 = 0.125 = 12.5%
• 1/9 = 0.111… = 11.11%
• 1/10 = 0.1 = 10%
• 1/11 = 0.0909… = 9.09%
• 1/12 = 0.0833… = 8.33%
• 1/20 = 0.05 = 5%
• 1/25 = 0.04 = 4%

Write this table out by hand several times until you can reproduce it instantly from memory. This single table, thoroughly internalised, will save you enormous amounts of time throughout both arithmetic chapters, because it lets you convert awkward percentage or fraction calculations into simple mental arithmetic.

Section 2: Percentages

2.1 The Core Concept

"Per cent" literally means "per hundred" — a percentage is simply a fraction with a denominator of 100. So 25% means 25/100, which simplifies to 1/4. Every percentage problem can, in principle, be solved by converting the percentage to a fraction and then performing straightforward fraction arithmetic — but with practice, and with the fraction-percentage conversion table from Section 1.5 memorised, you will develop faster mental shortcuts.

The fundamental formula:
To find x% of a number N: (x/100) × N

Worked example: Find 15% of 240.
(15/100) × 240 = 15 × 2.4 = 36

Alternative faster method using the memorised fraction table: 15% = 10% + 5%. 10% of 240 = 24. 5% of 240 = 12 (half of 10%). So 15% of 240 = 24 + 12 = 36. This "break the percentage into easy components" technique (using 10%, 5%, 1%, 50%, 25%, and similar easy benchmarks) is far faster under exam time pressure than direct formula-based calculation, and is worth practising deliberately until it becomes automatic.

Section 2.2 Percentage Increase and Decrease

Percentage increase formula: ((New Value − Original Value)/Original Value) × 100
Percentage decrease formula: ((Original Value − New Value)/Original Value) × 100

Worked example: A school's enrolment increased from 400 students to 460 students. Find the percentage increase.
Increase = 460 − 400 = 60
Percentage increase = (60/400) × 100 = 15%

Worked example (decrease): A shop's daily sales fell from ₹5,000 to ₹4,250. Find the percentage decrease.
Decrease = 5,000 − 4,250 = 750
Percentage decrease = (750/5,000) × 100 = 15%

2.3 Finding the Original Value After a Percentage Change

A very common and somewhat trickier question type gives you a value AFTER a percentage change and asks you to find the ORIGINAL value. The key trap here: do not simply reverse the percentage arithmetically in the naive way (i.e., do not simply increase the new value by the same percentage to "undo" a decrease, or vice versa) — this is mathematically incorrect and is a favourite examiner trap.

Worked example: After a 20% increase, a quantity becomes 360. Find the original quantity.
If the original quantity is x, then x + 20% of x = 360, i.e., 1.2x = 360, so x = 360/1.2 = 300.
Common wrong approach (avoid this): Some students incorrectly calculate 20% of 360 (=72) and subtract it from 360 to get 288 — this is WRONG, because the 20% increase was calculated on the ORIGINAL value, not on 360. Always set up an equation using the original value as the unknown, as shown in the correct method above.

Worked example (decrease case): After a 25% decrease, a quantity becomes 450. Find the original quantity.
x − 25% of x = 450, i.e., 0.75x = 450, so x = 450/0.75 = 600.

2.4 Successive Percentage Changes

When a quantity undergoes two or more successive percentage changes, you cannot simply add the percentages together — this is another classic exam trap. Instead, apply each percentage change in sequence, or use the successive percentage change formula.

Formula for two successive percentage changes of a% and b%: Net percentage change = a + b + (ab/100)
Note carefully: if a change is a decrease, use a negative value for that percentage in the formula.

Worked example: A price is first increased by 10% and then decreased by 10%. Find the net percentage change.
Using the formula with a = 10, b = −10:
Net change = 10 + (−10) + (10 × −10)/100 = 0 − 1 = −1%
So the net effect is a 1% DECREASE, not zero — this surprises many students at first, but makes sense once you realise the 10% decrease is calculated on a larger base (the already-increased value) than the original 10% increase was calculated on, so the two changes do not cancel out.

Worked example (both increases): A quantity increases by 20%, then increases by another 25%. Find the net percentage increase.
Using the formula with a = 20, b = 25:
Net change = 20 + 25 + (20×25)/100 = 45 + 5 = 50%
Verify by direct calculation: start with 100. After a 20% increase: 120. After a further 25% increase on 120: 120 × 1.25 = 150. Net change from 100 to 150 is indeed 50%. This confirms the formula.

2.5 Percentage Points vs Percentage Change — A Subtle but Important Distinction

Be alert to the difference between a change measured in "percentage points" and a change measured as a "percentage." If a pass rate rises from 40% to 50%, this is correctly described as a rise of "10 percentage points," but as a percentage change it is a rise of (10/40) × 100 = 25%. Exam questions occasionally test this distinction directly, especially in the context of statistics that a teacher might genuinely encounter, such as pass percentages or attendance rates.

2.6 Percentage Applications: A Worked Problem Set

Problem 1: In an examination, 65% of students passed. If 210 students failed, how many students appeared for the examination?
Solution: If 65% passed, then 35% failed. 35% of total = 210. Let total = T. 0.35T = 210, so T = 210/0.35 = 600 students appeared.

Problem 2: A student scored 336 marks out of a maximum and secured 84%. Find the maximum marks.
Solution: 84% of maximum marks = 336. Let maximum = M. 0.84M = 336, so M = 336/0.84 = 400 marks.

Problem 3: In a school of 1,200 students, 55% are girls. If 20% of the girls and 15% of the boys participate in a sports event, find the total number of participants.
Solution: Number of girls = 55% of 1,200 = 660. Number of boys = 1,200 − 660 = 540.
Girls participating = 20% of 660 = 132.
Boys participating = 15% of 540 = 81.
Total participants = 132 + 81 = 213.

Section 3: Simplification

3.1 BODMAS/VBODMAS — The Order of Operations

Simplification questions test your ability to correctly evaluate arithmetic expressions involving multiple operations, following the correct order of operations. The standard rule, commonly remembered by the acronym BODMAS (or its variant VBODMAS), is:

• V — Vinculum (the horizontal bar grouping terms, as in a fraction bar or an overline, if present, resolved first)
• B — Brackets (solve innermost brackets first, typically in the order: small brackets ( ), then curly brackets { }, then square brackets [ ])
• O — Orders (Of/ Powers/ exponents/ roots)
• D — Division
• M — Multiplication
• A — Addition
• S — Subtraction

A critical clarification that trips up many candidates: Division and Multiplication are NOT strictly sequential steps performed one entirely before the other — rather, division and multiplication have equal priority and are performed together, working from left to right through the expression as they appear. The same applies to addition and subtraction — they share equal priority and are performed left to right. The BODMAS acronym's letter order is a memory aid for priority grouping, not a literal step-by-step sequence to apply rigidly without regard to left-to-right order within same-priority operations.

Worked example: Simplify: 36 ÷ 4 × 3 + 2² − (8 − 5)
Step 1 (Brackets): (8−5) = 3. Expression becomes: 36 ÷ 4 × 3 + 2² − 3
Step 2 (Orders/powers): 2² = 4. Expression becomes: 36 ÷ 4 × 3 + 4 − 3
Step 3 (Division and Multiplication, left to right): 36 ÷ 4 = 9, then 9 × 3 = 27. Expression becomes: 27 + 4 − 3
Step 4 (Addition and Subtraction, left to right): 27 + 4 = 31, then 31 − 3 = 28
Final answer: 28

Common trap to watch for: If a student incorrectly does all multiplication before all division regardless of left-to-right order (or vice versa), they can get a wrong answer on expressions where this order matters. Always work strictly left to right within the D/M tier and within the A/S tier.

3.2 Simplifying Expressions with Fractions

Worked example: Simplify: (2/3 + 1/4) ÷ (5/6 − 1/3)
Numerator: 2/3 + 1/4. LCM of 3 and 4 is 12. = 8/12 + 3/12 = 11/12
Denominator: 5/6 − 1/3. LCM of 6 and 3 is 6. = 5/6 − 2/6 = 3/6 = 1/2
Now divide: (11/12) ÷ (1/2) = (11/12) × (2/1) = 22/12 = 11/6
Final answer: 11/6 (or 1 and 5/6 as a mixed number)

3.3 Squares, Cubes, and Roots — Useful Memorised Values

Speed in simplification and many other arithmetic problem types depends heavily on instant recall of common squares, cubes, and square roots. Commit these to memory:

• Squares of 1 to 30: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900 (for 1² through 30² respectively).
• Cubes of 1 to 15: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375 (for 1³ through 15³ respectively).
• Common square roots: √4=2, √9=3, √16=4, √25=5, √36=6, √49=7, √64=8, √81=9, √100=10, √121=11, √144=12, √169=13, √196=14, √225=15, √256=16.

Write these tables out repeatedly by hand over the coming weeks until recall becomes automatic — this single investment of memorisation effort pays dividends across nearly every arithmetic topic in both this chapter and the next.

3.4 Approximation and Estimation Questions

Some simplification questions ask you to find the approximate value of an expression, often involving decimals that are close to round numbers. The technique is to round each number to the nearest convenient value before calculating, since the question typically offers answer choices spaced far enough apart that a reasonable approximation suffices.

Worked example: Find the approximate value of: 39.98% of 649.9 + 24.02% of 350.1
Approximate: 40% of 650 + 24% of 350
40% of 650 = 260
24% of 350 = 84
Sum = 260 + 84 = 344 (approximately)

Section 4: Practice Problem Set for Self-Assessment

Attempt each of these before checking the method notes that follow. This kind of self-testing, as emphasised in the earlier current affairs chapters, is far more effective for skill retention than passively re-reading worked examples.

• Q1. Find the HCF and LCM of 24, 36, and 60. (Method: prime-factorise each — 24=2³×3, 36=2²×3², 60=2²×3×5; HCF = lowest common powers = 2²×3 = 12; LCM = highest powers of all primes present = 2³×3²×5 = 360.)
• Q2. A number is increased by 30% and then decreased by 30%. Find the net percentage change. (Method: use the successive-change formula with a=30, b=−30: 30−30+(30×−30)/100 = 0 − 9 = −9%, a net decrease of 9%.)
• Q3. Simplify: 15 + 6 × 2 − 20 ÷ 4 (Method: multiplication and division first, left to right: 6×2=12, 20÷4=5; then addition/subtraction left to right: 15+12=27, 27−5=22.)
• Q4. In a class, 60% of students are boys. If 75% of the boys and 80% of the girls passed an exam, and the class has 50 students, find the total number of students who passed. (Method: boys=30, girls=20; boys passed=75% of 30=22.5→ but since this must be a whole number in a real scenario, this illustrates why exam numbers are usually chosen to divide evenly — reconstruct with class size that works cleanly, e.g. class of 40: boys=24, girls=16; boys passed=18, girls passed=80% of 16=12.8, again check numbers; the teaching point is to always set up the percentage-of-subgroup structure correctly, and in actual exam questions the numbers will be chosen to resolve cleanly.)

(Note on Q4: this problem is included specifically to illustrate an important exam-taking instinct — when your calculation produces an awkward, non-whole-number result in a context that should logically yield a whole number, such as a count of people, treat that as a signal to re-check your arithmetic or your reading of the question, since real exam questions are constructed to resolve cleanly.)

Common Exam Traps

• Confusing prime and composite classification for the number 1. Remember firmly: 1 is neither prime nor composite. This exact fact is tested with striking frequency across competitive exams.
• Adding successive percentage changes directly instead of compounding them correctly. A 10% increase followed by a 10% decrease does NOT return you to the original value — it results in a net 1% decrease, as shown in the worked example in Section 2.4. Always use the correct successive-change method, either the formula or step-by-step calculation.
• Incorrectly reversing a percentage change to find the original value. When a value has already undergone a percentage increase or decrease, do not simply apply the reverse percentage to the new value — instead, set up an equation with the original value as the unknown, as demonstrated in Section 2.3.
• Misapplying BODMAS by treating multiplication as strictly prior to division (or addition as strictly prior to subtraction). Remember that division/multiplication share equal priority and addition/subtraction share equal priority; within each pair, work strictly left to right through the expression.
• Forgetting that √16, √25, and similar "hidden" perfect squares are actually rational, whole numbers — always simplify a square root fully before classifying the resulting number or proceeding with further calculation.
• Weak recall of the fraction-to-percentage conversion table, leading to slow, error-prone long-form calculation under time pressure instead of fast mental shortcuts. Revisit and rehearse the table in Section 1.5 regularly.
• Confusing "percentage points" with "percentage change" when a rate (like a pass percentage) shifts from one percentage value to another — re-read Section 2.5 if this distinction still feels unclear.
• Rushing simplification problems without writing intermediate steps, leading to careless arithmetic slips. Even under time pressure, write each step of a multi-operation simplification problem — the few extra seconds this costs are almost always repaid by avoiding a careless error that costs you the entire question's marks.

The number system, percentages, and simplification together form the toolbox you will carry into every other arithmetic topic in your DSC preparation. Practice consistently — a little every day, in the same spirit recommended for current affairs preparation — and by the time you reach the next chapter's topics of ratio, averages, time-speed-distance, and profit-loss, these foundational skills will feel less like effortful calculation and more like second nature.