Case Study Questions Mastery Guide for CBSE Class 10 Mathematics

In the modern CBSE Class 10 Mathematics board examination, Section E represents a significant pedagogical shift toward competence-based assessment. Worth 12 marks (comprising three 4-mark questions), case study problems test your ability to apply abstract mathematical principles to real-world engineering, architecture, sports, and economic situations.

Instead of directly asking you to "solve a quadratic equation" or "find the roots of an AP", case study questions embed these mathematical structures inside contextual narratives: the parabolic arch of an entrance gate, the seating arrangement of an auditorium, or the distance measurement of a shipping lighthouse.

This guide provides the definitive strategy for decoding case study scenarios, mapping physical stories to core mathematical models, and structuring step-by-step solutions to secure full marks.

What You Will Learn

• Anatomy and marking scheme of Section E Case Study Questions
• Strategic 4-step framework to decode real-world problem stems
• Case Archetype 1: Parabolic Arches & Highway Flyovers (Polynomials & Quadratic Equations)
• Case Archetype 2: Auditorium Seating & Production Tiers (Arithmetic Progressions)
• Case Archetype 3: Cartographic Grids & Sports Stadium Coordinates (Coordinate Geometry)
• Case Archetype 4: Multi-Tiered Lighthouses & Drone Navigation (Some Applications of Trigonometry)
• Common student pitfalls and presentation rubrics

1. Anatomy of a CBSE Case Study Question

Every Section E case study question carries 4 marks, organized into sub-parts:

• Part (i) [1 Mark]: A direct foundational question testing identification of parameters.
• Part (ii) [1 Mark]: A basic computational step or formula application.
• Part (iii) [2 Marks] (With Internal Choice): A multi-step analytical problem requiring algebraic manipulation or optimization.

2. The 4-Step Case Solving Framework

    Step 1: Skim the Narrative ───> Filter out background storytelling fluff.
    Step 2: Identify the Chapter ───> Parabola = Polynomials; Row progression = AP; Tower = Trig.
    Step 3: Extract Given Data ─────> Translate descriptive words into algebraic variables (a, d, h, θ).
    Step 4: Solve Sub-Questions ───> Treat each sub-part as an independent mini-problem!

Important: <u>Do not get overwhelmed by lengthy descriptive paragraphs! In almost every case study question, 70% of the text is purely descriptive context (history, geography, real-world context). The actual mathematical data is concentrated in 2 or 3 sentences and the accompanying diagram.</u>

3. High-Yield Solved Board Exam Case Studies

Case Study 1: The Parabolic Highway Overpass (Polynomials & Quadratic Equations)

Context: Highway engineering projects frequently utilize parabolic curves for flyovers and suspension cables because the parabolic shape evenly distributes mechanical load across bridge pillars. A highway engineer models the arch of a new overpass using the quadratic polynomial: $p(x) = -x^2 + 2x + 8$ where $x$ is the horizontal distance (in metres) and $p(x)$ represents the height of the arch above ground level.

                                    Height p(x)
                                         ^
                                       8 |        *  (Peak)
                                         |      /                                            |     /                                              |    /                                Ground Level ---+---*---------*---> Distance (x)
                                        -2   0         4

Questions & Step-by-Step Solutions:

(i) What is the shape of the curve represented by the quadratic polynomial? [1 Mark]

• Answer: The graph of any quadratic polynomial $ax^2 + bx + c$ ($a \ne 0$) is a Parabola. Since the coefficient of $x^2$ is negative ($a = -1 < 0$), the curve is a downward-opening parabola.

(ii) Find the horizontal distance between the two points where the arch touches the ground. [1 Mark]

• Solution: At ground level, the height is zero: $p(x) = 0$. $-x^2 + 2x + 8 = 0 \implies x^2 - 2x - 8 = 0$ Factorize by splitting the middle term: $(x - 4)(x + 2) = 0 \implies x = 4 \quad \text{or} \quad x = -2$
• The arch touches the ground at $x = -2\text{ m}$ and $x = 4\text{ m}$.
• The total horizontal span (distance) is: $\text{Distance} = 4 - (-2) = 4 + 2 = \mathbf{6\text{ metres}}$

(iii) Find the maximum height reached by the overpass arch above the ground. [2 Marks]

• Solution: In a symmetric parabola, the maximum height (vertex) occurs exactly midway between its zeroes: $x_{\text{mid}} = \frac{4 + (-2)}{2} = \frac{2}{2} = \mathbf{1\text{ metre}}$
• Substitute $x = 1$ into the polynomial $p(x)$: $p(1) = -(1)^2 + 2(1) + 8 = -1 + 2 + 8 = \mathbf{9\text{ metres}}$
• Therefore, <u>the maximum height reached by the overpass arch is $9\text{ metres}$</u>.

Case Study 2: Auditorium Seating Layout (Arithmetic Progressions)

Context: An architect is designing a semi-circular concert auditorium. To ensure optimal acoustics and sightlines, the rows are curved, with each successive row having more seats than the previous one. The first row contains $20$ seats, the second row contains $24$ seats, the third row contains $28$ seats, and so on. The auditorium has a total of $30$ rows.

Questions & Step-by-Step Solutions:

(i) Does the number of seats in successive rows form an Arithmetic Progression? Justify. [1 Mark]

• Answer: Yes. The difference between consecutive rows is constant: $d = 24 - 20 = 28 - 24 = \mathbf{4}$ Since the common difference $d = 4$ is uniform, the arrangement forms an Arithmetic Progression (AP) with first term $a = 20$ and common difference $d = 4$.

(ii) How many seats are there in the $15^{\text{th}}$ row? [1 Mark]

• Solution: Apply the $n^{\text{th}}$ term formula: $a_n = a + (n - 1)d$. $a_{15} = 20 + (15 - 1) \times 4 = 20 + (14 \times 4) = 20 + 56 = \mathbf{76\text{ seats}}$

(iii) What is the total seating capacity of the entire auditorium? [2 Marks]

• Solution: Apply the sum of first $n$ terms formula: $S_n = \frac{n}{2}[2a + (n - 1)d]$. $S_{30} = \frac{30}{2} [2(20) + (30 - 1) \times 4]$ $S_{30} = 15 [40 + (29 \times 4)] = 15 [40 + 116] = 15 \times 156 = \mathbf{2340\text{ seats}}$
• Therefore, <u>the total seating capacity of the auditorium is $2340$ seats</u>.

4. Summary and Examination Tips

Exam Tip: In 2-mark sub-questions with internal choices, read BOTH alternatives carefully before writing! Often, one option involves a direct single-line formula, while the other requires multi-step quadratic factorization. Choose the faster, cleaner option.

Common Mistake: Leaving units out of final answers in case study problems. In real-world problems, writing $9$ instead of $9\text{ metres}$ or $2340$ instead of $2340\text{ seats}$ loses half a mark!