Exercise Set 2.5 Class 9

Class 9 Mathematics Ganita Manjari Part 1 NCERT Solutions
Chapter-2: Introduction to Linear Polynomials

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 $\bm{y}$ depends on the number of modules accessed, $\bm{x}$ , according to the relation $\bm{y=ax+y}$ , find the values of $\bm{a}$ and $\bm{b}$ .

Solution:
Given linear relation
$y=ax+b$ ….. (1)
Here,
$y=\text{monthly bill (in ₹)}$
$x=\text{number of modules accessed}$
Given, when 10 modules are accessed, the bill is ₹400.
∴ $x=10,\quad y=400$ put these values in eq(1)
$400=10a+b$
⇒ $10a+b=400$ ….. (2)
Also, when 14 modules are accessed, the bill is ₹500.
∴ $x=14,\quad y=500$ put these values in eq(1)
$500=14a+b$
⇒ $14a+b=500$ ….. (3)
Subtract eq(2) from eq(3)
$(14a+b)-(10a+b)=500-400$
$14a+b-10a-b=100$
$4a=100$
$a=\dfrac{100}{4}$
$\boxed{a=25}$ put in eq(2)
$10(25)+b=400$
$250+b=400$
$b=400-250$
$\boxed{b=150}$

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 $\bm{y}$ depends on the hours of the use of the badminton court, $\bm{x}$ , according to the relation $\bm{y=ax+y}$ , find the values of $\bm{a}$ and $\bm{b}$ .

Solution:

Given linear relation
$y=ax+b$ ….. (1)
Here,
$y=\text{monthly bill (in ₹)}$
$x=\text{number of hours the badminton court used}$

Given, when the badminton court is used for 10 hours, the bill is ₹800.
∴ $x=10,\quad y=800$ put in eq(1)
$800=10a+b$
⇒ $10a+b=800$ ….. (2)

Aslo, when the badminton court is used for 15 hours, the bill is ₹1100.
∴ $x=15,\quad y=1100$ put in eq(1)
$1100=15a+b$
⇒ $15a+b=1100$ ….. (3)
Subtract eq(2) from eq(3)
$(15a+b)-(10a+b)=1100-800$
$15a+b-10a-b=300$
$5a=300$
$a=\dfrac{300}{5}$
$\boxed{a=60}$ put in eq(2)
$10(60)+b=800$
$600+b=800$
$b=800-600$
$\boxed{b=200}$

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

Solution:
Given linear relation
$\mathrm\degree C=a\mathrm\degree F+b$ ….. (1)

When $\mathrm\degree C=0$ and $\mathrm\degree F=32$ , then
$0=a(32)+b$
⇒ $32a+b=0$ ….. (2)

When $\mathrm\degree C=100$ and $\mathrm\degree F=212$ , then
$100=a(212)+b$
⇒ $212a+b=100$ ….. (3)
Subtract eq(2) from eq(3)
$(212a+b)-(32a+b)=100-0$
$212a+b-32a-b=100$
$180a=100$
$a=\dfrac{100}{180}$
$\boxed{a=\dfrac{5}{9}}$ put in eq(2)
$32\left(\dfrac{5}{9}\right)+b=0$
$\dfrac{160}{9}+b=0$
$\boxed{b=-\dfrac{160}{9}}$