Chapter 2

html_content = """

NCERT Class 9 Mathematics

Kalaimagal Academy — Chapter 2 Solutions

Exercise Set 2.1

1. Find the degrees of the following polynomials:

(i) 2x² - 5x + 3
Answer: Degree = 2 (Highest power of x is 2)

(ii) y³ + 2y - 1
Answer: Degree = 3 (Highest power of y is 3)

(iii) -9
Answer: Degree = 0 (-9 can be written as -9x⁰)

(iv) 4z - 3
Answer: Degree = 1 (Highest power of z is 1)

2. Write polynomials of degrees 1, 2 and 3.

Answer:

• Degree 1 (Linear): 3x + 5
• Degree 2 (Quadratic): x² + 4x + 4
• Degree 3 (Cubic): x³ - 2x + 1

3. What are the coefficients of x² and x³ in the polynomial x⁴ - 3x³ + 6x² - 2x + 7?

Answer:
Coefficient of x³ = -3
Coefficient of x² = 6

4. What is the coefficient of z in the polynomial 4z³ + 5z² - 11?

Answer: Coefficient of z = 0 (since the term containing z is missing)

5. What is the constant term of the polynomial 9x³ + 5x² - 8x - 10?

Answer: Constant term = -10

Exercise Set 2.2

1. Find the value of the linear polynomial 5x - 3 if:

(i) x = 0: 5(0) - 3 = -3
(ii) x = -1: 5(-1) - 3 = -5 - 3 = -8
(iii) x = 2: 5(2) - 3 = 10 - 3 = 7

2. Find the value of the quadratic polynomial 7s² - 4s + 6 if:

(i) s = 0: 7(0)² - 4(0) + 6 = 6
(ii) s = -3: 7(-3)² - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 18 = 81
(iii) s = 4: 7(4)² - 4(4) + 6 = 7(16) - 16 + 6 = 112 - 10 = 102

3. The present age of Salil's mother is three times Salil's present age. After 5 years, their ages will add up to 70 years. Find their present ages.

Solution:
Let Salil's present age be x years.
Mother's present age = 3x years.
After 5 years: Salil's age = x + 5, Mother's age = 3x + 5
Given sum = 70:
(x + 5) + (3x + 5) = 70
4x + 10 = 70 ⇒ 4x = 60 ⇒ x = 15
Answer: Salil's age = 15 years, Mother's age = 45 years.

4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.

Solution:
Let the integers be 2x and 5x.
5x - 2x = 63 ⇒ 3x = 63 ⇒ x = 21
Answer: First integer = 2(21) = 42; Second integer = 5(21) = 105.

5. Ruby has 3 times as many two-rupee coins as she has five-rupee coins. If she has a total of ₹88, how many coins does she have of each type?

Solution:
Let number of ₹5 coins be x.
Number of ₹2 coins = 3x
Total value = 5(x) + 2(3x) = 5x + 6x = 11x
11x = 88 ⇒ x = 8
Answer: ₹5 coins = 8, ₹2 coins = 24.

6. A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?

Solution:
Let shorter piece be x feet. Longer piece = 4x feet.
x + 4x = 300 ⇒ 5x = 300 ⇒ x = 60
Answer: Shorter piece = 60 ft, Longer piece = 240 ft.

7. If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?

Solution:
Let width = w cm. Length = 2w + 3 cm.
Perimeter = 2(Length + Width) = 24
2((2w + 3) + w) = 24 ⇒ 2(3w + 3) = 24 ⇒ 3w + 3 = 12 ⇒ 3w = 9 ⇒ w = 3
Answer: Width = 3 cm, Length = 2(3) + 3 = 9 cm.

Exercise Set 2.3

1. Bank savings problem

Initial amount = ₹500, Pocket money = ₹150/month
Amount at end of Month 1 = 500 + 150 = ₹650
Amount at end of Month 2 = 500 + 2(150) = ₹800
Linear expression for n-th month: A(n) = 500 + 150n

2. Rally dropout problem

Initial = 120 members. Drops 9 per hour.
After 1 hour = 111, 2 hours = 102, 3 hours = 93
Linear expression for n-th hour: M(n) = 120 - 9n

3. Area of rectangle with fixed length = 13 cm

(i) Breadth = 12 cm ⇒ Area = 13 × 12 = 156 cm²
(ii) Breadth = 10 cm ⇒ Area = 13 × 10 = 130 cm²
(iii) Breadth = 8 cm ⇒ Area = 13 × 8 = 104 cm²
Linear Pattern: Area = 13b (where b is breadth)

4. Volume of box (length = 7 cm, breadth = 11 cm)

Base Area = 7 × 11 = 77 cm²
(i) Height = 5 cm ⇒ Volume = 77 × 5 = 385 cm³
(ii) Height = 9 cm ⇒ Volume = 77 × 9 = 693 cm³
(iii) Height = 13 cm ⇒ Volume = 77 × 13 = 1001 cm³
Linear Pattern: V(h) = 77h

5. Sarita reading a 500-page book

Reads 20 pages/day. Pages left after 15 days = 500 - 20(15) = 500 - 300 = 200 pages.
Linear Pattern: P(d) = 500 - 20d

Exercise Set 2.4

1. Plant growth model

(i) Height after 7 months = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25 feet
(iii) Expression: h(t) = 1.75 + 0.5t. Represents linear growth because height increases by a constant amount (0.5 ft) each month.

2. Phone value depreciation

(i) Value after 3 years = 10,000 - 800(3) = ₹7,600
(iii) Expression: v(t) = 10,000 - 800t. Represents linear decay because value decreases by a constant amount (₹800) each year.

3. Village population growth

(i) Population after 6 years = 750 + 50(6) = 1,050
(iii) Expression: P(t) = 750 + 50t. Represents linear growth due to constant yearly increase.

4. Telecom balance decay

(i) Equation: b(x) = 600 - 15x. Represents linear decay as balance decreases linearly.
(ii) Balance runs out when b(x) = 0 ⇒ 600 - 15x = 0 ⇒ x = 40 days.

Exercise Set 2.5

1. Learning platform billing

10a + b = 400 and 14a + b = 500
Subtracting equations: 4a = 100 ⇒ a = 25
Substituting a: 10(25) + b = 400 ⇒ b = 150
Answer: a = 25, b = 150

2. Gym badminton court billing

10a + b = 800 and 15a + b = 1100
Subtracting equations: 5a = 300 ⇒ a = 60
Substituting a: 10(60) + b = 800 ⇒ b = 200
Answer: a = 60, b = 200

3. Celsius and Fahrenheit relationship: °C = a(°F) + b

At 0°C, °F = 32 ⇒ 0 = 32a + b ⇒ b = -32a
At 100°C, °F = 212 ⇒ 100 = 212a + b
Substitute b: 100 = 212a - 32a ⇒ 180a = 100 ⇒ a = 5/9
Then b = -32(5/9) = -160/9
Answer: a = 5/9, b = -160/9 (Relationship: °C = 5/9(°F - 32))

End-of-Chapter Exercises

1. Write a polynomial of degree 3 in variable x, where coefficient of x² term is -7.

Answer: x³ - 7x² + 2x + 5

2. Find values of polynomials:

(i) 5x² - 3x + 7 at x = 1: 5(1)² - 3(1) + 7 = 5 - 3 + 7 = 9
(ii) 4t³ - t² + 6 at t = a: 4a³ - a² + 6

3. Multiply a number by 5/2 and add 2/3 to product to get -7/12. Find number.

(5/2)x + 2/3 = -7/12
(5/2)x = -7/12 - 8/12 = -15/12 = -5/4
x = (-5/4) × (2/5) = -1/2
Answer: -1/2

4. Positive number is 5 times another. Adding 21 to both makes one twice the other.

Let numbers be x and 5x.
5x + 21 = 2(x + 21) ⇒ 5x + 21 = 2x + 42 ⇒ 3x = 21 ⇒ x = 7
Answer: 7 and 35

7. Identify slopes and y-intercepts:

(i) y = -3x + 4 ⇒ Slope = -3, y-intercept = 4, Cuts y-axis at (0, 4)
(ii) 2y = 4x + 7 ⇒ y = 2x + 3.5 ⇒ Slope = 2, y-intercept = 3.5, Cuts y-axis at (0, 3.5)
(iii) 5y = 6x - 10 ⇒ y = (6/5)x - 2 ⇒ Slope = 6/5, y-intercept = -2, Cuts y-axis at (0, -2)
(iv) 3y = 6x - 11 ⇒ y = 2x - 11/3 ⇒ Slope = 2, y-intercept = -11/3, Cuts y-axis at (0, -11/3)
Parallel lines: (ii) and (iv) are parallel because they have equal slopes (a = 2).

""" with open("ncert_class9_chapter2_solutions.html", "w", encoding="utf-8") as f: f.write(html_content) from weasyprint import HTML HTML("ncert_class9_chapter2_solutions.html").write_pdf("NCERT_Class_9_Maths_Chapter_2_Solutions.pdf") print("PDF generated successfully.")