In short: CBSE Class 12 Maths has 3 case study questions in Section E, each worth 4 marks, totalling 12 marks out of 80. Each case study presents a real-world scenario with 5 sub-questions — you answer any 4. Combined with assertion-reason MCQs and competency-based questions in other sections, 50% of the paper (40 marks) is competency-based. This post gives you 10 complete case studies in the exact board format.

Understanding Section E: Case Study Questions

The CBSE Class 12 Maths paper has 38 questions across 5 sections:

Section	Question Type	Marks
Section A	18 MCQs + 2 Assertion-Reason	20 × 1 = 20
Section B	5 Very Short Answer	5 × 2 = 10
Section C	6 Short Answer	6 × 3 = 18
Section D	4 Long Answer	4 × 5 = 20
Section E	3 Case Study Questions	3 × 4 = 12
Total	38 questions	80 marks

Case studies in Section E describe a real-world application — manufacturing, navigation, architecture, probability scenarios, or financial planning — followed by 5 MCQ sub-questions. You must answer any 4.

The Maths chapter-wise marks weightage shows that Calculus (35 marks), Algebra (10 marks), and Probability (8 marks) dominate the paper. Case studies are drawn from application-heavy chapters.

Case Study 1: Linear Programming — Manufacturing Optimisation

A small factory manufactures two products — chairs and tables. Each chair requires 2 hours of carpentry and 1 hour of painting. Each table requires 3 hours of carpentry and 2 hours of painting. The factory has 120 hours of carpentry time and 80 hours of painting time available per week. The profit on each chair is ₹500 and on each table is ₹800. The factory wants to maximise weekly profit.

(i) If x represents chairs and y represents tables, the objective function is: (a) Minimise Z = 500x + 800y (b) Maximise Z = 500x + 800y (c) Maximise Z = 2x + 3y (d) Minimise Z = x + y

(ii) The constraint for carpentry time is: (a) 2x + 3y ≤ 120 (b) x + 2y ≤ 120 (c) 2x + 3y ≥ 120 (d) 3x + 2y ≤ 120

(iii) The constraint for painting time is: (a) 2x + y ≤ 80 (b) x + 2y ≤ 80 (c) x + y ≤ 80 (d) 2x + 2y ≤ 80

(iv) Which of the following is a corner point of the feasible region? (a) (60, 0) (b) (0, 40) (c) (120, 80) (d) Both (a) and (b)

(v) The maximum profit is: (a) ₹30,000 (b) ₹32,000 (c) ₹36,000 (d) ₹40,000

Answers: (i) b (ii) a (iii) b (iv) d — x = 60, y = 0 satisfies both constraints; x = 0, y = 40 also satisfies both (v) b — Evaluating Z at corner points: (0,0) = 0, (60,0) = 30000, (0,40) = 32000, and the intersection of 2x+3y=120 and x+2y=80 gives (24,24) → Z = 12000+19200 = 31200. Maximum is ₹32,000 at (0,40).

Case Study 2: Probability — Medical Testing

A hospital uses a diagnostic test for a disease. The probability that a person has the disease is 0.01 (1% of the population). The test is 99% accurate for positive detection (sensitivity) and 95% accurate for negative detection (specificity). A patient tests positive.

(i) The probability of testing positive given the person has the disease, P(+|D), is: (a) 0.99 (b) 0.95 (c) 0.01 (d) 0.05

(ii) The probability of testing positive given the person does NOT have the disease, P(+|D'), is: (a) 0.99 (b) 0.95 (c) 0.01 (d) 0.05

(iii) Using Bayes' theorem, P(D|+) is approximately: (a) 0.99 (b) 0.17 (c) 0.50 (d) 0.01

(iv) The result P(D|+) ≈ 0.17 means: (a) The test is useless (b) Only about 17% of positive results actually have the disease (c) 99% of positives have the disease (d) The disease is very common

(v) To increase P(D|+), the hospital should: (a) Test more people (b) Improve specificity (reduce false positives) (c) Reduce sensitivity (d) Test fewer people

Case Study 3: Application of Integrals — Area Under Curves

An architect is designing a park. The boundary of one section of the park follows the curve y = x² from x = 0 to x = 3, and is bounded below by the x-axis. The architect needs to calculate the area for purchasing turf. Each square metre of turf costs ₹150.

(i) The area bounded by y = x², the x-axis, and x = 3 is: (a) 9 sq. units (b) 27 sq. units (c) 3 sq. units (d) 18 sq. units

(ii) The integral used to find this area is: (a) ∫₀³ x² dx (b) ∫₀³ x³ dx (c) ∫₀³ 2x dx (d) ∫₀³ √x dx

(iii) The value of ∫₀³ x² dx is: (a) [x³/3]₀³ = 9 (b) [x²/2]₀³ = 4.5 (c) [x⁴/4]₀³ (d) [2x]₀³ = 6

(iv) The cost of turf for this section is: (a) ₹1,350 (b) ₹4,050 (c) ₹2,700 (d) ₹900

(v) If the park boundary was y = √x instead, the area from x = 0 to x = 4 would be: (a) 8/3 sq. units (b) 16/3 sq. units (c) 4 sq. units (d) 2 sq. units

Answers: (i) a — ∫₀³ x² dx = [x³/3]₀³ = 27/3 = 9 (ii) a (iii) a (iv) a — 9 × 150 = ₹1,350 (v) b — ∫₀⁴ √x dx = [2x^(3/2)/3]₀⁴ = 2(8)/3 = 16/3

Case Study 4: Vectors and 3D Geometry — Navigation

A drone flies from point A(1, 2, 3) to point B(4, 6, 3) in a straight line. After reaching B, it changes direction and flies to point C(4, 6, 8). The drone's flight path is monitored by a control system that tracks its position vector at each point.

(i) The position vector of point A is: (a) î + 2ĵ + 3k̂ (b) 4î + 6ĵ + 3k̂ (c) î + 2ĵ (d) 3î + 2ĵ + k̂

(ii) The vector AB is: (a) 3î + 4ĵ (b) 3î + 4ĵ + 0k̂ (c) 5î + 8ĵ + 6k̂ (d) −3î − 4ĵ

(iii) The magnitude of vector AB is: (a) 5 units (b) 7 units (c) 3 units (d) 4 units

(iv) The vector BC is: (a) 0î + 0ĵ + 5k̂ (b) 8î + 12ĵ + 11k̂ (c) 4î + 6ĵ + 8k̂ (d) −4î − 6ĵ − 5k̂

(v) The angle between vectors AB and BC is: (a) 0° (b) 90° (c) 45° (d) 180°

Answers: (i) a (ii) b — AB = B − A = (4−1)î + (6−2)ĵ + (3−3)k̂ = 3î + 4ĵ + 0k̂ (iii) a — |AB| = √(9+16+0) = √25 = 5 (iv) a — BC = C − B = (4−4)î + (6−6)ĵ + (8−3)k̂ = 0î + 0ĵ + 5k̂ (v) b — AB·BC = 3(0) + 4(0) + 0(5) = 0, so cos θ = 0, θ = 90°

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Case Study 5: Matrices — Network Problems

A city has three bus routes connecting three zones. The number of buses running between zones is represented by the matrix:

A = | 0  5  3 |
    | 5  0  4 |
    | 3  4  0 |

The transport authority wants to double the frequency on all routes during a festival. They also want to add 2 extra buses on each route.

(i) The matrix representing doubled frequency is: (a) A + 2 (b) 2A (c) A² (d) A/2

(ii) 2A is equal to: (a) |0 10 6; 10 0 8; 6 8 0| (b) |2 7 5; 7 2 6; 5 6 2| (c) |0 25 9; 25 0 16; 9 16 0| (d) |1 6 4; 6 1 5; 4 5 1|

(iii) The matrix B representing 2 extra buses on each route is: (a) |2 2 2; 2 2 2; 2 2 2| (b) |0 2 2; 2 0 2; 2 2 0| (c) |2 0 0; 0 2 0; 0 0 2| (d) |1 1 1; 1 1 1; 1 1 1|

(iv) The total bus matrix during the festival (2A + B, where B has 2 extra on off-diagonal entries) is: (a) |0 12 8; 12 0 10; 8 10 0| (b) |2 12 8; 12 2 10; 8 10 2| (c) |0 10 6; 10 0 8; 6 8 0| (d) |4 14 10; 14 4 12; 10 12 4|

Answers: (i) b (ii) a (iii) b — since there are no buses from a zone to itself, the diagonal stays 0. Extra buses are added to off-diagonal entries. (iv) a — 2A + B = |0+0 10+2 6+2; 10+2 0+0 8+2; 6+2 8+2 0+0| = |0 12 8; 12 0 10; 8 10 0|

Case Study 6: Probability — Quality Control

A factory produces bolts using three machines A, B, and C, which produce 25%, 35%, and 40% of the total output respectively. The defective rates for machines A, B, and C are 5%, 4%, and 2% respectively. A bolt is selected at random and found to be defective.

(i) P(A) = 0.25, P(B) = 0.35, P(C) = 0.40. The probability of a defective bolt from machine A is: (a) 0.25 (b) 0.05 (c) 0.0125 (d) 0.50

(ii) The total probability of a defective bolt is: (a) 0.0125 + 0.014 + 0.008 = 0.0345 (b) 0.11 (c) 0.05 (d) 0.0125

(iii) Using Bayes' theorem, the probability that the defective bolt came from machine B is: (a) 0.362 (b) 0.406 (c) 0.232 (d) 0.500

(iv) Which machine is most likely to have produced the defective bolt? (a) Machine A (b) Machine B (c) Machine C (d) All equally likely

Answers: (i) c — P(A) × P(D|A) = 0.25 × 0.05 = 0.0125 (ii) a — P(D) = 0.0125 + 0.35×0.04 + 0.40×0.02 = 0.0125 + 0.014 + 0.008 = 0.0345 (iii) b — P(B|D) = P(B)P(D|B)/P(D) = 0.014/0.0345 ≈ 0.406 (iv) b — P(B|D) ≈ 0.406 is the highest among A(0.362), B(0.406), C(0.232)

Case Study 7: Differential Equations — Population Growth

A biologist studies a bacterial colony whose population P grows according to the equation dP/dt = kP, where k = 0.05 per hour. At time t = 0, the population is P₀ = 1000 bacteria.

(i) The type of this differential equation is: (a) Linear, first order (b) Non-linear, second order (c) Homogeneous (d) Partial differential equation

(ii) The solution of dP/dt = kP is: (a) P = P₀ + kt (b) P = P₀e^(kt) (c) P = P₀/t (d) P = kt²

(iii) The population after 10 hours is: (a) 1000 × e^(0.5) (b) 1000 × e^(0.05) (c) 1000 × 0.5 (d) 1000 + 500

(iv) The time taken for the population to double is: (a) 20 hours (b) ln2/0.05 ≈ 13.86 hours (c) 10 hours (d) 50 hours

Answers: (i) a (ii) b — separating variables and integrating gives P = P₀e^(kt) (iii) a — P = 1000 × e^(0.05×10) = 1000e^(0.5) (iv) b — 2P₀ = P₀e^(0.05t), so t = ln2/0.05 ≈ 13.86 hours

Case Study 8: Application of Derivatives — Maxima and Minima

A rectangular sheet of metal measuring 24 cm × 24 cm is used to make an open-top box by cutting equal squares of side x cm from each corner and folding up the flaps.

(i) The volume of the box as a function of x is: (a) V = x(24 − 2x)² (b) V = x(24 − x)² (c) V = (24 − 2x)³ (d) V = 24x²

(ii) The domain of x for a valid box is: (a) 0 < x < 24 (b) 0 < x < 12 (c) 0 < x < 6 (d) 0 < x < 8

(iii) To find the value of x that maximises volume, we set: (a) V = 0 (b) dV/dx = 0 (c) d²V/dx² = 0 (d) V = maximum constant

(iv) The value of x that maximises volume is: (a) 4 cm (b) 6 cm (c) 8 cm (d) 3 cm

(v) The maximum volume of the box is: (a) 1024 cm³ (b) 2048 cm³ (c) 512 cm³ (d) 768 cm³

Answers: (i) a — V = x(24−2x)(24−2x) = x(24−2x)² (ii) b — x must be positive and 24−2x > 0, so x < 12 (iii) b (iv) a — dV/dx = (24−2x)² + x·2(24−2x)(−2) = (24−2x)[(24−2x) − 4x] = (24−2x)(24−6x) = 0. So x = 12 (rejected) or x = 4. (v) a — V = 4(24−8)² = 4(16)² = 4×256 = 1024 cm³

Case Study 9: Continuity and Differentiability — Real-World Function

A water tank is being filled in two phases. For the first 3 hours, water flows in at a rate described by f(t) = t² litres per hour. After 3 hours, the flow rate changes to g(t) = 6t − 9 litres per hour. The water authority wants to know if the flow function is continuous and differentiable at t = 3.

(i) The value of f(3) is: (a) 6 (b) 9 (c) 3 (d) 12

(ii) The value of g(3) is: (a) 6 (b) 9 (c) 3 (d) 18

(iii) Is the combined function continuous at t = 3? (a) Yes, because f(3) = g(3) = 9 (b) No, because f(3) ≠ g(3) (c) Cannot be determined (d) Only from the left

(iv) The derivative of f(t) at t = 3 (from the left) is: (a) 6 (b) 9 (c) 3 (d) 0

(v) Is the combined function differentiable at t = 3? (a) Yes, because both left and right derivatives are equal to 6 (b) No, because the derivatives are different (c) Only from the right (d) Cannot be determined

Answers: (i) b — f(3) = 3² = 9 (ii) b — g(3) = 6(3) − 9 = 9 (iii) a — both give 9 at t = 3 (iv) a — f'(t) = 2t, f'(3) = 6 (v) a — g'(t) = 6, g'(3) = 6. Since left derivative (6) = right derivative (6) and function is continuous, it is differentiable.

Case Study 10: Relations and Functions — Encryption

A software company uses a function f: A → B for encrypting employee IDs, where A = {1, 2, 3, 4, 5} and B = {2, 5, 10, 17, 26}. The encryption function is f(x) = x² + 1.

(i) The value of f(3) is: (a) 9 (b) 10 (c) 7 (d) 4

(ii) Is the function f one-one (injective)? (a) Yes, because different inputs give different outputs (b) No, because two inputs give the same output (c) Cannot be determined (d) Only for even numbers

(iii) Is the function f onto (surjective)? (a) Yes, because every element of B has a pre-image in A (b) No, because some elements of B are unmapped (c) Only partially (d) Cannot be determined

(iv) Since f is both one-one and onto, f is: (a) Bijective (b) Surjective only (c) Injective only (d) Neither

(v) The inverse function f⁻¹(10) is: (a) 3 (b) 9 (c) 4 (d) 2

Answers: (i) b — f(3) = 9 + 1 = 10 (ii) a — f(1)=2, f(2)=5, f(3)=10, f(4)=17, f(5)=26 — all distinct (iii) a — B = {2,5,10,17,26} and each has a pre-image (iv) a (v) a — f(3) = 10, so f⁻¹(10) = 3

How to Approach Maths Case Studies

Time Allocation

Each case study is worth 4 marks. With 3 case studies, Section E totals 12 marks. Allocate 8-10 minutes per case study — that is 24-30 minutes for the entire section. Do not exceed this.

Step-by-Step Strategy

Read the scenario and identify the chapter. Linear programming, probability (Bayes' theorem), integration (area), vectors, matrices — the topic tells you which formulas to recall.
Set up the mathematical model. Convert the real-world problem into equations, constraints, or functions before looking at the sub-questions.
Answer the easier sub-questions first. In most case studies, (i) and (ii) are direct formula application. (iv) and (v) may require multi-step calculations.
Skip one sub-question. You only need 4 out of 5. If one involves complex calculation, skip it.
Check with options. After calculating, verify your answer matches an option. If not, recheck your setup.

For integration techniques specifically, see the integration tricks and substitutions guide.

Common Mistakes in Maths Case Studies

Misreading the scenario. In LP problems, students mix up constraints (≤ vs ≥) or confuse which quantity to maximise. Read twice before writing equations.
Forgetting non-negativity constraints. In linear programming, x ≥ 0 and y ≥ 0 are always required. Missing them changes the feasible region.
Bayes' theorem calculation errors. The denominator in Bayes' formula requires summing over all possibilities. Students often forget one term.
Integration limits. Area problems specify bounds. Swapping upper and lower limits gives a negative area — the magnitude is correct but the sign is wrong.
Not simplifying before comparing with options. Your calculation might give 16/3, but the option says 5.33. Convert fractions to decimals or vice versa to match.

Check the board exam answer sheet writing guide for structuring your Maths paper.

FAQs

How many case study questions are in CBSE Class 12 Maths 2027?

Three case study questions in Section E, each carrying 4 marks (12 marks total out of 80). Each has 5 MCQ sub-questions — answer any 4.

Which chapters do Maths case studies come from?

Linear Programming, Probability (Bayes' theorem), Application of Integrals (area under curves), Vectors and 3D Geometry, Application of Derivatives (maxima/minima), and Relations and Functions are the most common sources.

Are case study sub-questions connected to each other?

Sometimes loosely — for example, (i) may ask for the objective function and (ii) for the constraints in an LP problem. But you can usually answer (iii), (iv), (v) independently even if you get (i) wrong.

Can I score 12/12 in Section E?

Yes. Since you skip one sub-question per case study, you need 4/5 correct in each. With practice of 15-20 case studies, most students score 10-12 out of 12.

How is the Maths SQP 2027 different from 2025-26?

The pattern is the same: 38 questions, 80 marks, 5 sections with Section E containing 3 case studies of 4 marks each. See the SQP 2027 analysis for confirmation.

Should I solve case studies first or last in the exam?

Attempt them after Sections A and B (MCQs and 2-mark questions). By then you are warmed up. Do not leave case studies for the last 10 minutes — they require careful reading and calculation.

CBSE Class 12 Maths Case Study Questions — All Chapters Practice 2027