Real Numbers: CBSE Class 10 Previous Year Questions

203 different questions from Real Numbers (NCERT Chapter 1) asked in CBSE Class 10 Maths board exams 2022–2026. 4 of them came up in more than one year. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions104 questions 2 marks questions48 questions 3 marks questions49 questions 4 marks questions2 questions

Most asked Real Numbers questions

Q273 marksShort AnswerReal Numbers CBSE 2023 · Standard 30/1/1

Prove that $5$ is an irrational number.

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Answer: Proved.

Suppose $5$ is rational. Then $5 = \frac{a}{b}$ with integers a, b, $b \neq = 0$ , having no common factor other than 1.
Squaring, $a^{2} = 5 b^{2}$ , so 5 divides $a^{2}$ , hence 5 divides a (5 is prime).
Let $a = 5 c$ . Then $25 c^{2} = 5 b^{2}$ , so $b^{2} = 5 c^{2}$ ; 5 divides $b^{2}$ , hence 5 divides b.
So 5 is a common factor of a and b, a contradiction.
Hence $5$ is irrational.

Q303 marksShort AnswerReal Numbers CBSE 2023 · Standard 30/4/2

Prove that $3$ is an irrational number.

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Answer: Proved.

Assume $3$ is rational. Then $3 = \frac{a}{b}$ with $a$ , $b$ co-prime integers, $b \neq = 0$ .
Squaring: $a^{2} = 3 b^{2}$ , so 3 divides $a^{2}$ , hence 3 divides $a$ (3 is prime).
Let $a = 3 c$ . Then $9 c^{2} = 3 b^{2} \Rightarrow b^{2} = 3 c^{2}$ , so 3 divides $b^{2}$ and hence $b$ .
So 3 is a common factor of $a$ and $b$ , contradicting that they are co-prime.
Hence $3$ is irrational.

Also asked in: 2026 Standard 30/1/1, 2026 Standard 30/1/3, 2026 Standard 30/3/2, 2025 Standard 30/2/1, 2025 Standard 30/2/2, 2025 Standard 30/2/3, 2025 Standard 30/5/1, 2025 Standard 30/5/2, 2025 Standard 30/5/3, 2025 Basic 430/1/1, 2025 Basic 430/1/2, 2025 Basic 430/1/3, 2025 Basic 430/4/1, 2024 Standard 30/5/1, 2024 Standard 30/5/2, 2024 Standard 30/5/3, 2023 Standard 30/5/1, 2023 Standard 30/5/2, 2023 Standard 30/5/3

Q27 (OR) (OR)3 marksShort AnswerReal Numbers CBSE 2023 · Standard 30/6/1

Prove that $2$ is an irrational number.

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Answer: Proved.

Assume $2$ is rational, so $2 = \frac{a}{b}$ where a, b are co-prime integers and $b \neq = 0$ .
Squaring: $2 b^{2} = a^{2}$ , so 2 divides $a^{2}$ , hence 2 divides a.
Let a = 2c. Then $2 b^{2} = 4 c^{2}$ , so $b^{2} = 2 c^{2}$ ; 2 divides $b^{2}$ , hence 2 divides b.
So 2 is a common factor of a and b, contradicting that they are co-prime.
Hence $2$ is irrational.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/3/3, 2026 Basic 430/5/1, 2026 Basic 430/5/2, 2026 Basic 430/5/3, 2025 Standard 30/4/1, 2025 Standard 30/4/2, 2025 Standard 30/4/3, 2025 Basic 430/3/1, 2025 Basic 430/3/2, 2025 Basic 430/3/3, 2025 Basic 430/4/3, 2023 Standard 30/6/2, 2023 Standard 30/6/3

Q303 marksShort AnswerReal Numbers CBSE 2023 · Basic 430/6/1

Prove that $3 - 25$ is an irrational number, given that $5$ is an irrational number.

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Answer: Proved.

Suppose $3 - 25$ is rational, say $3 - 25 = r$ where $r$ is rational.
Then $5 = \frac{3 - r}{2}$ .
The right side is rational (rationals are closed under subtraction and division by a non-zero rational), so $5$ would be rational.
This contradicts the fact that $5$ is irrational. Hence $3 - 25$ is irrational.

Also asked in: 2024 Basic 430/5/2

Q171 markMCQReal Numbers CBSE 2026 · Basic 430/4/1

7 × 29 × 23 + 1 is :

(A)a prime number.
(B)divisible by 23.
(C)an odd number.
(D)a composite number.

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Answer: (D) a composite number.

$7 \times 29 \times 23 = 4669$ , so the number is 4670.
4670 is even, so it has factors other than 1 and itself.
Hence it is a composite number.

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q201 markAssertion–ReasonReal Numbers CBSE 2026 · Basic 430/4/1

Assertion (A) : $4^{n}$ can not end with the digit zero.
Reason (R) : Prime factorisation of $4^{n}$ is unique.

(A)Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
(B)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

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Answer: (A) Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).

A number ending in 0 must have both 2 and 5 as prime factors.
$4^{n} = 2^{2 n}$ , and by uniqueness of prime factorisation it has no other prime factor, so 5 never divides it.
So A is true, R is true and R explains A.

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q11 markMCQReal Numbers CBSE 2026 · Standard 30/1/1

The HCF of 960 and 432 is :

(A)48
(B)54
(C)72
(D)36

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Answer: (A) 48

$960 = 2^{6} \times 3 \times 5$ and $432 = 2^{4} \times 3^{3}$ .
HCF = product of the smallest powers of common primes = $2^{4} \times 3 = 48$ .

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q21 markMCQReal Numbers CBSE 2026 · Standard 30/1/1

The natural number 2 is :

(A)a prime number
(B)a composite number
(C)prime as well as composite
(D)neither prime nor composite

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Answer: (A) a prime number

2 has exactly two factors, 1 and 2.
So 2 is a prime number (the only even prime).

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q31 markMCQReal Numbers CBSE 2026 · Standard 30/1/1

For any natural number $n$ , $6^{n}$ ends with the digit :

(A)0
(B)6
(C)3
(D)2

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Answer: (B) 6

$6^{1} = 6$ , $6^{2} = 36$ , $6^{3} = 216$ , ...
Any power of a number ending in 6 also ends in 6, since $6 \times 6 = 36$ ends in 6.
So $6^{n}$ always ends with the digit 6.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q21 markMCQReal Numbers CBSE 2026 · Standard 30/2/1

The natural number 1 is :

(A)a prime number.
(B)a composite number.
(C)prime as well as composite.
(D)neither prime nor composite.

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Answer: (D) neither prime nor composite.

A prime number has exactly two factors and a composite number has more than two factors.
1 has only one factor (itself), so it is neither prime nor composite.

Also asked in: 2026 Standard 30/2/2, 2026 Standard 30/2/3

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