Chapter 4 : Class 10 : Quadratic Equations: Case Study Question Bank

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Chapter 4: Quadratic Equations

Class 10 Standard Mathematics (Case Study Bank)

Curated By Akash Srivastva

✨ Interactive TLMs, Worksheets & Digital Study Notes: www.akashmaths.online

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Case Study 1 Space Exploration 4 Marks

Asteroid Deflection Trajectory

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A space defense agency tests an Asteroid Deflection System to intercept an incoming asteroid. The relative distance d (in thousands of kilometers) between the probe and asteroid is modeled as a function of time t (in minutes) by:

$$d(t) = t^2 - 12t + k$$

where k is a parameter determined by the initial launching coordinates of the probe.

Q1. Conceptual Understanding 1 Mark

If the probe is launched such that k = 36, find the time t at which the relative distance between probe and asteroid becomes exactly 0 (interception).

(A) t = 4 minutes

(B) t = 6 minutes

(C) t = 8 minutes

(D) t = 12 minutes

Q2. Discriminant & Nature of Roots 1 Mark

If launching coordinates give k = 40, determine the nature of the roots of d(t) = 0 and state if interception occurs.

(A) D > 0; two interceptions occur

(B) D = 0; single tangent interception

(C) D = −16 < 0; no real roots, never intercepts

(D) D = −4 < 0; delayed interception occurs

Q3. Analytical Reasoning 2 Marks

For k = 27, find the times t at which the relative distance between probe and asteroid is exactly 16 thousand kilometers.

(A) t = 1 min and t = 11 min

(B) t = 2 min and t = 10 min

(C) t = 3 min and t = 9 min

(D) t = 4 min and t = 8 min

OR (Alternative Q3)

Find the range of values of k for which the probe will strictly pass by the asteroid without ever intercepting it (no real roots).

(A) k < 36

(B) k ≤ 36

(C) k = 36

(D) k > 36

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	For k = 36: t² − 12t + 36 = 0 ⇒ (t − 6)² = 0 ⇒ t = 6 minutes.	[0.5 Mark] for quadratic equation. [0.5 Mark] for t = 6 minutes.
Q2	For k = 40: D = b² − 4ac = (−12)² − 4(1)(40) = 144 − 160 = −16. Since D < 0, there are no real roots; probe never intercepts the asteroid.	[0.5 Mark] for D = −16. [0.5 Mark] for no real roots / conclusion.
Q3	t² − 12t + 27 = 16 ⇒ t² − 12t + 11 = 0. (t − 1)(t − 11) = 0 ⇒ t = 1 minute and t = 11 minutes.	[1 Mark] for equation setup. [1 Mark] for t = 1 and 11 minutes.
Q3 (OR)	No real roots requires D < 0: (−12)² − 4(1)(k) < 0 ⇒ 144 − 4k < 0 ⇒ 4k > 144 ⇒ k > 36.	[1 Mark] for D < 0 condition. [1 Mark] for range k > 36.

Case Study 2 Green Energy 4 Marks

EV Highway Efficiency

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An electric vehicle covers 360 km at uniform speed x km/h. Increasing speed by 10 km/h reduces travel time by 3 hours for the same distance:

• Leg 1: Distance = 360 km, Speed = x km/h
• Leg 2: Distance = 360 km, Speed = (x + 10) km/h

Q1. Equation Modeling 1 Mark

Write a quadratic equation in terms of speed x that models the scenario 360/x − 360/(x + 10) = 3.

(A) x² + 10x − 1200 = 0

(B) x² − 10x − 1200 = 0

(C) x² + 10x − 360 = 0

(D) x² + 20x − 1200 = 0

Q2. Speed Evaluation 1 Mark

Find the original uniform speed of the electric vehicle.

(A) 25 km/h

(B) 30 km/h

(C) 35 km/h

(D) 40 km/h

Q3. Analytical Calculations 2 Marks

If rain reduces original speed by 5 km/h, find the percentage increase in travel time of the first leg.

(A) 15%

(B) 18%

(C) 20%

(D) 25%

OR (Alternative Q3)

Find the ratio of time taken in Leg 1 to time taken in Leg 2 of the test journey.

(A) 4 : 3

(B) 3 : 2

(C) 5 : 4

(D) 6 : 5

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	360/x − 360/(x + 10) = 3 ⇒ 120[10/(x² + 10x)] = 1 ⇒ x² + 10x − 1200 = 0.	[0.5 Mark] for time equation. [0.5 Mark] for standard form.
Q2	(x + 40)(x − 30) = 0 ⇒ x = 30 or x = −40 (rejected) ⇒ Speed = 30 km/h.	[0.5 Mark] for factoring. [0.5 Mark] for x = 30 km/h.
Q3	Original time = 360/30 = 12 h; Rain time = 360/25 = 14.4 h. % increase = (2.4 / 12) × 100% = 20%.	[1 Mark] for both times. [1 Mark] for 20%.
Q3 (OR)	t₁ = 360/30 = 12 h; t₂ = 360/40 = 9 h. Ratio = 12 : 9 = 4 : 3.	[1 Mark] for times. [1 Mark] for ratio 4 : 3.

Case Study 3 Search & Rescue 4 Marks

Search & Rescue Drone

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A rescue drone flies at 15 km/h in calm air. It travels 30 km against wind (speed w) and returns 30 km with wind. Total round trip takes 4 hours 30 minutes (9/2 hours):

• Upstream speed: (15 − w) km/h
• Downstream speed: (15 + w) km/h

Q1. Velocity Expressions 1 Mark

Write expressions for drone speed during outward journey (upstream) and return journey (downstream).

(A) (15 − w) km/h and (15 + w) km/h

(B) (w − 15) km/h and (w + 15) km/h

(C) 15w km/h and (15/w) km/h

(D) (30 − w) km/h and (30 + w) km/h

Q2. Quadratic Equation 1 Mark

Construct the quadratic equation in terms of wind speed w representing 30/(15 − w) + 30/(15 + w) = 9/2.

(A) w² − 36 = 0

(B) w² − 25 = 0

(C) w² + 25 = 0

(D) w² − 16 = 0

Q3. Wind Speed & Flight Duration 2 Marks

Find the wind speed w in km/h by solving the quadratic equation.

(A) 4 km/h

(B) 6 km/h

(C) 5 km/h

(D) 7.5 km/h

OR (Alternative Q3)

Find return journey time. If wind speed doubles for return journey, calculate the new return travel time.

(A) Original = 1.5 h; New = 1.2 h (1 h 12 min)

(B) Original = 2 h; New = 1.5 h

(C) Original = 1.5 h; New = 1.0 h

(D) Original = 1.8 h; New = 1.4 h

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Upstream = (15 − w) km/h, Downstream = (15 + w) km/h.	[0.5 Mark] each expression.
Q2	30/(15 − w) + 30/(15 + w) = 9/2 ⇒ 10[30 / (225 − w²)] = 3/2 ⇒ 200 = 225 − w² ⇒ w² − 25 = 0.	[0.5 Mark] for setup. [0.5 Mark] for w² − 25 = 0.
Q3	w² = 25 ⇒ w = 5 or w = −5 (rejected) ⇒ Wind speed = 5 km/h.	[1 Mark] for solving roots. [1 Mark] for w = 5 km/h.
Q3 (OR)	Original return = 30 / (15 + 5) = 1.5 h. Doubled wind w' = 10 km/h ⇒ New speed = 15 + 10 = 25 km/h. New return time = 30 / 25 = 1.2 h (1 h 12 min).	[1 Mark] for 1.5 h. [1 Mark] for 1.2 h.

Case Study 4 Sustainable Architecture 4 Marks

Solar Farm Layout

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A clean energy cooperative designs a rectangular solar farm of area 1800 m². The length L is specified to be 10 meters more than twice its width W:

• Width: W meters
• Length: L = (2W + 10) meters
• Area: W(2W + 10) = 1800 m²

Q1. Standard Form 1 Mark

Formulate a quadratic equation in standard form aw² + bw + c = 0 representing the width W of the solar farm.

(A) W² + 10W − 1800 = 0

(B) W² + 5W − 900 = 0

(C) 2W² + 5W − 900 = 0

(D) W² + 10W − 900 = 0

Q2. Farm Dimensions 1 Mark

Find the width and length of this eco-friendly solar farm.

(A) Width = 25 m, Length = 60 m

(B) Width = 20 m, Length = 50 m

(C) Width = 30 m, Length = 70 m

(D) Width = 24 m, Length = 75 m

Q3. Perimeter Comparison 2 Marks

Compare perimeter of original rectangle (170 m) with alternative square of same area (1800 m²).

(A) Square perimeter is larger (172.5 m)

(B) Both perimeters are exactly equal

(C) Square perimeter is 160 m

(D) Rectangular perimeter (170 m) is strictly greater than square (≈169.68 m)

OR (Alternative Q3)

If a safety buffer of 2.5 m is cleared inside all around the farm boundary, find the remaining inner area for panels.

(A) 1000 m²

(B) 1100 m²

(C) 1200 m²

(D) 1250 m²

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	W(2W + 10) = 1800 ⇒ 2W² + 10W − 1800 = 0 ⇒ W² + 5W − 900 = 0.	[0.5 Mark] for area relation. [0.5 Mark] for standard form.
Q2	(W − 25)(W + 30) = 0 ⇒ W = 25 m (rejecting −30). Length L = 2(25) + 10 = 60 m.	[0.5 Mark] for W = 25 m. [0.5 Mark] for L = 60 m.
Q3	Rectangle Perimeter = 2(60 + 25) = 170 m. Square side = √1800 = 30√2 ≈ 42.42 m ⇒ Perimeter = 120√2 ≈ 169.68 m. 170 m > 169.68 m.	[1 Mark] for both perimeters. [1 Mark] for proof.
Q3 (OR)	Inner width = 25 − 2(2.5) = 20 m. Inner length = 60 − 2(2.5) = 55 m. Inner area = 55 × 20 = 1100 m².	[1 Mark] for inner dimensions. [1 Mark] for 1100 m².

Case Study 5 Fintech Algorithm 4 Marks

Algorithmic Trading Profit

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A fintech AI trading algorithm's daily profit P(x) (in ₹'000) depends on the number of executed contracts x:

$$P(x) = -x^2 + 50x - 400$$

The algorithm automatically pauses trading whenever net profit is negative (P(x) < 0).

Q1. Break-Even Analysis 1 Mark

Find the break-even points where the algorithm makes neither profit nor loss (P(x) = 0).

(A) 5 and 45 contracts

(B) 10 and 40 contracts

(C) 15 and 35 contracts

(D) 20 and 30 contracts

Q2. Maximum Profit 1 Mark

Determine the maximum daily profit (in INR) that this trading algorithm can achieve.

(A) ₹ 2,25,000 at x = 25

(B) ₹ 2,00,000 at x = 20

(C) ₹ 2,50,000 at x = 30

(D) ₹ 1,75,000 at x = 25

Q3. Target Interval 2 Marks

Calculate the range of daily contracts x executed to ensure a profit of at least ₹ 1,25,000 (P(x) ≥ 125).

(A) 10 ≤ x ≤ 40

(B) 12 ≤ x ≤ 38

(C) 15 ≤ x ≤ 35

(D) 20 ≤ x ≤ 30

OR (Alternative Q3)

For upgraded model Q(x) = −x² + 60x − 500, find new break-even points and compare operating range width.

(A) 10 and 50 contracts; Wider operating range (40 vs 30)

(B) 15 and 45 contracts; Same operating range

(C) 20 and 40 contracts; Narrower operating range

(D) 5 and 55 contracts; Wider operating range

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	−x² + 50x − 400 = 0 ⇒ x² − 50x + 400 = 0 ⇒ (x − 10)(x − 40) = 0 ⇒ x = 10 and 40 contracts.	[0.5 Mark] for setup. [0.5 Mark] for 10 and 40.
Q2	Vertex x = −50 / (2 × −1) = 25. P(25) = −625 + 1250 − 400 = 225 thousand INR = ₹ 2,25,000.	[0.5 Mark] for x = 25. [0.5 Mark] for ₹ 2,25,000.
Q3	−x² + 50x − 400 ≥ 125 ⇒ x² − 50x + 525 ≤ 0 ⇒ (x − 15)(x − 35) ≤ 0 ⇒ 15 ≤ x ≤ 35.	[1 Mark] for inequality setup. [1 Mark] for 15 ≤ x ≤ 35.
Q3 (OR)	x² − 60x + 500 = 0 ⇒ (x − 10)(x − 50) = 0 ⇒ x = 10, 50. Range = 50 − 10 = 40 contracts, which is wider than original (30 contracts).	[1 Mark] for new roots 10, 50. [1 Mark] for wider range comparison.

Case Study 6 Modern Architecture 4 Marks

Eco-Resort Amphitheater

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An outdoor eco-resort amphitheater accommodates 480 viewers in r semicircular rows. The seats per row are exactly 8 more than the number of rows:

• Number of rows: r
• Seats per row: r + 8
• Total capacity: r(r + 8) = 480

Q1. Equation Modeling 1 Mark

Write down the quadratic equation in terms of r that models this amphitheater layout.

(A) r² + 8r − 480 = 0

(B) r² − 8r − 480 = 0

(C) r² + 8r + 480 = 0

(D) 2r² + 8r − 480 = 0

Q2. Seating Details 1 Mark

Find the total number of rows and the number of seats in the outermost row under this design.

(A) 16 rows and 24 seats

(B) 20 rows and 28 seats

(C) 24 rows and 32 seats

(D) 18 rows and 26 seats

Q3. Layout Scaled Revision 2 Marks

If capacity is scaled down to 384 seats with same rule (seats per row = r + 8), calculate the new number of rows.

(A) 16 rows

(B) 14 rows

(C) 18 rows

(D) 12 rows

OR (Alternative Q3)

If seats per row is changed to twice the number of rows for 480 seats, determine if roots of new equation are real.

(A) D < 0; roots are not real

(B) D = 0; roots are real and equal

(C) r² − 240 = 0; D = 960 > 0, roots are real and distinct

(D) r² − 480 = 0; roots are irrational and imaginary

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Total seats = r(r + 8) = 480 ⇒ r² + 8r − 480 = 0.	[1 Mark] for quadratic formulation.
Q2	(r − 20)(r + 24) = 0 ⇒ r = 20 (rejecting −24). Outermost row seats = r + 8 = 20 + 8 = 28 seats.	[0.5 Mark] for r = 20 rows. [0.5 Mark] for 28 seats.
Q3	r(r + 8) = 384 ⇒ r² + 8r − 384 = 0 ⇒ (r − 16)(r + 24) = 0 ⇒ r = 16 rows.	[1 Mark] for equation. [1 Mark] for r = 16.
Q3 (OR)	r(2r) = 480 ⇒ 2r² − 480 = 0 ⇒ r² − 240 = 0. D = 0² − 4(1)(−240) = 960 > 0, hence roots are real and distinct.	[1 Mark] for equation. [1 Mark] for D > 0 and real roots conclusion.

Case Study 7 Warehouse Automation 4 Marks

Fulfillment Center Sorting

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Two robotic sorting systems (Alpha and Beta) complete a parcel shipment together in 6 hours. Working alone, Alpha takes 5 hours less than Beta:

• Beta alone: x hours
• Alpha alone: (x − 5) hours
• Together: 1/x + 1/(x − 5) = 1/6

Q1. Work-Rate Formulation 1 Mark

Write down the simplified quadratic equation in terms of x representing their combined work rate.

(A) x² − 17x + 30 = 0

(B) x² − 15x + 30 = 0

(C) x² − 12x + 25 = 0

(D) x² − 17x + 60 = 0

Q2. Machine Alpha's Time 1 Mark

Calculate the individual time (in hours) taken by Machine Alpha to complete the job alone.

(A) 8 hours

(B) 10 hours

(C) 12 hours

(D) 15 hours

Q3. Upgrades & Operational Delays 2 Marks

Machine Gamma is twice as fast as Alpha. Find Gamma's alone time and combined time of Alpha + Gamma.

(A) Gamma alone = 6 h; Together = 4 h

(B) Gamma alone = 4 h; Together = 3 h

(C) Gamma alone = 5 h; Together = 3 h 20 min (10/3 h)

(D) Gamma alone = 5 h; Together = 3 h 45 min

OR (Alternative Q3)

If Beta's alone time increases by 10 hours (to 25 h) while Alpha stays at 10 h, find new combined time.

(A) 50/7 hours (≈ 7.14 hours)

(B) 7.5 hours

(C) 8 hours

(D) 45/7 hours

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	1/x + 1/(x − 5) = 1/6 ⇒ (2x − 5)/(x² − 5x) = 1/6 ⇒ x² − 17x + 30 = 0.	[1 Mark] for quadratic derivation.
Q2	(x − 15)(x − 2) = 0 ⇒ x = 15 h for Beta (x > 5). Alpha alone = x − 5 = 15 − 5 = 10 hours.	[0.5 Mark] for Beta's time. [0.5 Mark] for Alpha = 10 h.
Q3	Gamma alone = 10/2 = 5 hours. Combined rate = 1/10 + 1/5 = 3/10 ⇒ Time = 10/3 h = 3 hours 20 minutes.	[1 Mark] for Gamma = 5 h. [1 Mark] for 3 h 20 min.
Q3 (OR)	Beta new = 15 + 10 = 25 h; Alpha = 10 h. Combined rate = 1/10 + 1/25 = 7/50 ⇒ Time = 50/7 hours ≈ 7.14 hours.	[0.5 Mark] for Beta = 25 h. [1.5 Marks] for 50/7 hours.

Case Study 8 Climate Infrastructure 4 Marks

Rainwater Harvesting Pond

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A rectangular rainwater reservoir measures 20 m by 15 m. A gravel path of uniform width x meters is constructed all around it, bringing total area to 500 m²:

• Reservoir: 20 m × 15 m (Area = 300 m²)
• Effective dimensions: (20 + 2x) m by (15 + 2x) m
• Total area: (20 + 2x)(15 + 2x) = 500 m²

Q1. Effective Dimensions 1 Mark

Express the total effective length and width of the installation in terms of path width x.

(A) (20 + 2x) m and (15 + 2x) m

(B) (20 + x) m and (15 + x) m

(C) (40 + 2x) m and (30 + 2x) m

(D) (20 − 2x) m and (15 − 2x) m

Q2. Simplified Equation 1 Mark

Formulate a quadratic equation in simplest form representing total effective area of 500 m².

(A) 4x² + 70x − 500 = 0

(B) 2x² + 35x − 100 = 0

(C) x² + 35x − 100 = 0

(D) 2x² + 70x − 200 = 0

Q3. Path Width & Construction Budget 2 Marks

Solve the quadratic equation to determine the exact width x of the gravel path in meters.

(A) 2.5 meters

(B) 3.0 meters

(C) 2.0 meters

(D) 3.5 meters

OR (Alternative Q3)

If laying the gravel path costs ₹250 per square meter, calculate total construction cost of the path.

(A) ₹ 45,000

(B) ₹ 60,000

(C) ₹ 50,000

(D) ₹ 55,000

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Effective length = (20 + 2x) m, Effective width = (15 + 2x) m.	[1 Mark] for dimensions.
Q2	(20 + 2x)(15 + 2x) = 500 ⇒ 4x² + 70x − 200 = 0 ⇒ 2x² + 35x − 100 = 0.	[1 Mark] for simplest quadratic equation.
Q3	(2x − 5)(x + 20) = 0 ⇒ x = 2.5 m (rejecting negative −20 m).	[1 Mark] for factoring. [1 Mark] for x = 2.5 m.
Q3 (OR)	Path Area = Total Area − Reservoir Area = 500 − (20 × 15) = 200 m². Cost = 200 × ₹250 = ₹ 50,000.	[1 Mark] for area = 200 m². [1 Mark] for total cost ₹ 50,000.

Live Practice: Chapter 4 Quadratic Equations

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Case Study 1

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