Mastering High-Frequency Word Problems in Class 10 Algebra

In the CBSE Class 10 Mathematics board examination, algebraic word problems represent the ultimate test of mathematical competence. Carrying 4 to 5 marks in Section D, these questions do not hand you an equation on a silver platter; you must extract variables, establish physical relationships, and construct quadratic or linear equations from real-world narratives.

For many students, translating English paragraphs into algebraic symbols feels intimidating. However, every algebraic word problem on the board exam belongs to one of four classic, highly predictable archetypes: Upstream and Downstream Boats, Two Water Taps Filling a Tank, Speed-Distance-Time Trains, and Two-Digit Number Reversals.

In this guide, we break down the exact mathematical blueprints for each archetype.

What You Will Learn

• The 4 foundational algebraic modeling archetypes
• Archetype 1: Upstream and Downstream Boat Problems ($v_{\text{up}} = x - y, \; v_{\text{down}} = x + y$)
• Archetype 2: Two Water Taps Filling a Tank Together (Work-Rate principle: $\frac{1}{x} + \frac{1}{x+10} = \frac{1}{T}$)
• Archetype 3: Speed-Distance-Time Train Problems ($T_1 - T_2 = \Delta t$)
• Archetype 4: Two-Digit Numbers and Reversal Equations ($10x + y$)
• Step-by-step quadratic factorization and rejecting extraneous roots

1. Archetype 1: Upstream and Downstream Boats

When a boat travels in moving water (a river or stream):

• Still Water Speed of Boat: Let this be $x\text{ km/h}$.
• Speed of River Stream: Let this be $y\text{ km/h}$ ($x > y$).

    Downstream (Moving WITH stream):        Upstream (Moving AGAINST stream):
    Water pushes boat faster!               Water opposes boat!
    Speed = (x + y) km/h                    Speed = (x - y) km/h

The Master Time Equation:

Since $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$: $\mathbf{T_{\text{upstream}} = \frac{d}{x - y}} \quad \text{and} \quad \mathbf{T_{\text{downstream}} = \frac{d}{x + y}}$

Solved Example: Upstream & Downstream (NCERT 5-Mark Classic)

Problem: A motor boat whose speed is $18\text{ km/h}$ in still water takes $1\text{ hour}$ more to go $24\text{ km}$ upstream than to return downstream to the same spot. Find the speed of the stream.

Solution:

1. Let the speed of the stream be $x\text{ km/h}$.
2. Given: Speed of boat in still water $= 18\text{ km/h}$.
   • Upstream speed $= (18 - x)\text{ km/h}$.
   • Downstream speed $= (18 + x)\text{ km/h}$.
3. Distance $d = 24\text{ km}$.
   • Time taken upstream: $T_1 = \frac{24}{18 - x}$.
   • Time taken downstream: $T_2 = \frac{24}{18 + x}$.
4. Formulate the Equation: Upstream takes $1\text{ hour}$ longer than downstream: $T_1 - T_2 = 1$ $\frac{24}{18 - x} - \frac{24}{18 + x} = 1$
5. Algebraic Simplification: $24 \left[ \frac{(18 + x) - (18 - x)}{(18 - x)(18 + x)} \right] = 1$ $24 \left[ \frac{18 + x - 18 + x}{18^2 - x^2} \right] = 1$ $24 \left[ \frac{2x}{324 - x^2} \right] = 1$ $48x = 324 - x^2$ $\mathbf{x^2 + 48x - 324 = 0}$
6. Factorize the Quadratic Equation: Find two numbers whose product is $-324$ and sum is $+48$ ($54$ and $-6$): $x^2 + 54x - 6x - 324 = 0$ $x(x + 54) - 6(x + 54) = 0$ $(x + 54)(x - 6) = 0$ $x = -54 \quad \text{or} \quad x = 6$
7. Reject Extraneous Root: Speed of a stream cannot be negative. We discard $x = -54$.
8. Therefore, <u>the speed of the stream is $6\text{ km/h}$</u>.

2. Archetype 2: Two Water Taps Filling a Tank

In work-rate problems, work done in 1 hour is the reciprocal of the total time taken:

• If Tap 1 fills the tank in $x\text{ hours}$, in $1\text{ hour}$ it fills $\frac{1}{x}$ of the tank.
• If Tap 2 takes $(x - 10)\text{ hours}$, in $1\text{ hour}$ it fills $\frac{1}{x - 10}$ of the tank.
• If together they fill the tank in $T\text{ hours}$, in $1\text{ hour}$ they fill $\frac{1}{T}$ of the tank:

The Work-Rate Equation: $\mathbf{\frac{1}{x} + \frac{1}{x - 10} = \frac{1}{T}}$

Solved Example: The Two Water Taps (NCERT Classic)

Problem: Two water taps together can fill a tank in $9\frac{3}{8}\text{ hours}$. The tap of larger diameter takes $10\text{ hours}$ less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solution:

1. Let the smaller tap take $x\text{ hours}$ to fill the tank.
   • The larger tap takes $(x - 10)\text{ hours}$.
2. Together they take: $T = 9\frac{3}{8} = \frac{75}{8}\text{ hours} \implies \frac{1}{T} = \mathbf{\frac{8}{75}}$
3. Formulate the Equation: $\frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75}$
4. Simplify Algebraically: $\frac{(x - 10) + x}{x(x - 10)} = \frac{8}{75}$ $\frac{2x - 10}{x^2 - 10x} = \frac{8}{75}$ $75(2x - 10) = 8(x^2 - 10x)$ $150x - 750 = 8x^2 - 80x$ $8x^2 - 230x + 750 = 0$ Divide the entire equation by $2$: $\mathbf{4x^2 - 115x + 375 = 0}$
5. Factorize: Product $= 4 \times 375 = 1500$. Sum $= -115$ ($-100$ and $-15$): $4x^2 - 100x - 15x + 375 = 0$ $4x(x - 25) - 15(x - 25) = 0$ $(4x - 15)(x - 25) = 0$ $x = \frac{15}{4} = 3.75 \quad \text{or} \quad x = 25$
6. Evaluate Feasibility:
   • If $x = 3.75\text{ hours}$, then the larger tap would take $3.75 - 10 = -6.25\text{ hours}$, which is physically impossible!
   • Therefore, $x = \mathbf{25\text{ hours}}$.
7. Conclusion:
   • Smaller tap takes: $25\text{ hours}$.
   • Larger tap takes: $25 - 10 = \mathbf{15\text{ hours}}$.
8. <u>The smaller tap takes $25 ext{ hours}$ and the larger tap takes $15 ext{ hours}$</u>.

3. Archetype 3: Speed-Distance-Time Train Problems

When a train's speed changes by $\Delta v$, the difference in transit times is given: $\mathbf{\frac{\text{Distance}}{\text{Slower Speed}} - \frac{\text{Distance}}{\text{Faster Speed}} = \text{Time Difference}}$

Example: A train travels $360\text{ km}$ at a uniform speed. If the speed had been $5\text{ km/h}$ more, it would have taken $1\text{ hour}$ less: $\frac{360}{x} - \frac{360}{x + 5} = 1 \implies x^2 + 5x - 1800 = 0 \implies x = 40\text{ km/h}$

4. Summary and Examination Tips

Exam Tip: Always write a concluding sentence explicitly rejecting extraneous negative roots! Evaluators award half a mark specifically for stating: "Since speed/time cannot be negative, we reject the negative root".

Common Mistake: In the boat problem, subtracting downstream time from upstream time incorrectly. Upstream is SLOWER, so upstream time is LARGER! The equation is always: $\text{Taller Time (Upstream)} - \text{Shorter Time (Downstream)} = \Delta t$.