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📐 Chapter 2: Introduction to Linear Polynomials • Exercise Set 2.5

Verified step-by-step solutions for your class notes.

Exercise Set 2.1 5Exercise Set 2.2 7Exercise Set 2.3 5Exercise Set 2.4 4Exercise Set 2.5 3Exercise Set 2.6 1End-of-Chapter Exercises 14All Exercises 39
Q1

Exercise Set 2.5

1. A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.

  1. We have two equations based on the information provided:
    400 = 10a + b
    500 = 14a + b
  2. Subtract the first equation from the second to eliminate b:
    (500 - 400) = (14a + b) - (10a + b)
    100 = 4a
    a = 25
  3. Substitute a = 25 back into the first equation to find b:
    400 = 10(25) + b
    400 = 250 + b
    b = 150
  4. The relationship is y = 25x + 150.
Q2

Exercise Set 2.5

2. A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.

  1. Establish the two linear equations:
    800 = 10a + b
    1100 = 15a + b
  2. Subtract the first from the second:
    300 = 5a
    a = 60
  3. Substitute a = 60 into the first equation:
    800 = 10(60) + b
    800 = 600 + b
    b = 200
  4. The relationship is y = 60x + 200.
Q3

Exercise Set 2.5

3. Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a°F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.

  1. Set up the equations based on the given points:
    0 = 32a + b
    100 = 212a + b
  2. Subtract the first equation from the second:
    100 = 180a
    a = 100 / 180 = 5/9
  3. Substitute a = 5/9 into the first equation:
    0 = 32(5/9) + b
    b = -160/9
  4. The relationship is °C = (5/9)°F - 160/9 (or °C = (5/9)(°F - 32)).