NCERT Exemplar Class 10 Maths Chapter 1 Real Numbers Solutions (CBSE 2026-27)
This post contains filtered NCERT Exemplar questions strictly based on the rationalised Class 10 syllabus. Deleted topics like Euclid’s Division Lemma/Algorithm and Decimal Expansions of Rational Numbers have been removed so you focus exclusively on scoring areas: the Fundamental Theorem of Arithmetic and Proofs of Irrationality.
Section A: Multiple Choice Questions (1 Mark Each)
Express both numbers as products of their prime factors:
\[65 = 5 \times 13\] \[117 = 3^2 \times 13\]The Highest Common Factor is the product of the smallest power of each common prime factor:
\[\operatorname{HCF}(65, 117) = 13\]According to the given linear expression:
\[65m – 117 = 13\] \[65m = 13 + 117\] \[65m = 130 \implies m = \frac{130}{65} = 2\]Subtract the required remainders from the given numbers:
\[70 – 5 = 65\] \[125 – 8 = 117\]The largest number dividing both numbers completely is \(\operatorname{HCF}(65, 117)\):
\[65 = 5 \times 13\] \[117 = 3^2 \times 13\] \[\operatorname{HCF}(65, 117) = 13\]The \(\operatorname{HCF}\) of algebraic monomials is the product of the smallest powers of each common variable:
- Smallest power of \(x\): \(x^{\min(3, 1)} = x^1\)
- Smallest power of \(y\): \(y^{\min(2, 3)} = y^2\)
The \(\operatorname{LCM}\) is the product of the greatest powers of each prime factor involved:
- Highest power of factor \(a\): \(a^{\max(1, 3)} = a^3\)
- Highest power of factor \(b\): \(b^{\max(2, 1)} = b^2\)
Let \(r \in \mathbb{Q}\) (\(r \neq 0\)) and let \(x \notin \mathbb{Q}\) (irrational).
Assume, for contradiction, that their product equals a rational number \(k \in \mathbb{Q}\):
\[r \times x = k \implies x = \frac{k}{r}\]Since the quotient of two rational numbers (\(r \neq 0\)) is always rational, \(\frac{k}{r}\) is rational. This implies \(x\) is rational, which contradicts the premise that \(x\) is irrational. Thus, the product is always irrational.
The required number is \(\operatorname{LCM}(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)\). Prime factorisation of each integer:
- \(1 = 1\), \(2 = 2^1\), \(3 = 3^1\), \(4 = 2^2\), \(5 = 5^1\)
- \(6 = 2^1 \times 3^1\), \(7 = 7^1\), \(8 = 2^3\), \(9 = 3^2\), \(10 = 2^1 \times 5^1\)
Taking the greatest power of each unique prime factor (\(2, 3, 5, 7\)):
\[\operatorname{LCM} = 2^3 \times 3^2 \times 5^1 \times 7^1 = 8 \times 9 \times 5 \times 7 = 2520\]Section B: Short Answer Questions with Reasoning (2 & 3 Marks)
No.
A natural number can end with the digit \(5\) only if \(5\) divides the number, meaning \(5\) must be a factor in its prime factorisation.
Writing \(6^n\) as prime factors:
\[6^n = (2 \times 3)^n = 2^n \times 3^n\]By the Fundamental Theorem of Arithmetic, prime factorisation is unique. The only prime factors of \(6^n\) are \(2\) and \(3\). Since \(5\) does not appear in its factorisation, \(6^n\) cannot end with the digit \(5\) for any \(n \in \mathbb{N}\).
Reason: The Highest Common Factor (\(\operatorname{HCF}\)) of two numbers is defined as the largest common divisor of both numbers. The set of common divisors of \(525\) and \(3000\) is given as \(\{3, 5, 15, 25, 75\}\). The greatest value in this set is \(75\).
Factoring the expression using the distributive property:
\[3 \times 5 \times 7 + 7 = 7 \times (3 \times 5 + 1)\] \[= 7 \times (15 + 1)\] \[= 7 \times 16\] \[= 7 \times 2^4\]Since the given expression can be written as a product of prime factors (\(2\) and \(7\)) and has positive factors other than \(1\) and itself, it is a composite number by the Fundamental Theorem of Arithmetic.
No.
The \(\operatorname{HCF}\) of two numbers must always divide their \(\operatorname{LCM}\) completely without any remainder.
\[\frac{\operatorname{LCM}}{\operatorname{HCF}} = \frac{380}{18} = \frac{190}{9} = 21.11\dots\]Dividing \(380\) by \(18\) leaves a remainder of \(2\). Since \(18\) does not divide \(380\) exactly, two numbers cannot have an \(\operatorname{HCF}\) of \(18\) and an \(\operatorname{LCM}\) of \(380\).
Section C: Proofs of Irrationality & Competency Word Problems
Assume, for contradiction, that \(\sqrt{2} + \sqrt{3}\) is a rational number \(a\) (\(a \in \mathbb{Q}, a \neq 0\)):
\[\sqrt{2} + \sqrt{3} = a \implies \sqrt{2} = a – \sqrt{3}\]Squaring both sides:
\[(\sqrt{2})^2 = (a – \sqrt{3})^2\] \[2 = a^2 – 2a\sqrt{3} + 3\]Rearranging the equation to isolate \(\sqrt{3}\):
\[2a\sqrt{3} = a^2 + 3 – 2\] \[2a\sqrt{3} = a^2 + 1\] \[\sqrt{3} = \frac{a^2 + 1}{2a}\]Because \(a\) is rational and \(a \neq 0\), the right-hand side \(\frac{a^2 + 1}{2a}\) is rational. This implies that \(\sqrt{3}\) is rational.
However, this contradicts the fact that \(\sqrt{3}\) is irrational. Therefore, our assumption is false, and \(\sqrt{2} + \sqrt{3}\) is irrational.
Assume, for contradiction, that \(\sqrt{3} + \sqrt{5}\) is rational. Let:
\[\sqrt{3} + \sqrt{5} = x \quad \text{where } x \in \mathbb{Q} \text{ and } x \neq 0\]Isolating \(\sqrt{5}\):
\[\sqrt{5} = x – \sqrt{3}\]Squaring both sides:
\[(\sqrt{5})^2 = (x – \sqrt{3})^2\] \[5 = x^2 – 2x\sqrt{3} + 3\]Rearranging terms to isolate \(\sqrt{3}\):
\[2x\sqrt{3} = x^2 + 3 – 5\] \[2x\sqrt{3} = x^2 – 2\] \[\sqrt{3} = \frac{x^2 – 2}{2x}\]Since \(x \in \mathbb{Q}\) and \(x \neq 0\), the quotient \(\frac{x^2 – 2}{2x}\) is rational. This implies that \(\sqrt{3}\) is rational, which contradicts the known theorem that \(\sqrt{3}\) is irrational.
Hence, our assumption is incorrect, and \(\sqrt{3} + \sqrt{5}\) is irrational.
A natural number ends with the digit \(0\) if and only if it is divisible by \(10 = 2 \times 5\). It ends with the digit \(5\) if it is divisible by \(5\). Thus, in both cases, \(5\) must be a prime factor.
Finding the prime factors of \(12^n\):
\[12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n\]The only prime factors present in the factorisation of \(12^n\) are \(2\) and \(3\). By the uniqueness property of the Fundamental Theorem of Arithmetic, there are no other prime factors in the expansion of \(12^n\).
Because \(5\) is not a factor of \(12^n\), it follows that \(12^n\) cannot end with the digit \(0\) or \(5\) for any \(n \in \mathbb{N}\).
The minimum common distance that can be covered in complete step counts is determined by the Least Common Multiple (\(\operatorname{LCM}\)) of the step lengths:
\[\text{Minimum Distance} = \operatorname{LCM}(40, 42, 45)\text{ cm}\]Prime factorisation of the given values:
\[40 = 2^3 \times 5^1\] \[42 = 2^1 \times 3^1 \times 7^1\] \[45 = 3^2 \times 5^1\]Selecting the greatest power of each factor:
\[\operatorname{LCM}(40, 42, 45) = 2^3 \times 3^2 \times 5^1 \times 7^1\] \[= 8 \times 9 \times 5 \times 7 = 2520\text{ cm}\]Converting the answer to metres:
\[2520\text{ cm} = \frac{2520}{100}\text{ m} = 25.2\text{ m}\]Answer: The minimum distance each person should walk is \(2520\text{ cm}\) (or \(25.2\text{ m}\)).
Assume to the contrary that \(\sqrt{p} + \sqrt{q}\) is rational. Let:
\[\sqrt{p} + \sqrt{q} = x \quad \text{where } x \in \mathbb{Q} \text{ and } x \neq 0\]Rearranging the equation:
\[\sqrt{p} = x – \sqrt{q}\]Squaring both sides:
\[(\sqrt{p})^2 = (x – \sqrt{q})^2\] \[p = x^2 – 2x\sqrt{q} + q\]Isolating \(\sqrt{q}\):
\[2x\sqrt{q} = x^2 + q – p\] \[\sqrt{q} = \frac{x^2 + q – p}{2x}\]Because \(x \in \mathbb{Q}\) (\(x \neq 0\)) and \(p, q\) are integers, the expression \(\frac{x^2 + q – p}{2x}\) is rational.
This implies that \(\sqrt{q}\) is rational. But this contradicts the theorem that the square root of any prime number is irrational. Hence, our assumption is false, and \(\sqrt{p} + \sqrt{q}\) is irrational.
