NCERT Solutions for Class 9 Maths Chapter 2 — Exercise Set 2.5#
To find the constants $a$ and $b$ in $y = a x + b$ from two known points $(x_1, y_1)$ and $(x_2, y_2)$, subtract the two equations to eliminate $b$: $a = \dfrac{y_2 - y_1}{x_2 - x_1}$, then substitute back to get $b$.
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. #
Solution
When she accessed 10 modules, her bill was ₹400, so:
$$10a + b = 400 \quad \cdots (1)$$When she accessed 14 modules, her bill was ₹500, so:
$$14a + b = 500 \quad \cdots (2)$$Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} 14a + b &=& 500 \\ 10a + b &=& 400 \\ \hline 4a &=& 100 \end{array}$$$$\Rightarrow a = 25$$
Substituting $a = 25$ in equation (1):
$$10a + b = 400$$$$10(25) + b = 400$$
$$\Rightarrow 250 + b = 400$$
$$\Rightarrow b = 150$$
∴ $a = 25$ and $b = 150$, so $y = 25x + 150$: a fixed fee of ₹150 a month and ₹25 for each module.
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. #
Solution
When she used the court for 10 hours, her bill was ₹800, so:
$$10a + b = 800 \quad \cdots (1)$$When she used it for 15 hours, her bill was ₹1100, so:
$$15a + b = 1100 \quad \cdots (2)$$Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} 15a + b &=& 1100 \\ 10a + b &=& 800 \\ \hline 5a &=& 300 \end{array}$$$$\Rightarrow a = 60$$
Substituting $a = 60$ in equation (1):
$$10a + b = 800$$$$10(60) + b = 800$$
$$\Rightarrow 600 + b = 800$$
$$\Rightarrow b = 200$$
∴ $a = 60$ and $b = 200$, so $y = 60x + 200$: a fixed fee of ₹200 a month and ₹60 for each hour on the court.
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. (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.) #
Solution
Ice melts at 0 °C and 32 °F, so:
$$32a + b = 0 \quad \cdots (1)$$Water boils at 100 °C and 212 °F, so:
$$212a + b = 100 \quad \cdots (2)$$Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} 212a + b &=& 100 \\ 32a + b &=& 0 \\ \hline 180a &=& 100 \end{array}$$$$\Rightarrow a = \dfrac{100}{180}$$
$$\Rightarrow a = \dfrac{5}{9}$$
Substituting $a = \dfrac{5}{9}$ in equation (1):
$$32a + b = 0$$$$32\left(\dfrac{5}{9}\right) + b = 0$$
$$\Rightarrow \dfrac{160}{9} + b = 0$$
$$\Rightarrow b = -\dfrac{160}{9}$$
∴ $a = \dfrac{5}{9}$ and $b = -\dfrac{160}{9}$, so the relationship is:
$$°\text{C} = \dfrac{5}{9}\,°\text{F} - \dfrac{160}{9}$$This is the familiar rule $°\text{C} = \dfrac{5}{9}(°\text{F} - 32)$.