Exercise 2.5 Practice
00:00
Overview
This page provides comprehensive Ch 2: Polynomials - Exercise 2.5 Practice. Practice solving systems of equations to find parameters of linear relationships, and converting temperatures between Celsius and Fahrenheit. Free step-by-step NCERT solutions and explanations.
Determining Linear Relation Parameters
Q1: Learning Platform Fees
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 $b$.
The given linear relation is:
$$y = ax + b$$
where:
$x$ = number of modules accessed,
$y$ = monthly bill (in ₹),
$a$ = cost per module,
$b$ = fixed monthly fee.
$x$ = number of modules accessed,
$y$ = monthly bill (in ₹),
$a$ = cost per module,
$b$ = fixed monthly fee.
Form two equations from the given observations:
1. When $x = 10$, $y = 400$: $$10a + b = 400 \quad \text{--- (Equation 1)}$$
2. When $x = 14$, $y = 500$: $$14a + b = 500 \quad \text{--- (Equation 2)}$$
1. When $x = 10$, $y = 400$: $$10a + b = 400 \quad \text{--- (Equation 1)}$$
2. When $x = 14$, $y = 500$: $$14a + b = 500 \quad \text{--- (Equation 2)}$$
Subtract Equation 1 from Equation 2 to eliminate $b$:
$$(14a + b) - (10a + b) = 500 - 400$$
$$4a = 100$$
$$a = 25$$
Substitute $a = 25$ back into Equation 1:
$$10(25) + b = 400$$
$$250 + b = 400$$
$$b = 150$$
$a = 25$ (cost per module), $b = 150$ (fixed monthly fee)
Q2: Gym Court Charges
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 $b$.
The given linear relation is:
$$y = ax + b$$
where:
$x$ = court usage hours,
$y$ = monthly gym bill (in ₹),
$a$ = hourly badminton court fee,
$b$ = fixed monthly gym fee.
$x$ = court usage hours,
$y$ = monthly gym bill (in ₹),
$a$ = hourly badminton court fee,
$b$ = fixed monthly gym fee.
Form two equations from the given observations:
1. When $x = 10$, $y = 800$: $$10a + b = 800 \quad \text{--- (Equation 1)}$$
2. When $x = 15$, $y = 1100$: $$15a + b = 1100 \quad \text{--- (Equation 2)}$$
1. When $x = 10$, $y = 800$: $$10a + b = 800 \quad \text{--- (Equation 1)}$$
2. When $x = 15$, $y = 1100$: $$15a + b = 1100 \quad \text{--- (Equation 2)}$$
Subtract Equation 1 from Equation 2 to eliminate $b$:
$$(15a + b) - (10a + b) = 1100 - 800$$
$$5a = 300$$
$$a = 60$$
Substitute $a = 60$ back into Equation 1:
$$10(60) + b = 800$$
$$600 + b = 800$$
$$b = 200$$
$a = 60$ (hourly fee), $b = 200$ (fixed monthly fee)
Q3: Celsius to Fahrenheit Relationship
Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by $\text{°C} = a\text{°F} + b$. 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.
The temperature relation is given by:
$$C = aF + b$$
We are given two pairs of values:
1. Melting point of ice: $C = 0$ when $F = 32$.
2. Boiling point of water: $C = 100$ when $F = 212$.
1. Melting point of ice: $C = 0$ when $F = 32$.
2. Boiling point of water: $C = 100$ when $F = 212$.
Form two equations using these values:
$$0 = 32a + b \quad \text{--- (Equation 1)}$$
$$100 = 212a + b \quad \text{--- (Equation 2)}$$
From Equation 1, express $b$ in terms of $a$:
$$b = -32a$$
Substitute $b = -32a$ into Equation 2:
$$100 = 212a - 32a$$
$$100 = 180a$$
$$a = \frac{100}{180} = \frac{5}{9}$$
Now substitute $a = \frac{5}{9}$ into the expression for $b$:
$$b = -32\left(\frac{5}{9}\right) = -\frac{160}{9}$$
Thus, the linear relationship is:
$$C = \frac{5}{9}F - \frac{160}{9} = \frac{5}{9}(F - 32)$$
$a = \frac{5}{9}$, $b = -\frac{160}{9}$