Maths Competency-Based Word Problem Practice Pack (NCF 2023 Aligned)

#Section A: Real-World Case Studies (4 Marks Each)  

Case Study 1: Smart City Solar Array Optimization (Engineering & Area Optimization)

Context: As part of a green energy initiative, a municipal corporation is installing a rectangular array of solar panels on the roof of a community hall. The total area allocated for the panels is $180 \text{ m}^2$.

To optimize energy capture during peak sunlight hours, engineers determined that the length of the array must be $3\text{ meters}$ more than twice its width. Furthermore, a border walkway of uniform width $1\text{ meter}$ must be left around all four sides of the roof surface outside the array.

Questions:

1. Formulate the quadratic equation that represents the dimensions of the solar panel array in terms of its width $x$. (1 Mark)

2. Find the exact dimensions (length and width) of the solar array. (2 Marks)

3. Calculate the total roof area required including the $1\text{-meter}$ surrounding walkway. (1 Mark)

Case Study 2: Vande Bharat Express Speed Delta (Kinematics & Transportation)

Context: The Vande Bharat Express covers a distance of $360\text{ km}$ between two major junctions at a uniform average speed. On a rainy day, due to track safety protocols, the train's average speed is reduced by $10\text{ km/h}$, causing the journey to take $1\text{ hour}$ longer than scheduled.

Questions:

1. Write the algebraic expression for the time taken for the trip under normal speed ($v$) and reduced speed. (1 Mark)

2. Set up the quadratic equation and solve for the original average speed ($v$) of the train. (2 Marks)

3. If the track safety speed limit on rainy days permits a maximum delay of $45\text{ minutes}$, what is the maximum speed reduction allowed for a $360\text{ km}$ trip assuming an original speed of $90\text{ km/h}$? (1 Mark)

Question 3: Financial Modeling & Shared Unit Economics

A group of Class 10 students organized a STEM exhibition and rented an advanced 3D printing module for a fixed total cost of $\text{₹ }7,200$, which was to be shared equally among all participating students.

Just before paying, $6\text{ students}$ withdrew from the project. As a result, each remaining student had to contribute $\text{₹ }60\text{ more}$ than originally planned to cover the rental fee.

• Task A: Derive the quadratic equation representing the initial number of participating students $n$.

• Task B: Solve for $n$ and determine the actual contribution paid by each student who remained in the project.

🗝️ Official Marking Schemes & Detailed Solutions

Solution to Case Study 1

Part 1: Equation Formulation (1 Mark)

• Let the width of the solar array be $x\text{ meters}$.

• Then, the length $L = (2x + 3)\text{ meters}$.

• $\text{Area} = \text{Length} \times \text{Width} \implies x(2x + 3) = 180$

• Standard Quadratic Form:

  $$2x^2 + 3x - 180 = 0 \quad \text{[1 Mark]}$$

Part 2: Solving for Array Dimensions (2 Marks)

• Calculate Discriminant ($D$) or Factorize by splitting the middle term:

  $$2x^2 + 20x - 17x - 180 = 0 \quad (\text{Product} = -360, \text{ Sum} = +3)$$
  $$2x(x + 10) - 17(x + 10) = 0$$
  $$(2x - 17)(x + 10) = 0 \quad \text{[1 Mark]}$$

• Roots: $x = \frac{17}{2} = 8.5$ or $x = -10$.

• Since width cannot be negative, reject $x = -10$.

• Width ($W$) $= 8.5\text{ m}$

• Length ($L$) $= 2(8.5) + 3 = 17 + 3 = 20\text{ m}$ $\quad \text{[1 Mark]}$

Part 3: Total Roof Area with Walkway (1 Mark)

• Walkway adds $1\text{ m}$ on both sides of width and length:

  $$\text{Total Roof Length} = 20 + 1 + 1 = 22\text{ m}$$
  $$\text{Total Roof Width} = 8.5 + 1 + 1 = 10.5\text{ m}$$

• $\text{Total Roof Area} = 22 \times 10.5 = 231\text{ m}^2 \quad \text{[1 Mark]}$

Solution to Case Study 2

Part 1: Time Expressions (1 Mark)

• Distance $S = 360\text{ km}$.

• Normal speed = $v\text{ km/h} \implies \text{Normal time } t_1 = \frac{360}{v}\text{ hours}$.

• Reduced speed = $(v - 10)\text{ km/h} \implies \text{Reduced time } t_2 = \frac{360}{v - 10}\text{ hours}$. $\quad \text{[1 Mark]}$

Part 2: Quadratic Equation and Solving (2 Marks)

• Given time difference $t_2 - t_1 = 1\text{ hour}$:

  $$\frac{360}{v - 10} - \frac{360}{v} = 1$$
  $$360 \left( \frac{v - (v - 10)}{v(v - 10)} \right) = 1 \implies \frac{360 \times 10}{v^2 - 10v} = 1$$
  $$v^2 - 10v - 3600 = 0 \quad \text{[1 Mark]}$$

• Factorize:

  $$v^2 - 60v + 50v - 3600 = 0$$
  $$(v - 60)(v + 50) = 0$$

• $v = 60$ or $v = -50$.

• Reject $v = -50$ as speed cannot be negative.

• Original Speed ($v$) $= 60\text{ km/h}$. $\quad \text{[1 Mark]}$

Part 3: Speed Reduction Threshold (1 Mark)

• Normal speed $v = 90\text{ km/h} \implies t_1 = \frac{360}{90} = 4\text{ hours}$.

• Max allowed time $t_2 = 4\text{ hours} + 45\text{ mins} = 4.75\text{ hours} = \frac{19}{4}\text{ hours}$.

• New speed $v' = \frac{360}{4.75} = \frac{360 \times 4}{19} \approx 75.79\text{ km/h}$.

• Max speed reduction allowed $= 90 - 75.79 = 14.21\text{ km/h}$. $\quad \text{[1 Mark]}$

Solution to Question 3

Task A: Equation Derivation (2.5 Marks)

• Let the original number of students be $n$.

• Original cost per student $= \frac{7200}{n}\text{ rupees}$. $\quad \text{[0.5 Mark]}$

• New number of students $= (n - 6)$.

• New cost per student $= \frac{7200}{n - 6}\text{ rupees}$. $\quad \text{[0.5 Mark]}$

• Difference equation:

  $$\frac{7200}{n - 6} - \frac{7200}{n} = 60$$
  $$7200 \left( \frac{n - (n - 6)}{n(n - 6)} \right) = 60 \implies \frac{7200 \times 6}{n^2 - 6n} = 60$$
  $$\frac{43200}{n^2 - 6n} = 60 \implies n^2 - 6n = 720$$

• Standard Form:

  $$n^2 - 6n - 720 = 0 \quad \text{[1.5 Marks]}$$

Task B: Solving for $n$ and Final Contributions (2.5 Marks)

• Factorize:

  $$n^2 - 30n + 24n - 720 = 0 \quad (\text{Product} = -720, \text{ Sum} = -6)$$
  $$(n - 30)(n + 24) = 0 \quad \text{[1 Mark]}$$

• $n = 30$ or $n = -24$.

• Reject $n = -24$ since the number of students must be a positive integer.

• Initial number of students ($n$) $= 30$. $\quad \text{[0.5 Mark]}$

• Remaining students $= 30 - 6 = 24$.

• Actual contribution paid per student $= \frac{7200}{24} = \text{₹ } 300$. $\quad \text{[1 Mark]}$