Number System & Simplification

Natural numbers (N): 1,2,3,... Whole numbers (W): 0,1,2,3,... (add zero to N). Integers (Z): ...-2,-1,0,1,2... Rational numbers: can be written as p/q (q not 0), e.g. 3/4, 0.25, 0.333. Irrational numbers: non-terminating non-repeating, e.g. root2, pi. Prime number: exactly two factors (1 and itself); 2 is the smallest and ONLY even prime. Composite: more than two factors. 1 is NEITHER prime nor composite. Even: divisible by 2; Odd: not. Memory aid: every Natural is Whole, every Whole is Integer, every Integer is Rational. There are 25 prime numbers below 100. Co-prime numbers share only 1 as common factor, e.g. (8,15).

By 2: last digit even (0,2,4,6,8). By 3: digit-sum divisible by 3. By 4: last two digits divisible by 4. By 5: last digit 0 or 5. By 6: divisible by 2 AND 3. By 8: last three digits divisible by 8. By 9: digit-sum divisible by 9. By 10: ends in 0. By 11: (sum of odd-position digits) minus (sum of even-position digits) is 0 or a multiple of 11. By 7 shortcut: double the last digit, subtract from rest; if result divisible by 7 the number is. Trick for 6,12,15: check the two co-prime factors (e.g. 12 = 3 and 4). Always test smaller co-prime factors rather than the big number itself.

Eleven is the trickiest of the school-age divisibility rules — and that is exactly why RPF Constable papers love to test it. Once you understand why the alternating-sum rule works, you can decide divisibility of any six- or seven-digit number in under fifteen seconds. Let us walk through the two worked examples you must internalise.

Definition: A number n is divisible by 11 if and only if the difference between the sum of digits in odd positions and the sum of digits in even positions is either 0 or a multiple of 11 (i.e., ±11, ±22, ±33, …). Position counting may start from the right or the left — only consistency matters.

Why the Rule Works (the Quick Why)

Powers of 10, taken mod 11, alternate between +1 and −1:

10⁰ = 1, 10¹ = 10 ≡ −1, 10² = 100 ≡ +1, 10³ = 1000 ≡ −1, …

So any number written in decimal as d₀ + d₁·10 + d₂·10² + … reduces, mod 11, to d₀ − d₁ + d₂ − d₃ + …, which is exactly the alternating sum of its digits. That alternating sum is what the textbook calls "odd-position sum minus even-position sum". This is why the rule is bulletproof — it is just modular arithmetic of 10 mod 11.

Worked Example 1: Is 4,83,771 Divisible by 11?

The number is 4 8 3 7 7 1 (six digits). We label positions starting from the rightmost digit.

Question: Check whether 483771 is divisible by 11.

Solution:
Step 1: Number the positions from the right.

:::compare

Digit	4	8	3	7	7	1
Position from right	6	5	4	3	2	1
Odd or even	even	odd	even	odd	even	odd
:::

Step 2: Sum the digits in odd positions: 1 + 7 + 8 = 16.
Step 3: Sum the digits in even positions: 7 + 3 + 4 = 14.
Step 4: Difference = 16 − 14 = 2.
Step 5: Is 2 equal to 0 or a multiple of 11? No.
Conclusion: 483771 is not divisible by 11. (For confirmation, 483771 ÷ 11 = 43,979.18…)

Worked Example 2: Is 4,83,769 Divisible by 11?

Same six-digit shape, last two digits changed. Watch how a tiny tweak flips the answer.

Question: Check whether 483769 is divisible by 11.

Solution:
Step 1: List positions from the right for 4 8 3 7 6 9.

:::compare

Digit	4	8	3	7	6	9
Position from right	6	5	4	3	2	1
Odd or even	even	odd	even	odd	even	odd
:::

Step 2: Sum the digits in odd positions: 9 + 7 + 8 = 24.
Step 3: Sum the digits in even positions: 6 + 3 + 4 = 13.
Step 4: Difference = 24 − 13 = 11.
Step 5: Is 11 a multiple of 11? Yes.
Conclusion: 483769 is divisible by 11. (Confirmation: 483769 ÷ 11 = 43,979.)

Why It Matters — and the One Trap to Avoid

The rule itself is simple, but every year candidates lose marks by changing the direction of counting in the middle of the calculation. The rule does not care whether you start from the right or the left — but you must stay consistent within a single problem. If you start position-1 from the left for one number and from the right for another, you may flip the sign of the difference. The magnitude stays the same, so the divisibility conclusion is unchanged (because we only care whether the answer is a multiple of 11), but you can confuse yourself by reading negative differences as wrong.

Why it matters: in RPF Constable / SSC arithmetic sections, divisibility by 7, 11 and 13 routinely appears as a "find the smallest number to be added/subtracted to make N divisible by 11" problem. Without a fast rule, you would do long division — and lose 30 seconds per question. With the alternating-sum rule, every such item should take you under 20 seconds.

Real-world example: bank account numbers, ISBN-10 book codes and many product barcodes use modulo-11 check digits to detect typing mistakes. The very same alternating-sum logic is what your phone uses to validate that you typed an ISBN correctly.

Common misconception: "Difference of 11 means the number is not divisible by 11." Wrong. Eleven is itself a multiple of 11, so a difference of 11, 22, 33, … all confirm divisibility by 11.

Common misconception: "Difference must equal exactly 0 to mean divisibility." Wrong. Zero or any multiple of 11 works.

Common misconception: "Position counting must start from the left." It can start from either end, but stay consistent. The magnitude of the difference does not change with direction.

A Faster Mental Shortcut

For exam speed, alternate +/− signs as you sweep left-to-right and sum:

For 483771: +4 − 8 + 3 − 7 + 7 − 1 = −2. |−2| = 2, not a multiple of 11. Not divisible.

For 483769: +4 − 8 + 3 − 7 + 6 − 9 = −11. |−11| = 11. Divisible by 11.

This is identical to the textbook rule, but you write fewer numbers and you finish under a heartbeat. Many toppers prefer it.

:::keypoints

• Divisibility by 11: take odd-position sum minus even-position sum; check whether the result is 0 or a multiple of 11.
• Position numbering can start from either end, but stay consistent.
• Equivalently: sweep digits left-to-right with alternating + and − signs and take absolute value.
• The rule works because 10 ≡ −1 (mod 11), so powers of 10 alternate ±1.
• 483771 → diff 2 → not divisible. 483769 → diff 11 → divisible.
• A "find the least number to add/subtract" problem reduces to making the alternating sum a multiple of 11.
• Same modular trick gives divisibility tests for 7 and 13 via grouping in threes.
  :::

:::memory
"Plus, minus, plus, minus — answer must be zero or eleven's family." Visualise the digits with alternating signs and add them up in your head.
:::

:::recap

• Sum digits at odd positions and at even positions separately; take the difference.
• A difference of 0, ±11, ±22 … means the number is divisible by 11.
• A difference of 2 (as in 483771) is not — but 11 (as in 483769) is.
• The shortcut is to alternate + and − signs while reading the number left to right.
  :::

Imagine you have to lay tiles on a floor that is 60 cm by 96 cm without cutting any tile. What is the largest square tile that will fit perfectly? Or imagine three bells in a temple that ring every 12, 18 and 30 minutes — when will they ring together again? Both puzzles boil down to the same two ideas: HCF and LCM. RPF Constable, RPF SI, SSC GD and RRB Group D ask 1–2 questions on these every paper, and the trap is almost always picking the wrong one between the two.

Definition: 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 (leaving zero remainder).

Definition: The Least Common Multiple (LCM) of two or more numbers is the smallest positive number that is exactly divisible by every one of them.

HCF: The Largest Common Divisor

If you list every divisor of two numbers, the HCF is the biggest one that appears in both lists.

Take 18 and 24.

• Divisors of 18: 1, 2, 3, 6, 9, 18.
• Divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24.
• Common divisors: 1, 2, 3, 6. The largest is 6.
• So HCF(18, 24) = 6.

The HCF is always smaller than or equal to the smallest of the given numbers — because a divisor of n cannot exceed n.

LCM: The Smallest Common Multiple

If you list multiples of each number, the LCM is the smallest number that appears in every list.

Same example, 18 and 24.

• Multiples of 18: 18, 36, 54, 72, 90, …
• Multiples of 24: 24, 48, 72, 96, …
• The smallest common multiple is 72.

The LCM is always greater than or equal to the largest of the given numbers — because any common multiple must contain each number.

Method 1: Prime Factorisation

Step-by-step: write each number as a product of prime powers. Then:

• HCF = product of every prime factor common to all numbers, raised to its lowest power.
• LCM = product of every prime factor that appears in any number, raised to its highest power.

Question: Find HCF and LCM of 72 and 120 by prime factorisation.

Solution:
Step 1: 72 = 2³ × 3².
Step 2: 120 = 2³ × 3 × 5.
Step 3: Common primes: 2 and 3. Lowest powers: 2³ and 3¹.
Step 4: HCF = 2³ × 3 = 8 × 3 = 24.
Step 5: All primes appearing: 2, 3, 5. Highest powers: 2³, 3², 5¹.
Step 6: LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360.
Conclusion: HCF = 24, LCM = 360.

Verification by the golden formula: HCF × LCM = 24 × 360 = 8640 = 72 × 120. ✓

Method 2: Division Method (Euclid's Algorithm)

For two large numbers where prime factorisation is slow, the division method is faster. It is essentially Euclid's algorithm: divide, take remainder, repeat with divisor and remainder, until remainder is 0. The last divisor is the HCF.

Question: Find HCF of 408 and 1032.

Solution:
Step 1: 1032 ÷ 408 = 2 remainder 216.
Step 2: 408 ÷ 216 = 1 remainder 192.
Step 3: 216 ÷ 192 = 1 remainder 24.
Step 4: 192 ÷ 24 = 8 remainder 0.
Conclusion: Last non-zero divisor = 24. HCF(408, 1032) = 24.

Then use HCF × LCM = product of the numbers to get LCM: 24 × LCM = 408 × 1032, so LCM = 17544.

The Golden Formula

For exactly two numbers a and b:

HCF(a, b) × LCM(a, b) = a × b

This single identity converts any HCF problem into an LCM problem (and vice versa) for two numbers. Memorise it — it shows up in 60% of HCF/LCM questions in RPF Constable.

Important caveat: the formula works only for two numbers. For three or more numbers, HCF × LCM is not equal to the product of the numbers. Example: 4, 6, 8 → HCF = 2, LCM = 24, product = 192, but 2 × 24 = 48 ≠ 192.

Co-prime Numbers — a Free Mark

Definition: Two numbers are co-prime (relatively prime) if their HCF = 1, i.e., they share no prime factor.

Two facts about co-primes that the exam exploits constantly:

• HCF of co-primes = 1 (by definition).
• LCM of co-primes = their product.

Examples: 8 and 15 are co-prime (8 = 2³, 15 = 3 × 5; no common prime), so HCF(8, 15) = 1 and LCM(8, 15) = 120. Note that co-prime numbers need not themselves be prime — 8 and 15 are both composite, yet co-prime.

HCF is Small, LCM is Large — Don't Get Them Mixed Up

This is the most common bleeding mistake in RPF aspirants' practice papers. A memory anchor that works:

• HCF = Highest Common Factor → it is the highest among the factors, but a factor cannot be bigger than the number itself, so HCF is at most the smallest number.
• LCM = Least Common Multiple → it is the least among the multiples, but a multiple cannot be smaller than the number, so LCM is at least the largest number.

In one line: HCF is small, LCM is large. If you ever get HCF > smallest number, or LCM < largest number, you have made an arithmetic mistake — recompute.

Choosing Between HCF and LCM in Word Problems

The exam pattern is predictable. The keywords in the question almost always tell you which to compute.

:::compare

Use HCF when the question says...	Use LCM when the question says...
Largest size of tile / largest cloth piece / longest tape / greatest length	Smallest length / smallest quantity that fits all
Maximum number of students with equal share	Minimum number of items, sweets, etc. that can be equally divided
Greatest number that divides A, B, C leaving the same remainder	Smallest number divisible by each
Make equal-sized groups from different totals	Bells/lights/buzzers/traffic-signals all ringing/blinking together
Cut things into equal pieces — no waste	Round trips / racing tracks / cyclical events meeting again
:::

The simple test: "largest / greatest / max" → HCF; "smallest / least / min / together again" → LCM.

Worked Example: The Bell Problem

Question: Three bells ring at intervals of 12, 18, and 30 minutes. If they ring together at 8:00 AM, when will they next ring together?

Solution:
Step 1: "Together again" is a together event — we want the smallest common multiple of the intervals → LCM.
Step 2: Prime factorise: 12 = 2² × 3; 18 = 2 × 3²; 30 = 2 × 3 × 5.
Step 3: LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180.
Step 4: 180 minutes = 3 hours.
Conclusion: The bells ring together again at 11:00 AM.

Worked Example: The Tile Problem

Question: What is the largest square tile that can cover a floor of 60 cm × 96 cm with no tile being cut?

Solution:
Step 1: "Largest" → HCF.
Step 2: HCF(60, 96). Prime factorise: 60 = 2² × 3 × 5; 96 = 2⁵ × 3.
Step 3: HCF = 2² × 3 = 12.
Conclusion: The largest square tile is 12 cm × 12 cm. (Floor needs (60/12) × (96/12) = 5 × 8 = 40 such tiles.)

Why It Matters

Number-system questions account for around 8–10 marks of every RPF Constable paper. HCF and LCM contribute at least 2 marks directly and another 2 indirectly (through ratio, time-and-work, and pipes-and-cisterns problems that use LCM-of-times to set total work). Mastery here lifts your overall maths score noticeably.

Real-World Example

The Indian Railways uses LCM logic to schedule simultaneous arrivals on different platforms. If train A arrives every 45 minutes and train B every 60 minutes, both will meet on a shared crossover at LCM(45, 60) = 180-minute intervals. Engineers use this to plan signal cycles so that the crossover does not block both trains at the same instant.

Common Misconception

"HCF × LCM = product, always."

False — only for two numbers. As shown above, three numbers can violate this. Always verify how many numbers the question gives you before reaching for the golden formula.

A second common slip: assuming co-prime numbers must be prime. 9 and 16 are co-prime (no shared prime), yet neither 9 nor 16 is prime. "Co-prime" only requires HCF = 1.

:::keypoints

• HCF = largest divisor common to all numbers; LCM = smallest multiple common to all.
• HCF ≤ smallest number; LCM ≥ largest number.
• Two methods: prime factorisation (works for any count); Euclidean division (fast for two big numbers).
• HCF × LCM = product of the two numbers — only for two numbers.
• Co-primes have HCF = 1 and LCM = product.
• Word problem rule: "largest / greatest" → HCF; "smallest / together again" → LCM.
• HCF of co-primes is always 1, even if the numbers themselves are composite.
  :::

:::memory
"HCF eats Factors (small), LCM eats Multiples (big)."

• Factors live inside the number, so HCF can never exceed the smallest input.
• Multiples live outside the number, so LCM can never be smaller than the largest input.
  :::

:::recap

• HCF is the biggest sharer; LCM is the smallest meeting point.
• Prime-factorisation method handles any number of inputs cleanly.
• Golden formula HCF × LCM = a × b is your fastest tool — for two numbers only.
• Read the word problem twice for the trigger word: largest / smallest decides which to compute.
  :::

For fractions, use these special formulas. HCF of fractions = HCF of numerators / LCM of denominators. LCM of fractions = LCM of numerators / HCF of denominators. Example: HCF of 2/3 and 4/9 = HCF(2,4)/LCM(3,9) = 2/9. LCM of 2/3 and 4/9 = LCM(2,4)/HCF(3,9) = 4/3. Memory hook: for HCF take the small route (HCF on top, LCM on bottom); for LCM take the big route (LCM on top, HCF on bottom). Always reduce fractions to lowest terms first. These appear regularly in RPF and other RRB Level-1 exams as one-mark direct questions.

One identity ends most HCF–LCM questions in the RPF Constable Mathematics paper before they begin: for any two positive integers, their product equals the product of their HCF and LCM. Use it and you save the time other candidates spend on prime factorisation.

Definition: The HCF (Highest Common Factor), also called GCD (Greatest Common Divisor), of two numbers is the largest positive integer that divides both of them without a remainder.

Definition: The LCM (Lowest Common Multiple) of two numbers is the smallest positive integer that is a multiple of both.

Definition: The Product Rule for two numbers states: a × b = HCF(a, b) × LCM(a, b). This identity holds for any two positive integers and is the single fastest tool in this chapter.

Why the product rule works (the intuition)

Write a = HCF × m and b = HCF × n, where m and n share no common factor (this is what it means to factor out the HCF). Then ab = HCF² × m × n, and LCM = HCF × m × n. Multiply HCF and LCM together: HCF × (HCF × m × n) = HCF² × m × n = ab. Done.

This is also why the rule only works for two numbers. For three or more, the simple product no longer equals HCF × LCM — extra correction factors creep in. If an RPF question gives three numbers and asks for one from HCF and LCM data, treat it differently (usually prime-factorise).

The set-up: find the missing number

Question: The HCF of two numbers is 12 and their LCM is 144. If one number is 36, find the other number.

Solution:
Step 1: Apply the product rule. Product of the two numbers = HCF × LCM = 12 × 144 = 1728.
Step 2: One number is given as 36, so the other number = 1728 / 36.
Step 3: 1728 / 36 = 48 (because 36 × 4 = 144, and 36 × 48 = 1728).
Step 4: Verify. Factorise 36 = 2² × 3² and 48 = 2⁴ × 3. Take the minimum power of each prime for HCF: HCF = 2² × 3 = 12. Take the maximum power for LCM: LCM = 2⁴ × 3² = 16 × 9 = 144. Both match.
Conclusion: The other number is 48.

Why the rule matters in the exam

Real-world example: Imagine the Indian Railways procurement team buying lengths of rail in two standard sizes that can be cut to fit a junction. If two slab sizes are 36 m and 48 m, the longest single piece that divides both evenly is HCF = 12 m, and the shortest length where both sizes line up perfectly is LCM = 144 m. The product rule reassures the planner that 36 × 48 = 12 × 144 = 1728 — a sanity check on the entire compatibility chart.

Why it matters: RPF and Railway exam setters re-use this template — "HCF = …, LCM = …, one number = …, find the other" — almost every year. Knowing the rule turns a 90-second problem into a 15-second one.

Common misconception: "The product rule works for any number of values." It does not. For three or more numbers the identity a × b × c = HCF × LCM is generally false. Use it only on pairs.

Common misconception: "Any HCF and LCM are compatible." Wrong — LCM must be a multiple of HCF. If a question says HCF = 12 and LCM = 100, those are inconsistent (12 does not divide 100). Spotting this lets you eliminate options instantly.

The bell-ringing / traffic-light family of LCM problems

These are RPF favourites. A signal blinks every 24 seconds, another every 36 seconds, a third every 54 seconds. They all start together — when do they next blink together?

Solution:
Step 1: They blink together at any common multiple of their periods.
Step 2: The first time after the start is the least common multiple: LCM(24, 36, 54).
Step 3: Prime-factorise. 24 = 2³ × 3, 36 = 2² × 3², 54 = 2 × 3³.
Step 4: LCM uses the highest power of each prime: 2³ × 3³ = 8 × 27 = 216 seconds.
Step 5: Convert: 216 seconds = 3 minutes 36 seconds.
Conclusion: The three signals blink together again 3 min 36 s after the start.

This template covers bells in a temple, traffic lights at a junction, runners on a circular track, and gear-meshing problems — all are LCM under different paint.

:::compare

Quantity	Built from	Use when…	Example for 24, 36
HCF	Lowest powers of common primes	Asking the biggest divisor that fits both	HCF = 12
LCM	Highest powers of all primes seen	Asking the smallest number both divide into	LCM = 72
Product rule	HCF × LCM = a × b	Only for two numbers, when missing one value	12 × 72 = 864 = 24 × 36
:::

:::keypoints

• For two numbers: a × b = HCF × LCM.
• Find HCF by taking minimum powers of common primes.
• Find LCM by taking maximum powers of all primes.
• LCM is always a multiple of HCF — useful as a sanity check.
• Product rule fails for three or more numbers; prime-factorise instead.
• "Together-again" timing problems are LCM problems in disguise.
• Always verify HCF and LCM consistency before solving.
• If HCF = 1, the two numbers are co-prime and LCM = their product.
  :::

:::memory
"Pair Product" — for any pair of numbers, the product equals HCF × LCM. Whisper "Pair Product" before HCF–LCM questions to lock in the rule.
:::

:::recap

• HCF × LCM = product of the two numbers; one identity, many marks.
• Use prime factorisation to verify or to handle three or more numbers.
• LCM is the natural answer to "when do they coincide again?".
• Watch the trap: the product rule is for two numbers only.
  :::

BODMAS gives the order in which operations are solved: B - Brackets, O - Of (powers/orders), D - Division, M - Multiplication, A - Addition, S - Subtraction. Division and Multiplication rank EQUAL - do them left to right. Same for Addition and Subtraction. Bracket order: solve innermost first - ( ) then { } then [ ]. The word 'of' means multiply but is done before division (e.g. 1/2 of 8 = 4 first). Memory aid: 'Brackets Open, Divide Multiply, Add Subtract'. A frequent trap: students do addition before multiplication - always finish multiply/divide first. Sign rule: +x+ = +, -x- = +, +x- = -, -x+ = -.

'Of' is treated as multiplication but evaluated immediately after brackets and before plain division. Example: 1/3 of 12 / 2 = (1/3 x 12) / 2 = 4 / 2 = 2 - do the 'of' first. When simplifying fractions, convert mixed numbers to improper fractions, then apply BODMAS. To divide by a fraction, multiply by its reciprocal: a / (b/c) = a x c/b. For a 'fraction bar' that acts as a bracket, simplify the numerator and denominator separately before dividing. Memory tip: vinculum (bar over numbers like a line) is solved FIRST, even before brackets. Keep signs careful when removing brackets preceded by a minus sign - every inside term flips sign.

"36 ÷ 4 × 3 + 8 − 5 × 2 — what is the answer?" In RPF Constable, SSC GD, and almost every banking prelims paper, one BODMAS expression sneaks in and a quarter of the candidates lose the mark. The maths is easy; the order is where everything goes wrong. This lesson works the full expression step by step so that the rule sinks in for life.

Definition: BODMAS is the agreed order of operations: Brackets → Orders (powers and roots) → Division and Multiplication (left to right) → Addition and Subtraction (left to right).

Definition: "Left to right" means within the D-M pair, you do whichever symbol appears first as you read the expression from left to right — not the one you find more familiar. The same rule applies to the A-S pair.

The question

Simplify: 36 ÷ 4 × 3 + 8 − 5 × 2

Apply BODMAS. There are no brackets, no powers — so we go straight to Division and Multiplication, left to right.

Step 1 — Division and Multiplication, strictly left to right

Read the expression from the left. The first D-or-M symbol you meet is the division 36 ÷ 4. Do it first:

36 ÷ 4 = 9

The expression becomes: 9 × 3 + 8 − 5 × 2

The next D-or-M symbol from the left is the multiplication 9 × 3:

9 × 3 = 27

The expression becomes: 27 + 8 − 5 × 2

Continue scanning. The next D-or-M symbol is 5 × 2:

5 × 2 = 10

The expression becomes: 27 + 8 − 10

All multiplications and divisions are now done.

Step 2 — Addition and Subtraction, strictly left to right

Again read from the left:

27 + 8 = 35

The expression becomes: 35 − 10

35 − 10 = 25

Conclusion: The value of 36 ÷ 4 × 3 + 8 − 5 × 2 is 25.

Why "left to right" is non-negotiable

Definition: Equal precedence. Division and multiplication share the same priority level; addition and subtraction share another, lower level. When two operators of equal precedence sit in the same expression, mathematicians long ago agreed to resolve them in reading order, i.e., left to right.

The catch is that BODMAS, as taught, lists "D" before "M". A surprising number of students read that as "always division first, then multiplication". It is not. The D and the M are tied; only their position in the expression decides which is done first. The same applies to A and S — addition is not automatically before subtraction.

The single most common mistake on this problem

Common misconception: Many candidates solve 36 ÷ 4 × 3 as 36 ÷ (4 × 3) = 36 ÷ 12 = 3, treating multiplication as automatically higher than division. The expression then collapses to 3 + 8 − 10 = 1 — completely wrong. The error is treating "BODMAS" as a fixed left-to-right list of priorities instead of grouping D-M as a single equal-precedence pair.

Another common error is to do 5 × 2 = 10 before 36 ÷ 4 = 9. That happens to give the same final answer here, but only by luck — change a sign and the trick fails. The safe habit is always left-to-right within the D-M tier.

A third error is treating subtraction as higher than addition. For 27 + 8 − 10, doing 8 − 10 = −2 first gives 27 − 2 = 25, which happens to be correct here, again by coincidence of sign. But the rule says: scan from the left, addition comes first, then subtraction. Build the habit before the exam, not in the exam.

Why it matters: BODMAS / order-of-operations questions appear in every quantitative aptitude paper — RPF Constable, RPF SI, SSC GD, SSC CHSL, SSC CGL, IBPS Clerk, SBI Clerk, Railway NTPC, state-level constabulary — usually as a quick one-mark question with negative marking. They are the easiest marks on the paper if you train the order, and a guaranteed loss if you do not.

Real-world example: When you key 36 ÷ 4 × 3 + 8 − 5 × 2 into a basic calculator at a railway counter (a simple, non-scientific calculator that processes left to right and does not respect order of operations), you may get a wrong number, because it adds 8 the moment you press +. The scientific calculator in your science classroom — and the one in your phone — does follow BODMAS and returns 25. The lesson: the order of operations is a convention humans agreed on; not every device follows it. The exam expects the BODMAS answer.

A second worked example to lock in the rule

Question: Simplify 48 ÷ 6 × 2 + 14 − 3 × 4.

Solution:
Step 1 (D-M left to right):
48 ÷ 6 = 8 → 8 × 2 + 14 − 3 × 4
8 × 2 = 16 → 16 + 14 − 3 × 4
3 × 4 = 12 → 16 + 14 − 12
Step 2 (A-S left to right):
16 + 14 = 30 → 30 − 12
30 − 12 = 18
Conclusion: 48 ÷ 6 × 2 + 14 − 3 × 4 = 18.

The procedure is identical to the main problem — proof that one drilled habit handles the entire genre.

:::compare

Stage	What to do	In our problem	Result
B	Solve brackets	None	—
O	Solve powers/roots	None	—
D & M	Left-to-right	36 ÷ 4, 9 × 3, 5 × 2	27 + 8 − 10
A & S	Left-to-right	27 + 8, then 35 − 10	25
:::

:::compare

Wrong path	Calculation	Wrong answer
Multiplication before division	4 × 3 = 12, then 36 ÷ 12 = 3, then 3 + 8 − 10 = 1	1 (wrong)
Mixing the order of additions and subtractions	27 + (8 − 10) = 25 — luck, not method	25 (right by accident)
Correct BODMAS path	9 × 3 = 27; 5 × 2 = 10; 27 + 8 − 10 = 25	25
:::

:::keypoints

• BODMAS gives the agreed order: Brackets, Orders, Division/Multiplication, Addition/Subtraction.
• D and M are at the same priority level; resolve them left to right in the expression.
• A and S are at the same lower level; resolve them left to right as well.
• 36 ÷ 4 × 3 = (36 ÷ 4) × 3 = 27, never 36 ÷ (4 × 3) = 3.
• Without brackets, you can never assume that the "later" operator is meant to bind tighter.
• The same rule covers any combination — division then multiplication, addition then subtraction, in any sequence.
• Train the eye to scan from left to right within each tier; speed comes from habit, not from cleverness.
  :::

:::memory
"Before Old Dogs, Many Are Sent" — the classic acronym; remember to pair D with M and A with S, and resolve each pair left to right.

A second mantra for the exam hall: "Same-tier, same-direction — left first."
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:::recap

• 36 ÷ 4 × 3 + 8 − 5 × 2 = 25.
• After D and M (left to right): 27 + 8 − 10.
• After A and S (left to right): 35 − 10 = 25.
• The trap is treating D-M (or A-S) as a strict left-to-right priority list; they are equal-precedence pairs.
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Proper fraction: numerator < denominator (3/5). Improper: numerator >= denominator (7/4). Mixed number: whole + fraction (1 3/4). To convert fraction to decimal, divide numerator by denominator. To convert a terminating decimal to a fraction, write digits over the place value (0.75 = 75/100 = 3/4). For comparing fractions, cross-multiply or make denominators equal. To add/subtract fractions, take LCM of denominators. To multiply, multiply across; to divide, multiply by the reciprocal. Memory aid for recurring decimals: 0.333... = 1/3, 0.666... = 2/3, 0.142857... = 1/7. Ascending order trick: smaller decimal value = smaller fraction. Always reduce final answers to lowest terms.

A surd is an irrational root like root2, root3, root5. Rules: root(a) x root(b) = root(ab); root(a)/root(b) = root(a/b); root(a^2 x b) = a root(b). Useful values to memorise: root2 = 1.414, root3 = 1.732, root5 = 2.236, root7 = 2.646. Rationalising the denominator: multiply numerator and denominator by the surd, e.g. 1/root2 = root2/2. For square roots of perfect squares, learn squares up to 30 (e.g. 25^2 = 625, 30^2 = 900). To find square root by factorisation, pair the prime factors: root324 = root(2^2 x 3^4) = 2 x 9 = 18. These shortcuts save crucial seconds in the exam.

Simplify 2/3 + 3/4 - 1/6. Step 1: LCM of 3, 4 and 6 is 12. Convert: 2/3 = 8/12, 3/4 = 9/12, 1/6 = 2/12. Step 2: 8/12 + 9/12 - 2/12 = (8 + 9 - 2)/12 = 15/12. Step 3: reduce 15/12 = 5/4 = 1 1/4. Answer = 5/4. For arranging in ascending order, e.g. 2/3, 3/5, 4/7: convert to decimals - 0.667, 0.600, 0.571 - so ascending order is 4/7 < 3/5 < 2/3. Decimal conversion is the fastest, error-free way to compare unlike fractions under exam time pressure.