Official Syllabus Item: 1, 28, 29, 30, 43 Arithmetic & Number Theory

Number System, Large Numbers & Whole Numbers

Master Indian and International numeral systems, face and place values, properties of whole numbers, large numbers in practice, and fundamental operations for RIMC Entrance Exam.

Overview & Fundamental Concept

Number System forms the foundational bedrock of the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). In the descriptive written exam, examiners evaluate candidates on their mathematical precision in expressing multi-digit numbers in the Indian (Lakhs, Crores) and International (Millions, Billions) systems, calculating place values and face values, applying axiomatic properties of whole numbers (closure, commutativity, associativity, and distributivity), deploying Euclid's division algorithm, and solving multi-step word problems involving real-world metric conversions. Full marks require formal variable declarations, theorem citations, step-by-step arithmetic reduction, and boxed answers with standard units.

Official Syllabus Coverage Breakdown
Syllabus Item Prescribed Scope Exam Relevance & Pedagogical Objective
Item 1 Natural numbers, Integers, Whole numbers Defines number sets $(\mathbb{N} \subset \mathbb{W} \subset \mathbb{Z})$, predecessor-successor properties, number line behavior, and the foundational role of zero.
Item 28 Indian and International System of Numerals Tests period demarcation, comma placement rules, cross-system numerical conversions ($1\text{ Million} = 10\text{ Lakhs}$, $1\text{ Billion} = 100\text{ Crores}$), and numeral-to-words representations.
Item 29 Larger numbers in operations Multi-digit arithmetic algorithms (addition, subtraction, long multiplication, division), estimation, and multi-unit conversions (length: km/m/cm/mm; mass: kg/g/mg; capacity: kL/L/mL).
Item 30 Properties of operations on numbers Closure, Commutative, Associative, and Distributive laws of whole numbers; special properties of $0$ and $1$; Division Algorithm ($Dividend = Divisor \times Quotient + Remainder$).
Item 43 Face value and place value of numbers Distinguishes intrinsic face value from position-dependent place value, expanded form in powers of 10, digit-reversal theorems, and place-value algebraic difference problems.
1. Classification of Numbers & Set Hierarchy (Syllabus Item 1)

In mathematics, numbers are grouped into nested collections according to their arithmetic definitions and properties. A clear conceptual grasp of these subsets is vital for RIMC theoretical reasoning questions.

1. Natural Numbers ($\mathbb{N}$): The set of counting numbers starting from 1: $$\mathbb{N} = \{1, 2, 3, 4, 5, \dots\}$$ The smallest natural number is $1$. There is no greatest natural number, as every natural number has a successor ($n + 1$).

2. Whole Numbers ($\mathbb{W}$): The set of natural numbers including the number zero ($0$): $$\mathbb{W} = \{0, 1, 2, 3, 4, \dots\}$$ The smallest whole number is $0$. Every natural number is a whole number, but $0$ is a whole number that is not a natural number. Hence, $\mathbb{N} \subset \mathbb{W}$.

3. Integers ($\mathbb{Z}$): The collection of all positive natural numbers, zero, and the negative opposites of natural numbers: $$\mathbb{Z} = \{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$$

Successor and Predecessor:

For any given whole number $n$:
• Successor: The number immediately following $n$, obtained by adding $1$: $\text{Successor}(n) = n + 1$.
• Predecessor: The number immediately preceding $n$, obtained by subtracting $1$: $\text{Predecessor}(n) = n - 1$.
Crucial RIMC Rule: The whole number $0$ has no predecessor in $\mathbb{W}$. However, in the set of integers $\mathbb{Z}$, the predecessor of $0$ is $-1$.

Exam Caution: Predecessor of Zero
If an exam question asks: 'Does every whole number have a predecessor?', the correct descriptive answer is: 'No. The whole number 0 has no predecessor in the set of whole numbers ($\mathbb{W}$), though its predecessor in the integer set ($\mathbb{Z}$) is -1.' Omitting the set qualification causes mark loss.
2. Indian & International Numeral Systems (Syllabus Item 28)

Numeration is the formal method of expressing numbers in words and figures. Both the Indian (Hindu-Arabic) and International systems use base-10 positional notation, but they group place values into fundamentally different periods.

Period Comma Separation Rules:
• Indian System: The first period ('Ones') contains 3 digits (Units, Tens, Hundreds). Thereafter, all subsequent periods ('Thousands', 'Lakhs', 'Crores', 'Arabs') contain exactly 2 digits each. Example: $7,65,43,210$.
• International System: Every single period ('Ones', 'Thousands', 'Millions', 'Billions', 'Trillions') contains exactly 3 digits. Example: $76,543,210$.

Numerical ValuePower of 10Indian System Period & PlaceInternational System Period & Place
1$10^0$Ones (Units)Ones (Units)
10$10^1$Ones (Tens)Ones (Tens)
100$10^2$Ones (Hundreds)Ones (Hundreds)
1,000$10^3$Thousands (Thousands)Thousands (One Thousands)
10,000$10^4$Thousands (Ten Thousands)Thousands (Ten Thousands)
100,000$10^5$Lakhs (One Lakh)Thousands (Hundred Thousands)
1,000,000$10^6$Lakhs (Ten Lakhs)Millions (One Million)
10,000,000$10^7$Crores (One Crore)Millions (Ten Millions)
100,000,000$10^8$Crores (Ten Crores)Millions (Hundred Millions)
1,000,000,000$10^9$Arabs (One Arab)Billions (One Billion)
10,000,000,000$10^{10}$Arabs (Ten Arabs)Billions (Ten Billions)
Golden Conversion Equivalence for RIMC
Commit these exact conversion identities to memory:
• $1 \text{ Million} = 10 \text{ Lakhs} = 1,000 \text{ Thousands}$
• $10 \text{ Millions} = 1 \text{ Crore} = 100 \text{ Lakhs}$
• $100 \text{ Millions} = 10 \text{ Crores}$
• $1 \text{ Billion} = 1,000 \text{ Millions} = 100 \text{ Crores} = 1 \text{ Arab}$
• $1 \text{ Trillion} = 1,000 \text{ Billions} = 100,000 \text{ Crores}$
3. Place Value, Face Value & Algebraic Number Theory (Syllabus Item 43)

Every digit in a numeral possesses two distinct values: an intrinsic face value and a contextual place value.

Face Value: The intrinsic value of the digit itself, independent of its position. In the numeral $9,48,251$, the face value of $4$ is simply $4$, and the face value of $8$ is $8$.

Place Value (Local Value): The value of a digit determined by its position in the place-value hierarchy: $$\text{Place Value of a Digit} = (\text{Face Value of the Digit}) \times (\text{Value of its Place Position})$$ In $9,48,251$:
• Place value of $4$ (Ten-Thousands place) = $4 \times 10,000 = 40,000$.
• Place value of $8$ (Thousands place) = $8 \times 1,000 = 8,000$.

Expanded Form: Expressing a number as the algebraic sum of the place values of all its constituent digits: $$8,45,29,316 = (8 \times 10^7) + (4 \times 10^6) + (5 \times 10^5) + (2 \times 10^4) + (9 \times 10^3) + (3 \times 10^2) + (1 \times 10^1) + (6 \times 1)$$

Algebraic Theorems on Digit Reversal:
• Two-Digit Number: Let tens digit $= t$ and units digit $= u$. The number is $N = 10t + u$. When digits are reversed, $N' = 10u + t$.
  1. Sum: $N + N' = (10t + u) + (10u + t) = 11(t + u)$ $\implies$ Always divisible by 11 and $(t + u)$.
  2. Difference: $|N - N'| = |(10t + u) - (10u + t)| = 9|t - u|$ $\implies$ Always divisible by 9 and $|t - u|$.
• Three-Digit Number: Let $N = 100h + 10t + u$ and $N' = 100u + 10t + h$.
  Difference: $|N - N'| = 99|h - u|$ $\implies$ Always divisible by 99, 9, 11, and $|h - u|$.

4. Axiomatic Properties of Operations on Whole Numbers (Syllabus Item 30)

Arithmetic operations on whole numbers follow strict structural laws. When solving descriptive questions in RIMC, you must explicitly state which property you are applying to earn complete step credit.

A. Closure Property

A set is closed under an operation if applying that operation to any two elements in the set always yields an element belonging to the same set.

  • Addition: For any $a, b \in \mathbb{W}$, $a + b \in \mathbb{W}$. Whole numbers are closed under addition (e.g., $4 + 9 = 13 \in \mathbb{W}$).
  • Multiplication: For any $a, b \in \mathbb{W}$, $a \times b \in \mathbb{W}$. Whole numbers are closed under multiplication (e.g., $6 \times 7 = 42 \in \mathbb{W}$).
  • Subtraction: Whole numbers are NOT closed under subtraction, because $4 - 9 = -5 \notin \mathbb{W}$.
  • Division: Whole numbers are NOT closed under division, because $7 \div 2 = 3.5 \notin \mathbb{W}$ and division by zero is undefined.

B. Commutative Property

The order of operands does not change the result.

  • Addition: $a + b = b + a$ (Valid for all whole numbers).
  • Multiplication: $a \times b = b \times a$ (Valid for all whole numbers).
  • Subtraction & Division: Commutative property does NOT hold: $a - b \ne b - a$ and $a \div b \ne b \div a$ (when $a \ne b$).

C. Associative Property

The grouping of three or more operands does not alter the final result.

  • Addition: $(a + b) + c = a + (b + c)$.
  • Multiplication: $(a \times b) \times c = a \times (b \times c)$.
  • Topper Regrouping Trick: Evaluate $(250 \times 17) \times 4$ by regrouping as $(250 \times 4) \times 17 = 1,000 \times 17 = 17,000$.

D. Distributive Property of Multiplication

Multiplication distributes across both addition and subtraction. This is the single most frequently tested simplification property in RIMC Paper I.

  • Over Addition: $a \times (b + c) = (a \times b) + (a \times c)$.
  • Over Subtraction: $a \times (b - c) = (a \times b) - (a \times c)$.
  • Application: $854 \times 102 = 854 \times (100 + 2) = (854 \times 100) + (854 \times 2) = 85,400 + 1,708 = 87,108$.

E. Identity Elements & Special Properties of Zero and One

Crucial identity elements and arithmetic constraints:

  • Additive Identity: Zero ($0$) is the additive identity because $a + 0 = 0 + a = a$.
  • Multiplicative Identity: One ($1$) is the multiplicative identity because $a \times 1 = 1 \times a = a$.
  • Multiplication by Zero: $a \times 0 = 0 \times a = 0$.
  • Division by Zero is Undefined: If $a \div 0 = k$, then $k \times 0 = a$. For any non-zero $a$, no such number $k$ exists. If $a = 0$, $0 \div 0$ is indeterminate.
  • Zero Divided by Non-Zero: For any $a \ne 0$, $0 \div a = 0$.

F. Euclid's Division Algorithm

For any dividend $D$ and non-zero divisor $d$, there exist unique whole numbers $q$ (quotient) and $r$ (remainder) such that: $$\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder} \quad \text{where } 0 \le \text{Remainder} < \text{Divisor}$$

  • If $\text{Remainder} = 0$, then Dividend is exactly divisible by Divisor.
  • Least number to be subtracted from $N$ to make it exactly divisible by $d$: Subtract $\text{Remainder} (r)$.
  • Least number to be added to $N$ to make it exactly divisible by $d$: Add $(d - r)$, where $r$ is the non-zero remainder.
5. Real-World Practical Units & Metric Measurements (Syllabus Item 29)

Part B (10-mark questions) frequently presents multi-step word problems involving commercial transactions, medicine dosages, sheet printing, or container filling requiring multiple unit conversions. Always convert all quantities into the lowest common unit before carrying out multiplication or division.

Measurement CategoryBase UnitKey Multipliers & Standard EquivalencesSmallest Practical Unit
LengthMeter (m)• $1\text{ km} = 1,000\text{ m}$
• $1\text{ m} = 100\text{ cm} = 1,000\text{ mm}$
• $1\text{ km} = 100,000\text{ cm} = 1,000,000\text{ mm}$		Millimeter (mm)
Mass (Weight)Gram (g) / Kilogram (kg)• $1\text{ metric ton} = 10\text{ quintals} = 1,000\text{ kg}$
• $1\text{ quintal} = 100\text{ kg}$
• $1\text{ kg} = 1,000\text{ g} = 1,000,000\text{ mg}$
• $1\text{ g} = 1,000\text{ mg}$		Milligram (mg)
Capacity (Volume)Liter (L)• $1\text{ kiloliter (kL)} = 1,000\text{ liters (L)}$
• $1\text{ L} = 1,000\text{ milliliters (mL)}$
• $1\text{ kL} = 1,000,000\text{ mL}$		Milliliter (mL)
6. Estimation, Rounding Off Rules & The General Estimation Principle (Syllabus Item 29)

Estimation gives a quick, reasonable approximation of large computations without performing tedious paper-and-pencil operations. In RIMC Stage-I, questions frequently ask candidates to estimate sums, differences, or products using the General Rule of Estimation.

1. Rounding to Nearest Ten: Examine the ones (units) digit. If it is $0, 1, 2, 3,$ or $4$, round down to $0$. If it is $5, 6, 7, 8,$ or $9$, round up by adding $1$ to the tens digit and setting the ones digit to $0$. Examples: $73 \to 70$; $86 \to 90$.

2. Rounding to Nearest Hundred: Examine the tens digit. If $< 5$, round down. If $\ge 5$, round up. Examples: $438 \to 400$; $782 \to 800$.

3. Rounding to Nearest Thousand: Examine the hundreds digit. If $< 5$, round down. If $\ge 5$, round up. Examples: $2,468 \to 2,000$; $6,840 \to 7,000$.

The General Rule of Estimation (NCERT & RIMC Standard): Round each contributing number to its greatest place value position, and then carry out the arithmetic operation.

• Sum Estimation Example: Estimate $730 + 998$. Both numbers have Hundreds as their greatest place. $730$ rounds to $700$; $998$ rounds to $1,000$. Estimated sum $= 700 + 1,000 = 1,700$.

• Product Estimation Example 1: Estimate $578 \times 161$. Round each factor to nearest hundred (their greatest place): $578 \to 600$, $161 \to 200$. Estimated product $= 600 \times 200 = 1,20,000$.

• Product Estimation Example 2: Estimate $9,250 \times 29$. Round $9,250$ to nearest thousand ($9,000$) and $29$ to nearest ten ($30$). Estimated product $= 9,000 \times 30 = 2,70,000$.

Cadet Estimation Strategy
Never round off intermediate steps to different decimal places unless specified. If the question states 'Estimate using general rule', round every factor to its highest non-zero place before multiplying.
Illustrated Conceptual Diagrams & Visual Schematics
Number Sets Nested Hierarchy & Place Value Structure
Sets of Numbers Hierarchy (ℝ) Real Numbers (ℝ) Rational (ℚ): p/q, q ≠ 0 (e.g. -3/4, 0.5, 7) Integers (ℤ): { ...-2, -1, 0, 1, 2... } Whole Numbers (𝕎): { 0, 1, 2, 3... } Natural (ℕ): { 1, 2, 3, 4... } Counting Numbers (Excludes 0) Place Value Period Comparison Indian System (Lakhs & Crores) Grouping: 3 digits, then every 2 digits 8,76,54,321 8 Crore 76 Lakh 54 Thousand 321 1 Lakh = 100,000 | 1 Crore = 10,000,000 International System (Millions) Grouping: Every 3 digits uniformly 87,654,321 87 Million 654 Thousand 321 1 Million = 10 Lakh | 10 Million = 1 Crore
Number System Architecture: All natural numbers are whole numbers, integers, and rationals ($mathbb{N} subset mathbb{W} subset mathbb{Z} subset mathbb{Q} subset mathbb{R}$). Indian system groups into Lakhs and Crores ($3:2:2:2$ pattern) whereas the International system groups into Thousands and Millions ($3:3:3$ pattern).

Topper Strategy & Descriptive Step Breakdown

Mastering RIMC Mathematics Standard

In RIMC, clear step-by-step layout is essential. Write down all given values, declare formulas before substitution, and conclude with full sentences with proper units.

Official Descriptive Marking Scheme & Step Allocation

In RIMC Stage-I, answers without step justification receive only 20-30% of total marks even if the numeric answer is correct. Follow this strict 4-tier layout on every question:

Step 1: Given Statement & Let Declaration 1 Mark (Part A) / 2 Marks (Part B)

Explicitly list all numerical quantities given in the question with their original units. Declare variables clearly (e.g., 'Let the required number be $x$').

Booklet Layout:
Given:
Total cloth purchased = 2,450 m = 2,45,000 cm.
Cloth required per uniform = 2 m 75 cm = 275 cm.
Step 2: Formula, Law & Theorem Citation 1 Mark (Part A) / 2 Marks (Part B)

State the mathematical identity, property, or division algorithm in standard algebraic notation before inserting numerical values.

Booklet Layout:
By Euclid's Division Algorithm:
Dividend = (Divisor × Quotient) + Remainder, where 0 ≤ Remainder < Divisor.
Step 3: Systematic Calculation & Margin Working 2 Marks (Part A) / 4 Marks (Part B)

Carry out each arithmetic reduction on a separate line. Keep all scratch computations in a dedicated 'ROUGH WORK' column drawn with a pencil on the right-hand side of your page.

Booklet Layout:
2,45,000 = (275 × 890) + 250
Quotient = 890 uniforms, Remainder = 250 cm.
Step 4: Concluding Statement with Proper Units 1 Mark (Part A) / 2 Marks (Part B)

State the final answer in a full English sentence with appropriate physical units (₹, kg, m, mL). Highlight or box your final numerical answer.

Booklet Layout:
∴ Exactly 890 uniforms can be stitched and 2 m 50 cm cloth remains.
Ans: [890 Uniforms, 2 m 50 cm left over]
Examiner Traps & Common Blunders (Marks Penalties)
Trap 1: Placing '0' at the Leading Position in Number Formation Deduction: -2 to -3 Marks
Common Mistake: When asked to form the smallest 6-digit number using digits $5, 0, 8, 2, 9, 3$, writing $023589$. A number starting with $0$ is a 5-digit number, NOT a 6-digit number!
Cadet Antidote: Place the smallest non-zero digit first, followed immediately by $0$: write $203589$.
Trap 2: Mixing Indian and International Period Commas Deduction: -1 to -2 Marks
Common Mistake: Writing hybrid commas such as $45,678,92$ or $1,234,567$ when Indian periods were asked for.
Cadet Antidote: Indian: 3 digits from right, then groups of 2 ($12,34,567$). International: Groups of 3 throughout ($1,234,567$).
Trap 3: Omitting the Property Name in Simplifications Deduction: -2 Marks
Common Mistake: Evaluating $738 \times 103 = 76014$ directly via long multiplication without declaring the Distributive Law.
Cadet Antidote: Always write: 'Using Distributive Property of Multiplication over Addition: $a \times (b + c) = (a \times b) + (a \times c)$'.
Trap 4: Subtracting instead of Adding in 'Least Number to be Added' Deduction: -3 to -4 Marks
Common Mistake: To find the smallest 6-digit number divisible by 43, subtracting the remainder from 1,00,000, which reduces it to a 5-digit number (99,975)!
Cadet Antidote: To find the least number to add to reach a higher multiple, add $(d - r)$: $1,00,000 + (43 - 25) = 1,00,018$.
Trap 5: Calculating Operations without Unifying Units Deduction: -3 to -5 Marks
Common Mistake: Dividing $4\text{ L } 500\text{ mL}$ by $25\text{ mL}$ as $4.5 \div 25$ without formal metric conversion.
Cadet Antidote: Convert everything to milliliters first: $4\text{ L } 500\text{ mL} = (4 \times 1000) + 500 = 4,500\text{ mL}$, then divide: $4500 \div 25 = 180$.
Trap 6: Calculating Exact Answer When Estimation is Explicitly Ordered Deduction: -3 to -4 Marks
Common Mistake: When asked to 'Estimate the product $5,281 \times 349$ using the general rule', computing the exact long product ($18,43,069$) instead of rounding.
Cadet Antidote: Round each factor to its greatest place ($5,281 \to 5,000$; $349 \to 300$), then evaluate: $5,000 \times 300 = 15,00,000$.
Cadet Topper Mental Verification Shortcuts
Total Count of n-digit Numbers Formula
The total number of $n$-digit whole numbers in existence is given by: $$N_n = (\text{Greatest } n\text{-digit number}) - (\text{Smallest } n\text{-digit number}) + 1 = 9 \times 10^{n-1}$$
• Total 5-digit numbers = $9 \times 10^4 = 90,000$
• Total 6-digit numbers = $9 \times 10^5 = 900,000$
• Total 7-digit numbers = $9 \times 10^6 = 90,00,000$
Digital Root / Casting Out Nines Verification
Check multi-digit multiplications in 5 seconds without re-multiplying: the product of digital roots of factors must equal the digital root of the product (modulo 9).
Check: $738 \times 103 = 76,014$.
• Digital root of $738 = 7+3+8 = 18 \to 1+8 = 9 \equiv 0$.
• Product digital root must be $0 \times 4 = 0$ (or 9).
• Digital root of $76,014 = 7+6+0+1+4 = 18 \to 1+8 = 9$. Matches perfectly!
Remainder Transformation in Division Algorithm
Given number $N$ and divisor $d$ with remainder $r$:
• Least number to be subtracted from $N$ to yield exact divisibility = $r$
• Least number to be added to $N$ to yield exact divisibility = $d - r$ (if $r > 0$)
When $1,00,000$ is divided by $348$, quotient is $287$ and remainder is $124$.
• Least number to subtract = $124$ (yields $99,876$).
• Least number to add = $348 - 124 = 224$ (yields $100,224$).
Fast Distributive Multipliers (Splitting Powers)
Replace difficult multipliers with expansions of base $10, 100, 1000$ or friendly factors ($25, 125$):
• $\times 99 = \times (100 - 1)$
• $\times 105 = \times (100 + 5)$
• $\times 125 = \times (1000 \div 8)$
$4,275 \times 125 = (4,275 \times 1000) \div 8 = 4,275,000 \div 8 = 534,375$.
Product Digit Cardinality Rule (Sanity Check for Multiplication)
If an $m$-digit number is multiplied by an $n$-digit number, their product $P = A \times B$ will ALWAYS have either $(m + n - 1)$ or $(m + n)$ digits. It can never have fewer than $(m + n - 1)$ or more than $(m + n)$ digits.
A 4-digit number multiplied by a 3-digit number has either $(4+3-1)=6$ digits or $(4+3)=7$ digits.
Example: $1,000 \times 100 = 1,00,000$ (6 digits); $9,999 \times 999 = 9,989,001$ (7 digits). Immediate sanity check for long multiplication in RIMC!

Key Formulas & Conversion Cheat Sheet

Indian Place Value Periods
$$\text{Ones } (O, T, H) \; | \; \text{Thousands } (Th, T\text{-}Th) \; | \; \text{Lakhs } (L, T\text{-}L) \; | \; \text{Crores } (C, T\text{-}C)$$

First group contains 3 digits; all subsequent groups contain 2 digits.

International Place Value Periods
$$\text{Ones } (O, T, H) \; | \; \text{Thousands } (Th, T\text{-}Th, H\text{-}Th) \; | \; \text{Millions } (M, T\text{-}M, H\text{-}M) \; | \; \text{Billions}$$

All periods consist of exactly 3 digits separated by commas.

Cross-System Equivalence Conversions
$$1 \text{ Million} = 10 \text{ Lakhs}; \quad 10 \text{ Millions} = 1 \text{ Crore}; \quad 1 \text{ Billion} = 100 \text{ Crores} = 1 \text{ Arab}$$

Fundamental conversion benchmark for currency and population figures.

Euclid's Division Algorithm
$$\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}, \quad 0 \le \text{Remainder} < \text{Divisor}$$

Determines quotient and remainder with non-negative bounded remainder.

Distributive Property of Multiplication
$$a \times (b \pm c) = (a \times b) \pm (a \times c)$$

Used to simplify large multi-digit products without direct long multiplication.

Total Count of n-Digit Numbers
$$N_n = (10^n - 1) - 10^{n-1} + 1 = 9 \times 10^{n-1}$$

Counts the exact cardinality of all whole numbers having exactly n digits.

Metric Unit Multipliers (Length, Mass, Volume)
$$1\text{ km} = 10^6\text{ mm}; \quad 1\text{ kg} = 10^6\text{ mg}; \quad 1\text{ kL} = 10^6\text{ mL}; \quad 1\text{ ton} = 10^3\text{ kg}$$

Standard decimal multipliers for dimensional reduction in word problems.

Digit Reversal Difference Theorem
$$|(10a + b) - (10b + a)| = 9|a - b|; \quad |(100a+10b+c) - (100c+10b+a)| = 99|a - c|$$

Two-digit reversal difference is divisible by 9; three-digit difference by 99.

Product Digit Bound Inequality
$$10^{m+n-2} \le A \times B < 10^{m+n} \implies \text{Digits in } (A \times B) \in \{m+n-1, \; m+n\}$$

Limits the exact number of digits produced by multiplying an m-digit and n-digit number.

Solved Model Examples (Descriptive Format with Step Mark Distribution)
Part A (5 Marks) 5 Marks
Find the difference between the place values of the two 7s occurring in the numeral 5,73,27,419. Express the resulting difference in words in the International System of numeration.
Step-by-Step Model Solution:
Step 1: Identify positions of both digit 7s 1 Mark
In the given numeral $5,73,27,419$:
• The first $7$ is at the Ten-Lakhs position ($10,00,000$).
• The second $7$ is at the Thousands position ($1,000$).
Step 2: Calculate individual place values 1 Mark
• $\text{Place value of first 7} = 7 \times 10,00,000 = 70,00,000$
• $\text{Place value of second 7} = 7 \times 1,000 = 7,000$
Step 3: Compute the difference 2 Marks
$$\text{Difference} = 70,00,000 - 7,000 = 69,93,000$$
Rough Column Working: 7000000 - 7000 = 6993000
Step 4: Convert and write in the International System 1 Mark
In the International System, insert commas after every 3 digits from the right: $6,993,000$.
In words: Six million nine hundred ninety-three thousand.
Final Answer: Difference = $6,993,000$ (Six million nine hundred ninety-three thousand).
Examiner Scoring Key: Candidates often write the number name in the Indian system ('Sixty-nine lakh ninety-three thousand'). RIMC specifically penalizes this with -1 mark when the International system is requested.
Part A (5 Marks) 5 Marks
Using the digits 4, 0, 7, 2, 8, 5 each only once, find: (i) the greatest 6-digit number, (ii) the smallest 6-digit number, and (iii) their difference.
Step-by-Step Model Solution:
Step 1: Arrange digits in descending and ascending order 1 Mark
Given digits: $4, 0, 7, 2, 8, 5$. Number of digits $= 6$.
Descending order: $8, 7, 5, 4, 2, 0$.
Ascending order: $0, 2, 4, 5, 7, 8$.
Step 2: Form greatest 6-digit number 1 Mark
Place the greatest digits in the highest place-value positions from left to right:
$$\text{Greatest 6-digit number} = 8,75,420$$
Step 3: Form smallest 6-digit number (accounting for zero) 1 Mark
Since a 6-digit number cannot begin with $0$ (which would make it a 5-digit number), the smallest non-zero digit $2$ must be placed at the highest place (Lakhs), followed immediately by $0$:
$$\text{Smallest 6-digit number} = 2,04,578$$
Step 4: Compute difference & conclude 2 Marks
$$\text{Difference} = 8,75,420 - 2,04,578 = 6,70,842$$
Final Answer: (i) Greatest = $8,75,420$, (ii) Smallest = $2,04,578$, (iii) Difference = $6,70,842$.
Rough Column Working: 875420 - 204578 = 670842. Verification: 670842 + 204578 = 875420.
Part A (5 Marks) 5 Marks
Find the product using suitable properties of whole numbers: (a) 738 × 103, (b) 81265 × 169 - 81265 × 69.
Step-by-Step Model Solution:
Step 1: Solve part (a) using Distributive Property over Addition 2.5 Marks
Rewrite $103$ as $(100 + 3)$:
$$738 \times 103 = 738 \times (100 + 3)$$
By Distributive Property of Multiplication over Addition [$a \times (b + c) = a \times b + a \times c$]:
$$= (738 \times 100) + (738 \times 3) = 73,800 + 2,214 = 76,014$$
Step 2: Solve part (b) using Distributive Property over Subtraction 2.5 Marks
Notice the common factor $81,265$:
$$81265 \times 169 - 81265 \times 69 = 81265 \times (169 - 69)$$
By Distributive Property of Multiplication over Subtraction [$a \times b - a \times c = a \times (b - c)$]:
$$= 81,265 \times 100 = 81,26,500$$
Final Answer: (a) $76,014$, (b) $81,26,500$.
Examiner Scoring Key: Full credit requires citing the property name by theorem. Candidates who multiply directly without using the distributive law receive only 1 out of 5 marks.
Part B (10 Marks) 10 Marks
Find the smallest 6-digit number which is exactly divisible by 348. Also, find the greatest 6-digit number divisible by 348. Show complete long division workings.
Step-by-Step Model Solution:
Step 1: Smallest 6-digit number setup & long division 2 Marks
The smallest 6-digit number is $1,00,000$.
Divide $1,00,000$ by $348$:
$$1,00,000 = (348 \times 287) + 124$$
Quotient = $287$, Remainder = $124$.
Rough Column Working: 348 x 2 = 696; 1000 - 696 = 304 -> 3040. 348 x 8 = 2784; 3040 - 2784 = 256 -> 2560. 348 x 7 = 2436; 2560 - 2436 = 124.
Step 2: Determine least number to add for smallest 6-digit number 3 Marks
Since remainder $124 \ne 0$, $1,00,000$ is not divisible by $348$.
Subtracting $124$ would yield $99,876$ (a 5-digit number).
Therefore, we must add the difference between the divisor and remainder:
$$\text{Number to be added} = \text{Divisor} - \text{Remainder} = 348 - 124 = 224$$
$$\text{Smallest 6-digit number divisible by 348} = 1,00,000 + 224 = 1,00,224$$
Verification: $1,00,224 \div 348 = 288$ (exact).
Step 3: Greatest 6-digit number setup & long division 2 Marks
The greatest 6-digit number is $9,99,999$.
Divide $9,99,999$ by $348$:
$$9,99,999 = (348 \times 2873) + 195$$
Quotient = $2,873$, Remainder = $195$.
Rough Column Working: 999999 / 348: 348 x 2873 = 999804. 999999 - 999804 = 195.
Step 4: Subtract remainder to obtain greatest 6-digit multiple & conclude 3 Marks
To remain within 6 digits, we must subtract the remainder $195$:
$$\text{Greatest 6-digit number divisible by 348} = 9,99,999 - 195 = 9,99,804$$
Verification: $9,99,804 \div 348 = 2,873$ (exact).
Final Answer: Smallest 6-digit number = $1,00,224$; Greatest 6-digit number = $9,99,804$.
Part B (10 Marks) 10 Marks
A pharmaceutical company manufactured 18 kg 750 g of medicinal powder. It is to be filled into capsules each of capacity 250 mg. The filled capsules are then packed into boxes containing 150 capsules each. (a) How many total capsules can be manufactured? (b) How many complete boxes can be packed, and how many capsules remain unpacked? (c) If each box is sold for ₹ 375, find the total revenue generated from the complete boxes.
Step-by-Step Model Solution:
Step 1: Declare given data & convert total powder to milligrams 2 Marks
Given:
Total medicinal powder = $18\text{ kg } 750\text{ g}$.
Since $1\text{ kg} = 1,000\text{ g}$ and $1\text{ g} = 1,000\text{ mg}$:
$$1\text{ kg} = 1,000,000\text{ mg}$$
$$\text{Total powder in mg} = (18 \times 1,000,000) + (750 \times 1,000) = 18,000,000 + 750,000 = 18,750,000\text{ mg}$$
Step 2: Calculate total number of capsules manufactured 3 Marks
Capacity of 1 capsule = $250\text{ mg}$.
$$\text{Total number of capsules} = \frac{\text{Total Powder in mg}}{\text{Capacity of 1 Capsule in mg}} = \frac{18,750,000}{250}$$
$$= \frac{1875000}{25} = 75,000\text{ capsules}$$
Rough Column Working: 1875 / 25 = 75, so 18,750,000 / 250 = 75,000.
Step 3: Calculate complete boxes and remaining capsules 3 Marks
Capacity of 1 packaging box = $150\text{ capsules}$.
$$\text{Number of boxes} = 75,000 \div 150$$
$$75,000 = (150 \times 500) + 0$$
• Number of complete boxes $= 500\text{ boxes}$.
• Number of capsules left unpacked $= 0\text{ capsules}$.
Step 4: Compute total revenue and state final answers 2 Marks
Selling price of 1 box = ₹ $375$.
$$\text{Total Revenue} = 500 \times 375 = \frac{1,000 \times 375}{2} = \frac{3,75,000}{2} = ₹ 1,87,500$$
Final Answer:
(a) Total capsules manufactured = $75,000$
(b) Complete boxes = $500$, Leftover capsules = $0$
(c) Total revenue = ₹ $1,87,500$.
Part B (10 Marks) 10 Marks
A uniform contractor purchased 2,450 meters of cloth at ₹ 145 per meter. To stitch one cadet uniform, 2 m 75 cm of cloth is required. (a) How many cadet uniforms can be stitched from the cloth purchased? (b) How much cloth will be left over? (c) If stitching and button charges cost ₹ 85 per uniform, and each finished uniform is supplied to the academy at ₹ 650, calculate the contractor's net profit.
Step-by-Step Model Solution:
Step 1: Convert cloth measurements into centimeters 2 Marks
Since $1\text{ m} = 100\text{ cm}$:
• Total cloth purchased = $2,450\text{ m} = 2,450 \times 100 = 2,45,000\text{ cm}$.
• Cloth required for 1 uniform = $2\text{ m } 75\text{ cm} = (2 \times 100) + 75 = 275\text{ cm}$.
Step 2: Divide total cloth by cloth per uniform 3 Marks
Divide $2,45,000$ by $275$:
By Euclid's Division Algorithm:
$$2,45,000 = (275 \times 890) + 250$$
• Quotient = $890$ (Number of uniforms stitched).
• Remainder = $250\text{ cm} = 2\text{ m } 50\text{ cm}$ (Cloth left over).
Rough Column Working: 245000 / 275: 275 x 8 = 2200; 2450 - 2200 = 250 -> 2500. 275 x 9 = 2475; 2500 - 2475 = 25 -> 250. 275 x 0 = 0; Remainder = 250 cm.
Step 3: Calculate total cost incurred by contractor 3 Marks
• Cost of cloth purchased = $2,450 \times 145 = ₹ 3,55,250$.
• Stitching and button charges for 890 uniforms = $890 \times 85 = ₹ 75,650$.
$$\text{Total Cost Incurred} = 3,55,250 + 75,650 = ₹ 4,30,900$$
Rough Column Working: 2450 x 145: 2450 x 100 = 245000; 2450 x 40 = 98000; 2450 x 5 = 12250. Sum = 355250. 890 x 85 = 75650.
Step 4: Compute revenue, net profit & conclude 2 Marks
Revenue from supplying 890 uniforms at ₹ 650 each:
$$\text{Total Revenue} = 890 \times 650 = ₹ 5,78,500$$
$$\text{Net Profit} = \text{Total Revenue} - \text{Total Cost} = 5,78,500 - 4,30,900 = ₹ 1,47,600$$
Final Answer: (a) $890$ uniforms, (b) $2\text{ m } 50\text{ cm}$ cloth left over, (c) Net profit = ₹ $1,47,600$.
Part B (10 Marks) 10 Marks
The state disaster management authority sanctioned a relief package of ₹ 45 Crores 75 Lakhs. An international relief agency made an additional contribution of 28 Million US Dollars, converted at the rate of ₹ 82 per Dollar. (a) Express both amounts in figures using commas in their respective systems. (b) Convert the international contribution into Indian Rupees in Crores. (c) Determine the grand total of the relief fund in Indian Rupees, stating the final sum in figures and in words.
Step-by-Step Model Solution:
Step 1: Express both sums in official figures with system commas 2 Marks
• State package: ₹ 45 Crores 75 Lakhs = ₹ 45,75,00,000 (Indian System).
• International aid: 28 Million Dollars = $ 28,000,000 (International System).
Step 2: Convert international aid to Indian Rupees 3 Marks
Given exchange rate: $1\text{ USD} = ₹ 82$.
$$\text{Aid in INR} = 28,000,000 \times 82$$
$$28 \times 82 = 28 \times (80 + 2) = 2,240 + 56 = 2,296$$
$$\text{Aid in INR} = ₹ 2,296,000,000$$
In Indian System: $₹ 2,29,60,00,000$ (₹ 229 Crores 60 Lakhs).
Rough Column Working: 28 x 82 = 2296. With 6 zeros: 2,29,60,00,000.
Step 3: Calculate grand total relief fund 3 Marks
$$\text{Total Relief Fund} = \text{State Package} + \text{International Contribution}$$
$$= 45,75,00,000 + 2,29,60,00,000$$
$$= ₹ 2,75,35,00,000$$
Rough Column Working: 45,75,00,000 + 2,29,60,00,000 = 2,75,35,00,000.
Step 4: Express final amount in words & conclude 2 Marks
In figures: ₹ 2,75,35,00,000.
In words (Indian System): Two hundred seventy-five crore thirty-five lakh rupees.
In words (International System): Two billion seven hundred fifty-three million five hundred thousand rupees.
Final Answer: Grand Total Relief Fund = ₹ 2,75,35,00,000 (Two hundred seventy-five crore thirty-five lakh rupees).

RIMC Practice & Drill Exercises

Attempt these questions on paper following proper step-wise conventions, then click Reveal Solution to verify your answers.

Part A: 5-Mark Descriptive Questions RIMC Descriptive Pattern
Question 1 5 Marks
Insert commas suitably and write the numeral names according to both Indian and International Systems of Numeration for the number: 85462831.
Model Descriptive Solution:
Step 1: Indian System of Numeration 2.5 Marks
Place commas after the first 3 digits from right, then after every 2 digits:
Numeral with commas: 8,54,62,831
Number Name: Eight crore fifty-four lakh sixty-two thousand eight hundred thirty-one.
Step 2: International System of Numeration 2.5 Marks
Place commas after every 3 digits from right:
Numeral with commas: 85,462,831
Number Name: Eighty-five million four hundred sixty-two thousand eight hundred thirty-one.
Final Answer: Indian: 8,54,62,831; International: 85,462,831
Question 2 5 Marks
Find the value of the following product using the Distributive Property of Whole Numbers: 4275 × 125.
Model Descriptive Solution:
Step 1: Split 125 using place value expansion 1 Mark
Write $125$ as $(100 + 20 + 5)$:
$$4275 \times 125 = 4275 \times (100 + 20 + 5)$$
Step 2: Apply Distributive Property of Multiplication over Addition 1 Mark
By Distributive Law: $a \times (b + c + d) = (a \times b) + (a \times c) + (a \times d)$:
$$= (4275 \times 100) + (4275 \times 20) + (4275 \times 5)$$
Step 3: Carry out individual multiplications and sum 3 Marks
• $4275 \times 100 = 4,27,500$
• $4275 \times 20 = 85,500$
• $4275 \times 5 = 21,375$
$$\text{Sum} = 4,27,500 + 85,500 + 21,375 = 5,34,375$$
Rough Column Working: 427500 + 85500 = 513000; 513000 + 21375 = 534375.
Final Answer: 4275 × 125 = 5,34,375
Question 3 5 Marks
Find the difference between the successor of the greatest 5-digit number and the predecessor of the smallest 6-digit number. Show formal justification.
Model Descriptive Solution:
Step 1: Identify greatest 5-digit number and its successor 2 Marks
• Greatest 5-digit number $= 99,999$.
• $\text{Successor}(99,999) = 99,999 + 1 = 1,00,000$.
Step 2: Identify smallest 6-digit number and its predecessor 2 Marks
• Smallest 6-digit number $= 1,00,000$.
• $\text{Predecessor}(1,00,000) = 1,00,000 - 1 = 99,999$.
Step 3: Compute difference 1 Mark
$$\text{Difference} = \text{Successor} - \text{Predecessor} = 1,00,000 - 99,999 = 1$$
Final Answer: Difference = 1
Question 4 5 Marks
How many 6-digit whole numbers exist in all? Provide the complete mathematical proof using the greatest and smallest 6-digit numbers.
Model Descriptive Solution:
Step 1: State smallest and greatest 6-digit numbers 1.5 Marks
• Smallest 6-digit whole number $= 1,00,000$.
• Greatest 6-digit whole number $= 9,99,999$.
Step 2: Apply formula for number of integers in a closed interval 2 Marks
The count of integers in the closed range $[a, b]$ is given by $(b - a) + 1$:
$$\text{Total 6-digit numbers} = (9,99,999 - 1,00,000) + 1$$
Step 3: Evaluate and conclude 1.5 Marks
$$= 8,99,999 + 1 = 9,00,000$$
Alternatively: By formula $9 \times 10^{n-1} = 9 \times 10^{6-1} = 9 \times 10^5 = 9,00,000$.
Final Answer: 9,00,000 (Nine Lakh numbers)
Question 5 5 Marks
In an operation of division, the divisor is 12 times the quotient and 5 times the remainder. If the remainder is 48, find the dividend.
Model Descriptive Solution:
Step 1: Calculate divisor from given remainder 1.5 Marks
Given:
Remainder ($r$) = $48$.
Divisor ($d$) = $5 \times \text{Remainder} = 5 \times 48 = 240$.
Step 2: Calculate quotient 1.5 Marks
Given that Divisor is 12 times Quotient ($q$):
$$d = 12 \times q \implies 240 = 12 \times q \implies q = \frac{240}{12} = 20$$
Step 3: Apply Euclid's Division Algorithm to find dividend 2 Marks
$$\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$$
$$= (240 \times 20) + 48 = 4,800 + 48 = 4,848$$
Final Answer: Dividend = 4,848
Part B: 10-Mark Descriptive Questions RIMC Descriptive Pattern
Question 6 10 Marks
A vessel contains 13 liters and 500 mL of water. (a) In how many glasses, each of 35 mL capacity, can it be filled, and how much water will remain? (b) If the capacity of each glass is increased to 50 mL, how many fewer glasses will be required?
Model Descriptive Solution:
Step 1: Convert total water capacity to milliliters 2 Marks
Given:
Total water in vessel = $13\text{ L } 500\text{ mL}$.
Since $1\text{ L} = 1,000\text{ mL}$:
$$\text{Total capacity} = (13 \times 1,000) + 500 = 13,000 + 500 = 13,500\text{ mL}$$
Step 2: Solve part (a) for 35 mL glasses 3 Marks
Capacity of 1 glass = $35\text{ mL}$.
Divide $13,500$ by $35$:
$$13,500 = (35 \times 385) + 25$$
• Number of glasses filled $= 385\text{ glasses}$.
• Water remaining $= 25\text{ mL}$.
Rough Column Working: 13500 / 35: 35 x 385 = 13475; 13500 - 13475 = 25 mL.
Step 3: Solve part (b) for 50 mL glasses 3 Marks
Capacity of upgraded glass = $50\text{ mL}$.
Divide $13,500$ by $50$:
$$\text{Number of glasses} = \frac{13,500}{50} = 270\text{ glasses}$$ (Remainder = $0\text{ mL}$).
Step 4: Calculate difference in glasses & state final answers 2 Marks
$$\text{Difference in glasses} = 385 - 270 = 115\text{ fewer glasses}$$
Final Answer:
(a) $385$ glasses filled, $25\text{ mL}$ water remains.
(b) $115$ fewer glasses required.
Final Answer: (a) 385 glasses, 25 mL left; (b) 115 fewer glasses
Question 7 10 Marks
The population of a city was 9,87,654 in the year 2021. In 2022, it increased by 65,420 due to migration. In 2023, it decreased by 28,975. Find: (i) The population at the end of 2022. (ii) The final population at the end of 2023. (iii) Express the final population in the International System in figures and in words.
Model Descriptive Solution:
Step 1: Calculate population at the end of 2022 2.5 Marks
Population at end of 2021 = $9,87,654$.
Increase in 2022 = $+65,420$.
$$\text{Population (end 2022)} = 9,87,654 + 65,420 = 10,53,074$$
Rough Column Working: 987654 + 65420 = 1053074.
Step 2: Calculate population at the end of 2023 2.5 Marks
Population at end of 2022 = $10,53,074$.
Decrease in 2023 = $-28,975$.
$$\text{Population (end 2023)} = 10,53,074 - 28,975 = 10,24,099$$
Rough Column Working: 1053074 - 28975 = 1024099.
Step 3: Convert to International System figures 2.5 Marks
Insert commas after every 3 digits from right:
$$\text{Figures in International System} = 1,024,099$$
Step 4: Write in words in International System & conclude 2.5 Marks
In words: One million twenty-four thousand ninety-nine.
Final Answer: (i) $10,53,074$; (ii) $10,24,099$; (iii) $1,024,099$ (One million twenty-four thousand ninety-nine).
Final Answer: (i) 10,53,074; (ii) 10,24,099; (iii) 1,024,099 (One million twenty-four thousand ninety-nine)
Question 8 10 Marks
Find the least number of 5 digits which when divided by 47 leaves a remainder of 12. Also, find the greatest 5-digit number satisfying the same condition.
Model Descriptive Solution:
Step 1: Smallest 5-digit number division analysis 2.5 Marks
The smallest 5-digit number is $10,000$.
Divide $10,000$ by $47$:
$$10,000 = (47 \times 212) + 36$$
Quotient = $212$, Remainder = $36$.
Rough Column Working: 47 x 212 = 9964; 10000 - 9964 = 36.
Step 2: Determine least 5-digit number leaving remainder 12 2.5 Marks
The smallest 5-digit multiple of $47$ is:
$$10,000 + (47 - 36) = 10,000 + 11 = 10,011$$
To leave a remainder of $12$, add $12$ to this multiple:
$$\text{Required smallest number} = 10,011 + 12 = 10,023$$
Verification: $10,023 = (47 \times 213) + 12$.
Step 3: Greatest 5-digit number division analysis 2.5 Marks
The greatest 5-digit number is $99,999$.
Divide $99,999$ by $47$:
$$99,999 = (47 \times 2127) + 30$$
Quotient = $2,127$, Remainder = $30$.
Rough Column Working: 47 x 2127 = 99969; 99999 - 99969 = 30.
Step 4: Determine greatest 5-digit number leaving remainder 12 & conclude 2.5 Marks
The greatest 5-digit multiple of $47$ is:
$$99,999 - 30 = 99,969$$
To leave a remainder of $12$, add $12$ to this multiple:
$$\text{Required greatest number} = 99,969 + 12 = 99,981$$
Verification: $99,981 = (47 \times 2127) + 12$.
Final Answer: Least number = $10,023$; Greatest number = $99,981$.
Final Answer: Least = 10,023; Greatest = 99,981
Question 9 10 Marks
A printing company manufactured 75,000 sheets of paper. Each sheet yields 8 pages of a notebook. A standard cadet notebook requires 250 pages. (a) How many total pages are available? (b) How many complete notebooks can be prepared? (c) How many pages remain unused? (d) If each notebook is sold for ₹ 45, calculate the total revenue generated.
Model Descriptive Solution:
Step 1: Calculate total number of pages manufactured 2 Marks
Given:
Total sheets = $75,000$.
Pages yielded per sheet = $8$.
$$\text{Total pages} = 75,000 \times 8 = 6,00,000\text{ pages}$$
Rough Column Working: 75 x 8 = 600, so 75,000 x 8 = 6,00,000.
Step 2: Calculate number of complete notebooks prepared 3 Marks
Pages required per notebook = $250\text{ pages}$.
Divide $6,00,000$ by $250$:
$$\text{Number of notebooks} = \frac{6,00,000}{250} = \frac{60,000}{25} = 2,400\text{ notebooks}$$
Step 3: Determine unused leftover pages 2 Marks
By Euclid's Algorithm:
$$6,00,000 = (250 \times 2400) + 0$$
Therefore, Remainder = $0\text{ pages}$ (Zero unused pages).
Step 4: Compute total revenue and state answers 3 Marks
Selling price per notebook = ₹ $45$.
$$\text{Total Revenue} = 2,400 \times 45 = 2,400 \times \frac{90}{2} = 1,200 \times 90 = ₹ 1,08,000$$
Final Answer:
(a) Total pages = $6,00,000$
(b) Complete notebooks = $2,400$
(c) Leftover pages = $0$
(d) Total revenue = ₹ $1,08,000$.
Final Answer: (a) 6,00,000 pages; (b) 2,400 notebooks; (c) 0 pages left; (d) ₹ 1,08,000
Question 10 10 Marks
An officer mess contractor placed an order for 65 commercial refrigerators at ₹ 18,750 each and 45 heavy-duty washing machines at ₹ 14,250 each. (a) Calculate the total expenditure using suitable properties of whole numbers. (b) If the appliance manufacturer offered a special defence discount of ₹ 850 on every individual appliance supplied, determine the net amount payable by the mess.
Model Descriptive Solution:
Step 1: Set up total expenditure equation using properties 2 Marks
Cost of 65 refrigerators = $65 \times 18,750$.
Cost of 45 washing machines = $45 \times 14,250$.
$$\text{Total Cost} = (65 \times 18,750) + (45 \times 14,250)$$
Step 2: Evaluate using Distributive Law 3 Marks
• $65 \times 18,750 = 65 \times (18,000 + 750) = (65 \times 18,000) + (65 \times 750) = 11,70,000 + 48,750 = ₹ 12,18,750$.
• $45 \times 14,250 = 45 \times (14,000 + 250) = (45 \times 14,000) + (45 \times 250) = 6,30,000 + 11,250 = ₹ 6,41,250$.
$$\text{Gross Total Cost} = 12,18,750 + 6,41,250 = ₹ 18,60,000$$
Rough Column Working: 1218750 + 641250 = 1860000.
Step 3: Calculate total discount across all appliances 3 Marks
Total appliances ordered = $65 + 45 = 110\text{ appliances}$.
Discount per appliance = ₹ $850$.
$$\text{Total Discount} = 110 \times 850 = 11 \times 8,500 = ₹ 93,500$$
Rough Column Working: 110 x 850: 100 x 850 = 85000; 10 x 850 = 8500. 85000 + 8500 = 93500.
Step 4: Compute net amount payable & state final answers 2 Marks
$$\text{Net Amount Payable} = \text{Gross Total} - \text{Total Discount}$$
$$= 18,60,000 - 93,500 = ₹ 17,66,500$$
Final Answer: (a) Gross expenditure = ₹ $18,60,000$; (b) Net amount payable = ₹ $17,66,500$.
Final Answer: (a) Gross Cost = ₹ 18,60,000; (b) Net Payable = ₹ 17,66,500

High-Yield FAQs & Revision Notes

What is the general formula to find the total count of n-digit numbers?
The total number of $n$-digit whole numbers in base-10 numeration is given by: $$\text{Total } n\text{-digit numbers} = 9 \times 10^{n-1}$$ For example: 3-digit numbers = $9 \times 10^2 = 900$; 5-digit numbers = $9 \times 10^4 = 90,000$; 6-digit numbers = $9 \times 10^5 = 9,00,000$.
Why is division of a whole number by zero mathematically undefined?
Division is defined as the inverse operation of multiplication. If $a \div 0 = k$, it requires that $k \times 0 = a$. For any non-zero whole number $a$, multiplying any number by zero always equals zero ($k \times 0 = 0$), so no such number $k$ can exist. When $a = 0$, $0 \div 0 = k$ means $k \times 0 = 0$, which is satisfied by every number, making it indeterminate. Therefore, division by zero is strictly undefined.
What is the key difference between Hindu-Arabic numerals and Roman numerals?
The Hindu-Arabic system utilizes a positional place-value mechanism with a dedicated symbol for zero ($0$), allowing representation of infinitely large numbers. The Roman numeral system is an additive-subtractive system without place value and without any symbol for zero ($0$), making complex arithmetic division or multiplication intractable.
Does the RIMC Mathematics Paper award step marks if my final answer has a minor arithmetic slip?
Yes! RIMC Paper I is evaluated on a strict step-marking rubric. If you declare variables, state the relevant algebraic property or division algorithm, and show systematic steps, you receive 70% to 80% of the total marks even if a minor addition slip occurs in the final step. Conversely, writing only the final numeric answer without steps yields at most 1 mark out of 10.
How should rough calculations be presented on the answer booklet?
Never perform rough work on loose paper or erase it. Draw a neat vertical margin of approximately 5 to 6 cm on the right-hand side of your ruled answer sheet using a pencil. Title the column 'ROUGH WORK', carry out all long division and multiplications there, and leave them visible so the examiner can verify your authenticity.
What is the difference between the predecessor of a whole number and an integer?
In the set of whole numbers ($\mathbb{W} = \{0, 1, 2, \dots\}$), the number $0$ has no predecessor because no whole number is smaller than $0$. However, in the set of integers ($\mathbb{Z}$), the predecessor of $0$ exists and is $-1$ ($0 - 1 = -1$).
What are the general rules for estimating and rounding off numbers in RIMC?
To round off a number to a given place (e.g., nearest ten, hundred, thousand): examine the digit immediately to its right. If that digit is $5, 6, 7, 8,$ or $9$, round up (add 1 to target digit and replace trailing digits with zeros). If the digit is $0, 1, 2, 3,$ or $4$, round down (leave target digit unchanged and replace trailing digits with zeros).
How is the distributive property of multiplication tested in RIMC questions?
Examiners test it either by asking candidates to simplify products like $854 \times 102 = 854 \times (100 + 2)$ or $987 \times 99 = 987 \times (100 - 1)$, or by grouping common factors such as $81265 \times 169 - 81265 \times 69 = 81265 \times (169 - 69) = 81,26,500$.
Can commas be placed according to both Indian and International systems in the same paper?
You must follow the system specified in the question. If a question asks for the Indian system, place commas as $3, 2, 2$ ($12,34,567$). If it asks for the International system, place commas as $3, 3, 3$ ($1,234,567$). If no system is specified, use the Indian system by default.
What is the golden rule when solving word problems involving capacity, length, or mass?
Always convert all given measurements to the smallest common unit before carrying out long division or multiplication. For example, convert $4\text{ L } 500\text{ mL}$ into $4,500\text{ mL}$, or $18\text{ kg } 750\text{ g}$ into $18,750,000\text{ mg}$.
How does this topic relate to the official RIMC syllabus?
This module directly covers Serial No. 1, 28, 29, 30, 43 of the broader syllabus published in Chapter 8 of the official RIMC admission brochure, ensuring 100% syllabus alignment.
Last-Minute Cadet High-Yield Revision Matrix
Topic / Concept High-Yield Exam Rules & Memory Anchor
Natural Numbers ($\mathbb{N}$) Counting numbers $\{1, 2, 3, \dots\}$; smallest is 1; no greatest.
Whole Numbers ($\mathbb{W}$) Natural numbers plus zero $\{0, 1, 2, \dots\}$; smallest is 0. 0 has no predecessor in $\mathbb{W}$.
Indian Periods Ones (3 digits), Thousands (2 digits), Lakhs (2 digits), Crores (2 digits). Separator: 3:2:2:2.
International Periods Ones (3 digits), Thousands (3 digits), Millions (3 digits), Billions (3 digits). Separator: 3:3:3:3.
Cross-System Identities $1\text{ Million} = 10\text{ Lakhs}$; $1\text{ Crore} = 10\text{ Millions}$; $1\text{ Billion} = 100\text{ Crores} = 1\text{ Arab}$.
Euclid's Division Algorithm $\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$, with $0 \le \text{Remainder} < \text{Divisor}$.
Least to Add / Subtract To make divisible by $d$: Subtract Remainder ($r$); Add $(d - r)$ (if $r \ne 0$).
Distributive Property $a \times (b \pm c) = (a \times b) \pm (a \times c)$. Essential for mental math & step-marking.