NCERT Exemplar Class 10 Maths Chapter 1: Real Numbers
NCERT Exemplar Class 10 Maths Chapter 1: Real Numbers

NCERT Exemplar Class 10 Maths Chapter 1: Real Numbers. NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers prepare students for their Class 10 exams thoroughly.

Maths problems and solutions for the Class 10 pdf are provided here which are similar to the questions being asked in the previous year’s board.

NCERT Exemplar Class 10 Maths Chapter 1: Real Numbers

Class 10 : Maths Chapter 1 solutions. Complete Class 10 Maths Chapter 1 Notes.

Main Concepts and Results
Euclid’s Division Lemma : Given two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.
Euclid’s Division Algorithm to obtain the HCF of two positive integers, say c and dc > d.
Step 1: Apply Euclid’s division lemma to c and d, to find whole numbers q and r, such that c = dq + r, 0 ≤ r < d.
Step 2: If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
Fundamental Theorem of Arithmetic : Every composite number can be expressed as a product of primes, and this expression (factorisation) is unique, apart from the order in which the prime factors occur.


The sum or difference of a rational and an irrational number is irrational.
The product or quotient of a non-zero rational number and an irrational number is irrational.
For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.

Multiple Choice Questions (Solved Examples)

Choose the correct answer from the given four options:
Sample Question 1: The decimal expansion of the rational number  will terminate after
(A) one decimal place (B) two decimal places (C) three decimal places (D) more than 3 decimal places
Solution: Answer (B)
Sample Question 2: Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy
(A) 1 < r < b (B) 0 < r ≤ b (C) 0 ≤ r < b (D) 0 < r < b
Solution: Answer (C)

Multiple Choice Questions

Real Numbers/

Short Answer Questions with Reasoning

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Short Answer Type Questions

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Long Answer Type Questions (Solved Examples)

Sample Question 1: Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.
Solution: We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5, for some integer m.
Thus, an odd positive integer can be of the form 6m + 1, 6m + 3, or 6m + 5
Thus we have:
(6m+1)² = 36  + 12m + 1= 6 ( 6m² + 2m) + 1= 6+ 1, q is an integer
(6m+3)² = 36m² + 36+ 9 = 6 (6+6+ 1) + 3= 6+ 3, q is an integer
(6m + 5)² = 36 + 60 m + 25 = 6 (6+ 10 m + 4) + 1 = 6q + 1, q is an integer.
Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.

Long Answer Type Questions

Real Numbers/

Answers