Official Syllabus Item: 42 Arithmetic & Number Theory

Rational Numbers: Properties, Operations & Insertion

Master rational numbers for RIMC Entrance Exam: definition p/q (q ≠ 0), positive and negative rationals, standard irreducible form, number line plotting, density property, insertion techniques, distributive/associative laws, and descriptive word problems.

Overview & Fundamental Concept

A rational number is formally defined as any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and the denominator $q \neq 0$. In the descriptive RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes), rational numbers are rigorously tested across both Part A (5 Marks) and Part B (10 Marks). The examination evaluates candidates on their mathematical precision in converting fractions into standard irreducible forms (with strictly positive denominators and co-prime numerators), plotting proper and improper rationals on a calibrated number line, inserting a specified number of rational numbers using common denominator scaling or the mean method, formally verifying axiomatic laws (Closure, Commutativity, Associativity, and Distributivity), and solving multi-stage descriptive word problems involving geometric perimeters, fractional remnants, and cost rates. Scoring full credit demands step-wise algebraic notation, explicit property citations, neat arithmetic working, and boxed final answers with appropriate physical units.

Official Syllabus Coverage Breakdown

Syllabus Item	Prescribed Scope	Exam Relevance & Pedagogical Objective
Item 42	Rational Numbers: Definition, Representation, Axiomatic Properties & Density	Directly prescribed in Chapter 8 of the official RIMC brochure. Assesses set-theoretic relations ($\mathbb{N} \subset \mathbb{W} \subset \mathbb{Z} \subset \mathbb{Q}$), standard form conversions, number line partitioning, insertion between bounds, and algebraic property verifications.

1. Formal Definition, Set Structure & Classification of Rational Numbers

The word rational arises from the mathematical term ratio. A number is termed a rational number if it can be written in the quotient form $\frac{p}{q}$, where $p$ (the numerator) and $q$ (the denominator) are integers $(\mathbb{Z})$ and $q \neq 0$. The set of all rational numbers is denoted by the blackboard bold symbol $\mathbb{Q}$ (from quotient):

$$\mathbb{Q} = \left\{ \frac{p}{q} \;\middle|\; p \in \mathbb{Z},\; q \in \mathbb{Z},\; q \neq 0 \right\}$$

Why must $q \neq 0$? In mathematical analysis, division by zero is strictly undefined. If $\frac{p}{0} = k$, then $p = k \times 0 = 0$. If $p \neq 0$, no such real number $k$ exists; if $p = 0$, $k$ can be any arbitrary number (indeterminate). Thus, zero can never serve as a denominator in the rational number system.

Hierarchical Number Hierarchy & Universal Inclusions

Every natural number, whole number, and integer is an inherent member of the rational number set $\mathbb{Q}$:

Natural Numbers $(\mathbb{N})$: Any $n \in \mathbb{N}$ can be written as $\frac{n}{1}$. For example, $7 = \frac{7}{1} \in \mathbb{Q}$.
Whole Numbers $(\mathbb{W})$: The number $0$ is a rational number because $0 = \frac{0}{1} = \frac{0}{5} = \frac{0}{-9}$. Here $p = 0 \in \mathbb{Z}$ and $q = 1 \in \mathbb{Z}$ with $q \neq 0$.
Integers $(\mathbb{Z})$: Any negative integer $-k$ can be written as $\frac{-k}{1}$. For example, $-14 = \frac{-14}{1} \in \mathbb{Q}$.
Fractions: All positive and mixed fractions are positive rational numbers (e.g., $3\frac{1}{4} = \frac{13}{4} \in \mathbb{Q}$).

Classification: Positive, Negative & Neutral Rationals

Positive Rational Numbers: A rational number $\frac{p}{q}$ is positive if its numerator and denominator have the same sign (both positive or both negative). For example, $\frac{3}{7} > 0$ and $\frac{-5}{-8} = \frac{5}{8} > 0$.
Negative Rational Numbers: A rational number $\frac{p}{q}$ is negative if its numerator and denominator have opposite signs. For example, $\frac{-4}{9} < 0$ and $\frac{7}{-11} = -\frac{7}{11} < 0$.
Neutral Rational Number: The number $0$ is neither positive nor negative.

2. Standard (Irreducible) Form of a Rational Number

A rational number $\frac{p}{q}$ is said to be in its Standard Form (or Canonical Irreducible Form) if and only if it satisfies two mandatory criteria:

1. Its denominator $q$ is a strictly positive integer ($q > 0$).
2. The greatest common divisor (HCF) of $|p|$ and $q$ is exactly $1$, meaning $p$ and $q$ are co-prime: $\gcd(|p|, q) = 1$.

Two-Step Algorithm to Reduce Any Rational Number to Standard Form

In RIMC descriptive answers, never leave a negative sign in the denominator or an unreduced fraction. Apply this exact protocol:

Step I (Make Denominator Positive): If the denominator is negative, multiply both the numerator and denominator by $-1$:
$$\frac{a}{-b} = \frac{a \times (-1)}{-b \times (-1)} = \frac{-a}{b}$$
Step II (Cancel Common Factors via HCF): Find $H = \gcd(|a|, b)$. Divide both numerator and denominator by $H$:
$$\frac{-a \div H}{b \div H}$$
Example: Express $\frac{36}{-84}$ in standard form:
Multiply by $-1$: $\frac{-36}{84}$. Compute $\gcd(36, 84) = 12$. Divide by $12$: $\frac{-36 \div 12}{84 \div 12} = \mathbf{-\frac{3}{7}}$.

Examiner Penalty Trap: Negative Denominators in Booklets

Leaving a final answer as $\frac{5}{-13}$ or $\frac{-15}{-25}$ will instantly lose 1 mark in RIMC marking schemes. You must write $\mathbf{-\frac{5}{13}}$ and $\mathbf{\frac{3}{5}}$.

3. Representation of Rational Numbers on the Number Line

Every rational number corresponds to a unique point on the real number line. The method of geometric division depends on whether the rational number is proper ($|p| < q$) or improper ($|p| > q$).

Case I: Proper Rational Numbers $(|p| < q)$

Proper rationals lie entirely within the primary unit segment between $0$ and $1$ (for positive) or between $-1$ and $0$ (for negative):

To represent $\frac{p}{q}$, divide the unit segment from $0$ to $1$ into $q$ equal parts using $(q - 1)$ equidistant marking ticks.
Count $p$ parts starting from $0$ toward the right. Mark this point as $P = \frac{p}{q}$.
To represent $-\frac{p}{q}$, divide the segment between $0$ and $-1$ into $q$ equal parts and count $p$ units to the left of $0$.

Case II: Improper Rational Numbers $(|p| > q)$

Convert the improper rational into a mixed numeral $W\frac{r}{q}$, where $W$ is the quotient (whole integer) and $r$ is the remainder ($r < q$):

Identify the consecutive integers $W$ and $W + 1$ between which the number lies.
Divide the interval between $W$ and $W + 1$ into $q$ equal segments.
Advance $r$ divisions from $W$ toward $W + 1$.
Example: To plot $\frac{11}{4} = 2\frac{3}{4}$, locate the interval $[2, 3]$. Divide $[2, 3]$ into $4$ equal intervals of length $0.25$ each. The third tick mark represents $2\frac{3}{4} = \frac{11}{4}$.

4. Comprehensive Axiomatic Properties of Operations on $\mathbb{Q}$

In RIMC Part B (10-mark questions), examiners frequently require students to state and verify algebraic properties with given numerical values. The table below consolidates the four primary operations across the rational number system:

Key Axiomatic Nuances Tested in RIMC

Additive Inverse (Opposite): For any rational number $\frac{p}{q}$, its unique additive inverse is $-\frac{p}{q}$ because $\frac{p}{q} + \left(-\frac{p}{q}\right) = 0$. Note that $-(-x) = x$.
Multiplicative Inverse (Reciprocal): For any non-zero rational number $\frac{p}{q}$, its reciprocal is $\frac{q}{p}$ because $\frac{p}{q} \times \frac{q}{p} = 1$. Zero has no reciprocal because $\frac{1}{0}$ is undefined.
Numbers that are their own reciprocals: Exactly two rational numbers satisfy $x = \frac{1}{x}$, namely $\mathbf{1}$ and $\mathbf{-1}$.

Property	Addition $(+)$	Subtraction $(-)$	Multiplication $(\times)$	Division $(\div)$
Closure	Closed: $a+b \in \mathbb{Q}$	Closed: $a-b \in \mathbb{Q}$	Closed: $a \cdot b \in \mathbb{Q}$	Not Closed (division by $0$ undefined)
Commutativity	Valid: $a+b = b+a$	Fails: $a-b \neq b-a$	Valid: $a \cdot b = b \cdot a$	Fails: $a \div b \neq b \div a$
Associativity	Valid: $(a+b)+c = a+(b+c)$	Fails: $(a-b)-c \neq a-(b-c)$	Valid: $(a \cdot b) \cdot c = a \cdot (b \cdot c)$	Fails: $(a \div b) \div c \neq a \div (b \div c)$
Identity Element	Identity is $0$: $a + 0 = a$	No two-sided identity	Identity is $1$: $a \times 1 = a$	No two-sided identity
Inverse Element	Additive Inverse: $-a$ $a + (-a) = 0$	No standard inverse	Reciprocal: $\frac{1}{a}$ ($a \neq 0$) $a \times \frac{1}{a} = 1$	No standard inverse
Distributivity	—	—	Over $+$: $a(b+c) = ab+ac$ Over $-$: $a(b-c) = ab-ac$	Fails to distribute over $+/-$ from right

5. The Density Property & Systematic Insertion of $n$ Rational Numbers

Unlike natural numbers or integers, where there are a finite number of integers between any two values (e.g., exactly two integers between $3$ and $6$), the set of rational numbers is dense: between any two distinct rational numbers $a$ and $b$ ($a < b$), there exist infinitely many rational numbers.

Method A: The Mean (Midpoint) Algorithm

For two rational numbers $a$ and $b$ with $a < b$:

The arithmetic mean $q_1 = \frac{a + b}{2}$ lies strictly between $a$ and $b$: $a < q_1 < b$.
A second rational number can be generated between $a$ and $q_1$: $q_2 = \frac{a + q_1}{2}$.
A third rational number can be found between $q_1$ and $b$: $q_3 = \frac{q_1 + b}{2}$.
Note: The mean method is theoretically elegant and ideal for inserting $1$ or $2$ numbers, but computationally tedious for inserting $5$ to $10$ numbers.

Method B: Common Denominator & $(n+1)$ Scaling Multiplier (Recommended for RIMC)

To insert $n$ rational numbers between $\frac{a}{b}$ and $\frac{c}{d}$:

Step 1: Find $L = \text{LCM}(b, d)$. Express both numbers with denominator $L$: $\frac{a'}{L}$ and $\frac{c'}{L}$.
Step 2: Check if the difference between numerators $(c' - a')$ provides at least $n$ integer gaps. If yes, simply pick the intermediate numerators.
Step 3: If the gap is insufficient, multiply both numerator and denominator of each fraction by $(n + 1)$ (or a convenient multiplier like $10$):
$$\frac{a' \times (n+1)}{L \times (n+1)} \quad \text{and} \quad \frac{c' \times (n+1)}{L \times (n+1)}$$
Step 4: Write down $n$ consecutive fractions by incrementing the numerator by $1$ while keeping the denominator constant.

Illustrated Conceptual Diagrams & Visual Schematics

The Rational Number Line & Density Property Continuum

Rational Number Definition: Any number expressible as $\frac{p}{q}$ where $p, q \in \mathbb{Z}$ and $q \neq 0$. Standard form mandates $q > 0$ and $\gcd(|p|, q) = 1$. Density Property: Between any two distinct rational numbers, there exist infinitely many rational numbers, found using the mean method $\frac{a + b}{2}$.

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: Parameter Declaration & Common Denominator Setup 1 - 2 Marks

State the given numbers, identify whether denominators are negative, convert to standard form, and compute the LCM of denominators explicitly.

Booklet Layout:
Given rational numbers: $-\frac{3}{5}$ and $\frac{2}{3}$.\nLCM of denominators $5$ and $3 = 15$.\nConverting to like fractions: $-\frac{3 \times 3}{5 \times 3} = -\frac{9}{15}$ and $\frac{2 \times 5}{3 \times 5} = \frac{10}{15}$.

Step 2: Property Statement & Algebraic Formulation 1 - 2 Marks

Formally cite the mathematical law being applied (e.g., Distributive Law: $a(b+c) = ab+ac$) before substituting numeric values.

Booklet Layout:
By the Distributive Property of Multiplication over Addition:\n$a \times (b + c) = (a \times b) + (a \times c)$

Step 3: Step-by-Step Arithmetic Calculation & Sign Handling 2 - 4 Marks

Carry out bracket evaluation, fractional addition/subtraction with explicit numerators, and multiplication of signs step-by-step. Keep rough work neat.

Booklet Layout:
$$\text{LHS} = -\frac{3}{4} \times \left[ \frac{2 \times 2 + (-5) \times 1}{6} \right] = -\frac{3}{4} \times \left( -\frac{1}{6} \right) = \frac{3}{24} = \frac{1}{8}$$

Step 4: Standard Irreducible Form & Boxed Answer 1 Mark

Simplify the final result to lowest terms with a positive denominator. Conclude with a clear statement and box the final value.

Booklet Layout:
$$\therefore \text{LHS} = \text{RHS} = \frac{1}{8}. \quad \text{[Hence Verified]}$$

Examiner Traps & Common Blunders (Marks Penalties)

Leaving Negative Signs in Denominators -1 Mark

Common Mistake: Writing the final answer as $\frac{7}{-12}$ or leaving double negatives like $\frac{-9}{-15}$.

Cadet Antidote: Always multiply numerator and denominator by $-1$ to force the denominator positive: $\frac{7}{-12} = -\frac{7}{12}$ and $\frac{-9}{-15} = \frac{3}{5}$.

Claiming Division is Closed over Rational Numbers -2 Marks

Common Mistake: Writing in true/false or conceptual questions that 'Rational numbers are closed under division'.

Cadet Antidote: Rational numbers are closed under division ONLY if zero is excluded. Since $\frac{a}{0}$ is undefined, the set $\mathbb{Q}$ as a whole is NOT closed under division.

Sign Errors in Distributive Expansion with Negative Multipliers -2 to -3 Marks

Common Mistake: Expanding $-\frac{2}{3} \times \left(\frac{4}{5} - \frac{7}{10}\right)$ as $-\frac{8}{15} - \frac{14}{30}$ instead of $-\frac{8}{15} + \frac{14}{30}$.

Cadet Antidote: Explicitly treat $- \times - = +$: write $-\frac{2}{3} \times \left(-\frac{7}{10}\right) = +\frac{14}{30}$ before combining.

Confusing Additive Inverse with Multiplicative Inverse -2 Marks

Common Mistake: Writing the additive inverse of $-\frac{4}{7}$ as $-\frac{7}{4}$ (mixing reciprocal with opposite).

Cadet Antidote: Additive inverse changes the sign only ($-\frac{4}{7} \to +\frac{4}{7}$). Multiplicative inverse flips the fraction without changing the sign ($-\frac{4}{7} \to -\frac{7}{4}$).

Cadet Topper Mental Verification Shortcuts

The $(n+1)$ Scaling Multiplier for Flawless Insertion

When asked to insert $n$ rational numbers between two fractions with common denominator $D$, multiply both numerator and denominator by $(n+1)$. This guarantees exactly $(n+1) - 1 = n$ intermediate integers between the two numerators with zero guesswork.

Insert 5 rationals between $\frac{1}{3}$ and $\frac{2}{3}$. Common denominator is 3. Here $n = 5$, so multiply by $n+1 = 6$: $\frac{1 \times 6}{3 \times 6} = \frac{6}{18}$ and $\frac{2 \times 6}{3 \times 6} = \frac{12}{18}$. The 5 intermediate numbers are immediately: $\frac{7}{18}, \frac{8}{18}, \frac{9}{18}, \frac{10}{18}, \frac{11}{18}$.

Cross-Multiplication Shortcut for Comparing Two Rationals

To compare $\frac{a}{b}$ and $\frac{c}{d}$ where $b, d > 0$, compute cross-products $a \times d$ and $b \times c$. If $ad > bc$, then $\frac{a}{b} > \frac{c}{d}$. If $ad < bc$, then $\frac{a}{b} < \frac{c}{d}$.

Compare $-\frac{4}{7}$ and $-\frac{5}{9}$: $(-4) \times 9 = -36$ and $7 \times (-5) = -35$. Since $-36 < -35$, we have $-\frac{4}{7} < -\frac{5}{9}$.

Regrouping Like Denominators to Minimize Arithmetic

In long multi-term sums involving 4 or more rationals, use commutativity and associativity to group terms with identical or multiple denominators together before taking a global LCM.

For $\frac{3}{7} + \left(-\frac{6}{11}\right) + \left(-\frac{8}{21}\right) + \frac{5}{22}$, group: $\left[\frac{3}{7} - \frac{8}{21}\right] + \left[-\frac{6}{11} + \frac{5}{22}\right] = \left[\frac{9-8}{21}\right] + \left[\frac{-12+5}{22}\right] = \frac{1}{21} - \frac{7}{22}$.

Key Formulas & Conversion Cheat Sheet

Definition of Rational Number Set

$$\mathbb{Q} = \left\{ \frac{p}{q} \;\middle|\; p, q \in \mathbb{Z},\; q \neq 0 \right\}$$

Every rational number is the quotient of two integers where the divisor is non-zero.

Standard Irreducible Form

$$\text{Standard Form} = \frac{p}{q} \quad (q > 0,\; \gcd(|p|, q) = 1)$$

Denominator must be strictly positive and numerator and denominator must be co-prime.

Additive Inverse (Opposite)

$$\frac{p}{q} + \left(-\frac{p}{q}\right) = 0$$

The sum of a number and its additive inverse is always zero (the additive identity).

Multiplicative Inverse (Reciprocal)

$$\frac{p}{q} \times \frac{q}{p} = 1 \quad (p \neq 0,\; q \neq 0)$$

The product of a non-zero rational and its reciprocal is always 1 (the multiplicative identity).

Mean Method for Rational Insertion

$$q_{\text{mid}} = \frac{a + b}{2} = \frac{1}{2}(a + b)$$

The arithmetic average of two rationals always lies strictly between them: $a < q_{\text{mid}} < b$.

Density Scaling Formula for $n$ Numbers

$$\frac{a \times (n+1)}{b \times (n+1)} < \frac{a(n+1) + 1}{b(n+1)} < \dots < \frac{c \times (n+1)}{b \times (n+1)}$$

Multiplying like fractions by $(n+1)$ creates exactly $n$ integer spaces between numerators.

Distributive Property of Multiplication

$$a \times (b \pm c) = (a \times b) \pm (a \times c)$$

Multiplication distributes over both addition and subtraction for all rational numbers.

Rational Division via Reciprocal

$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc} \quad (b, c, d \neq 0)$$

Dividing by a fraction is equivalent to multiplying by its reciprocal.

Solved Model Examples (Descriptive Format with Step Mark Distribution)

Model 1: Standard Form & Number Line 5 Marks

Express $-\frac{36}{84}$ in standard irreducible form, and represent the resulting rational number on a neatly labeled number line.

Step-by-Step Model Solution:

Step 1: Check Denominator & Compute HCF 2 Marks

The given rational number is $-\frac{36}{84}$. The denominator is $84 > 0$, so the positive denominator condition is already satisfied.
Now find $\gcd(36, 84)$ using prime factorization:
$36 = 2^2 \times 3^2$
$84 = 2^2 \times 3 \times 7$
$\gcd(36, 84) = 2^2 \times 3 = 12$.

Rough Column Working: 84 = 36 × 2 + 12; 36 = 12 × 3 + 0 => HCF = 12

Step 2: Divide Numerator & Denominator by HCF 1 Mark

$$-\frac{36}{84} = \frac{-36 \div 12}{84 \div 12} = -\frac{3}{7}$$
Since $7 > 0$ and $\gcd(3, 7) = 1$, the standard form is $\mathbf{-\frac{3}{7}}$.

Step 3: Number Line Representation 2 Marks

Since $-\frac{3}{7}$ is a proper negative fraction, it lies strictly between $0$ and $-1$ on the number line:
1. Draw a horizontal line and mark origin $O(0)$ and point $A(-1)$.
2. Divide the segment $OA$ into $7$ equal sub-intervals of length $\frac{1}{7}$ unit each.
3. Moving $3$ units to the left from $0$, mark the third division point as $P$.
Point $P$ represents $-\frac{3}{7}$.

Examiner Scoring Key: Award 2 marks for HCF calculation and standard reduction, 2 marks for correct number line plotting between 0 and -1 with 7 equal intervals, and 1 mark for clear labeling.

Model 2: Inserting 5 Rational Numbers 5 Marks

Find five rational numbers strictly lying between $-\frac{3}{5}$ and $\frac{2}{3}$. Show complete mathematical working.

Step-by-Step Model Solution:

Step 1: Convert to Like Fractions via LCM 2 Marks

The given rational numbers are $-\frac{3}{5}$ and $\frac{2}{3}$.
The denominators are $5$ and $3$.
$\text{LCM}(5, 3) = 15$.
Converting both numbers to equivalent fractions with denominator $15$:
$$-\frac{3}{5} = -\frac{3 \times 3}{5 \times 3} = -\frac{9}{15}$$
$$\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}$$

Step 2: Identify Intermediate Numerators 2 Marks

The numerators are $-9$ and $10$.
Integers between $-9$ and $10$ include: $-8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, \dots, 9$.
Since there are $18$ available integers, we can directly select any five distinct integers between $-9$ and $10$.
Choosing five convenient integers: $-5, -3, 0, 5, 8$.

Step 3: Form Fractions & Reduce to Lowest Terms 1 Mark

The corresponding five rational numbers are:
$$-\frac{5}{15} = -\frac{1}{3}, \quad -\frac{3}{15} = -\frac{1}{5}, \quad \frac{0}{15} = 0, \quad \frac{5}{15} = \frac{1}{3}, \quad \frac{8}{15}$$
Arranged in ascending order:
$$-\frac{9}{15} < -\frac{1}{3} < -\frac{1}{5} < 0 < \frac{1}{3} < \frac{8}{15} < \frac{10}{15}$$

Examiner Scoring Key: Full marks require explicit statement of LCM = 15 and writing the final 5 numbers within the correct bounds.

Model 3: Verification of Distributive Property 7 Marks

Verify the Distributive Property of Multiplication over Subtraction: $a \times (b - c) = (a \times b) - (a \times c)$ for $a = -\frac{3}{4}$, $b = \frac{2}{3}$, and $c = -\frac{5}{6}$.

Step-by-Step Model Solution:

Step 1: State Property & Evaluate LHS 3 Marks

We need to verify: $\text{LHS} = a \times (b - c)$ and $\text{RHS} = (a \times b) - (a \times c)$.
Given: $a = -\frac{3}{4}$, $b = \frac{2}{3}$, $c = -\frac{5}{6}$.
Evaluating LHS:
$$\text{LHS} = -\frac{3}{4} \times \left[ \frac{2}{3} - \left(-\frac{5}{6}\right) \right] = -\frac{3}{4} \times \left[ \frac{2}{3} + \frac{5}{6} \right]$$
LCM of $3$ and $6 = 6$.
$$\frac{2}{3} + \frac{5}{6} = \frac{2 \times 2 + 5}{6} = \frac{4 + 5}{6} = \frac{9}{6} = \frac{3}{2}$$
Now multiply by $a$:
$$\text{LHS} = -\frac{3}{4} \times \frac{3}{2} = -\frac{9}{8}$$

Step 2: Evaluate RHS Separately 3 Marks

Evaluating RHS: $(a \times b) - (a \times c)$
$$a \times b = \left(-\frac{3}{4}\right) \times \frac{2}{3} = -\frac{6}{12} = -\frac{1}{2}$$
$$a \times c = \left(-\frac{3}{4}\right) \times \left(-\frac{5}{6}\right) = \frac{(-3) \times (-5)}{4 \times 6} = \frac{15}{24} = \frac{5}{8}$$
Now subtract $(a \times c)$ from $(a \times b)$:
$$\text{RHS} = -\frac{1}{2} - \frac{5}{8} = \frac{-1 \times 4 - 5}{8} = \frac{-4 - 5}{8} = -\frac{9}{8}$$

Step 3: Compare LHS and RHS & Conclude 1 Mark

Since $\text{LHS} = -\frac{9}{8}$ and $\text{RHS} = -\frac{9}{8}$, we have:
$$\text{LHS} = \text{RHS} = -\frac{9}{8}$$
Hence, the Distributive Property of Multiplication over Subtraction is verified.

Examiner Scoring Key: Deduct 1 mark if students do not show separate evaluation of LHS and RHS. Signs in (-3/4) * (-5/6) = +5/8 must be explicitly correct.

Model 4: Strategic Evaluation using Axiomatic Laws 7 Marks

Evaluate the following expression using suitable properties of rational numbers:
$$\frac{3}{7} + \left(-\frac{6}{11}\right) + \left(-\frac{8}{21}\right) + \frac{5}{22}$$

Step-by-Step Model Solution:

Step 1: Regroup Terms using Commutative & Associative Properties 2 Marks

Notice that $7$ and $21$ share a common factor ($21 = 7 \times 3$), and $11$ and $22$ share a common factor ($22 = 11 \times 2$).
By the Commutative and Associative Properties of Addition, we can rearrange and group the terms:
$$\left[ \frac{3}{7} + \left(-\frac{8}{21}\right) \right] + \left[ \left(-\frac{6}{11}\right) + \frac{5}{22} \right]$$

Step 2: Simplify Each Bracket Independently 3 Marks

First group (LCM of $7$ and $21 = 21$):
$$\frac{3}{7} - \frac{8}{21} = \frac{3 \times 3 - 8}{21} = \frac{9 - 8}{21} = \frac{1}{21}$$
Second group (LCM of $11$ and $22 = 22$):
$$-\frac{6}{11} + \frac{5}{22} = \frac{-6 \times 2 + 5}{22} = \frac{-12 + 5}{22} = -\frac{7}{22}$$

Step 3: Combine Results via Global LCM 2 Marks

Now add the two simplified results:
$$\frac{1}{21} + \left(-\frac{7}{22}\right) = \frac{1}{21} - \frac{7}{22}$$
Since $21$ and $22$ are consecutive integers, they are co-prime: $\text{LCM}(21, 22) = 21 \times 22 = 462$.
$$= \frac{1 \times 22 - 7 \times 21}{462} = \frac{22 - 147}{462} = -\frac{125}{462}$$
Check for common factors: $125 = 5^3$. $462$ is not divisible by $5$ ($462 = 2 \times 3 \times 7 \times 11$).
Thus, $\gcd(125, 462) = 1$. The fraction is in standard form:
$$\mathbf{-\frac{125}{462}}$$

Examiner Scoring Key: Students taking the direct LCM of 7, 11, 21, 22 in one huge step without grouping receive only partial marks because the question specifically instructed 'using suitable properties'.

Model 5: Multi-Stage Descriptive Word Problem 10 Marks

A roll of electric cable of total length $33\frac{3}{5}\text{ m}$ is cut into pieces of equal length $1\frac{3}{5}\text{ m}$ each. How many complete pieces can be obtained? If the remaining piece is sold at the rate of ₹ $45\frac{1}{2}$ per metre, what is the value of the leftover cable?

Step-by-Step Model Solution:

Step 1: Declare Given Parameters & Convert to Improper Fractions 2 Marks

Total length of the electric cable $= 33\frac{3}{5}\text{ m} = \frac{33 \times 5 + 3}{5} = \frac{165 + 3}{5} = \frac{168}{5}\text{ m}$.
Length of each required piece $= 1\frac{3}{5}\text{ m} = \frac{1 \times 5 + 3}{5} = \frac{8}{5}\text{ m}$.

Step 2: Calculate Number of Complete Pieces via Division 3 Marks

Number of pieces $= \text{Total Length} \div \text{Length of One Piece}$
$$= \frac{168}{5} \div \frac{8}{5} = \frac{168}{5} \times \frac{5}{8}$$
Canceling common factor $5$:
$$= \frac{168}{8}$$
Dividing $168$ by $8$:
$168 = 8 \times 21$.
Therefore, exactly $21$ complete pieces can be obtained with $0\text{ m}$ remainder from this exact division.

Step 3: Analyze Remainder & Practical Extension 3 Marks

Since $168$ is exactly divisible by $8$ ($168 \div 8 = 21$), the remainder length is $0\text{ m}$.
$$\text{Remaining cable length} = 0\text{ m}$$
Rate per metre = ₹ $45\frac{1}{2} = ₹ \frac{91}{2}$.
Value of leftover cable $= 0 \times ₹ \frac{91}{2} = ₹ 0$.

Step 4: Formal Concluding Statement 2 Marks

Conclusion:
1. Total number of complete pieces of cable obtained = 21 pieces.
2. Value of leftover cable = ₹ 0 (Nil).

Examiner Scoring Key: Writing the formula 'Total Length / Length of each piece' gets 1 mark. Correct reciprocal multiplication gets 2 marks.

Model 6: Algebraic Equation with Rational Numbers 10 Marks

The product of two rational numbers is $-\frac{28}{81}$. If one of the numbers is $\frac{14}{27}$, find the other number. Also, determine by what rational number this second number must be increased to equal $\frac{5}{9}$.

Step-by-Step Model Solution:

Step 1: Set Up Algebraic Equation for Unknown Number 2 Marks

Let the required second rational number be $x$.
Given:
First rational number $= \frac{14}{27}$
Product of the two numbers $= -\frac{28}{81}$
According to the question:
$$\frac{14}{27} \times x = -\frac{28}{81}$$

Step 2: Solve for $x$ using Multiplicative Inverse 3 Marks

Multiplying both sides by the reciprocal of $\frac{14}{27}$, which is $\frac{27}{14}$:
$$x = -\frac{28}{81} \div \frac{14}{27} = -\frac{28}{81} \times \frac{27}{14}$$
Simplify by canceling cross-terms:
$\frac{-28}{14} = -2$
$\frac{27}{81} = \frac{1}{3}$
$$x = (-2) \times \frac{1}{3} = -\frac{2}{3}$$
Thus, the other rational number is $-\frac{2}{3}$.

Step 3: Set Up Equation for Second Question Part 3 Marks

Let the number to be added to $x$ be $y$.
According to the given condition:
$$x + y = \frac{5}{9}$$
Substitute $x = -\frac{2}{3}$:
$$-\frac{2}{3} + y = \frac{5}{9}$$
Transpose $-\frac{2}{3}$ to the RHS:
$$y = \frac{5}{9} - \left(-\frac{2}{3}\right) = \frac{5}{9} + \frac{2}{3}$$

Step 4: Compute Final Value & Box Answer 2 Marks

LCM of $9$ and $3 = 9$.
$$y = \frac{5 + 2 \times 3}{9} = \frac{5 + 6}{9} = \frac{11}{9} = 1\frac{2}{9}$$
Final Statement:
$$\text{The second rational number is } \mathbf{-\frac{2}{3}} \quad \text{and it must be increased by } \mathbf{\frac{11}{9}} \text{ to equal } \frac{5}{9}.$$

Examiner Scoring Key: Award 5 marks for finding x = -2/3 with proper reciprocal multiplication steps, and 5 marks for finding y = 11/9.

RIMC Practice & Drill Exercises

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

Question 1 5 Marks

Express the rational number $-\frac{48}{60}$ in equivalent standard form with denominator $5$. Furthermore, state its additive inverse and its multiplicative inverse.

Model Descriptive Solution:

Step 1: Reduce to Denominator 5 2 Marks

Given fraction $= -\frac{48}{60}$. To convert denominator from $60$ to $5$, divide both numerator and denominator by $60 \div 5 = 12$:
$$\frac{-48 \div 12}{60 \div 12} = -\frac{4}{5}$$

Step 2: Additive & Multiplicative Inverses 3 Marks

Additive Inverse of $-\frac{4}{5}$ is $+\frac{4}{5}$ because $-\frac{4}{5} + \frac{4}{5} = 0$.
Multiplicative Inverse (Reciprocal) of $-\frac{4}{5}$ is $-\frac{5}{4}$ because $\left(-\frac{4}{5}\right) \times \left(-\frac{5}{4}\right) = 1$.

Final Answer: Standard form: $-\frac{4}{5}$; Additive Inverse: $\frac{4}{5}$; Multiplicative Inverse: $-\frac{5}{4}$

Question 2 5 Marks

Explain the geometric construction steps to represent the rational numbers $-\frac{5}{6}$ and $\frac{7}{4}$ on separate number lines.

Model Descriptive Solution:

Step 1: Number Line for -5/6 2.5 Marks

$-\frac{5}{6}$ is a negative proper fraction ($0 > -\frac{5}{6} > -1$).
1. Draw a straight line and mark origin $0$ and $-1$.
2. Divide the unit interval $[0, -1]$ into $6$ equal parts of width $\frac{1}{6}$.
3. Move $5$ segments left from $0$. Mark the 5th tick as point $P = -\frac{5}{6}$.

Step 2: Number Line for 7/4 2.5 Marks

$\frac{7}{4} = 1\frac{3}{4}$, which lies between $1$ and $2$.
1. Mark points $1$ and $2$ on the positive side of the number line.
2. Divide the interval $[1, 2]$ into $4$ equal segments of width $\frac{1}{4}$.
3. Count $3$ divisions to the right of $1$. Mark this point as $Q = 1\frac{3}{4} = \frac{7}{4}$.

Final Answer: Point P located at 5th tick left of 0 in [0, -1]; Point Q located at 3rd tick right of 1 in [1, 2].

Question 3 5 Marks

Find six rational numbers lying strictly between $-\frac{1}{3}$ and $\frac{1}{2}$ using the common denominator scaling method.

Model Descriptive Solution:

Step 1: Convert to Like Denominators 2 Marks

$\text{LCM}(3, 2) = 6$.
$$-\frac{1}{3} = -\frac{1 \times 2}{3 \times 2} = -\frac{2}{6}, \quad \frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}$$

Step 2: Scale Denominators by Multiplier 2 2 Marks

Integers between $-2$ and $3$ are: $-1, 0, 1, 2$ (only 4 numbers). We need 6 numbers.
Multiply both fractions by $2$:
$$-\frac{2 \times 2}{6 \times 2} = -\frac{4}{12}, \quad \frac{3 \times 2}{6 \times 2} = \frac{6}{12}$$
Integers between $-4$ and $6$ are: $-3, -2, -1, 0, 1, 2, 3, 4, 5$ (9 numbers available).

Step 3: Select 6 Rational Numbers 1 Mark

Pick six numbers: $-\frac{3}{12}, -\frac{2}{12}, -\frac{1}{12}, \frac{1}{12}, \frac{2}{12}, \frac{3}{12}$.
Simplifying: $-\frac{1}{4}, -\frac{1}{6}, -\frac{1}{12}, \frac{1}{12}, \frac{1}{6}, \frac{1}{4}$.

Final Answer: $-\frac{1}{4}, -\frac{1}{6}, -\frac{1}{12}, \frac{1}{12}, \frac{1}{6}, \frac{1}{4}$

Question 4 7 Marks

Using the Distributive Property of Multiplication over Addition, simplify the following expression in stepwise detail:
$$\left(-\frac{3}{5}\right) \times \frac{5}{12} + \left(-\frac{3}{5}\right) \times \frac{7}{12}$$

Model Descriptive Solution:

Step 1: State Distributive Property 2 Marks

Notice the common factor $\left(-\frac{3}{5}\right)$ in both terms.
By the Distributive Property: $a \times b + a \times c = a \times (b + c)$, where $a = -\frac{3}{5}$, $b = \frac{5}{12}$, and $c = \frac{7}{12}$.

Step 2: Factor Out Common Rational Number 3 Marks

$$\left(-\frac{3}{5}\right) \times \left[ \frac{5}{12} + \frac{7}{12} \right] = \left(-\frac{3}{5}\right) \times \left[ \frac{5 + 7}{12} \right] = \left(-\frac{3}{5}\right) \times \frac{12}{12}$$

Step 3: Compute Final Result 2 Marks

Since $\frac{12}{12} = 1$:
$$\left(-\frac{3}{5}\right) \times 1 = -\frac{3}{5}$$

Final Answer: $-\frac{3}{5}$

Question 5 7 Marks

The sum of two rational numbers is $-\frac{1}{2}$. If one of the numbers is $\frac{5}{6}$, find the other number. Also, compute the product of these two rational numbers.

Model Descriptive Solution:

Step 1: Set Up Addition Equation 2 Marks

Let the required rational number be $x$.
Given: $x + \frac{5}{6} = -\frac{1}{2}$.

Step 2: Solve for x 3 Marks

$$x = -\frac{1}{2} - \frac{5}{6}$$
$\text{LCM}(2, 6) = 6$.
$$x = \frac{-1 \times 3 - 5}{6} = \frac{-3 - 5}{6} = -\frac{8}{6} = -\frac{4}{3}$$

Step 3: Compute Their Product 2 Marks

$$\text{Product} = x \times \frac{5}{6} = \left(-\frac{4}{3}\right) \times \frac{5}{6} = \frac{-4 \times 5}{3 \times 6} = -\frac{20}{18} = -\frac{10}{9}$$

Final Answer: Other rational number = $-\frac{4}{3}$; Product = $-\frac{10}{9}$

Question 6 7 Marks

By what rational number should $-\frac{33}{8}$ be multiplied to obtain $-\frac{11}{2}$? Show the full algebraic solution with cancellation steps.

Model Descriptive Solution:

Step 1: Formulate Algebraic Equation 2 Marks

Let the unknown multiplier be $y$.
According to the question:
$$\left(-\frac{33}{8}\right) \times y = -\frac{11}{2}$$

Step 2: Divide by First Rational Number 3 Marks

$$y = \left(-\frac{11}{2}\right) \div \left(-\frac{33}{8}\right) = \left(-\frac{11}{2}\right) \times \left(-\frac{8}{33}\right)$$
Multiplying two negative fractions gives a positive result.

Step 3: Simplify by Cross Cancellation 2 Marks

$$\frac{11}{33} = \frac{1}{3}, \quad \frac{8}{2} = 4$$
$$y = \frac{1 \times 4}{1 \times 3} = \frac{4}{3} = 1\frac{1}{3}$$

Final Answer: $\frac{4}{3}$ or $1\frac{1}{3}$

Question 7 7 Marks

A rectangular sports ground has length $15\frac{2}{3}\text{ m}$ and perimeter $50\frac{2}{9}\text{ m}$. Find the breadth of the ground as a rational number in lowest terms, and subsequently calculate its area.

Model Descriptive Solution:

Step 1: Parameter Declaration & Conversion 2 Marks

Length $L = 15\frac{2}{3}\text{ m} = \frac{47}{3}\text{ m}$.
Perimeter $P = 50\frac{2}{9}\text{ m} = \frac{452}{9}\text{ m}$.
Formula: $P = 2(L + B) \implies L + B = \frac{P}{2}$.

Step 2: Solve for Breadth B 3 Marks

$$L + B = \frac{452}{9} \div 2 = \frac{452}{18} = \frac{226}{9}\text{ m}$$
$$B = \frac{226}{9} - L = \frac{226}{9} - \frac{47}{3} = \frac{226 - 47 \times 3}{9} = \frac{226 - 141}{9} = \frac{85}{9}\text{ m}$$
$B = 9\frac{4}{9}\text{ m}$.

Step 3: Calculate Area of Ground 2 Marks

$$\text{Area} = L \times B = \frac{47}{3} \times \frac{85}{9} = \frac{47 \times 85}{27}$$
Rough work: $47 \times 85 = 3995$.
$$\text{Area} = \frac{3995}{27}\text{ m}^2 = 147\frac{26}{27}\text{ m}^2$$

Final Answer: Breadth = $\frac{85}{9}\text{ m}$ ($9\frac{4}{9}\text{ m}$); Area = $\frac{3995}{27}\text{ m}^2$ ($147\frac{26}{27}\text{ m}^2$)

Question 8 10 Marks

If $\frac{3}{5}$ of a tin of mustard oil weighs $16\frac{1}{2}\text{ kg}$, find the total weight of the completely filled tin. Further, calculate the weight of $\frac{4}{11}$ part of such a tin.

Model Descriptive Solution:

Step 1: Let Total Weight be W & Form Equation 2 Marks

Let the total weight of the full tin be $W\text{ kg}$.
Given: $\frac{3}{5} \times W = 16\frac{1}{2}\text{ kg} = \frac{33}{2}\text{ kg}$.

Step 2: Solve for Full Tin Weight W 3 Marks

$$W = \frac{33}{2} \div \frac{3}{5} = \frac{33}{2} \times \frac{5}{3}$$
Cancel $33 \div 3 = 11$:
$$W = \frac{11 \times 5}{2} = \frac{55}{2}\text{ kg} = 27\frac{1}{2}\text{ kg}$$

Step 3: Compute Weight of 4/11 Part 3 Marks

$$\text{Weight of } \frac{4}{11} \text{ part} = \frac{4}{11} \times W = \frac{4}{11} \times \frac{55}{2}$$
Simplify cross-terms: $\frac{4}{2} = 2$ and $\frac{55}{11} = 5$:
$$= 2 \times 5 = 10\text{ kg}$$

Step 4: Formal Concluding Statement 2 Marks

Full weight of tin = $27\frac{1}{2}\text{ kg}$ ($55/2\text{ kg}$).
Weight of $\frac{4}{11}$ part = $10\text{ kg}$.

Final Answer: Full tin = $27\frac{1}{2}\text{ kg}$; $\frac{4}{11}$ part = $10\text{ kg}$

Question 9 10 Marks

Verify the Associative Property of Multiplication: $(x \times y) \times z = x \times (y \times z)$ for the rational numbers $x = -\frac{7}{3}$, $y = \frac{12}{5}$, and $z = -\frac{4}{9}$.

Model Descriptive Solution:

Step 1: Compute LHS = (x * y) * z 4 Marks

$$x \times y = \left(-\frac{7}{3}\right) \times \frac{12}{5} = \frac{-7 \times 12}{3 \times 5} = \frac{-84}{15} = -\frac{28}{5}$$
Now multiply by $z$:
$$\text{LHS} = \left(-\frac{28}{5}\right) \times \left(-\frac{4}{9}\right) = \frac{(-28) \times (-4)}{5 \times 9} = \frac{112}{45}$$

Step 2: Compute RHS = x * (y * z) 4 Marks

$$y \times z = \frac{12}{5} \times \left(-\frac{4}{9}\right) = \frac{12 \times (-4)}{5 \times 9} = \frac{-48}{45} = -\frac{16}{15}$$
Now multiply by $x$:
$$\text{RHS} = \left(-\frac{7}{3}\right) \times \left(-\frac{16}{15}\right) = \frac{(-7) \times (-16)}{3 \times 15} = \frac{112}{45}$$

Step 3: Verification Statement 2 Marks

Since $\text{LHS} = \frac{112}{45}$ and $\text{RHS} = \frac{112}{45}$, we conclude:
$$\text{LHS} = \text{RHS} = \frac{112}{45} = 2\frac{22}{45} \quad \text{[Verified]}$$

Final Answer: $\text{LHS} = \text{RHS} = \frac{112}{45}$ [Verified]

Question 10 10 Marks

In a library, a cadet reads $\frac{1}{4}$ of an encyclopaedia on Monday, and $\frac{2}{5}$ of the remaining part on Tuesday. If there are still $180$ unread pages left in the book, calculate the total number of pages in the encyclopaedia, and determine how many pages the cadet read on Tuesday.

Model Descriptive Solution:

Step 1: Define Total Pages as X & Analyze Monday 2 Marks

Let total pages in encyclopaedia $= X$.
Pages read on Monday $= \frac{1}{4}X$.
Remaining pages after Monday $= X - \frac{1}{4}X = \frac{3}{4}X$.

Step 2: Analyze Tuesday Reading & Remaining Fraction 3 Marks

Pages read on Tuesday $= \frac{2}{5} \times \left(\text{Remaining}\right) = \frac{2}{5} \times \frac{3}{4}X = \frac{6}{20}X = \frac{3}{10}X$.
Total fraction read over Monday and Tuesday:
$$= \frac{1}{4}X + \frac{3}{10}X = \frac{5 + 6}{20}X = \frac{11}{20}X$$
Fraction remaining unread $= 1 - \frac{11}{20} = \frac{9}{20}X$.

Step 3: Solve for Total Pages X 3 Marks

Given unread pages $= 180$.
$$\frac{9}{20}X = 180 \implies X = 180 \times \frac{20}{9} = 20 \times 20 = 400$$
Total number of pages $= 400$.

Step 4: Compute Tuesday Pages & Conclude 2 Marks

Pages read on Tuesday $= \frac{3}{10}X = \frac{3}{10} \times 400 = 120\text{ pages}$.
Check: Monday $= 100$, Tuesday $= 120$, Left $= 180$. $100 + 120 + 180 = 400$ (Matches perfectly!).

Final Answer: Total pages = 400 pages; Pages read on Tuesday = 120 pages

High-Yield FAQs & Revision Notes

Is zero a rational number in RIMC syllabus?

Yes. Zero can be expressed as $\frac{0}{q}$ where $q$ is any non-zero integer (e.g. $\frac{0}{1}, \frac{0}{5}$). Here $p = 0 \in \mathbb{Z}$ and $q = 1 \in \mathbb{Z}$ with $q \neq 0$.

Why is division by zero undefined in mathematics?

If $\frac{a}{0} = k$, then $a = k \times 0 = 0$. For any $a \neq 0$, no number multiplied by $0$ can equal $a$. If $a = 0$, $k$ could be any number (indeterminate). Thus division by zero has no consistent mathematical meaning.

What is the density property of rational numbers?

The density property states that between any two distinct rational numbers, there are infinitely many rational numbers. Integers do not possess this property.

What is the difference between additive inverse and reciprocal?

Additive inverse changes the sign so their sum is $0$ ($x + (-x) = 0$). Reciprocal (multiplicative inverse) flips the numerator and denominator without changing sign so their product is $1$ ($x \times \frac{1}{x} = 1$).

Which rational numbers are equal to their own reciprocals?

Exactly $1$ and $-1$. For $x = 1$, $\frac{1}{1} = 1$. For $x = -1$, $\frac{1}{-1} = -1$.

Does zero have a reciprocal?

No. The reciprocal of $0$ would be $\frac{1}{0}$, which is undefined. Zero is the only rational number with no reciprocal.

Are rational numbers closed under division?

No. The set $\mathbb{Q}$ is not closed under division because division by zero is undefined. However, the set of non-zero rational numbers $\mathbb{Q} \setminus \{0\}$ is closed under division.

What is the standard irreducible form of a rational number?

A rational number $\frac{p}{q}$ is in standard form when $q > 0$ and $\gcd(|p|, q) = 1$. The negative sign must always be on the numerator or in front of the fraction.

When inserting rational numbers, why is multiplying by (n + 1) recommended?

Multiplying like fractions by $(n+1)$ guarantees that the numerators differ by at least $(n+1)$, creating exactly $n$ consecutive integer slots between them with zero guesswork.

How does this module map to Chapter 8 of the official RIMC brochure?

It directly corresponds to Serial No. 42 ('Rational') of Chapter 8 broader syllabus, comprehensively addressing arithmetic, algebraic verifications, and descriptive problem solving.

How does this topic relate to the official RIMC syllabus?

This module directly covers Serial No. 42 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
Set Definition	$\mathbb{Q} = \{p/q \mid p, q \in \mathbb{Z},\; q \neq 0\}$. Zero is rational ($0 = 0/1$).
Standard Form	Ensure denominator $q > 0$ and $\gcd(\|p\|, q) = 1$. Multiply by $-1$ if denominator is negative.
Axiomatic Operations	Closed under $+$, $-$, $\times$; NOT closed under $\div$. Distributive: $a(b \pm c) = ab \pm ac$.
Inverses	Additive inverse $= -a/b$ (sum $= 0$). Multiplicative inverse $= b/a$ (product $= 1$, $a \neq 0$).
Density Property	Between $a$ and $b$, infinitely many rationals exist. Insert using $(n+1)$ scaling on like denominators.
Booklet Presentation	Cite properties by name, show step-wise cancellations, reduce to lowest terms, and box final answers.