Word Problems on Pair of Linear Equations for CBSE Class 10

Word problems represent the ultimate test of algebraic mastery in CBSE Class 10 Mathematics. Rather than handing you pre-formulated equations to solve mechanically, word problems require you to decode real-world scenarios, identify unknown quantities, establish algebraic constraints, and solve the resulting system of equations.

In CBSE board examinations, word problems appear consistently in Section C and Section D, carrying 3, 4, or 5 marks. Most students struggle not with the equation solving, but with the initial mathematical translation. This guide provides step-by-step structural templates for all five major categories of board exam word problems.

What You Will Learn

• Universal 5-step framework for converting word statements into linear equations
• Category 1: Age-related problems (past, present, and future relationships)
• Category 2: Fraction problems (modifications to numerator and denominator)
• Category 3: Two-digit number problems (place value expansion $10x + y$)
• Category 4: Speed, Distance, Time & Upstream/Downstream boat problems
• Category 5: Fixed and variable cost problems (taxi fares, library fees, hostel food)
• Board exam presentation templates and common algebraic traps

1. The Universal 5-Step Translation Framework

Whenever you encounter a word problem, follow this structured process:

1. Identify the Two Unknowns: Assign variable $x$ to the first unknown quantity and $y$ to the second unknown quantity. Explicitly declare their units (e.g., Let the speed of the boat be $x$ km/h).
2. Identify the Two Independent Conditions: Every solvable two-variable word problem provides exactly two distinct relational facts.
3. Formulate Equation 1: Translate Condition 1 into algebraic terms using $x$ and $y$.
4. Formulate Equation 2: Translate Condition 2 into algebraic terms using $x$ and $y$.
5. Solve and Verify: Solve using elimination or substitution, and verify that your solutions make physical sense (e.g., age or speed cannot be negative).

2. Category 1: Age-Related Problems

The Structural Template

• Let present age of person A be $x$ years and person B be $y$ years.
• Age $n$ years ago: $\text{A} = (x - n)$, $\text{B} = (y - n)$.
• Age $m$ years hence (in future): $\text{A} = (x + m)$, $\text{B} = (y + m)$.

Solved Example: Aftab and His Daughter (NCERT Classic)

Problem: Seven years ago, Aftab was seven times as old as his daughter was then. Three years from now, he will be three times as old as his daughter will be. Find their present ages.

Solution:

1. Let Aftab's present age be $x$ years and daughter's present age be $y$ years.
2. Condition 1 (Seven years ago):
   • Aftab's age $= x - 7$; Daughter's age $= y - 7$.
   • Relation: $x - 7 = 7(y - 7) \implies x - 7 = 7y - 49 \implies x - 7y = -42 \quad \text{--- (1)}$
3. Condition 2 (Three years hence):
   • Aftab's age $= x + 3$; Daughter's age $= y + 3$.
   • Relation: $x + 3 = 3(y + 3) \implies x + 3 = 3y + 9 \implies x - 3y = 6 \quad \text{--- (2)}$
4. Subtract Equation (1) from Equation (2): $(x - 3y) - (x - 7y) = 6 - (-42)$ $4y = 48 \implies y = 12$
5. Substitute $y = 12$ into (2): $x - 3(12) = 6 \implies x - 36 = 6 \implies x = 42$
6. Therefore, <u>Aftab's present age is $42$ years and his daughter's present age is $12$ years</u>.

3. Category 2: Two-Digit Number Problems

The Structural Template

In a two-digit number, digits have place values:

• Let the tens digit be $x$ and the units (ones) digit be $y$.
• Original Number: $10x + y$
• When digits are reversed, tens digit becomes $y$ and units digit becomes $x$: Reversed Number: $10y + x$

Important: <u>A two-digit number with digits $x$ and $y$ is NEVER written as $xy$ (which means multiplication $x \times y$). You must always use the place-value expansion: $10x + y$!</u>

Solved Example: Sum of Digits

Problem: The sum of the digits of a two-digit number is $9$. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution:

1. Let the tens digit be $x$ and units digit be $y$. Original number $= 10x + y$; Reversed number $= 10y + x$.
2. Condition 1 (Sum of digits is 9): $x + y = 9 \quad \text{--- (1)}$
3. Condition 2 (9 times original = 2 times reversed): $9(10x + y) = 2(10y + x)$ $90x + 9y = 20y + 2x$ $90x - 2x + 9y - 20y = 0 \implies 88x - 11y = 0$ Divide by 11: $8x - y = 0 \implies y = 8x \quad \text{--- (2)}$
4. Substitute $y = 8x$ into Equation (1): $x + 8x = 9 \implies 9x = 9 \implies x = 1$
5. Find $y$: $y = 8(1) = 8$
6. Original Number: $10x + y = 10(1) + 8 = 18$
7. Therefore, <u>the number is $18$</u>.

4. Category 3: Upstream and Downstream Boat Problems

The Structural Template

• Let speed of boat in still water be $x\text{ km/h}$.
• Let speed of water stream/current be $y\text{ km/h}$ (with $x > y$).
• Downstream Speed (with current): $(x + y)\text{ km/h}$
• Upstream Speed (against current): $(x - y)\text{ km/h}$
• Formula: $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$

Solved Example: Boat Travel

Problem: A boat goes $30\text{ km}$ upstream and $44\text{ km}$ downstream in $10\text{ hours}$. In $13\text{ hours}$, it can go $40\text{ km}$ upstream and $55\text{ km}$ downstream. Determine the speed of the stream and that of the boat in still water.

Solution:

1. Let boat speed in still water be $x\text{ km/h}$ and stream speed be $y\text{ km/h}$. Upstream speed $= x - y$; Downstream speed $= x + y$.
2. Condition 1: $\frac{30}{x - y} + \frac{44}{x + y} = 10 \quad \text{--- (1)}$
3. Condition 2: $\frac{40}{x - y} + \frac{55}{x + y} = 13 \quad \text{--- (2)}$
4. Let $\frac{1}{x - y} = u$ and $\frac{1}{x + y} = v$: $30u + 44v = 10 \quad \text{--- (3)}$ $40u + 55v = 13 \quad \text{--- (4)}$
5. Multiply (3) by $4$ and (4) by $3$: $120u + 176v = 40$ $120u + 165v = 39$ Subtracting gives: $11v = 1 \implies v = \frac{1}{11}$.
6. Substitute $v = \frac{1}{11}$ into (3): $30u + 44\left(\frac{1}{11}\right) = 10 \implies 30u + 4 = 10 \implies 30u = 6 \implies u = \frac{1}{5}$
7. Solve for $x$ and $y$: $x - y = 5 \quad \text{and} \quad x + y = 11$ Adding: $2x = 16 \implies x = 8\text{ km/h}$. Subtracting: $2y = 6 \implies y = 3\text{ km/h}$.
8. Therefore, <u>speed of the boat in still water is $8\text{ km/h}$ and speed of the stream is $3\text{ km/h}$</u>.

5. Category 4: Fixed and Variable Cost Problems

Solved Example: Taxi Charges

Problem: The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of $10\text{ km}$, the charge paid is $\text{₹}105$, and for a journey of $15\text{ km}$, the charge paid is $\text{₹}155$. What are the fixed charges and the charge per km?

Solution:

1. Let the fixed charge be $\text{₹}x$ and charge per km be $\text{₹}y$.
2. For 10 km: $x + 10y = 105 \quad \text{--- (1)}$
3. For 15 km: $x + 15y = 155 \quad \text{--- (2)}$
4. Subtract (1) from (2): $(x + 15y) - (x + 10y) = 155 - 105 \implies 5y = 50 \implies y = 10$
5. Substitute $y = 10$ into (1): $x + 10(10) = 105 \implies x + 100 = 105 \implies x = 5$
6. Therefore, <u>fixed charge is $\text{₹}5$ and charge per km is $\text{₹}10$</u>.

6. Summary of Core Word Problem Formulas

Exam Tip: Always write a final concluding sentence stating the requested quantities along with their appropriate physical units (e.g., years, km/h, ₹, cm). Never leave your answer as just "$x = 42, y = 12$"!

Common Mistake: Mixing up upstream and downstream speeds. Upstream is against the stream, so it is ALWAYS $(x - y)$, where $x$ is boat speed and $y$ is stream speed. Never write $y - x$, as boat speed must be greater than stream speed for the boat to move forward!