Worksheet Solutions-Real numbers-10th class-Mathematics-English Medium-2026-2027

WORKSHEET - 1 Class X — Mathematics

Chapter: 1. Real Numbers Topic: Division Algorithm & Euclid's Lemma

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Q1. In a = bq + r (Division algorithm):
a is _______, q is _______, r is _______, b is _______ and condition for "r" is _______.

a = Dividend,    b = Divisor

q = Quotient,    r = Remainder

Condition for r: 0 ≤ r < b

Q2. Find the HCF of 56 and 98, using Euclid's Division algorithm.

Applying Euclid's Division Lemma (a = bq + r):

98 = 56 × 1 + 42    (r = 42 ≠ 0)

56 = 42 × 1 + 14    (r = 14 ≠ 0)

42 = 14 × 3 + 0     (r = 0)

Since remainder is 0, HCF(56, 98) = 14.

Q3. Use Euclid's Division Algorithm to find the HCF of 135 and 225.

Here 225 > 135. Using a = bq + r:

225 = 135 × 1 + 90    (r = 90 ≠ 0)

135 = 90 × 1 + 45      (r = 45 ≠ 0)

90 = 45 × 2 + 0        (r = 0)

Therefore, HCF(135, 225) = 45.

Q4. If the HCF of 12 and 84 is expressed as 7m - 2, then find the value of "m".

First, find HCF of 12 and 84:

84 = 12 × 7 + 0  ⇒  HCF(12, 84) = 12

Given HCF = 7m - 2

⇒ 7m - 2 = 12 ⇒ 7m = 14 ⇒ m = 2

Q5. Find the greatest number that divides 245 and 1029 leaving the remainder 5 in each case.

Subtract remainder 5 from given numbers:

245 - 5 = 240  and  1029 - 5 = 1024

Now, find HCF(240, 1024) using Euclid's Algorithm:

1024 = 240 × 4 + 64  |  240 = 64 × 3 + 48

64 = 48 × 1 + 16       |  48 = 16 × 3 + 0

The required greatest number is 16.

WORKSHEET - 2 Class X — Mathematics

Chapter: 1. Real Numbers Topic: Prime Factorisation

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Q1. Find the prime factorisation of 60.

Dividing 60 by prime factors step-by-step:

60 = 2 × 30 = 2 × 2 × 15 = 2 × 2 × 3 × 5

Prime factorisation of 60 = 2² × 3 × 5

Q2. Find the prime factorisation of the following:

(a) 84
84 = 2 × 42
= 2 × 2 × 21
= 2² × 3 × 7

(b) 180
180 = 2 × 90
= 2 × 2 × 45 = 2 × 2 × 3 × 15
= 2² × 3² × 5

(c) 360
360 = 2 × 180
= 2 × 2 × 2 × 3² × 5
= 2³ × 3² × 5

(d) 945
945 = 3 × 315
= 3 × 3 × 105 = 3 × 3 × 3 × 35
= 3³ × 5 × 7

Q3. Find the smallest number that must be multiplied to 540 to make it a perfect square.

Prime factorisation of 540:

540 = 2 × 2 × 3 × 3 × 3 × 5 = (2²) × (3²) × 3 × 5

Here, prime factors 3 and 5 do not have pairs.

To make it a perfect square, we must multiply by 3 × 5 = 15.

Required smallest number = 15.

Q4. Check whether 8ⁿ can end with digit '0' for any natural number n. Justify your answer.

If a number ends with digit 0, its prime factorisation must contain both 2 and 5.

Now, 8ⁿ = (2³)ⁿ = 2³ⁿ

The only prime factor of 8ⁿ is 2. It does not contain 5 as a prime factor.

By Fundamental Theorem of Arithmetic, 8ⁿ can NEVER end with digit 0 for any n ∈ N.

WORKSHEET - 3 Class X — Mathematics

Chapter: 1. Real Numbers Topic: Fundamental Theorem of Arithmetic - HCF & LCM

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Q1. Two numbers have prime factorizations: A = 2² × 3 × 5 and B = 2 × 3² × 7.
(a) HCF = _______      (b) LCM = _______

(a) HCF = Product of smallest powers of common prime factors

         = 2¹ × 3¹ = 6

(b) LCM = Product of greatest powers of all prime factors

         = 2² × 3² × 5 × 7 = 4 × 9 × 5 × 7 = 1260

Q2. Find HCF of 24 and 36 by prime factorisation.

24 = 2 × 2 × 2 × 3 = 2³ × 3¹

36 = 2 × 2 × 3 × 3 = 2² × 3²

HCF(24, 36) = 2² × 3¹ = 4 × 3 = 12

Q3. Using prime factorisation, find HCF and LCM of 84 and 180. Also verify HCF × LCM = product of numbers.

84 = 2² × 3 × 7  |  180 = 2² × 3² × 5

HCF = 2² × 3 = 12  |  LCM = 2² × 3² × 5 × 7 = 1260

Verification: HCF × LCM = 12 × 1260 = 15,120

Product of numbers = 84 × 180 = 15,120. (Verified!)

Q4. Determine the least number that is divisible by 15, 20 and 24.

The required least number is the LCM(15, 20, 24):

15 = 3 × 5  |  20 = 2² × 5  |  24 = 2³ × 3

LCM = 2³ × 3 × 5 = 8 × 3 × 5 = 120

The least divisible number is 120.

Q5. Find the greatest number dividing 306 and 657 exactly.

The required greatest number is HCF(306, 657):

306 = 2 × 3² × 17  |  657 = 3² × 73

HCF(306, 657) = 3² = 9.

WORKSHEET - 4 Class X — Mathematics

Chapter: 1. Real Numbers Topic: Rational Numbers & Decimal Expansions

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Q1. (a) When does a rational number have a terminating decimal expansion?
(b) What type of decimal expansion does a rational number have if its denominator (in lowest form) contains primes other than 2 or 5?

(a) A rational number p/q has a terminating decimal expansion when prime factorisation of denominator q is of the form 2ⁿ × 5^m (n, m ≥ 0).

(b) It will have a Non-terminating recurring (repeating) decimal expansion.

Q2. Without actual division, state whether the decimal expansion of the following is terminating or non-terminating recurring?
(a) 7/8          (b) 9/15

(a) 7/8: Denominator = 8 = 2³ = 2³ × 5⁰. Form 2ⁿ × 5^m ⇒ Terminating.

(b) 9/15: In simplest form = 3/5. Denominator = 5 = 2⁰ × 5¹. Form 2ⁿ × 5^m ⇒ Terminating.

Q3. Write the decimal expansion of the following without actual division:
(a) 1/8          (b) 13/30

(a) 1/8 = 1 / 2³ = (1 × 5³) / (2³ × 5³) = 125 / 1000 = 0.125

(b) 13/30 = 13 / (3 × 10) = 0.4333... = 0.43̄ (Non-terminating recurring)

Q4. Express 0.375 as a rational form (p/q).

0.375 = 375 / 1000

Dividing numerator and denominator by 125:

0.375 = 3 / 8

Q5. Express 0.27 (0.2727...) as a rational number.

Let x = 0.2727... — (1)

Multiplying by 100: 100x = 27.2727... — (2)

Subtract (1) from (2): 99x = 27

x = 27 / 99 = 3 / 11

WORKSHEET - 5 Class X — Mathematics

Chapter: 1. Real Numbers Topic: Proofs of Irrational Numbers

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Q1. Prove that √3 is irrational.

Assume to the contrary that √3 is rational. Then √3 = a/b (where a, b are co-prime integers, b ≠ 0).

⇒ a = b√3 ⇒ Squaring both sides: a² = 3b² — (1)

Thus, 3 divides a², which means 3 divides a. Let a = 3c.

Substitute in (1): (3c)² = 3b² ⇒ 9c² = 3b² ⇒ b² = 3c²

Thus, 3 divides b², so 3 divides b.

Hence, 3 is a common factor of both a and b, contradicting that a, b are co-prime. Therefore, √3 is irrational.

Q2. Prove that 2 + √3 is irrational.

Let us assume 2 + √3 is rational.

Then 2 + √3 = a/b (where a, b are integers, b ≠ 0).

⇒ √3 = (a/b) - 2 ⇒ √3 = (a - 2b) / b

Since a, b are integers, (a - 2b)/b is rational.

This implies √3 is rational, which contradicts the known fact that √3 is irrational.

Hence, 2 + √3 is irrational.

Q3. Prove that √3 + 5√2 is irrational.

Assume x = √3 + 5√2 is rational.

⇒ x - 5√2 = √3 ⇒ Squaring both sides: (x - 5√2)² = (√3)²

⇒ x² - 10√2x + 50 = 3 ⇒ x² + 47 = 10√2x

⇒ √2 = (x² + 47) / (10x)

Since x is rational, (x² + 47)/(10x) is rational, implying √2 is rational.

This is a contradiction. Thus, √3 + 5√2 is irrational.

WORKSHEET - 6 (Part 1) Class X — Mathematics

Chapter: 1. Real Numbers Topic: Logarithms — Forms, Evaluations & Laws

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Q1. Convert into logarithmic form:
(i) 10³ = 1000      (ii) 2⁴ = 16      (iii) 3² = 9

(i) 10³ = 1000 ⇒ log₁₀ 1000 = 3

(ii) 2⁴ = 16 ⇒ log₂ 16 = 4

(iii) 3² = 9 ⇒ log₃ 9 = 2

Q2. Convert into exponential form:
(i) log₂ 8 = 3      (ii) log₅ 25 = 2      (iii) log₁₀ 100 = 2

(i) log₂ 8 = 3 ⇒ 2³ = 8

(ii) log₅ 25 = 2 ⇒ 5² = 25

(iii) log₁₀ 100 = 2 ⇒ 10² = 100

Q3. Evaluate:
(i) log₃ 3    (ii) log₃ 81    (iii) log₁₀ 1    (iv) log₅ 125    (v) log₁₀ 1000    (vi) log₁₀ 0.01

(i) log₃ 3 = 1

(ii) log₃ 3⁴ = 4

(iii) log₁₀ 1 = 0

(iv) log₅ 5³ = 3

(v) log₁₀ 10³ = 3

(vi) log₁₀ 10^-2 = -2

Q4. Write the laws of logarithms:

a) Product Law: log_a (mn) = log_a m + log_a n

b) Quotient Law: log_a (m/n) = log_a m - log_a n

c) Power Law: log_a (m^k) = k · log_a m

Q5. Expand using laws of logarithms:

a) log(5x) = log 5 + log x

b) log(3y/2) = log 3 + log y - log 2

c) log(7a³) = log 7 + 3 log a

d) log(4x²/y) = log 4 + 2 log x - log y

WORKSHEET - 6 (Part 2) Class X — Mathematics

Chapter: 1. Real Numbers Topic: Logarithms — Simplification & Proofs

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Q6. Simplify:

a) log 8 + log 5 = log(40)

b) log 12 - log 3 = log(4)

c) 3 log 2 = log(2³) = log 8

d) 2 log x - log y = log(x²/y)

e) log₁₀ 8 - log₁₀ 2 = log₁₀ 4

f) log₁₀ 50 + log₁₀ 2 = log₁₀ 100 = 2

Q7. Evaluate (Given: log 2 = 0.3010, log 3 = 0.4771):
(a) log 6     (b) log 9     (c) log(8/3)     (d) log₁₀ 20

a) log 6 = log(2 × 3) = log 2 + log 3 = 0.3010 + 0.4771 = 0.7781

b) log 9 = log(3²) = 2 log 3 = 2 × 0.4771 = 0.9542

c) log(8/3) = 3 log 2 - log 3 = 3(0.3010) - 0.4771 = 0.9030 - 0.4771 = 0.4259

d) log₁₀ 20 = log₁₀ (2 × 10) = log₁₀ 2 + 1 = 0.3010 + 1 = 1.3010

Q8. If log a = 2 and log b = 3, find the values of:
(a) log(ab)          (b) log(a²/b)

a) log(ab) = log a + log b = 2 + 3 = 5

b) log(a²/b) = 2 log a - log b = 2(2) - 3 = 4 - 3 = 1

Q9. If x² + y² = 18xy, then prove that 2 log(x - y) = 4 log 2 + log x + log y.

Given: x² + y² = 18xy

Subtract 2xy from both sides:

x² + y² - 2xy = 18xy - 2xy  ⇒  (x - y)² = 16xy

Taking log on both sides:

log(x - y)² = log(16xy)

⇒ 2 log(x - y) = log 16 + log x + log y

⇒ 2 log(x - y) = log(2⁴) + log x + log y

⇒ 2 log(x - y) = 4 log 2 + log x + log y. (Hence Proved!)