Number System — Andhra Pradesh (SSC) Class 9 Mathematics Solutions (Free)

Free step-by-step Andhra Pradesh (SSC) Class 9 Mathematics solutions for "Number System" — important questions with detailed answers, download PDF for board exam preparation.

Number System — Andhra Pradesh (SSC) Class 9 Mathematics Solutions (Free)

Free step-by-step Andhra Pradesh (SSC) Class 9 Mathematics solutions for "Number System" — important questions with detailed answers, download PDF for board exam preparation.

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TL;DR: Free step-by-step Andhra Pradesh (SSC) Class 9 Mathematics solutions for "Number System" — important questions with detailed answers, download PDF for…

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Number System — Textbook Questions Answered (8)

Q1: Express 0.3̄ (where 3 repeats) as a rational number in the form p/q.

Step 1: Let x = 0.3̄ = 0.3333... Step 2: Multiply both sides by 10: 10x = 3.3333... Step 3: Subtract equation 1 from equation 2: 10x - x = 3.3333... - 0.3333... 9x = 3. Step 4: x = 3/9 = 1/3. Step 5: Verify: 1/3 = 0.3̄ (divide 1 by 3). Final Answer: 0.3̄ = 1/3.

Q2: Simplify: (2√3 + 3√2)(2√3 - 3√2).

Step 1: Use the identity (a + b)(a - b) = a² - b². Step 2: Here a = 2√3 and b = 3√2. Step 3: (2√3 + 3√2)(2√3 - 3√2) = (2√3)² - (3√2)². Step 4: = 4 × 3 - 9 × 2. Step 5: = 12 - 18 = -6. Final Answer: (2√3 + 3√2)(2√3 - 3√2) = -6.

Q3: Prove that √2 is irrational.

Step 1: We prove by contradiction. Assume √2 is rational. Step 2: Then √2 = p/q where p and q are integers with no common factors (p and q are coprime). Step 3: Squaring both sides: 2 = p²/q². Step 4: Therefore: 2q² = p² ... (equation 1). Step 5: From equation 1, p² is even (since 2q² is even). Therefore p must be even. Let p = 2m. Step 6: Substituting p = 2m in equation 1: 2q² = (2m)² = 4m². Step 7: Dividing by 2: q² = 2m². Step 8: This means q² is even, so q must be even. Let q = 2n. Step 9: But if both p and q are even, they have common factor 2, contradicting our assumption that p and q are coprime. Step 10: Therefore our assumption is false. √2 is irrational. Final Answer: √2 is irrational (proved by contradiction).

Q4: Simplify: (2 × 3⁻¹) / (4 × 2⁻³).

Step 1: Simplify numerator: 2 × 3⁻¹ = 2/3. Step 2: Simplify denominator: 4 × 2⁻³ = 4/2³ = 4/8 = 1/2. Step 3: Divide: (2/3) ÷ (1/2) = (2/3) × (2/1) = 4/3. Final Answer: (2 × 3⁻¹) / (4 × 2⁻³) = 4/3.

Q5: Express 2.1̄7̄ (where 17 repeats) as a rational number p/q in lowest terms.

Step 1: Let x = 2.1̄7̄ = 2.171717... Step 2: Multiply by 100 (since 2 digits repeat): 100x = 217.171717... Step 3: Subtract: 100x - x = 217.171717... - 2.171717... 99x = 215. Step 4: x = 215/99. Step 5: Check if fraction can be reduced: GCD(215, 99) = 1 (since 215 = 5 × 43 and 99 = 9 × 11, no common factors). Final Answer: 2.1̄7̄ = 215/99.

Q6: If a = 2 + √3, find the value of a² - 4a.

Step 1: a = 2 + √3. Step 2: Calculate a²: a² = (2 + √3)² = 4 + 4√3 + 3 = 7 + 4√3. Step 3: Calculate 4a: 4a = 4(2 + √3) = 8 + 4√3. Step 4: Find a² - 4a: (7 + 4√3) - (8 + 4√3) = 7 + 4√3 - 8 - 4√3 = -1. Final Answer: a² - 4a = -1.

Q7: Rationalize the denominator: 1 / (√5 - √3).

Step 1: Multiply numerator and denominator by the conjugate (√5 + √3). Step 2: = 1/(√5 - √3) × (√5 + √3)/(√5 + √3). Step 3: Numerator: 1 × (√5 + √3) = √5 + √3. Step 4: Denominator: (√5 - √3)(√5 + √3) = (√5)² - (√3)² = 5 - 3 = 2. Step 5: = (√5 + √3) / 2. Final Answer: 1/(√5 - √3) = (√5 + √3) / 2.

Q8: Simplify: (3⁴ × 3⁻²) / 3⁻³.

Step 1: Using law of exponents: 3ᵃ × 3ᵇ = 3^(a+b). Step 2: Numerator: 3⁴ × 3⁻² = 3^(4-2) = 3². Step 3: Divide: 3² / 3⁻³ = 3² × 3³ = 3^(2+3) = 3⁵. Step 4: Calculate: 3⁵ = 243. Final Answer: (3⁴ × 3⁻²) / 3⁻³ = 243.

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