ROUTERA

Chapter 4 Quadratic Equations

Class 10th Maths Chapter Case Study Questions

Case Study 1: Profit Maximization for a Product Launch

Case Description:
A company plans to launch a new product and estimates that the profit, P(x)P(x), they will make depends on the selling price xx (in hundreds of rupees). The profit function is given by a quadratic equation:

P(x)=−2x2+12x+50P(x) = -2x^2 + 12x + 50

where xx represents the selling price in hundreds of rupees. The company wants to find the selling price that will maximize the profit, as well as the maximum profit achievable. Additionally, they need to determine the selling prices at which they would make a specific profit level, such as zero profit or a profit of ₹100.

MCQs:

  1. At what selling price xx will the company achieve the maximum profit?

    • A) ₹200
    • B) ₹300
    • C) ₹600
    • D) ₹1200

  2. What is the maximum profit the company can achieve?

    • A) ₹100
    • B) ₹122
    • C) ₹86
    • D) ₹128

  3. What are the selling prices at which the company will make no profit (i.e., P(x)=0P(x) = 0)?

    • A) ₹0 and ₹400
    • B) ₹200 and ₹400
    • C) ₹0 and ₹600
    • D) ₹100 and ₹300

  4. If the company wants to achieve a profit of ₹100, what selling price(s) would satisfy this condition?

    • A) ₹100 and ₹300
    • B) ₹250 and ₹350
    • C) ₹50 and ₹300
    • D) ₹100 and ₹200

Case Study 2: Design of a Flower Bed

Case Description:
A garden designer is planning a rectangular flower bed surrounded by a uniform pathway. The length of the flower bed is 10 meters more than its width. The designer wants the area of the flower bed itself to be 300 square meters. If the width of the flower bed is xx meters, then the length will be x+10x + 10 meters. The designer sets up a quadratic equation based on the area requirement to determine the dimensions of the flower bed.

MCQs:

  1. Which of the following represents the equation for the area of the flower bed?

    • A) x(x+10)=300x(x + 10) = 300
    • B) x2+10x=300x^2 + 10x = 300
    • C) x(x+10)=600x(x + 10) = 600
    • D) x(x+10)=100x(x + 10) = 100

  2. What is the width of the flower bed?

    • A) 10 meters
    • B) 15 meters
    • C) 20 meters
    • D) 25 meters

  3. What is the length of the flower bed?

    • A) 25 meters
    • B) 20 meters
    • C) 30 meters
    • D) 15 meters

  4. If the designer decides to increase both the length and width by 2 meters, what will be the new area of the flower bed?

    • A) 336 square meters
    • B) 360 square meters
    • C) 400 square meters
    • D) 340 square meters

Case Study 3: Projectile Motion

Case Description:
A ball is thrown from a height of 45 meters above the ground, and its height hh (in meters) after tt seconds is given by the equation:

h=−5t2+20t+45h = -5t^2 + 20t + 45

The player who throws the ball wants to know when it will hit the ground, how long it will take to reach the maximum height, and what that maximum height will be. By solving this quadratic equation, they can answer these questions and make adjustments to the throw as needed.

MCQs:

  1. What is the initial height from which the ball is thrown?

    • A) 20 meters
    • B) 30 meters
    • C) 45 meters
    • D) 50 meters

  2. After how many seconds will the ball hit the ground?

    • A) 1 second
    • B) 2 seconds
    • C) 3 seconds
    • D) 5 seconds

  3. After how many seconds will the ball reach its maximum height?

    • A) 2 seconds
    • B) 4 seconds
    • C) 1 second
    • D) 3 seconds

  4. What is the maximum height reached by the ball?

    • A) 45 meters
    • B) 50 meters
    • C) 65 meters
    • D) 60 meters

Case Study 4: Fencing a Playground

Case Description:
A school wants to fence a rectangular playground that has an area of 400 square meters. The length of the playground is 20 meters longer than the width. To calculate the dimensions of the playground and the cost of fencing, the school administration sets up a quadratic equation to find the width and length of the playground.

MCQs:

  1. Which equation represents the area of the playground?

    • A) x(x+20)=400x(x + 20) = 400
    • B) x(x+20)=800x(x + 20) = 800
    • C) x(x+10)=400x(x + 10) = 400
    • D) x(x−20)=400x(x - 20) = 400

  2. What is the width of the playground?

    • A) 10 meters
    • B) 20 meters
    • C) 30 meters
    • D) 40 meters

  3. What is the length of the playground?

    • A) 40 meters
    • B) 50 meters
    • C) 60 meters
    • D) 70 meters

  4. If the cost of fencing is ₹5 per meter, what will be the total cost of fencing the playground?

    • A) ₹800
    • B) ₹900
    • C) ₹1,200
    • D) ₹1,500

Case Study 5: Revenue Optimization for a Show

Case Description:
A theater owner organizes a play and sells tickets at two price points, ₹100 and ₹150. The number of ₹100 tickets sold is 10 more than twice the number of ₹150 tickets sold. To maximize revenue, the owner wants to find the number of tickets to sell at each price point. If the total revenue from ticket sales is ₹5,000, the owner uses a quadratic equation to calculate how many tickets were sold at each price.

MCQs:

  1. Which equation represents the revenue constraint based on ticket prices?

    • A) 100x+150y=5,000100x + 150y = 5,000
    • B) 150x+100y=5,000150x + 100y = 5,000
    • C) 100(2y+10)+150y=5,000100(2y + 10) + 150y = 5,000
    • D) 100y+150x=5,000100y + 150x = 5,000

  2. If yy represents the number of ₹150 tickets sold, which equation can represent the condition involving the number of ₹100 tickets sold?

    • A) x=2yx = 2y
    • B) x=2y+10x = 2y + 10
    • C) x=y+10x = y + 10
    • D) x=3y+5x = 3y + 5

  3. How many ₹150 tickets were sold?

    • A) 10
    • B) 15
    • C) 20
    • D) 25

  4. How many ₹100 tickets were sold?

    • A) 40
    • B) 50
    • C) 30
    • D) 45