Applications and Word Problems on Quadratic Equations for CBSE Class 10

Algebra reveals its true power when applied to real-world situations. While solving purely numerical equations tests computational mechanics, word problems on quadratic equations require you to analyze a physical situation, set up algebraic variables, translate verbal descriptions into quadratic models, and interpret the physical validity of the resulting roots.

In CBSE Class 10 Mathematics, word problems in Chapter 4 carry substantial weightage (typically 4 or 5 marks in Section D). Whether it is calculating the speed of a stream, the time taken by water taps to fill a pool, or the dimensions of a rectangular garden, mastering these structural problem templates is essential for scoring top marks.

What You Will Learn

• Universal 5-step framework for formulating and solving quadratic word problems
• Category 1: Geometry and mensuration problems (rectangles, Pythagoras theorem)
• Category 2: Number theory problems (consecutive positive integers, reciprocal sums)
• Category 3: Uniform speed, distance, and time problems (trains and flights)
• Category 4: Work and time problems (two water taps filling a tank)
• The rule of rejecting extraneous and physically impossible roots (negative lengths or speeds)
• Step-by-step solved CBSE board exam questions and presentation templates

1. The 5-Step Quadratic Problem-Solving Framework

1. Step 1 (Variable Declaration): Assign variable $x$ to the fundamental unknown quantity and clearly state its physical units (e.g., Let the uniform speed of the train be $x$ km/h).
2. Step 2 (Formulate Expressions): Express all other related quantities in terms of $x$ using the relationships described in the prompt.
3. Step 3 (Formulate Quadratic Equation): Set up the governing equation based on the given physical condition (e.g., $\text{Time}_1 - \text{Time}_2 = \text{Time Difference}$).
4. Step 4 (Simplify to Standard Form): Clear fractions, expand brackets, and arrange in standard form $ax^2 + bx + c = 0$.
5. Step 5 (Solve and Filter Roots): Solve using factorisation or the quadratic formula. <u>Always reject extraneous roots that lack physical meaning (such as negative speed, negative length, or negative time), and clearly state the reason for rejection in your final written answer.</u>

2. Category 1: Geometry and Mensuration Problems

Solved Example: Right-Angled Triangle Dimensions

Problem: The altitude of a right triangle is $7\text{ cm}$ less than its base. If the hypotenuse is $13\text{ cm}$, find the other two sides.

Solution:

1. Let the base of the right triangle be $x\text{ cm}$.
2. Then the altitude (height) is $(x - 7)\text{ cm}$.
3. Hypotenuse is given as $13\text{ cm}$.
4. By the Pythagoras Theorem: $\text{Base}^2 + \text{Altitude}^2 = \text{Hypotenuse}^2$ $x^2 + (x - 7)^2 = 13^2$
5. Expand and simplify: $x^2 + (x^2 - 14x + 49) = 169$ $2x^2 - 14x + 49 - 169 = 0$ $2x^2 - 14x - 120 = 0$ Divide the entire equation by 2: $x^2 - 7x - 60 = 0$
6. Factor by splitting the middle term (product $=-60$, sum $=-7$ $\implies -12$ and $+5$): $(x - 12)(x + 5) = 0$ $x = 12 \quad \text{or} \quad x = -5$
7. Since length cannot be negative, $x = -5$ is rejected.
8. Therefore, base $= 12\text{ cm}$ and altitude $= 12 - 7 = 5\text{ cm}$.
9. Conclusion: <u>The other two sides are $12\text{ cm}$ and $5\text{ cm}$</u>.

3. Category 2: Speed, Distance, and Time Problems (CBSE Board Classic)

The Structural Template

$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$ If speed increases, time decreases. The difference between original time and new time gives the equation: $\frac{\text{Distance}}{\text{Original Speed}} - \frac{\text{Distance}}{\text{Increased Speed}} = \text{Time Difference}$

Solved Example: Uniform Speed of a Train

Problem: 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 for the same journey. Find the original speed of the train.

Solution:

1. Let the original uniform speed of the train be $x\text{ km/h}$.
2. Increased speed $= (x + 5)\text{ km/h}$.
3. Total distance $= 360\text{ km}$.
4. Time taken at original speed: $T_1 = \frac{360}{x}\text{ hours}$
5. Time taken at increased speed: $T_2 = \frac{360}{x + 5}\text{ hours}$
6. According to the problem, the time difference is $1\text{ hour}$ ($T_1 - T_2 = 1$): $\frac{360}{x} - \frac{360}{x + 5} = 1$
7. Combine fractions: $360\left(\frac{1}{x} - \frac{1}{x + 5}\right) = 1$ $360\left(\frac{(x + 5) - x}{x(x + 5)}\right) = 1$ $360\left(\frac{5}{x^2 + 5x}\right) = 1$ $\frac{1800}{x^2 + 5x} = 1 \implies x^2 + 5x - 1800 = 0$
8. Factorize (product $=-1800$, sum $=5$ $\implies +45$ and $-40$): $(x + 45)(x - 40) = 0$ $x = 40 \quad \text{or} \quad x = -45$
9. Speed cannot be negative, so $x = -45$ is rejected.
10. Therefore, <u>the original speed of the train is $40\text{ km/h}$</u>.

4. Category 3: Work and Time (Two Water Taps Problem)

Solved Example: Filling a Tank Together (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 time taken by the smaller tap to fill the tank be $x\text{ hours}$.
2. Then the time taken by the larger tap is $(x - 10)\text{ hours}$.
3. In $1\text{ hour}$:
   • Portion filled by smaller tap $= \frac{1}{x}$
   • Portion filled by larger tap $= \frac{1}{x - 10}$
   • Portion filled by both taps together $= \frac{1}{9\frac{3}{8}} = \frac{1}{\frac{75}{8}} = \frac{8}{75}$
4. Formulate the equation: $\frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75}$
5. Simplify the LHS: $\frac{(x - 10) + x}{x(x - 10)} = \frac{8}{75}$ $\frac{2x - 10}{x^2 - 10x} = \frac{8}{75}$ $\frac{2(x - 5)}{x^2 - 10x} = \frac{8}{75} \implies \frac{x - 5}{x^2 - 10x} = \frac{4}{75}$
6. Cross-multiply: $75(x - 5) = 4(x^2 - 10x)$ $75x - 375 = 4x^2 - 40x$ $4x^2 - 40x - 75x + 375 = 0 \implies 4x^2 - 115x + 375 = 0$
7. Factorize (product $= 4 \times 375 = 1500$, sum $= -115$ $\implies -100$ and $-15$): $4x^2 - 100x - 15x + 375 = 0$ $4x(x - 25) - 15(x - 25) = 0$ $(x - 25)(4x - 15) = 0$ $x = 25 \quad \text{or} \quad x = \frac{15}{4} = 3.75$
8. Evaluate Roots: If $x = 3.75\text{ hours}$, then the larger tap would take $x - 10 = 3.75 - 10 = -6.25\text{ hours}$, which is physically impossible. Hence, $x = 3.75$ is rejected.
9. Thus, $x = 25\text{ hours}$.
   • Time taken by smaller tap $= 25\text{ hours}$.
   • Time taken by larger tap $= 25 - 10 = 15\text{ hours}$.
10. Conclusion: <u>The smaller tap takes $25\text{ hours}$ and the larger tap takes $15\text{ hours}$ separately</u>.

5. Summary and Examination Tips

Exam Tip: Always write the physical units in your concluding statement (e.g., km/h, hours, cm). Failing to write units loses half a mark in CBSE board exams!

Common Mistake: Forgetting to test both roots against physical constraints. In the two-taps problem, both roots ($25$ and $3.75$) were positive numbers, but $3.75$ had to be rejected because $3.75 - 10 = -6.25$ was negative!