Real Numbers — Class 10 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Real Numbers" — 9 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Real Numbers" — 9 important questions with detailed answers for CBSE board exam pr…
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NCERT Solutions for Class 10 Mathematics — Real Numbers. Step-by-step answers to all 9 textbook questions from this chapter, written for CBSE board preparation and free to use.
Real Numbers — All 9 Questions Solved
Q1. State the Fundamental Theorem of Arithmetic.
The Fundamental Theorem of Arithmetic states that every composite number can be expressed (factorised) as a product of prime numbers, and this factorisation is unique, apart from the order in which the prime factors occur.
For example:
- 60 = 2² × 3 × 5
- 84 = 2² × 3 × 7
This means that if we write a composite number as a product of primes in two different ways, the prime factors (with their powers) must be exactly the same.
Q2. Prove that √2 is irrational.
Proof by contradiction:
Assume √2 is rational. Then √2 = p/q, where p and q are coprime integers (HCF(p,q) = 1) and q ≠ 0.
Squaring both sides: 2 = p²/q²
Therefore: 2q² = p² ... (1)
From equation (1), p² is even, which means p must be even.
Let p = 2m for some integer m.
Substituting in equation (1):
2q² = (2m)²
2q² = 4m²
q² = 2m²
This shows q² is even, so q must be even.
But if both p and q are even, then HCF(p,q) ≥ 2, which contradicts our assumption that p and q are coprime.
Therefore, our assumption is wrong, and √2 is irrational.
Q3. Find the HCF of 96 and 404 using Euclid's division algorithm.
Using Euclid's division algorithm:
Step 1: Divide 404 by 96
404 = 96 × 4 + 20
Step 2: Divide 96 by 20
96 = 20 × 4 + 16
Step 3: Divide 20 by 16
20 = 16 × 1 + 4
Step 4: Divide 16 by 4
16 = 4 × 4 + 0
Since the remainder is 0, the HCF is 4.
Therefore, HCF(96, 404) = 4
Q4. Find the LCM and HCF of 12 and 18 using prime factorisation method.
Step 1: Find prime factorisation of both numbers
12 = 2² × 3
18 = 2 × 3²
Step 2: Find HCF (take lowest powers of common prime factors)
Common prime factors: 2 and 3
HCF = 2¹ × 3¹ = 6
Step 3: Find LCM (take highest powers of all prime factors)
LCM = 2² × 3² = 4 × 9 = 36
Verification: HCF × LCM = 6 × 36 = 216
and 12 × 18 = 216 ✓
Therefore, HCF(12, 18) = 6 and LCM(12, 18) = 36
Q5. Express 0.3̄5̄ (0.353535...) as a fraction in simplest form.
Let x = 0.353535...
Since two digits repeat, multiply both sides by 100:
100x = 35.353535...
Subtract the original equation:
100x - x = 35.353535... - 0.353535...
99x = 35
x = 35/99
Check if this can be simplified:
HCF(35, 99) = HCF(5 × 7, 9 × 11) = 1
Therefore, 0.3̄5̄ = 35/99
Q6. Find the HCF of 84, 120, and 138.
Using prime factorisation:
84 = 2² × 3 × 7
120 = 2³ × 3 × 5
138 = 2 × 3 × 23
HCF = Product of lowest powers of common prime factors
Common prime factors: 2 and 3
HCF = 2¹ × 3¹ = 6
Therefore, HCF(84, 120, 138) = 6
Q7. A number when divided by 407 leaves a remainder of 4. When the same number is divided by 11, what is the remainder? (Assume the number is divisible by 37)
Let the number be N.
Given: N = 407k + 4 for some integer k
Also given: N is divisible by 37
Step 1: Express 407 in terms of 37
407 = 11 × 37
Step 2: Substitute in the equation
N = 11 × 37 × k + 4
N = 37(11k) + 4
Since N is divisible by 37:
37(11k) + 4 ≡ 0 (mod 37)
4 ≡ 0 (mod 37)
This is not possible. Let me recalculate:
N = 407k + 4
Since N is divisible by 37:
407k + 4 ≡ 0 (mod 37)
(11 × 37)k + 4 ≡ 0 (mod 37)
4 ≡ 0 (mod 37) — contradiction
Let's check: if 407 = 11 × 37, and N = 407k + 4:
N mod 11 = (407k + 4) mod 11 = (0 + 4) mod 11 = 4
Therefore, remainder when N is divided by 11 is 4
Q8. If LCM(a,b) = 180 and HCF(a,b) = 12, and one number is 36, find the other number.
Given:
LCM(a,b) = 180
HCF(a,b) = 12
One number (say a) = 36
Find the other number (say b) = ?
Using the relation: HCF(a,b) × LCM(a,b) = a × b
12 × 180 = 36 × b
2160 = 36 × b
b = 2160/36
b = 60
Verification:
60 = 2² × 3 × 5
36 = 2² × 3²
HCF = 2² × 3 = 12 ✓
LCM = 2² × 3² × 5 = 180 ✓
Therefore, the other number is 60
Q9. Prove that 5 + 3√2 is irrational.
Proof by contradiction:
Assume 5 + 3√2 is rational.
Then 5 + 3√2 = p/q, where p and q are coprime integers and q ≠ 0.
Rearranging:
3√2 = p/q - 5
3√2 = (p - 5q)/q
√2 = (p - 5q)/(3q)
Since p and q are integers, (p - 5q)/(3q) is a rational number.
This implies √2 is rational, which contradicts the fact that √2 is irrational.
Therefore, our assumption is wrong, and 5 + 3√2 is irrational.
Real Numbers — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the HCF of 96 and 404 using prime factorization?
- A. 4
- B. 8
- C. 12
- D. 16
Answer: A. 4 — Prime factorization of 96 = 2^5 * 3, and 404 = 2^2 * 101. The HCF is the product of the lowest powers of common prime factors, which is 2^2 = 4.
MCQ 2. If two positive integers 'a' and 'b' are written as a = pq^2 and b = p^3q, where p and q are prime numbers, then HCF(a, b) is:
- A. pq
- B. p^3q^2
- C. p^2q^2
- D. pq^2
Answer: A. pq — HCF is the product of the lowest powers of common prime factors. For 'p', the lowest power is p^1, and for 'q', it's q^1. So, HCF(a, b) = pq.
MCQ 3. A sweet seller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the maximum number of barfis that can be placed in each stack?
- A. 10
- B. 20
- C. 30
- D. 40
Answer: A. 10 — To find the maximum number of barfis in each stack, we need to find the HCF of 420 and 130. Using Euclid's algorithm: 420 = 3 * 130 + 30; 130 = 4 * 30 + 10; 30 = 3 * 10 + 0. The HCF is 10.
MCQ 4. The prime factorization of 156 is:
- A. 2 * 3 * 13
- B. 2^2 * 3 * 13
- C. 2 * 3^2 * 13
- D. 2^2 * 3^2 * 13
Answer: B. 2^2 * 3 * 13 — 156 ÷ 2 = 78; 78 ÷ 2 = 39; 39 ÷ 3 = 13; 13 ÷ 13 = 1. So, 156 = 2^2 * 3 * 13.
MCQ 5. If HCF(26, 91) = 13, then LCM(26, 91) is:
- A. 182
- B. 91
- C. 26
- D. 2366
Answer: A. 182 — For any two positive integers a and b, HCF(a, b) * LCM(a, b) = a * b. So, LCM(26, 91) = (26 * 91) / 13 = 2366 / 13 = 182.
MCQ 6. The product of a non-zero rational number and an irrational number is:
- A. Always rational
- B. Always irrational
- C. Rational or irrational
- D. Always an integer
Answer: B. Always irrational — The product of a non-zero rational number and an irrational number is always irrational. For example, 2 * √3 = 2√3, which is irrational.
MCQ 7. The decimal expansion of the rational number 33/(2^2 * 5) will terminate after how many decimal places?
- A. 1
- B. 2
- C. 3
- D. 4
Answer: B. 2 — For a rational number p/q to have a terminating decimal expansion, the prime factorization of q must be of the form 2^m * 5^n. Here, the maximum power is 2, so it terminates after 2 decimal places.
MCQ 8. Which of the following rational numbers has a non-terminating repeating decimal expansion?
- A. 15/1600
- B. 23/(2^3 * 5^2)
- C. 17/6
- D. 35/50
Answer: C. 17/6 — A rational number has a non-terminating repeating decimal expansion if its denominator (in simplest form) has prime factors other than 2 or 5. For 17/6, the denominator 6 = 2 * 3, which includes a factor of 3.
MCQ 9. Which of the following is an irrational number?
- A. √16
- B. √12/√3
- C. 3 + √5
- D. √100
Answer: C. 3 + √5 — √16 = 4, √12/√3 = √4 = 2, and √100 = 10, all are rational. 3 + √5 is irrational because the sum of a rational and an irrational number is always irrational.
MCQ 10. The sum of two irrational numbers is:
- A. Always rational
- B. Always irrational
- C. Rational or irrational
- D. Always an integer
Answer: C. Rational or irrational — The sum of two irrational numbers can be rational (e.g., (2+√3) + (2-√3) = 4) or irrational (e.g., √2 + √3).
MCQ 11. Every composite number can be expressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur. This statement is known as:
- A. Euclid's Division Lemma
- B. Fundamental Theorem of Arithmetic
- C. Division Algorithm
- D. Irrationality Theorem
Answer: B. Fundamental Theorem of Arithmetic — This is the precise statement of the Fundamental Theorem of Arithmetic.
MCQ 12. If HCF(a, 8) = 4 and LCM(a, 8) = 24, then the value of a is:
- A. 12
- B. 16
- C. 24
- D. 8
Answer: A. 12 — Using the property HCF(a, b) * LCM(a, b) = a * b, we have 4 * 24 = a * 8. This simplifies to 96 = 8a, so a = 12.
MCQ 13. The largest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is:
- A. 13
- B. 65
- C. 875
- D. 1750
Answer: A. 13 — The number must divide (70-5) = 65 and (125-8) = 117 exactly. We need to find the HCF of 65 and 117. Prime factors of 65 = 5 * 13, and 117 = 3^2 * 13. The HCF is 13.
MCQ 14. Can two numbers have 18 as their HCF and 380 as their LCM?
- A. Yes
- B. No
- C. Can't say
- D. Only if one number is a multiple of the other
Answer: B. No — For any two positive integers, their HCF must always be a factor of their LCM. Since 18 is not a factor of 380 (380 divided by 18 leaves a remainder), it is not possible.
MCQ 15. Which of the following is a rational number?
- A. π
- B. √7
- C. 0.333...
- D. 2 + √2
Answer: C. 0.333... — π is irrational, √7 is irrational, and 2 + √2 is irrational. 0.333... can be expressed as 1/3, which is a rational number.
MCQ 16. The decimal expansion of the number 147/1200 will terminate after how many decimal places?
- A. 1
- B. 2
- C. 3
- D. 4
Answer: D. 4 — First, simplify the fraction: 147/1200 = 49/400. Factorize the denominator: 400 = 2^4 * 5^2. The highest power is 4, so it terminates after 4 decimal places.
MCQ 17. If p is a prime number, then √p is:
- A. Always rational
- B. Always irrational
- C. Always an integer
- D. Sometimes rational, sometimes irrational
Answer: B. Always irrational — According to a fundamental theorem, the square root of any prime number is always irrational.
MCQ 18. Find the least number that is divisible by all numbers from 1 to 10 (inclusive).
- A. 100
- B. 2520
- C. 5040
- D. 720
Answer: B. 2520 — The least number divisible by all numbers from 1 to 10 is their LCM. LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) = 2^3 * 3^2 * 5 * 7 = 8 * 9 * 5 * 7 = 2520.
MCQ 19. A number which can be expressed in the form p/q, where p and q are integers and q ≠ 0, is called a:
- A. Irrational number
- B. Real number
- C. Integer
- D. Rational number
Answer: D. Rational number — This is the definition of a rational number.
MCQ 20. The number 7 * 11 * 13 + 13 is a:
- A. Prime number
- B. Composite number
- C. Rational number
- D. Irrational number
Answer: B. Composite number — 7 * 11 * 13 + 13 = 13 * (7 * 11 + 1) = 13 * (77 + 1) = 13 * 78 = 13 * (2 * 3 * 13). Since it has more than two factors (1, itself, and others like 2, 3), it is a composite number.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.