Chapter 2 - Introduction to Linear Polynomials

Exercise Set 2.2

Q1. Find the value of the linear polynomial 5x – 3 if: (i) x = 0 (ii) x = –1 (iii) x = 2

Solution: Substitute the given values of x into the polynomial.

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

Q2. Find the value of the quadratic polynomial 7s² – 4s + 6 if: (i) s = 0 (ii) s = –3 (iii) s = 4

Solution: Substitute the values of s into the polynomial.

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

Q3. 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.
Salil's mother's present age = 3x years.
After 5 years, Salil's age = x + 5 and his mother's age = 3x + 5.
According to the problem: (x + 5) + (3x + 5) = 70.

4x + 10 = 70
4x = 60
x = 15

Salil's present age is 15 years, and his mother's present age is 3 × 15 = 45 years.

Q6. 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 the length of the shorter piece be x feet.
The length of the longer piece = 4x feet.
Their total length is 300 feet, so x + 4x = 300.
5x = 300 ⇒ x = 60.
The shorter piece is 60 feet, and the longer piece is 4 × 60 = 240 feet.

Exercise Set 2.5

Q1. A student observes that when she accessed 10 learning 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: We are given the linear relationship y = ax + b.
Substitute the given values into the equation:
Equation 1 (for 10 modules): 400 = 10a + b.
Equation 2 (for 14 modules): 500 = 14a + b.
Subtracting Equation 1 from Equation 2:

(500 - 400) = (14a + b) - (10a + b)
100 = 4a ⇒ a = 25

Substitute a = 25 into Equation 1:
400 = 10(25) + b ⇒ 400 = 250 + b ⇒ b = 150.
Therefore, a = 25 and b = 150.

Chapter 3 - The World of Numbers

Non-terminating Repeating Decimals

Example: Convert 0.666... (or 0.6̅) into the form p/q.

Solution: Let x = 0.666....
Step 1: Since one digit repeats, multiply both sides by 10¹ = 10.
10x = 6.666....
Step 2: Subtract the original equation from this new equation:

10x - x = 6.666... - 0.666...
9x = 6

Step 3: Solve for x: x = 6 / 9, which simplifies to 2/3.

Proving Irrationality

Proof: Prove that √2 is an irrational number.

Solution: We prove this by contradiction.
Step 1: Assume √2 is rational. Thus, it can be written as √2 = p/q, where p and q are integers that share no common factors other than 1 (they are co-prime) and q ≠ 0.
Step 2: Square both sides: 2 = p² / q².
Step 3: Multiply by q²: 2q² = p².
Step 4: Because p² is equal to 2 times an integer, p² must be an even number. Therefore, p is also an even integer. Let p = 2k.
Step 5: Substitute p = 2k back into the equation: 2q² = (2k)² ⇒ 2q² = 4k².
Step 6: Divide by 2: q² = 2k².
Step 7: This shows q² is even, so q must also be even.
Step 8: Contradiction! We deduced that both p and q are even, meaning they share a common factor of 2. This contradicts our initial assumption that they are co-prime. Thus, √2 is irrational.

Chapter 4 - Exploring Algebraic Identities

Example 3: Expand (5x + 2y)² using algebraic identities.

Solution: We use the identity (a + b)² = a² + 2ab + b².
Here, a = 5x and b = 2y.

(5x + 2y)² = (5x)² + 2(5x)(2y) + (2y)²
= 25x² + 20xy + 4y²

Thus, the expanded form is 25x² + 20xy + 4y².

Example 4: Calculate 43² without direct multiplication, using identities.

Solution: We can write 43 as (40 + 3).
Using the identity (a + b)² = a² + 2ab + b²:

43² = (40 + 3)²
= 40² + 2(40)(3) + 3²
= 1600 + 240 + 9
= 1849

The answer is 1849.

Example 7: Factorise the expression 50p² + 60pq + 18q².

Solution: First, observe that 2 is a common factor of all the terms: 50p², 60pq, 18q².
50p² + 60pq + 18q² = 2(25p² + 30pq + 9q²).
Now focus on the inner expression. We can rewrite it using the identity a² + 2ab + b² = (a + b)².

25p² = (5p)²
9q² = (3q)²
30pq = 2(5p)(3q)

Therefore, 25p² + 30pq + 9q² = (5p + 3q)².
The completely factorised expression is 2(5p + 3q)².

Example: What is the side of a cube whose volume is p³ + 6p²q + 12pq² + 8q³ cubic units?

Solution: We need to factorise the volume expression. Compare it with the identity (a + b)³ = a³ + 3a²b + 3ab² + b³.
Rewriting the expression:
p³ + 3(p)²(2q) + 3(p)(2q)² + (2q)³.
This matches the identity with a = p and b = 2q.
Hence, the volume is (p + 2q)³.
The side of the cube is (p + 2q) units.