CHAPTER 1 TELANGANA CLASS X MATHEMATICS - CHAPTER 1: REAL NUMBERS (EXERCISE 1.1 TO 1.5 FULL SOLUTIONS)

BOARD OF SECONDARY EDUCATION, TELANGANA

CLASS 10 MATHEMATICS - CHAPTER 1: REAL NUMBERS

Complete Exercise-Wise Questions & Step-by-Step Solutions (Ex 1.1 - 1.5) | Muqeem Tutorial

SUBJECT: MATHEMATICS CHAPTER 1 (FULL EXERCISES) WEIGHTAGE: 6 - 8 MARKS

EXERCISE 1.1 — Euclid's Division Lemma & Applications

Q1. Use Euclid's algorithm to find the HCF of: (i) 900 and 270, (ii) 196 and 38220, (iii) 1651 and 2032.

Solution:
By Euclid's Division Lemma: a = bq + r (0 ≤ r < b)

• (i) 900 and 270:
1. 900 = 270 × 3 + 90
2. 270 = 90 × 3 + 0
Since remainder is 0, the divisor is 90. 👉 HCF(900, 270) = 90

• (ii) 196 and 38220:
1. 38220 = 196 × 195 + 0
Remainder is 0 at first step. 👉 HCF(196, 38220) = 196

• (iii) 1651 and 2032:
1. 2032 = 1651 × 1 + 381
2. 1651 = 381 × 4 + 127
3. 381 = 127 × 3 + 0
Remainder is 0, divisor is 127. 👉 HCF(1651, 2032) = 127

Q2. Use division algorithm to show that any positive odd integer is of the form 6q + 1, or 6q + 3 or 6q + 5, where q is some integer.

Solution:
• Let a be any positive odd integer, and take divisor b = 6.
• By division algorithm: a = 6q + r where 0 ≤ r < 6 (i.e., r = 0, 1, 2, 3, 4, 5).
• If r = 0, 2, 4: a = 6q, 6q+2, 6q+4 which are all divisible by 2 (even numbers).
• Since a is odd, r must be 1, 3, or 5.
👉 Therefore, any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5.

Q3. Use division algorithm to show that the square of any positive integer is of the form 3p or 3p + 1.

Solution:
• Let a be any positive integer, and take b = 3. Then a = 3q + r (r = 0, 1, 2).
• Case 1 (r = 0): a = 3q ⇒ a² = 9q² = 3(3q²) = 3p (where p = 3q²).
• Case 2 (r = 1): a = 3q + 1 ⇒ a² = (3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3p + 1.
• Case 3 (r = 2): a = 3q + 2 ⇒ a² = (3q + 2)² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1 = 3p + 1.
👉 Hence, square of any positive integer is of the form 3p or 3p + 1.

Q4. Use division algorithm to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution:
• Let a = 3q + r (r = 0, 1, 2).
• Case 1: a = 3q ⇒ a³ = 27q³ = 9(3q³) = 9m.
• Case 2: a = 3q + 1 ⇒ a³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1 = 9m + 1.
• Case 3: a = 3q + 2 ⇒ a³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8 = 9m + 8.
👉 Hence proved.

Q5. Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer.

Solution:
• Any positive integer n can be expressed as 3q, 3q+1, or 3q+2.
• If n = 3q: n is divisible by 3. (n+2 = 3q+2, n+4 = 3q+4 not divisible by 3).
• If n = 3q+1: n+2 = 3q+3 = 3(q+1) is divisible by 3.
• If n = 3q+2: n+4 = 3q+6 = 3(q+2) is divisible by 3.
👉 Therefore, exactly one out of n, n+2, n+4 is divisible by 3.

EXERCISE 1.2 — Fundamental Theorem of Arithmetic

Q1. Express each of the following numbers as a product of its prime factors:

• (i) 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7
• (ii) 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13
• (iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17
• (iv) 5005 = 5 × 7 × 11 × 13
• (v) 7429 = 17 × 19 × 23

Q2. Find the LCM and HCF of the following integers by the prime factorization method:

• (i) 12, 15, 21: 12 = 2²×3, 15 = 3×5, 21 = 3×7 ⇒ HCF = 3, LCM = 420
• (ii) 17, 23, 29: All primes ⇒ HCF = 1, LCM = 1139
• (iii) 8, 9, 25: Coprime numbers ⇒ HCF = 1, LCM = 1800
• (iv) 72, 108: 72 = 2³×3², 108 = 2²×3³ ⇒ HCF = 36, LCM = 216
• (v) 306, 657: HCF = 9 ⇒ LCM = (306×657)/9 = 22338

Q3. Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution:
• If a number ends with 0, its prime factorization must contain both 2 and 5.
• Prime factorization of 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ.
• Since 5 is not a prime factor of 6ⁿ, by the Fundamental Theorem of Arithmetic,
👉 6ⁿ can NEVER end with digit 0 for any natural number n.

Q4. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution:
• 1st Number = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78. Since it has factors other than 1 and itself, it is Composite.
• 2nd Number = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009. Having factors 5 and 1009 makes it Composite.

Q5. How will you show that (17 × 11 × 2) + (17 × 11 × 5) is a composite number? Explain.

Solution:
• Taking common terms: (17 × 11 × 2) + (17 × 11 × 5) = 17 × 11 × (2 + 5) = 17 × 11 × 7.
👉 Since the number can be expressed as a product of prime factors (7, 11, 17), it is a Composite number.

Q6. Which digit would occupy the units place of resulting number of 6¹⁰⁰?

Solution:
• Any positive power of 6 always ends with 6 (6¹=6, 6²=36, 6³=216...).
👉 The digit in the units place of 6¹⁰⁰ is 6.

EXERCISE 1.3 — Rational Numbers & Decimal Expansions

Q1. Write the following rational numbers in their decimal form and state whether they are terminating or non-terminating repeating:

• (i) 3/8 = 3/2³ = 0.375 (Terminating)
• (ii) 229/400 = 229/(2⁴ × 5²) = 0.5725 (Terminating)
• (iii) 4 1/5 = 21/5 = 4.2 (Terminating)
• (iv) 2/11 = 0.1818... (Non-terminating repeating)
• (v) 8/125 = 8/5³ = 0.064 (Terminating)

Q2. Without performing division, state whether the following rational numbers will have a terminating or non-terminating repeating decimal form:

• (i) 13/3125 = 13/5⁵ ⇒ Terminating
• (ii) 11/12 = 11/(2² × 3) ⇒ Non-terminating repeating (contains factor 3)
• (iii) 64/455 = 64/(5 × 7 × 13) ⇒ Non-terminating repeating
• (iv) 15/1600 = 3/320 = 3/(2⁶ × 5) ⇒ Terminating
• (v) 29/343 = 29/7³ ⇒ Non-terminating repeating
• (vi) 23 / (2³ × 5²) ⇒ Terminating
• (vii) 129 / (2² × 5⁷ × 7⁵) ⇒ Non-terminating repeating
• (viii) 9/15 = 3/5 ⇒ Terminating
• (ix) 36/100 = 9/25 = 9/5² ⇒ Terminating
• (x) 77/210 = 11/30 = 11/(2 × 3 × 5) ⇒ Non-terminating repeating

Q3. Write the following rational numbers in decimal form using Theorem 1.4:

• (i) 13/25 = (13 × 2²)/(5² × 2²) = 52/100 = 0.52
• (ii) 15/16 = (15 × 5⁴)/(2⁴ × 5⁴) = 9375/10000 = 0.9375
• (iii) 23/(2³ × 5²) = (23 × 5)/(2³ × 5³) = 115/1000 = 0.115
• (iv) 7218/(3² × 5²) = 802/25 = (802 × 4)/100 = 32.08
• (v) 143/110 = 13/10 = 1.3

Q4. Express the following decimal numbers in p/q form and analyze the prime factors of q:

• (i) 43.123: 43123 / 1000 = 43123 / (2³ × 5³) ⇒ Prime factors of q are 2 and 5.
• (ii) 0.120112001120001...: Non-terminating & non-repeating ⇒ Irrational Number (Cannot be written as p/q).
• (iii) 43.1212...: x = 43.1212... ⇒ 99x = 4269 ⇒ x = 4269/99 = 1423/33 ⇒ q = 33 = 3 × 11 (has factors other than 2 or 5).
• (iv) 0.6363...: x = 0.6363... ⇒ 99x = 63 ⇒ x = 63/99 = 7/11 ⇒ q = 11 (factor other than 2 or 5).

EXERCISE 1.4 — Proofs of Irrationality

Q1. Prove that the following are irrational: (i) 1/√2, (ii) √3 + √5, (iii) 6 + √2, (iv) √5, (v) 3 + 2√5.

• (i) 1/√2: Assume 1/√2 = a/b (rational) ⇒ √2 = b/a. Since b/a is rational, √2 must be rational, which is a contradiction. Hence 1/√2 is Irrational.

• (ii) √3 + √5: Let x = √3 + √5 (rational) ⇒ x² = 3 + 5 + 2√15 = 8 + 2√15 ⇒ √15 = (x² - 8)/2. RHS is rational, implying √15 is rational, contradiction. Hence √3 + √5 is Irrational.

• (iii) 6 + √2: Let 6 + √2 = a/b ⇒ √2 = (a - 6b)/b. Rational RHS contradicts √2 being irrational. Hence 6 + √2 is Irrational.

• (iv) √5: Let √5 = a/b (a, b coprime) ⇒ 5b² = a² ⇒ 5 divides a. Let a = 5c ⇒ 5b² = 25c² ⇒ b² = 5c² ⇒ 5 divides b. Contradicts co-prime assumption. Hence √5 is Irrational.

• (v) 3 + 2√5: Let 3 + 2√5 = a/b ⇒ √5 = (a - 3b)/2b. Rational RHS contradicts √5 being irrational. Hence 3 + 2√5 is Irrational.

Q2. Prove that √p + √q is an irrational number, where p and q are primes.

Solution:
1. Let x = √p + √q be a rational number.
2. Squaring both sides: x² = p + q + 2√(pq) ⇒ 2√(pq) = x² - (p + q) ⇒ √(pq) = [x² - (p + q)] / 2.
3. Since x, p, q are rational, the RHS is rational, which implies √(pq) is rational.
4. But p and q are distinct prime numbers, so pq is not a perfect square, making √(pq) irrational.
👉 Contradiction! Hence, √p + √q is Irrational. (Proved)

EXERCISE 1.5 — Logarithms & Laws of Logarithms

Q1. Determine the value of the following:

• (i) log₂₅ 5 = 1/2 (since 25^1/2 = 5)
• (ii) log₈₁ 3 = 1/4 (since 81^1/4 = 3)
• (iii) log₂ (1/16) = log₂ (2^-4) = -4
• (iv) log₇ 1 = 0
• (v) log_x x = 1
• (vi) log₂ 512 = log₂ (2⁹) = 9
• (vii) log₁₀ 0.01 = log₁₀ (10^-2) = -2
• (viii) log_2/3 (8/27) = log_2/3 ((2/3)³) = 3
• (ix) 2^{2 + log₂ 3} = 2² × 2^{log₂ 3} = 4 × 3 = 12

Q2. Write the following expressions as log N and find their values:

• (i) log 2 + log 5 = log(2 × 5) = log 10 = 1
• (ii) log₂ 16 - log₂ 2 = log₂(16/2) = log₂ 8 = 3
• (iii) 3 log₆₄ 4 = log₆₄(4³) = log₆₄ 64 = 1
• (iv) 2 log 3 - 3 log 2 = log(3²) - log(2³) = log(9/8)
• (v) log 10 + 2 log 3 - log 2 = log 10 + log 9 - log 2 = log(90/2) = log 45

Q3. Evaluate each of the following in terms of x and y, if x = log₂ 3 and y = log₂ 5:

• (i) log₂ 15 = log₂(3 × 5) = x + y
• (ii) log₂ 7.5 = log₂(15/2) = log₂ 15 - log₂ 2 = x + y - 1
• (iii) log₂ 60 = log₂(4 × 15) = log₂ 4 + log₂ 15 = x + y + 2
• (iv) log₂ 6750 = log₂(2 × 3³ × 5³) = 1 + 3 log₂ 3 + 3 log₂ 5 = 1 + 3x + 3y

Q4. Expand the following:

• (i) log 1000 = log(10³) = 3
• (ii) log(128 / 625) = log 128 - log 625 = 7 log 2 - 4 log 5
• (iii) log(x² y³ z⁴) = 2 log x + 3 log y + 4 log z
• (iv) log(p² q³ / r⁴) = 2 log p + 3 log q - 4 log r
• (v) log √(x³ / y²) = log(x^3/2 / y) = (3/2) log x - log y

Q5. If x² + y² = 25xy, then prove that 2 log(x + y) = 3 log 3 + log x + log y.

Solution:
• Adding 2xy on both sides: x² + 2xy + y² = 25xy + 2xy ⇒ (x + y)² = 27xy.
• Taking log on both sides: log(x + y)² = log(27xy).
• 2 log(x + y) = log(3³) + log x + log y.
👉 2 log(x + y) = 3 log 3 + log x + log y. (Proved)

Q6. If log((x + y)/3) = (1/2)(log x + log y), find the value of (x/y + y/x).

Solution:
• log((x + y)/3) = log √(xy) ⇒ (x + y)/3 = √(xy).
• Squaring both sides: (x + y)² / 9 = xy ⇒ x² + 2xy + y² = 9xy ⇒ x² + y² = 7xy.
• Dividing by xy: (x²/xy) + (y²/xy) = 7.
👉 x/y + y/x = 7

Q7. If (2.3)^x = (0.23)^y = 1000, find the value of (1/x - 1/y).

Solution:
• (2.3)^x = 10³ ⇒ 2.3 = 10^3/x.
• (0.23)^y = 10³ ⇒ 0.23 = 10^3/y.
• Dividing equations: 2.3 / 0.23 = 10^3/x / 10^3/y ⇒ 10¹ = 10^{3(1/x - 1/y)}.
• Equating exponents: 3(1/x - 1/y) = 1.
👉 1/x - 1/y = 1/3

Q8. If 2^x+1 = 3^1-x, find the value of x.

Solution:
• Taking log on both sides: log(2^x+1) = log(3^1-x).
• (x + 1) log 2 = (1 - x) log 3 ⇒ x log 2 + log 2 = log 3 - x log 3.
• x(log 2 + log 3) = log 3 - log 2 ⇒ x log 6 = log(3/2).
👉 x = (log 3 - log 2) / (log 2 + log 3)

Q9. Is (i) log 2 rational or irrational? Justify. (ii) log 100 rational or irrational? Justify.

• (i) log 2: Let log₁₀ 2 = p/q (rational) ⇒ 10^p/q = 2 ⇒ 10^p = 2^q ⇒ 2^p × 5^p = 2^q ⇒ 5^p = 2^q-p. Left side is odd while right side is even, which is impossible. Hence log 2 is Irrational.

• (ii) log 100: log₁₀ 100 = log₁₀ (10²) = 2 = 2/1. Since 2 is an integer ratio, log 100 is Rational.