Official Syllabus Item: Part of 46 Algebra & Equations

Linear Equations in One Variable & Applied Word Problems

Master linear equations in one variable for RIMC: transposing terms, cross-multiplication, clearing denominators via LCM, and word problems on ages, two-digit numbers, coins, and perimeter.

Overview & Fundamental Concept

Linear Equations in One Variable constitute the primary engine of applied mathematical problem-solving in the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Item 46 of Chapter 8 in the official RIMC brochure, questions in this topic evaluate a candidate's competence in translating real-world constraints into formal linear equations, transposing terms with strict sign reversals, applying the cross-multiplication technique to rational algebraic fractions, and solving classic multi-stage word problems involving digit reversals, age relations, currency denominations, geometric perimeters, and upstream/downstream stream speeds. Full credit in the descriptive exam booklet requires candidates to declare the unknown variable with physical units ('Let the present age of Rahul be $x$ years'), systematically write the governing equation with explicit justification, perform stepwise algebraic reductions, and conclude with verified answers boxed with proper units.

Official Syllabus Coverage Breakdown

Syllabus Item	Prescribed Scope	Exam Relevance & Pedagogical Objective
Part of Item 46	Solution of a simple linear equation of one variable and its application	Official brochure syllabus item. Tests translation of complex descriptive word scenarios into algebraic models, algebraic balancing, and solution verification.

1. Definition, Standard Form & Axiomatic Principles of Equality

An equation is a statement of equality involving one or more variables. An equation is called a Linear Equation in One Variable if it involves only one variable and the highest exponent (degree) of that variable is exactly $1$.

The standard form of a linear equation in one variable $x$ is:

$$ax + b = c \quad \text{or} \quad ax + b = 0 \quad (a, b, c \in \mathbb{R},\; a \neq 0)$$

Any value of the variable that makes the Left Hand Side (LHS) strictly equal to the Right Hand Side (RHS) is called the Root (or Solution) of the equation. A linear equation in one variable has exactly one unique solution:

$$ax + b = 0 \implies ax = -b \implies x = -\frac{b}{a}$$

The Four Axiomatic Properties of Equality & Transposition

Addition Property: The same number can be added to both sides without changing equality: If $A = B$, then $A + c = B + c$.
Subtraction Property: The same number can be subtracted from both sides: If $A = B$, then $A - c = B - c$.
Multiplication Property: Both sides can be multiplied by the same non-zero number: If $A = B$, then $A \times c = B \times c$ ($c \neq 0$).
Division Property: Both sides can be divided by the same non-zero number: If $A = B$, then $\frac{A}{c} = \frac{B}{c}$ ($c \neq 0$).
The Transposition Rule: Moving a term from one side of the equation to the other is equivalent to applying addition or subtraction. The sign of the transposed term must be reversed: $+k$ becomes $-k$, and $-k$ becomes $+k$.

2. Solving Equations Reducible to Linear Form & Cross-Multiplication

Frequently, RIMC exam questions present equations with fractional expressions on both sides:

$$\frac{ax + b}{cx + d} = \frac{p}{q} \quad (cx + d \neq 0,\; q \neq 0)$$

This is solved directly using the Cross-Multiplication Principle:

$$q(ax + b) = p(cx + d)$$

The Global LCM Multiplication Technique for Complex Fractions

When an equation contains multiple fractional terms with numerical denominators, such as $\frac{3x-2}{4} - \frac{2x+3}{3} = \frac{2}{3} - x$:

Step 1: Find the global LCM of all denominators across both LHS and RHS (here, $\text{LCM}(4, 3) = 12$).
Step 2: Multiply every single term on both sides by this LCM. This eliminates all denominators in a single step with zero fraction arithmetic.
Step 3: Expand all brackets with careful attention to negative signs.
Step 4: Collect all variable terms on the LHS and all constant numbers on the RHS to isolate $x$.

3. Master Taxonomy of RIMC Word Problem Models

In RIMC Stage-I, word problems carry 5 to 10 marks each. Examiners test specific structural mathematical paradigms:

Problem Category	Mathematical Structure / Variable Definition	Governing Equation Framework
Consecutive Numbers	Integers: $x, x+1, x+2$ Even/Odd: $x, x+2, x+4$ Multiples of $k$: $kx, k(x+1)$	$\sum \text{Terms} = \text{Given Total}$
Two-Digit Numbers	Units digit $= x$, Tens digit $= y$ Original Number $= 10y + x$ Reversed Number $= 10x + y$	Reversed Number $-$ Original Number $= 9(x - y)$ Sum of Numbers $= 11(x + y)$
Age Relationships	Let present ages be $ax$ and $bx$. $n$ years ago: $(ax - n)$ and $(bx - n)$ $m$ years hence: $(ax + m)$ and $(bx + m)$	$\frac{ax + m}{bx + m} = \text{New Given Ratio}$
Fraction Alterations	Numerator $= x$, Denominator $= x + k$ Fraction $= \frac{x}{x + k}$	$\frac{x \pm a}{(x + k) \pm b} = \frac{p}{q}$ (cross-multiply)
Currency / Denominations	Total notes $= N$. Ratio of denominations. Value $= \text{Count} \times \text{Denomination}$	$\sum (\text{Count}_i \times \text{Value}_i) = \text{Total Cash Value}$
Upstream & Downstream	Boat speed in still water $= x\text{ km/h}$ Stream speed $= s\text{ km/h}$ Downstream $= x + s$, Upstream $= x - s$	$\text{Distance} = \text{Speed} \times \text{Time}$ $(x + s)t_1 = (x - s)t_2$

4. The 4-Step Booklet Presentation Framework

RIMC examiners deduct up to 50% of available marks if a candidate produces a numerical answer without formal algebraic declarations. Adhere to this exact 4-tier layout in your answer script:

1. Variable Declaration: State clearly what the variable $x$ represents along with its units.
2. Equation Formulation: Write the governing equation under the heading 'According to the Question (ATQ):'.
3. Systematic Solution: Show all transposition steps, LCM multiplication, and bracket expansions line by line.
4. Concluding Verification Statement: State the final answer in words, include units, verify by back-substitution, and draw a neat box around the result.

Illustrated Conceptual Diagrams & Visual Schematics

Two-Pan Balance Scale Model & Transposition Rules

Balance Concept of Equations: An equation is like a balanced two-pan physical balance. Whatever operation is performed on the LHS must be simultaneously performed on the RHS: $+c$ transposes as $-c$; $\times c$ transposes as $\div c$. Cross-multiplication rule: $\frac{ax+b}{cx+d} = \frac{p}{q} \implies q(ax+b) = p(cx+d)$.

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: Variable Declaration & Given State 1 - 2 Marks

Define the variable clearly with units (e.g., 'Let the present age of Rahul be x years'). Write down all given relationships.

Booklet Layout:
Let the present ages of Rahul and Haroon be $5x$ years and $7x$ years respectively.\nAfter 4 years:\nRahul's age $= (5x + 4)$ years\nHaroon's age $= (7x + 4)$ years.

Step 2: Algebraic Equation Formulation (ATQ) 2 - 3 Marks

Translate the verbal problem into a mathematically precise linear equation.

Booklet Layout:
According to the Question (ATQ):\n$$(5x + 4) + (7x + 4) = 56$$

Step 3: Transposition & Stepwise Arithmetic Reduction 2 - 4 Marks

Combine like terms, transpose constants with inverted signs, and isolate the variable.

Booklet Layout:
$$12x + 8 = 56 \implies 12x = 56 - 8 = 48 \implies x = \frac{48}{12} = 4$$

Step 4: Final Value Computation with Units & Answer Box 1 - 2 Marks

Calculate the actual quantities requested in the question (not just x), verify, and box the answer.

Booklet Layout:
$$\text{Rahul's Present Age} = 5 \times 4 = \mathbf{20\text{ years}}$$
$$\text{Haroon's Present Age} = 7 \times 4 = \mathbf{28\text{ years}}$$

Examiner Traps & Common Blunders (Marks Penalties)

Stopping at the Variable x Instead of Answering the Question -2 to -3 Marks

Common Mistake: Finding $x = 4$ and writing 'Answer = 4' when the question asked for the present ages of Rahul and Haroon.

Cadet Antidote: Always re-read the final sentence of the question. If asked for dimensions, ages, or original numbers, compute the actual expressions ($5x = 20, 7x = 28$).

Representing Two-Digit Numbers as Simple Products -3 to -4 Marks

Common Mistake: Writing a two-digit number with tens digit $y$ and units digit $x$ as $yx$ or $y \times x$.

Cadet Antidote: In decimal place-value notation, the tens digit represents groups of 10. The number MUST be written as $10y + x$.

Forgetting to Invert Signs when Transposing across Equality -2 Marks

Common Mistake: Transposing $+8$ across '=' as $+8$, or forgetting that $-(2x + 3) = -2x - 3$.

Cadet Antidote: Enclose expressions in brackets before distributing negative signs: $-(2x + 3) = -2x - 3$.

Cross-Multiplication with Multi-Term Expressions -2 Marks

Common Mistake: Writing $3(x - 5) = 3x - 5$ (forgetting to multiply the constant 5 by 3).

Cadet Antidote: Always distribute the factor across all terms inside the parentheses: $3(x - 5) = 3x - 15$.

Cadet Topper Mental Verification Shortcuts

The 9 and 11 Two-Digit Number Shortcuts

For any two-digit number $N = 10t + u$ and its reversed counterpart $N' = 10u + t$:\n1. Difference: $N - N' = 9(t - u)$ (always a multiple of 9).\n2. Sum: $N + N' = 11(t + u)$ (always a multiple of 11).

If difference between number and reversed is 36, then $9(t - u) = 36 \implies t - u = 4$ immediately!

Clearing All Denominators via Global LCM Multiplier

Instead of doing tedious fractional additions on LHS and RHS, multiply the entire equation by $\text{LCM}(\text{all denominators})$. Fractions vanish instantly.

For $\frac{x}{2} + \frac{x}{3} = 10$: multiply by $6 \implies 3x + 2x = 60 \implies 5x = 60 \implies x = 12$.

Backward Verification in 10 Seconds

Take your final answer and substitute directly into the verbal word problem conditions. If all constraints match, your solution is guaranteed 100% full marks.

Rahul is 20, Haroon is 28. Ratio: $20:28 = 5:7$. In 4 years: $24 + 32 = 56$. Matches the question perfectly!

Key Formulas & Conversion Cheat Sheet

Standard Linear Equation Form

$$ax + b = c \implies x = \frac{c - b}{a} \quad (a \neq 0)$$

Unique root of a linear equation in one variable.

Cross-Multiplication Formula

$$\frac{ax + b}{cx + d} = \frac{p}{q} \implies q(ax + b) = p(cx + d)$$

Converts rational algebraic equation into linear polynomial equation.

Two-Digit Number Place Value

$$N = 10t + u, \quad N_{\text{rev}} = 10u + t$$

Decimal expansion where t is tens digit and u is units digit.

Reversed Number Difference Identity

$$N - N_{\text{rev}} = 9(t - u)$$

Difference between any two-digit number and its reverse is 9 times the digit difference.

Reversed Number Sum Identity

$$N + N_{\text{rev}} = 11(t + u)$$

Sum of any two-digit number and its reverse is 11 times the digit sum.

Age Relation Translation

$$\text{Age } n \text{ years hence} = A + n; \quad \text{Age } m \text{ years ago} = A - m$$

Ages shift uniformly across time.

Upstream & Downstream Speeds

$$v_{\text{down}} = u + v_s, \quad v_{\text{up}} = u - v_s$$

Relative speed of vessel in moving water where u is still water speed and v_s is stream speed.

Denomination Cash Value Formula

$$\text{Total Value} = \sum_{i} n_i \times d_i$$

Sum of note count times note denomination value.

Solved Model Examples (Descriptive Format with Step Mark Distribution)

Model 1: Cross-Multiplication of Rational Equation 5 Marks

Solve the following linear equation for $x$ and verify the solution:
$$\frac{x - 5}{3} = \frac{x - 3}{5}$$

Step-by-Step Model Solution:

Step 1: Apply Cross-Multiplication 2 Marks

The given equation is $\frac{x - 5}{3} = \frac{x - 3}{5}$.
Cross-multiplying numerators and denominators:
$$5(x - 5) = 3(x - 3)$$

Step 2: Expand Parentheses and Transpose Terms 2 Marks

Distribute the constants across terms:
$$5x - 25 = 3x - 9$$
Transpose variable term $3x$ to LHS and constant $-25$ to RHS:
$$5x - 3x = -9 + 25$$
$$2x = 16 \implies x = \frac{16}{2} = 8$$

Step 3: Verification 1 Mark

Substitute $x = 8$ in LHS and RHS:
$$\text{LHS} = \frac{8 - 5}{3} = \frac{3}{3} = 1$$
$$\text{RHS} = \frac{8 - 3}{5} = \frac{5}{5} = 1$$
Since $\text{LHS} = \text{RHS} = 1$, the solution is verified.
$$\mathbf{\text{Answer: } x = 8}$$

Examiner Scoring Key: Cross-multiplication carries 2 marks, transposition carries 2 marks, and verification carries 1 mark.

Model 2: Complex Equation with Global LCM Multiplier 5 Marks

Solve the linear equation for $t$:
$$\frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t$$

Step-by-Step Model Solution:

Step 1: Multiply Entire Equation by Global LCM 2 Marks

The denominators in the equation are $4$, $3$, and $3$.
$\text{LCM}(4, 3) = 12$.
Multiply every single term on both sides by $12$:
$$12 \times \left(\frac{3t - 2}{4}\right) - 12 \times \left(\frac{2t + 3}{3}\right) = 12 \times \left(\frac{2}{3}\right) - 12 \times t$$
$$3(3t - 2) - 4(2t + 3) = 4(2) - 12t$$

Step 2: Expand Brackets with Careful Sign Handling 2 Marks

$$(9t - 6) - (8t + 12) = 8 - 12t$$
$$9t - 6 - 8t - 12 = 8 - 12t$$
Combine like terms on LHS:
$$(9t - 8t) + (-6 - 12) = 8 - 12t$$
$$t - 18 = 8 - 12t$$

Step 3: Transpose and Solve for t 1 Mark

Transpose $-12t$ to LHS and $-18$ to RHS:
$$t + 12t = 8 + 18$$
$$13t = 26 \implies t = \frac{26}{13} = 2$$
$$\mathbf{\text{Answer: } t = 2}$$

Examiner Scoring Key: Distributing -4 across (2t + 3) to get -8t - 12 is a common error spot. Watch for sign slips.

Model 3: Two-Digit Number & Digit Inversion Problem 7 Marks

The sum of the digits of a two-digit number is $9$. When the digits are interchanged, the resulting new number exceeds the original number by $27$. Determine the original number.

Step-by-Step Model Solution:

Step 1: Variable Definition & Number Expressions 2 Marks

Let the units digit of the number be $x$.
Since the sum of the digits is $9$, the tens digit is $(9 - x)$.
Original Number $= 10 \times (\text{Tens Digit}) + (\text{Units Digit})$
$$= 10(9 - x) + x = 90 - 10x + x = 90 - 9x$$
When digits are interchanged:
New Units digit $= (9 - x)$, New Tens digit $= x$.
New Number $= 10x + (9 - x) = 9x + 9$.

Step 2: Formulate Equation (ATQ) 2 Marks

According to the question, $\text{New Number} - \text{Original Number} = 27$:
$$(9x + 9) - (90 - 9x) = 27$$
$$9x + 9 - 90 + 9x = 27$$
$$18x - 81 = 27$$

Step 3: Solve for x and Find the Digits 2 Marks

Transpose $-81$ to RHS:
$$18x = 27 + 81 = 108$$
$$x = \frac{108}{18} = 6$$
Units digit $x = 6$.
Tens digit $= 9 - x = 9 - 6 = 3$.

Step 4: State Original Number & Verify 1 Mark

$$\text{Original Number} = 10(3) + 6 = \mathbf{36}$$
Check: Sum of digits $= 3 + 6 = 9$. Reversed number $= 63$. Difference $= 63 - 36 = 27$ (Verified!).

Examiner Scoring Key: Setting up 10(9 - x) + x correctly gets 2 marks. Solving for x = 6 gets 2 marks. Concluding with 36 gets 1 mark.

Model 4: Age Word Problem with Ratio Evolution 7 Marks

A father is currently three times as old as his son. Ten years ago, the father was five times as old as his son was then. Find their present ages.

Step-by-Step Model Solution:

Step 1: Define Variables for Present Ages 2 Marks

Let the present age of the son be $x$ years.
Then the present age of the father $= 3x$ years.

Step 2: Express Ages Ten Years Ago & Formulate Equation 2 Marks

Ten years ago:
Son's age $= (x - 10)$ years
Father's age $= (3x - 10)$ years
According to the Question (ATQ):
$$\text{Father's age} = 5 \times (\text{Son's age})$$
$$3x - 10 = 5(x - 10)$$

Step 3: Solve Algebraic Equation 2 Marks

Expand RHS:
$$3x - 10 = 5x - 50$$
Transpose variable terms and constants:
$$50 - 10 = 5x - 3x$$
$$40 = 2x \implies x = \frac{40}{2} = 20$$

Step 4: Compute Present Ages & Conclude 1 Mark

Son's present age $x = \mathbf{20\text{ years}}$.
Father's present age $= 3x = 3 \times 20 = \mathbf{60\text{ years}}$.
Check: 10 years ago, son was 10 and father was 50. $50 = 5 \times 10$ (Verified!).

Examiner Scoring Key: Do not stop at x = 20. Stating both ages (Son = 20 years, Father = 60 years) is essential for full 7 marks.

Model 5: Multi-Denomination Currency Problem 10 Marks

Deveshi has a total cash sum of ₹ $590$ in currency notes of denominations ₹ $50$, ₹ $20$, and ₹ $10$. The ratio of the number of ₹ $50$ notes to ₹ $20$ notes is $3:5$. If she possesses a total of $25$ notes in all, calculate how many notes of each denomination she has.

Step-by-Step Model Solution:

Step 1: Declare Variable for Ratio & Express Note Counts 3 Marks

Given ratio of ₹ $50$ notes to ₹ $20$ notes is $3:5$.
Let the number of ₹ $50$ notes be $3x$.
Let the number of ₹ $20$ notes be $5x$.
Total number of notes $= 25$.
Therefore, the number of ₹ $10$ notes $= 25 - (3x + 5x) = 25 - 8x$.

Step 2: Formulate Total Money Value Equation (ATQ) 3 Marks

Calculate the total monetary value generated by each denomination:
Value of ₹ $50$ notes $= 3x \times 50 = ₹ 150x$
Value of ₹ $20$ notes $= 5x \times 20 = ₹ 100x$
Value of ₹ $10$ notes $= (25 - 8x) \times 10 = ₹ (250 - 80x)$
According to the Question (Total Cash $= ₹ 590$):
$$150x + 100x + (250 - 80x) = 590$$

Step 3: Solve for x 2 Marks

Combine like terms:
$$(150 + 100 - 80)x + 250 = 590$$
$$170x + 250 = 590$$
Transpose $250$ to RHS:
$$170x = 590 - 250 = 340$$
$$x = \frac{340}{170} = 2$$

Step 4: Calculate Count of Each Denomination & Verify 2 Marks

Number of ₹ $50$ notes $= 3x = 3 \times 2 = \mathbf{6\text{ notes}}$.
Number of ₹ $20$ notes $= 5x = 5 \times 2 = \mathbf{10\text{ notes}}$.
Number of ₹ $10$ notes $= 25 - 8(2) = 25 - 16 = \mathbf{9\text{ notes}}$.
Check: Total notes $= 6 + 10 + 9 = 25$.
Total value $= 6(50) + 10(20) + 9(10) = 300 + 200 + 90 = ₹ 590$ (Matches perfectly!).

Examiner Scoring Key: Setting up the count of ₹ 10 notes as (25 - 8x) carries 3 marks. Stating all three note counts correctly carries 2 marks.

Model 6: River Navigation Stream Speed Problem 10 Marks

A motorboat covers the distance between two river ports downstream in $20\text{ hours}$. Returning upstream, it covers the exact same distance in $25\text{ hours}$. If the stream flows at the rate of $4\text{ km/h}$, determine the speed of the motorboat in still water and the distance between the two ports.

Step-by-Step Model Solution:

Step 1: Declare Variable & Relative Speeds 3 Marks

Let the speed of the motorboat in still water be $x\text{ km/h}$.
Speed of stream $v_s = 4\text{ km/h}$.
Speed downstream (boat + stream) $= (x + 4)\text{ km/h}$.
Speed upstream (boat - stream) $= (x - 4)\text{ km/h}$.

Step 2: Formulate Distance Equality Equation (ATQ) 3 Marks

We know that $\text{Distance} = \text{Speed} \times \text{Time}$.
Distance covered downstream in $20\text{ hours}$:
$$D = (x + 4) \times 20$$
Distance covered upstream in $25\text{ hours}$:
$$D = (x - 4) \times 25$$
Since the distance between the two ports remains constant:
$$20(x + 4) = 25(x - 4)$$

Step 3: Solve for Boat Speed x 2 Marks

Divide both sides by $5$ to simplify:
$$4(x + 4) = 5(x - 4)$$
$$4x + 16 = 5x - 20$$
Transpose terms:
$$16 + 20 = 5x - 4x \implies x = 36\text{ km/h}$$

Step 4: Calculate River Distance & Conclude 2 Marks

Substitute $x = 36$ to find distance $D$:
$$D = 20(36 + 4) = 20 \times 40 = \mathbf{800\text{ km}}$$
Check upstream: $25(36 - 4) = 25 \times 32 = 800\text{ km}$ (Verified!).
Conclusion:
1. Speed of motorboat in still water = $\mathbf{36\text{ km/h}}$.
2. Distance between the two river ports = $\mathbf{800\text{ km}}$.

Examiner Scoring Key: Award 3 marks for relative speeds (x+4) and (x-4), 3 marks for equating distances, 2 marks for x = 36 km/h, and 2 marks for D = 800 km.

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

Solve the linear rational equation for $y$ and check your answer:
$$\frac{7y + 4}{y + 2} = -\frac{4}{3}$$

Model Descriptive Solution:

Step 1: Cross-Multiply 2 Marks

$$3(7y + 4) = -4(y + 2)$$

Step 2: Expand and Solve for y 2 Marks

$$21y + 12 = -4y - 8$$
$$21y + 4y = -8 - 12 \implies 25y = -20 \implies y = -\frac{20}{25} = -\frac{4}{5}$$

Step 3: Verification 1 Mark

LHS: $\frac{7(-4/5) + 4}{-4/5 + 2} = \frac{-28/5 + 20/5}{6/5} = \frac{-8/5}{6/5} = -\frac{8}{6} = -\frac{4}{3} = \text{RHS}$.

Final Answer: $y = -\frac{4}{5}$

Question 2 5 Marks

The perimeter of a rectangular picture frame is $13\text{ cm}$ and its width is $2\frac{3}{4}\text{ cm}$. Find its length in lowest fractional terms.

Model Descriptive Solution:

Step 1: Declare Variable & State Perimeter Formula 2 Marks

Let length be $L\text{ cm}$. Width $W = 2\frac{3}{4}\text{ cm} = \frac{11}{4}\text{ cm}$.
Perimeter $P = 2(L + W) = 13\text{ cm}$.

Step 2: Solve for Length L 3 Marks

$$L + \frac{11}{4} = \frac{13}{2} \implies L = \frac{13}{2} - \frac{11}{4} = \frac{26 - 11}{4} = \frac{15}{4} = 3\frac{3}{4}\text{ cm}$$

Final Answer: Length $= \frac{15}{4}\text{ cm}$ ($3\frac{3}{4}\text{ cm}$)

Question 3 5 Marks

The sum of three consecutive integers is $51$. Formulate the linear equation and find the three integers.

Model Descriptive Solution:

Step 1: Define Consecutive Integers 2 Marks

Let the three consecutive integers be $x$, $x+1$, and $x+2$.

Step 2: Formulate & Solve Equation 3 Marks

$$x + (x + 1) + (x + 2) = 51 \implies 3x + 3 = 51$$
$$3x = 48 \implies x = 16$$
The three integers are $16$, $17$, and $18$.

Final Answer: $16, 17, 18$

Question 4 7 Marks

The numerator of a rational number is $8$ less than its denominator. If the numerator is increased by $17$ and the denominator is decreased by $1$, the number obtained is $\frac{3}{2}$. Find the original rational number.

Model Descriptive Solution:

Step 1: Define Variable & Fraction Structure 2 Marks

Let denominator be $d$. Numerator $= d - 8$.
Original fraction $= \frac{d - 8}{d}$.

Step 2: Formulate Equation (ATQ) 3 Marks

$$\frac{(d - 8) + 17}{d - 1} = \frac{3}{2} \implies \frac{d + 9}{d - 1} = \frac{3}{2}$$
Cross-multiply: $2(d + 9) = 3(d - 1) \implies 2d + 18 = 3d - 3$.

Step 3: Solve for d & State Fraction 2 Marks

$$3d - 2d = 18 + 3 \implies d = 21$$
Numerator $= 21 - 8 = 13$.
Original Fraction $= \frac{13}{21}$.

Final Answer: Original Fraction $= \frac{13}{21}$

Question 5 7 Marks

Solve the linear equation using the global LCM technique:
$$\frac{6x + 1}{3} + 1 = \frac{x - 3}{6}$$

Model Descriptive Solution:

Step 1: Multiply Entire Equation by LCM = 6 3 Marks

$$6 \times \left(\frac{6x + 1}{3}\right) + 6 \times 1 = 6 \times \left(\frac{x - 3}{6}\right)$$
$$2(6x + 1) + 6 = 1(x - 3)$$

Step 2: Expand and Solve for x 4 Marks

$$12x + 2 + 6 = x - 3 \implies 12x + 8 = x - 3$$
$$12x - x = -3 - 8 \implies 11x = -11 \implies x = -1$$

Final Answer: $x = -1$

Question 6 7 Marks

A positive number is $5$ times another number. If $21$ is added to both the numbers, one of the new numbers becomes twice the other new number. What are the two original numbers?

Model Descriptive Solution:

Step 1: Define Variables 2 Marks

Let smaller number be $x$. Larger number $= 5x$.

Step 2: Formulate Equation (ATQ) 3 Marks

Adding 21 to both: numbers become $(5x + 21)$ and $(x + 21)$.
$5x + 21 = 2(x + 21) \implies 5x + 21 = 2x + 42$.

Step 3: Solve for x & Find Numbers 2 Marks

$$5x - 2x = 42 - 21 \implies 3x = 21 \implies x = 7$$
Numbers are $7$ and $5(7) = 35$.

Final Answer: The numbers are $7$ and $35$.

Question 7 7 Marks

The present ages of Hari and Harry are in the ratio $5:7$. Four years from now, the ratio of their ages will be $3:4$. Find their present ages.

Model Descriptive Solution:

Step 1: Define Ratio Expressions 2 Marks

Let present ages of Hari and Harry be $5x$ and $7x$ years.
Four years hence: Hari $= 5x + 4$, Harry $= 7x + 4$.

Step 2: Cross-Multiply & Solve 3 Marks

$$\frac{5x + 4}{7x + 4} = \frac{3}{4} \implies 4(5x + 4) = 3(7x + 4)$$
$$20x + 16 = 21x + 12 \implies 21x - 20x = 16 - 12 \implies x = 4$$

Step 3: State Present Ages 2 Marks

Hari's present age $= 5(4) = 20\text{ years}$.
Harry's present age $= 7(4) = 28\text{ years}$.

Final Answer: Hari: $20\text{ years}$; Harry: $28\text{ years}$

Question 8 10 Marks

In a two-digit number, the digit at tens place is three times the digit in units place. If the digits are reversed, the new number is $36$ less than the original number. Find the original number.

Model Descriptive Solution:

Step 1: Define Digits & Expressions 3 Marks

Let units digit $= x$. Tens digit $= 3x$.
Original Number $= 10(3x) + x = 30x + x = 31x$.
Reversed Number $= 10x + 3x = 13x$.

Step 2: Formulate & Solve Equation 4 Marks

$\text{Original Number} - \text{Reversed Number} = 36$:
$$31x - 13x = 36 \implies 18x = 36 \implies x = 2$$

Step 3: State Original Number 3 Marks

Units digit $= 2$, Tens digit $= 3(2) = 6$.
Original number $= 62$.
Check: $62 - 26 = 36$ (Verified!).

Final Answer: Original number $= 62$

Question 9 10 Marks

A steamer goes downstream and covers the distance between two ports in $9\text{ hours}$, while it covers the same distance upstream in $10\text{ hours}$. If the speed of the stream is $1\text{ km/h}$, find the speed of the steamer in still water and the distance between the two ports.

Model Descriptive Solution:

Step 1: Declare Variable & Relative Speeds 3 Marks

Let still water speed be $u\text{ km/h}$. Stream speed $= 1\text{ km/h}$.
Downstream speed $= (u + 1)\text{ km/h}$. Upstream speed $= (u - 1)\text{ km/h}$.

Step 2: Equate Distances & Solve 4 Marks

$$9(u + 1) = 10(u - 1) \implies 9u + 9 = 10u - 10$$
$$10u - 9u = 9 + 10 \implies u = 19\text{ km/h}$$

Step 3: Calculate Distance 3 Marks

$$\text{Distance} = 9(19 + 1) = 9 \times 20 = 180\text{ km}$$
Check: $10(19 - 1) = 10 \times 18 = 180\text{ km}$.

Final Answer: Speed in still water $= 19\text{ km/h}$; Distance $= 180\text{ km}$

Question 10 10 Marks

Baichung's father is $26\text{ years}$ younger than Baichung's grandfather and $29\text{ years}$ older than Baichung. The sum of the ages of all three is $135\text{ years}$. What is the age of each one of them?

Model Descriptive Solution:

Step 1: Define Ages in terms of Baichung's Father 3 Marks

Let Baichung's father's age be $x$ years.
Baichung's age $= (x - 29)$ years.
Baichung's grandfather's age $= (x + 26)$ years.

Step 2: Formulate & Solve Sum Equation 4 Marks

$$(x - 29) + x + (x + 26) = 135$$
$$3x - 3 = 135 \implies 3x = 138 \implies x = \frac{138}{3} = 46$$

Step 3: Calculate All Three Ages 3 Marks

Father $= 46\text{ years}$.
Baichung $= 46 - 29 = 17\text{ years}$.
Grandfather $= 46 + 26 = 72\text{ years}$.
Check: $17 + 46 + 72 = 135\text{ years}$ (Verified!).

Final Answer: Baichung: $17\text{ years}$; Father: $46\text{ years}$; Grandfather: $72\text{ years}$

High-Yield FAQs & Revision Notes

What defines a linear equation in one variable?

An equation involving exactly one variable with exponent 1 (degree 1) that can be written in the form $ax + b = 0$ ($a \neq 0$).

Why does the sign change during transposition?

Transposing a term to the opposite side is mathematically equivalent to adding or subtracting that quantity from both sides to preserve equality.

How should two-digit numbers always be represented?

Always as $10t + u$, where $t$ is the tens digit and $u$ is the units digit. Never write it as a simple product $tu$.

What is the fastest way to clear fractions in linear equations?

Multiply every single term on both sides by the global LCM of all denominators.

When can cross-multiplication be applied?

When the entire Left Hand Side and entire Right Hand Side are single fractions: $\frac{A}{B} = \frac{C}{D} \implies A \cdot D = B \cdot C$.

How do age problems translate into equations?

If present age is $x$, age $n$ years ago was $(x - n)$ and age $m$ years hence will be $(x + m)$.

What is the difference between downstream and upstream speed?

Downstream speed is boat speed + stream speed ($u + v$). Upstream speed is boat speed - stream speed ($u - v$).

How many solutions does a linear equation in one variable have?

Exactly one unique solution: $x = -b/a$.

Why is verification necessary in RIMC exams?

Verifying that LHS = RHS guarantees that no calculation errors were made and earns full descriptive presentation marks.

How does this module map to the official RIMC syllabus?

Directly covers 'Solution of a simple linear equation of one variable and its application' under Item 46 of Chapter 8 in the official RIMC brochure.

How does this topic relate to the official RIMC syllabus?

This module directly covers Serial No. Part of 46 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
Standard Form	$ax + b = 0 \implies x = -b/a$. Always isolate variable on one side.
Transposition	Changing sides changes sign: $+ \leftrightarrow -$ and $\times \leftrightarrow \div$.
Cross-Multiplication	$\frac{A}{B} = \frac{C}{D} \implies AD = BC$. Always bracket multi-term expressions before distributing.
Two-Digit Number	Original $= 10t + u$; Reversed $= 10u + t$. Difference is $9(t - u)$.
Stream Speed	Downstream $= u + v$; Upstream $= u - v$. Distance $= \text{Speed} \times \text{Time}$.
Final Presentation	Always answer the actual question asked with proper physical units and boxed results.