Chapter 3 : Class 10 : Linear Equations: Case Study Question Bank

✦ NCF 2023 • COMPETENCY-FOCUSED QUESTION BANK ✦

Chapter 3: Linear Equations in Two Variables

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 Smart City Mobility 4 Marks

EV Cab Fare Structure

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In a newly designed eco-friendly smart city, a fleet of high-efficiency Electric Cabs (EV Cabs) operates under a specialized pricing algorithm. The fare structure is designed with two components: a fixed base charge (₹ x) and a variable charge per kilometer (₹ y).

• Ananya traveled a distance of 15 km and paid a total fare of ₹ 350.
• Kabir traveled a distance of 25 km and paid a total fare of ₹ 550.

Q1. Formulation of Equations 1 Mark

Formulate a pair of linear equations in two variables representing the fare structure for Ananya and Kabir.

(A) x + 15y = 350 and x + 25y = 550

(B) 15x + y = 350 and 25x + y = 550

(C) x + 15y = 550 and x + 25y = 350

(D) x − 15y = 350 and x − 25y = 550

Q2. Algebraic Solution 1 Mark

Find the value of the fixed booking charge (x) and the rate charged per kilometer (y).

(A) x = ₹40, y = ₹22/km

(B) x = ₹50, y = ₹20/km

(C) x = ₹60, y = ₹18/km

(D) x = ₹30, y = ₹25/km

Q3. Analytical Reasoning 2 Marks

Calculate the total fare a passenger has to pay for traveling a distance of 32 km in this smart city.

(A) ₹640

(B) ₹670

(C) ₹690

(D) ₹720

OR (Alternative Q3)

If the fixed booking charge is reduced by 20% and the per-km rate is increased by ₹2, find the new total fare for Ananya's 15 km journey.

(A) ₹370

(B) ₹360

(C) ₹385

(D) ₹340

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Let fixed charge be ₹ x and per-km charge be ₹ y. Ananya: x + 15y = 350. Kabir: x + 25y = 550.	[0.5 Mark] for each correct equation.
Q2	Subtracting equations: 10y = 200 ⇒ y = 20. Substituting y = 20: x + 15(20) = 350 ⇒ x = 50. Fixed charge = ₹ 50, Per-km charge = ₹ 20/km.	[0.5 Mark] for y = 20. [0.5 Mark] for x = 50.
Q3	T = x + 32y = 50 + 32(20) = 50 + 640 = ₹ 690.	[1 Mark] for formula setup. [1 Mark] for evaluating ₹ 690.
Q3 (OR)	New base charge = 50 − 20%(50) = ₹ 40. New per-km rate = 20 + 2 = ₹ 22/km. New fare for 15 km = 40 + 15(22) = 40 + 330 = ₹ 370.	[1 Mark] for new rates. [1 Mark] for final fare ₹ 370.

Case Study 2 Flight Operations 4 Marks

Drone Collision Avoidance

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An automated e-commerce fulfillment warehouse uses AI-guided quadcopter drones to sort packages in a 2D floor grid. Two drones are tracked along linear flight trajectories:

• Drone Alpha: 2x + 3y = 12
• Drone Beta: 3x − 2y = 5

Q1. Trajectory Nature 1 Mark

Determine whether the flight paths of Drone Alpha and Drone Beta are intersecting, parallel, or coincident using coefficient ratios.

(A) Intersecting lines (a₁/a₂ ≠ b₁/b₂)

(B) Parallel lines (a₁/a₂ = b₁/b₂ ≠ c₁/c₂)

(C) Coincident lines (a₁/a₂ = b₁/b₂ = c₁/c₂)

(D) Perpendicular but non-intersecting

Q2. Intersection Point 1 Mark

Calculate the coordinates of the potential collision point where the trajectories intersect.

(A) (2, 3)

(B) (3, 2)

(C) (4, 1)

(D) (1, 3)

Q3. Geometric Consistency 2 Marks

Drone Gamma flies along kx + 6y = 18. Find the value of k such that Drone Gamma's path is strictly parallel to Drone Alpha's path (2x + 3y = 12).

(A) k = 2

(B) k = 3

(C) k = 4

(D) k = 6

OR (Alternative Q3)

Drone Delta flies along 4x + my − 24 = 0. Find the value of m for which Drone Delta's path is coincident with Drone Alpha's path.

(A) m = 3

(B) m = 4

(C) m = 8

(D) m = 6

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	a₁/a₂ = 2/3, b₁/b₂ = 3/(−2) = −3/2. Since a₁/a₂ ≠ b₁/b₂, the lines intersect at a unique point.	[0.5 Mark] for ratios. [0.5 Mark] for conclusion.
Q2	2(2x + 3y = 12) ⇒ 4x + 6y = 24. 3(3x − 2y = 5) ⇒ 9x − 6y = 15. Adding: 13x = 39 ⇒ x = 3. 2(3) + 3y = 12 ⇒ 3y = 6 ⇒ y = 2. Intersection point = (3, 2).	[0.5 Mark] for elimination. [0.5 Mark] for point (3, 2).
Q3	Parallel condition: a₁/a₃ = b₁/b₃ ≠ c₁/c₃. 2/k = 3/6 ≠ (−12)/(−18) ⇒ 2/k = 1/2 ≠ 2/3 ⇒ k = 4.	[1 Mark] for condition. [1 Mark] for k = 4.
Q3 (OR)	Coincident condition: a₁/a₄ = b₁/b₄ = c₁/c₄. 2/4 = 3/m = (−12)/(−24) ⇒ 1/2 = 3/m = 1/2 ⇒ m = 6.	[1 Mark] for condition. [1 Mark] for m = 6.

Case Study 3 Renewable Energy 4 Marks

Hybrid Solar & Wind Grid

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A smart village community installs a hybrid power micro-grid featuring solar panels (x kW each) and wind turbines (y kW each). The maintenance team logs two equipment configurations:

• Config 1: 5 Solar Panels + 3 Wind Turbines = 29 kW
• Config 2: 3 Solar Panels + 4 Wind Turbines = 24 kW

Q1. System Formulation 1 Mark

Write down the linear system representing the peak power generation of solar panels (x) and wind turbines (y).

(A) 3x + 5y = 29 and 4x + 3y = 24

(B) 5x + 3y = 29 and 3x + 4y = 24

(C) 5x + 4y = 29 and 3x + 3y = 24

(D) 5x + 3y = 24 and 3x + 4y = 29

Q2. Unit Capacities 1 Mark

Determine the individual peak power generation capacity of a single Solar Panel and a single Wind Turbine in kW.

(A) Solar = 4 kW, Wind = 3 kW

(B) Solar = 5 kW, Wind = 2 kW

(C) Solar = 3 kW, Wind = 4 kW

(D) Solar = 4.5 kW, Wind = 2.5 kW

Q3. Grid Optimization 2 Marks

If the village scales up the hybrid grid to 8 Solar Panels and 6 Wind Turbines, calculate the total expected peak power output in kW.

(A) 46 kW

(B) 48 kW

(C) 54 kW

(D) 50 kW

OR (Alternative Q3)

Determine the ratio of the peak power produced by 4 Solar Panels to that produced by 3 Wind Turbines.

(A) 4 : 3

(B) 12 : 7

(C) 16 : 9

(D) 9 : 16

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Config 1: 5x + 3y = 29. Config 2: 3x + 4y = 24.	[0.5 Mark] each equation.
Q2	4(5x + 3y = 29) ⇒ 20x + 12y = 116. 3(3x + 4y = 24) ⇒ 9x + 12y = 72. Subtracting: 11x = 44 ⇒ x = 4 kW. 5(4) + 3y = 29 ⇒ 3y = 9 ⇒ y = 3 kW.	[0.5 Mark] for x = 4 kW. [0.5 Mark] for y = 3 kW.
Q3	P = 8x + 6y = 8(4) + 6(3) = 32 + 18 = 50 kW.	[1 Mark] for expression. [1 Mark] for 50 kW.
Q3 (OR)	4 Solar Panels = 4(4) = 16 kW. 3 Wind Turbines = 3(3) = 9 kW. Ratio = 16 : 9.	[1 Mark] for individual outputs. [1 Mark] for ratio 16:9.

Case Study 4 Marine Exploration 4 Marks

Deep-Sea Submersible

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An oceanographic institute deploys an autonomous robotic submersible in a marine trench with current flow:

• Upstream (against current): Covers 24 km in 4 hours.
• Downstream (with current): Covers 36 km in 3 hours.

Q1. Velocity Equations 1 Mark

Formulate the equations representing the speed of the submersible in still water (x) and the ocean current (y) upstream and downstream.

(A) x − y = 6 and x + y = 12

(B) x − y = 4 and x + y = 3

(C) x + y = 6 and x − y = 12

(D) 2x − y = 6 and 2x + y = 12

Q2. Speeds Evaluation 1 Mark

Determine the speed of the submersible in still water (x) and the velocity of the ocean current (y).

(A) x = 8 km/h, y = 4 km/h

(B) x = 9 km/h, y = 3 km/h

(C) x = 10 km/h, y = 2 km/h

(D) x = 7.5 km/h, y = 4.5 km/h

Q3. Time & Storm Conditions 2 Marks

How much time will the submersible take to travel a distance of 45 km in still water?

(A) 4 hours

(B) 4.5 hours

(C) 5 hours

(D) 6 hours

OR (Alternative Q3)

If a storm increases the ocean current velocity to double its original speed, calculate the new downstream speed.

(A) 15 km/h

(B) 18 km/h

(C) 12 km/h

(D) 16 km/h

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Upstream: x − y = 24/4 = 6. Downstream: x + y = 36/3 = 12.	[0.5 Mark] each equation.
Q2	Adding: 2x = 18 ⇒ x = 9 km/h. Substituting: 9 + y = 12 ⇒ y = 3 km/h.	[0.5 Mark] for x = 9. [0.5 Mark] for y = 3.
Q3	Time = Distance / Speed = 45 / 9 = 5 hours.	[1 Mark] for formula. [1 Mark] for 5 hours.
Q3 (OR)	New current y_new = 2 × 3 = 6 km/h. New downstream speed = x + y_new = 9 + 6 = 15 km/h.	[1 Mark] for new current. [1 Mark] for 15 km/h.

Case Study 5 Sports Nutrition 4 Marks

Sports Nutrition Plan

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A sports nutritionist designs a recovery diet combining Food A (x units) and Food B (y units):

• Food A: 4 g protein, 8 g carbs per unit.
• Food B: 6 g protein, 4 g carbs per unit.
• Target: Exactly 44 g protein and 48 g carbohydrates.

Q1. Constraint Equations 1 Mark

Formulate the simplified pair of linear equations representing the protein and carbohydrate constraints.

(A) 4x + 6y = 22 and 2x + y = 24

(B) 2x + 3y = 22 and 2x + y = 12

(C) 2x + 3y = 44 and 4x + 2y = 48

(D) 3x + 2y = 22 and x + 2y = 12

Q2. Consistency Verification 1 Mark

Check whether the system of equations is consistent and has a unique solution using coefficient ratios.

(A) Consistent with unique solution (a₁/a₂ ≠ b₁/b₂)

(B) Inconsistent with no solution

(C) Dependent with infinitely many solutions

(D) Coincident system

Q3. Dietary Solution 2 Marks

Determine the exact number of units of Food A and Food B that must be blended to meet the target.

(A) Food A = 4, Food B = 4

(B) Food A = 3, Food B = 6

(C) Food A = 3.5, Food B = 5

(D) Food A = 5, Food B = 3.5

OR (Alternative Q3)

If Food A costs ₹20/unit and Food B costs ₹30/unit, calculate the total cost of the recovery meal.

(A) ₹200

(B) ₹220

(C) ₹240

(D) ₹250

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Protein: 4x + 6y = 44 ⇒ 2x + 3y = 22. Carbs: 8x + 4y = 48 ⇒ 2x + y = 12.	[0.5 Mark] each equation.
Q2	a₁/a₂ = 2/2 = 1, b₁/b₂ = 3/1 = 3. Since a₁/a₂ ≠ b₁/b₂, the system is consistent with a unique solution.	[0.5 Mark] for ratios. [0.5 Mark] for consistency conclusion.
Q3	Subtracting (2x + y = 12) from (2x + 3y = 22): 2y = 10 ⇒ y = 5 units. 2x + 5 = 12 ⇒ 2x = 7 ⇒ x = 3.5 units. Food A = 3.5 units, Food B = 5 units.	[1 Mark] for finding y = 5. [1 Mark] for finding x = 3.5.
Q3 (OR)	Cost = 20(3.5) + 30(5) = 70 + 150 = ₹ 220.	[1 Mark] for substitution. [1 Mark] for evaluating ₹ 220.

Case Study 6 Warehouse Robotics 4 Marks

Warehouse AGV Sorters

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An automated sorting hub operates Type 1 robots (x packages/min) and Type 2 robots (y packages/min):

• Scenario A: 3 Type 1 + 2 Type 2 = 120 pkgs/min.
• Scenario B: 4 Type 1 + 5 Type 2 = 230 pkgs/min.

Q1. System Setup 1 Mark

Write down the system of two linear equations that model the sorting rates of both robots.

(A) 3x + 2y = 120 and 4x + 5y = 230

(B) 2x + 3y = 120 and 5x + 4y = 230

(C) 3x + 2y = 230 and 4x + 5y = 120

(D) 4x + 2y = 120 and 3x + 5y = 230

Q2. Fleet Output 1 Mark

How many packages can a fleet of 5 Type 2 AGVs sort in 10 minutes?

(A) 1200 packages

(B) 1000 packages

(C) 1800 packages

(D) 1500 packages

Q3. Rates Resolution 2 Marks

Find the individual sorting rates (packages sorted per minute) of a single Type 1 AGV and a single Type 2 AGV.

(A) Type 1 = 25, Type 2 = 25

(B) Type 1 = 20, Type 2 = 30

(C) Type 1 = 30, Type 2 = 20

(D) Type 1 = 15, Type 2 = 35

OR (Alternative Q3)

Can a fleet of 7 Type 1 AGVs and 7 Type 2 AGVs achieve a daily target rate of exactly 350 packages per minute?

(A) No, rate is 320 pkgs/min

(B) No, rate is 340 pkgs/min

(C) Yes, combined rate is exactly 350 pkgs/min

(D) Yes, exceeds target with 380 pkgs/min

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	3x + 2y = 120 and 4x + 5y = 230.	[0.5 Mark] each equation.
Q2	Using y = 30 pkgs/min: Rate = 5(30) = 150 pkgs/min. Volume in 10 min = 150 × 10 = 1500 packages.	[0.5 Mark] for rate. [0.5 Mark] for 1500 pkgs.
Q3	5(3x + 2y = 120) ⇒ 15x + 10y = 600. 2(4x + 5y = 230) ⇒ 8x + 10y = 460. Subtracting: 7x = 140 ⇒ x = 20 pkgs/min. 3(20) + 2y = 120 ⇒ 2y = 60 ⇒ y = 30 pkgs/min.	[1 Mark] for x = 20. [1 Mark] for y = 30.
Q3 (OR)	Rate = 7x + 7y = 7(20) + 7(30) = 140 + 210 = 350 pkgs/min. Yes, the proposed fleet can achieve the target.	[1 Mark] for evaluation. [1 Mark] for logical conclusion.

Case Study 7 Green Architecture 4 Marks

Eco-Friendly Community Park

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An architect plans a rectangular green zone of length x meters and width y meters (area = xy):

• Condition I: Length reduced by 5 m, width increased by 3 m ⇒ Area reduced by 9 m².
• Condition II: Length increased by 3 m, width increased by 2 m ⇒ Area increased by 67 m².

Q1. System Equations 1 Mark

Formulate the simplified pair of linear equations representing both dimensional alterations.

(A) 5x − 3y = 6 and 3x + 2y = 61

(B) 3x − 5y = 6 and 2x + 3y = 61

(C) 3x − 5y = 9 and 2x + 3y = 67

(D) 3x + 5y = 6 and 2x − 3y = 61

Q2. Park Perimeter 1 Mark

Find the perimeter of the original planned rectangular community park.

(A) 48 meters

(B) 50 meters

(C) 52 meters

(D) 56 meters

Q3. Dimensions & Budget 2 Marks

Determine the original planned dimensions (length and width) of the park.

(A) Length = 17 m, Width = 9 m

(B) Length = 18 m, Width = 8 m

(C) Length = 16 m, Width = 10 m

(D) Length = 19 m, Width = 7 m

OR (Alternative Q3)

Find the original area and calculate the budget required to sod it with grass at ₹50 per sq meter.

(A) 144 m² and ₹7,200

(B) 160 m² and ₹8,000

(C) 150 m² and ₹7,500

(D) 153 m² and ₹7,650

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	(x − 5)(y + 3) = xy − 9 ⇒ 3x − 5y = 6. (x + 3)(y + 2) = xy + 67 ⇒ 2x + 3y = 61.	[0.5 Mark] each equation.
Q2	Using x = 17 m, y = 9 m: Perimeter = 2(x + y) = 2(17 + 9) = 2(26) = 52 m.	[0.5 Mark] for formula. [0.5 Mark] for 52 m.
Q3	3(3x − 5y = 6) ⇒ 9x − 15y = 18. 5(2x + 3y = 61) ⇒ 10x + 15y = 305. Adding: 19x = 323 ⇒ x = 17 m. 3(17) − 5y = 6 ⇒ 5y = 45 ⇒ y = 9 m.	[1 Mark] for x = 17 m. [1 Mark] for y = 9 m.
Q3 (OR)	Area = 17 × 9 = 153 m². Budget = 153 × ₹50 = ₹ 7,650.	[1 Mark] for 153 m². [1 Mark] for ₹ 7,650.

Case Study 8 High-Speed Transit 4 Marks

Hyperloop Two-Way Travel

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An experimental Hyperloop connects stations Alpha and Beta 150 km apart. Pod speeds are u km/h (faster) and v km/h (slower):

• Same direction: Meet in 5 hours.
• Opposite directions: Meet in 1 hour.

Q1. Relative Speed System 1 Mark

Formulate the pair of linear equations representing both test run cases.

(A) u − v = 50 and u + v = 150

(B) u − v = 30 and u + v = 150

(C) u − v = 25 and u + v = 100

(D) 5u − v = 150 and u + v = 30

Q2. Distance Covered 1 Mark

If the slower pod travels continuously for 2.5 hours at its speed, how much distance will it cover?

(A) 120 km

(B) 140 km

(C) 150 km

(D) 175 km

Q3. Pod Speeds 2 Marks

Determine the respective speeds of both hyperloop pods in km/h under testing.

(A) Faster = 90 km/h, Slower = 60 km/h

(B) Faster = 100 km/h, Slower = 50 km/h

(C) Faster = 85 km/h, Slower = 65 km/h

(D) Faster = 95 km/h, Slower = 55 km/h

OR (Alternative Q3)

If distance is 180 km and pods meet in 6 hours (same direction) and 1.5 hours (towards each other), find the speed of the faster pod.

(A) 70 km/h

(B) 75 km/h

(C) 80 km/h

(D) 85 km/h

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Same direction: 5(u − v) = 150 ⇒ u − v = 30. Opposite: 1(u + v) = 150 ⇒ u + v = 150.	[0.5 Mark] each equation.
Q2	Using v = 60 km/h: Distance = 60 × 2.5 = 150 km.	[0.5 Mark] for formula. [0.5 Mark] for 150 km.
Q3	Adding equations: 2u = 180 ⇒ u = 90 km/h. 90 + v = 150 ⇒ v = 60 km/h. Faster = 90 km/h, Slower = 60 km/h.	[1 Mark] for u = 90 km/h. [1 Mark] for v = 60 km/h.
Q3 (OR)	6(u − v) = 180 ⇒ u − v = 30. 1.5(u + v) = 180 ⇒ u + v = 120. Adding: 2u = 150 ⇒ u = 75 km/h.	[1 Mark] for equations. [1 Mark] for u = 75 km/h.

Live Practice: Chapter 3 Linear Equations

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Case Study 1

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