Q1. Page 27: 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.
Answer:
We are given two points: (x=10, y=400) and (x=14, y=500).
Substituting into y = ax + b:
400 = 10a + b ...(1)
500 = 14a + b ...(2)
Subtracting (1) from (2):
100 = 4a
a = 25
Substitute a = 25 in (1):
400 = 10(25) + b
400 = 250 + b
b = 150
Therefore, the linear relationship is y = 25x + 150.
a = 25 (cost per module), b = 150 (fixed monthly fee).
Q2. Page 27: 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.
Answer:
We are given two points: (x=10, y=800) and (x=15, y=1100).
Substituting into y = ax + b:
800 = 10a + b ...(1)
1100 = 15a + b ...(2)
Subtracting (1) from (2):
300 = 5a
a = 60
Substitute a = 60 in (1):
800 = 10(60) + b
800 = 600 + b
b = 200
Therefore, the linear relationship is y = 60x + 200.
a = 60 (cost per hour for badminton court), b = 200 (fixed monthly fee).
Q3. Page 27: 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. (Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)
Answer:
We are given two points: (°F=32, °C=0) and (°F=212, °C=100).
Substituting into °C = a°F + b:
0 = 32a + b ...(1)
100 = 212a + b ...(2)
Subtracting (1) from (2):
100 = 180a
a = 100/180 = 5/9
Substitute a = 5/9 in (1):
0 = 32(5/9) + b
0 = 160/9 + b
b = -160/9
Therefore, the linear relationship is °C = (5/9)°F - 160/9, which can also be written as °C = (5/9)(°F - 32).
a = 5/9, b = -160/9.