This MCQ module is based on: Real Numbers – Chapter Exercises
TOPIC 3 OF 35
Real Numbers – Chapter Exercises
🎓 Class 10
Mathematics
CBSE
Theory
Ch 1 — Real Numbers
⏱ ~40 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Real Numbers – Chapter Exercises
Targeting Class 10 level in Number Theory, with Intermediate difficulty.
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Exercise 1.1
Q1. Express each number as a product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
(i) 140:
\(140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7\)
\(\boxed{140 = 2^2 \times 5 \times 7}\)
(ii) 156:
\(156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13\)
\(\boxed{156 = 2^2 \times 3 \times 13}\)
(iii) 3825:
\(3825 = 3 \times 1275 = 3 \times 3 \times 425 = 3^2 \times 425\)
\(425 = 5 \times 85 = 5 \times 5 \times 17\)
\(\boxed{3825 = 3^2 \times 5^2 \times 17}\)
(iv) 5005:
\(5005 = 5 \times 1001 = 5 \times 7 \times 143 = 5 \times 7 \times 11 \times 13\)
\(\boxed{5005 = 5 \times 7 \times 11 \times 13}\)
(v) 7429:
\(7429 = 17 \times 437 = 17 \times 19 \times 23\)
\(\boxed{7429 = 17 \times 19 \times 23}\)
\(140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7\)
\(\boxed{140 = 2^2 \times 5 \times 7}\)
(ii) 156:
\(156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13\)
\(\boxed{156 = 2^2 \times 3 \times 13}\)
(iii) 3825:
\(3825 = 3 \times 1275 = 3 \times 3 \times 425 = 3^2 \times 425\)
\(425 = 5 \times 85 = 5 \times 5 \times 17\)
\(\boxed{3825 = 3^2 \times 5^2 \times 17}\)
(iv) 5005:
\(5005 = 5 \times 1001 = 5 \times 7 \times 143 = 5 \times 7 \times 11 \times 13\)
\(\boxed{5005 = 5 \times 7 \times 11 \times 13}\)
(v) 7429:
\(7429 = 17 \times 437 = 17 \times 19 \times 23\)
\(\boxed{7429 = 17 \times 19 \times 23}\)
Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers:
(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
(i) 26 and 91:
\(26 = 2 \times 13\)
\(91 = 7 \times 13\)
\(\text{HCF} = 13\) (common prime factor with smallest power)
\(\text{LCM} = 2 \times 7 \times 13 = 182\) (all primes with greatest powers)
Verification: \(\text{HCF} \times \text{LCM} = 13 \times 182 = 2366\). Also \(26 \times 91 = 2366\). ✓
(ii) 510 and 92:
\(510 = 2 \times 3 \times 5 \times 17\)
\(92 = 2^2 \times 23\)
\(\text{HCF} = 2\)
\(\text{LCM} = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460\)
Verification: \(2 \times 23460 = 46920\). Also \(510 \times 92 = 46920\). ✓
(iii) 336 and 54:
\(336 = 2^4 \times 3 \times 7\)
\(54 = 2 \times 3^3\)
\(\text{HCF} = 2^1 \times 3^1 = 6\)
\(\text{LCM} = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024\)
Verification: \(6 \times 3024 = 18144\). Also \(336 \times 54 = 18144\). ✓
\(26 = 2 \times 13\)
\(91 = 7 \times 13\)
\(\text{HCF} = 13\) (common prime factor with smallest power)
\(\text{LCM} = 2 \times 7 \times 13 = 182\) (all primes with greatest powers)
Verification: \(\text{HCF} \times \text{LCM} = 13 \times 182 = 2366\). Also \(26 \times 91 = 2366\). ✓
(ii) 510 and 92:
\(510 = 2 \times 3 \times 5 \times 17\)
\(92 = 2^2 \times 23\)
\(\text{HCF} = 2\)
\(\text{LCM} = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460\)
Verification: \(2 \times 23460 = 46920\). Also \(510 \times 92 = 46920\). ✓
(iii) 336 and 54:
\(336 = 2^4 \times 3 \times 7\)
\(54 = 2 \times 3^3\)
\(\text{HCF} = 2^1 \times 3^1 = 6\)
\(\text{LCM} = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024\)
Verification: \(6 \times 3024 = 18144\). Also \(336 \times 54 = 18144\). ✓
Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method:
(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
(i) 12, 15 and 21:
\(12 = 2^2 \times 3\)
\(15 = 3 \times 5\)
\(21 = 3 \times 7\)
\(\text{HCF} = 3\) (only common prime factor is 3, smallest power \(3^1\))
\(\text{LCM} = 2^2 \times 3 \times 5 \times 7 = 420\)
(ii) 17, 23 and 29:
All three are prime numbers.
\(\text{HCF} = 1\) (no common prime factors)
\(\text{LCM} = 17 \times 23 \times 29 = 11339\)
(iii) 8, 9 and 25:
\(8 = 2^3\)
\(9 = 3^2\)
\(25 = 5^2\)
\(\text{HCF} = 1\) (no common prime factors)
\(\text{LCM} = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800\)
\(12 = 2^2 \times 3\)
\(15 = 3 \times 5\)
\(21 = 3 \times 7\)
\(\text{HCF} = 3\) (only common prime factor is 3, smallest power \(3^1\))
\(\text{LCM} = 2^2 \times 3 \times 5 \times 7 = 420\)
(ii) 17, 23 and 29:
All three are prime numbers.
\(\text{HCF} = 1\) (no common prime factors)
\(\text{LCM} = 17 \times 23 \times 29 = 11339\)
(iii) 8, 9 and 25:
\(8 = 2^3\)
\(9 = 3^2\)
\(25 = 5^2\)
\(\text{HCF} = 1\) (no common prime factors)
\(\text{LCM} = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800\)
Q4. Given that HCF(306, 657) = 9, find LCM(306, 657).
Step 1: We know that for two positive integers \(a\) and \(b\):
\[\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b\]
Step 2: Substitute the known values:
\(9 \times \text{LCM}(306, 657) = 306 \times 657\)
Step 3: Calculate:
\(306 \times 657 = 201042\)
Step 4: Solve for LCM:
\(\text{LCM}(306, 657) = \frac{201042}{9} = \boxed{22338}\)
\[\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b\]
Step 2: Substitute the known values:
\(9 \times \text{LCM}(306, 657) = 306 \times 657\)
Step 3: Calculate:
\(306 \times 657 = 201042\)
Step 4: Solve for LCM:
\(\text{LCM}(306, 657) = \frac{201042}{9} = \boxed{22338}\)
Q5. Check whether \(6^n\) can end with the digit 0 for any natural number \(n\).
Step 1: If a number ends with the digit 0, it must be divisible by 10, and hence by both 2 and 5.
Step 2: Prime factorisation of \(6^n\):
\(6^n = (2 \times 3)^n = 2^n \times 3^n\)
Step 3: The only prime factors of \(6^n\) are 2 and 3. The prime 5 does not appear in the factorisation.
Step 4: By the uniqueness of the Fundamental Theorem of Arithmetic, no other primes can divide \(6^n\). Since 5 does not divide \(6^n\), the number \(6^n\) cannot be divisible by 10.
Conclusion: \(\boxed{6^n \text{ can never end with the digit 0}}\) for any natural number \(n\).
Step 2: Prime factorisation of \(6^n\):
\(6^n = (2 \times 3)^n = 2^n \times 3^n\)
Step 3: The only prime factors of \(6^n\) are 2 and 3. The prime 5 does not appear in the factorisation.
Step 4: By the uniqueness of the Fundamental Theorem of Arithmetic, no other primes can divide \(6^n\). Since 5 does not divide \(6^n\), the number \(6^n\) cannot be divisible by 10.
Conclusion: \(\boxed{6^n \text{ can never end with the digit 0}}\) for any natural number \(n\).
Q6. Explain why \(7 \times 11 \times 13 + 13\) and \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\) are composite numbers.
First number: \(7 \times 11 \times 13 + 13\)
Factor out 13:
\(= 13(7 \times 11 + 1) = 13(77 + 1) = 13 \times 78 = 13 \times 78\)
Since \(78 = 2 \times 3 \times 13\), the number equals \(2 \times 3 \times 13^2 = 1014\).
Since it has factors other than 1 and itself, it is composite.
Second number: \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\)
\(= 5040 + 5\)
Factor out 5:
\(= 5(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5(1008 + 1) = 5 \times 1009\)
Since it is expressible as a product of 5 and 1009 (both greater than 1), it is composite.
Factor out 13:
\(= 13(7 \times 11 + 1) = 13(77 + 1) = 13 \times 78 = 13 \times 78\)
Since \(78 = 2 \times 3 \times 13\), the number equals \(2 \times 3 \times 13^2 = 1014\).
Since it has factors other than 1 and itself, it is composite.
Second number: \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\)
\(= 5040 + 5\)
Factor out 5:
\(= 5(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5(1008 + 1) = 5 \times 1009\)
Since it is expressible as a product of 5 and 1009 (both greater than 1), it is composite.
Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Step 1: To find when both will be at the starting point together again, we need the LCM of their round times.
Step 2: Prime factorisations:
\(18 = 2 \times 3^2\)
\(12 = 2^2 \times 3\)
Step 3: \(\text{LCM}(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36\)
Conclusion: They will meet again at the starting point after \(\boxed{36 \text{ minutes}}\).
Step 2: Prime factorisations:
\(18 = 2 \times 3^2\)
\(12 = 2^2 \times 3\)
Step 3: \(\text{LCM}(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36\)
Conclusion: They will meet again at the starting point after \(\boxed{36 \text{ minutes}}\).
Exercise 1.2
Q1. Prove that \(\sqrt{5}\) is irrational.
Step 1: Assume, to the contrary, that \(\sqrt{5}\) is rational. Then we can find coprime integers \(a\) and \(b\) (\(b \neq 0\)) such that \(\sqrt{5} = \frac{a}{b}\).
Step 2: Squaring both sides: \(5b^2 = a^2\).
Step 3: Therefore, 5 divides \(a^2\). Since 5 is prime, by Theorem 1.2, 5 divides \(a\). Let \(a = 5c\) for some integer \(c\).
Step 4: Substituting: \(5b^2 = 25c^2\), which gives \(b^2 = 5c^2\).
Step 5: This means 5 divides \(b^2\), and by Theorem 1.2, 5 divides \(b\).
Step 6 -- Contradiction: Both \(a\) and \(b\) are divisible by 5. But we assumed \(a\) and \(b\) are coprime (no common factors other than 1). This is a contradiction.
Conclusion: Our assumption was wrong. Therefore, \(\boxed{\sqrt{5} \text{ is irrational}}\). ■
Step 2: Squaring both sides: \(5b^2 = a^2\).
Step 3: Therefore, 5 divides \(a^2\). Since 5 is prime, by Theorem 1.2, 5 divides \(a\). Let \(a = 5c\) for some integer \(c\).
Step 4: Substituting: \(5b^2 = 25c^2\), which gives \(b^2 = 5c^2\).
Step 5: This means 5 divides \(b^2\), and by Theorem 1.2, 5 divides \(b\).
Step 6 -- Contradiction: Both \(a\) and \(b\) are divisible by 5. But we assumed \(a\) and \(b\) are coprime (no common factors other than 1). This is a contradiction.
Conclusion: Our assumption was wrong. Therefore, \(\boxed{\sqrt{5} \text{ is irrational}}\). ■
Q2. Prove that \(3 + 2\sqrt{5}\) is irrational.
Step 1: Assume, to the contrary, that \(3 + 2\sqrt{5}\) is rational. Then we can write \(3 + 2\sqrt{5} = \frac{a}{b}\), where \(a\) and \(b\) are coprime integers, \(b \neq 0\).
Step 2: Rearranging: \(2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}\).
Step 3: Therefore: \(\sqrt{5} = \frac{a - 3b}{2b}\).
Step 4: Since \(a\), \(b\), 3, and 2 are all integers (and \(b \neq 0\)), the expression \(\frac{a - 3b}{2b}\) is rational.
Step 5: This would mean \(\sqrt{5}\) is rational. But we proved in Q1 above that \(\sqrt{5}\) is irrational. This is a contradiction.
Conclusion: Our assumption was wrong. Therefore, \(\boxed{3 + 2\sqrt{5} \text{ is irrational}}\). ■
Step 2: Rearranging: \(2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}\).
Step 3: Therefore: \(\sqrt{5} = \frac{a - 3b}{2b}\).
Step 4: Since \(a\), \(b\), 3, and 2 are all integers (and \(b \neq 0\)), the expression \(\frac{a - 3b}{2b}\) is rational.
Step 5: This would mean \(\sqrt{5}\) is rational. But we proved in Q1 above that \(\sqrt{5}\) is irrational. This is a contradiction.
Conclusion: Our assumption was wrong. Therefore, \(\boxed{3 + 2\sqrt{5} \text{ is irrational}}\). ■
Q3. Prove that the following are irrationals:
(i) \(\frac{1}{\sqrt{2}}\) (ii) \(7\sqrt{5}\) (iii) \(6 + \sqrt{2}\)
(i) \(\frac{1}{\sqrt{2}}\) (ii) \(7\sqrt{5}\) (iii) \(6 + \sqrt{2}\)
(i) Prove \(\frac{1}{\sqrt{2}}\) is irrational:
Step 1: Assume \(\frac{1}{\sqrt{2}}\) is rational. Then \(\frac{1}{\sqrt{2}} = \frac{a}{b}\) for coprime integers \(a, b\) (\(b \neq 0\)).
Step 2: Rearranging: \(\sqrt{2} = \frac{b}{a}\).
Step 3: Since \(a\) and \(b\) are integers and \(a \neq 0\), \(\frac{b}{a}\) is rational, and so \(\sqrt{2}\) is rational.
Step 4: But this contradicts the fact that \(\sqrt{2}\) is irrational (Theorem 1.3).
Conclusion: \(\boxed{\frac{1}{\sqrt{2}} \text{ is irrational}}\). ■
(ii) Prove \(7\sqrt{5}\) is irrational:
Step 1: Assume \(7\sqrt{5}\) is rational. Then \(7\sqrt{5} = \frac{a}{b}\) for coprime integers \(a, b\).
Step 2: Rearranging: \(\sqrt{5} = \frac{a}{7b}\).
Step 3: Since \(a, b, 7\) are integers, \(\frac{a}{7b}\) is rational, making \(\sqrt{5}\) rational.
Step 4: This contradicts the irrationality of \(\sqrt{5}\).
Conclusion: \(\boxed{7\sqrt{5} \text{ is irrational}}\). ■
(iii) Prove \(6 + \sqrt{2}\) is irrational:
Step 1: Assume \(6 + \sqrt{2}\) is rational. Then \(6 + \sqrt{2} = \frac{a}{b}\) for coprime integers \(a, b\).
Step 2: Rearranging: \(\sqrt{2} = \frac{a}{b} - 6 = \frac{a - 6b}{b}\).
Step 3: Since \(a, b, 6\) are integers, \(\frac{a - 6b}{b}\) is rational, making \(\sqrt{2}\) rational.
Step 4: This contradicts Theorem 1.3.
Conclusion: \(\boxed{6 + \sqrt{2} \text{ is irrational}}\). ■
Step 1: Assume \(\frac{1}{\sqrt{2}}\) is rational. Then \(\frac{1}{\sqrt{2}} = \frac{a}{b}\) for coprime integers \(a, b\) (\(b \neq 0\)).
Step 2: Rearranging: \(\sqrt{2} = \frac{b}{a}\).
Step 3: Since \(a\) and \(b\) are integers and \(a \neq 0\), \(\frac{b}{a}\) is rational, and so \(\sqrt{2}\) is rational.
Step 4: But this contradicts the fact that \(\sqrt{2}\) is irrational (Theorem 1.3).
Conclusion: \(\boxed{\frac{1}{\sqrt{2}} \text{ is irrational}}\). ■
(ii) Prove \(7\sqrt{5}\) is irrational:
Step 1: Assume \(7\sqrt{5}\) is rational. Then \(7\sqrt{5} = \frac{a}{b}\) for coprime integers \(a, b\).
Step 2: Rearranging: \(\sqrt{5} = \frac{a}{7b}\).
Step 3: Since \(a, b, 7\) are integers, \(\frac{a}{7b}\) is rational, making \(\sqrt{5}\) rational.
Step 4: This contradicts the irrationality of \(\sqrt{5}\).
Conclusion: \(\boxed{7\sqrt{5} \text{ is irrational}}\). ■
(iii) Prove \(6 + \sqrt{2}\) is irrational:
Step 1: Assume \(6 + \sqrt{2}\) is rational. Then \(6 + \sqrt{2} = \frac{a}{b}\) for coprime integers \(a, b\).
Step 2: Rearranging: \(\sqrt{2} = \frac{a}{b} - 6 = \frac{a - 6b}{b}\).
Step 3: Since \(a, b, 6\) are integers, \(\frac{a - 6b}{b}\) is rational, making \(\sqrt{2}\) rational.
Step 4: This contradicts Theorem 1.3.
Conclusion: \(\boxed{6 + \sqrt{2} \text{ is irrational}}\). ■
1.4 Summary
Key Points from Chapter 1 -- Real Numbers
- The Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
- If \(p\) is a prime and \(p\) divides \(a^2\), then \(p\) divides \(a\), where \(a\) is a positive integer. (Theorem 1.2)
- To prove that \(\sqrt{p}\) is irrational for a prime \(p\), we use proof by contradiction along with Theorem 1.2.
- HCF (Highest Common Factor) of two or more integers is found by taking the product of the smallest power of each common prime factor.
- LCM (Least Common Multiple) of two or more integers is found by taking the product of the greatest power of each prime factor involved in the numbers.
- For any two positive integers \(a\) and \(b\): \(\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b\). (This does NOT extend to three or more numbers.)
A Note to the Reader
You have seen that:
\[\text{HCF}(p, q) \times \text{LCM}(p, q) = p \times q\]
where \(p\) and \(q\) are positive integers (see Example 3). However, the following results hold good for three positive integers \(p\), \(q\) and \(r\):
- \(\text{LCM}(p, q, r) = \dfrac{p \times q \times r \times \text{HCF}(p, q, r)}{\text{HCF}(p, q) \times \text{HCF}(q, r) \times \text{HCF}(p, r)}\)
- \(\text{HCF}(p, q, r) = \dfrac{p \times q \times r \times \text{LCM}(p, q, r)}{\text{LCM}(p, q) \times \text{LCM}(q, r) \times \text{LCM}(p, r)}\)
Activity: Self-Assessment Checklist
L5 Evaluate
Reflect: Go through each topic of this chapter and evaluate your understanding.
- Can you find the prime factorisation of any given number using a factor tree?
- Can you compute the HCF and LCM of two numbers using prime factorisation?
- Can you verify that HCF × LCM = product of two numbers?
- Can you prove that \(\sqrt{p}\) is irrational for any prime \(p\)?
- Can you prove that expressions like \(a + b\sqrt{p}\) are irrational (when \(a\) is rational, \(b \neq 0\) is rational, and \(p\) is prime)?
Revision Tips: If you answered "no" to any question above, revisit that section. The proofs by contradiction follow a standard pattern: assume the opposite, derive that both numerator and denominator share a common factor, which contradicts coprimality. Practice writing out the full proof for different primes (2, 3, 5, 7, 11) until the structure becomes automatic.
Competency-Based Questions
Scenario: A farmer has three rectangular fields with areas 336 m², 54 m² and 180 m². He wants to divide each field into the largest possible square plots of equal size. He also plans to fence the boundary of a combined rectangular area that has a length equal to the LCM of the three field numbers and a width of 1 metre.
Q1. What is the side length (in metres) of the largest square plot that can divide each field exactly?
L3 ApplyAnswer: (c) 6 m. \(336 = 2^4 \times 3 \times 7\), \(54 = 2 \times 3^3\), \(180 = 2^2 \times 3^2 \times 5\). HCF = \(2^1 \times 3^1 = 6\). The largest square plot side is 6 m.
Q2. How many square plots can the farmer get from the 336 m² field?
L3 ApplyAnswer: Each square plot has area \(6^2 = 36\) m². Number of plots from 336 m² field = \(\frac{336}{36} = 9.\overline{3}\). Since 336 is not perfectly divisible by 36, there would be 9 full square plots with some area remaining. (Note: If the field dimensions allow exact tiling, 336/36 rounds to 9 with 12 m² remainder.)
Q3. Analyse: For three numbers, does the relationship HCF × LCM = product of numbers hold?
L4 AnalyseAnswer: No. For three numbers, HCF × LCM does NOT equal the product of the three numbers. The formula \(\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b\) only works for exactly two numbers. For three numbers, a more complex formula involving pairwise HCFs and LCMs is needed.
Q4. Design a problem where finding HCF is practically useful in daily life, and write its solution using prime factorisation.
L6 CreateSample Problem: "A confectioner has 120 chocolates and 84 lollipops. She wants to make gift boxes with the same number of chocolates and the same number of lollipops in each box, using all sweets. What is the maximum number of boxes she can make?"
Solution: \(120 = 2^3 \times 3 \times 5\), \(84 = 2^2 \times 3 \times 7\). HCF = \(2^2 \times 3 = 12\). She can make 12 boxes (each with 10 chocolates and 7 lollipops).
Solution: \(120 = 2^3 \times 3 \times 5\), \(84 = 2^2 \times 3 \times 7\). HCF = \(2^2 \times 3 = 12\). She can make 12 boxes (each with 10 chocolates and 7 lollipops).
Assertion--Reason Questions
Assertion (A): \(\text{LCM}(306, 657) = 22338\).
Reason (R): \(\text{LCM}(a, b) = \dfrac{a \times b}{\text{HCF}(a, b)}\) for any two positive integers \(a, b\).
Reason (R): \(\text{LCM}(a, b) = \dfrac{a \times b}{\text{HCF}(a, b)}\) for any two positive integers \(a, b\).
Answer: (a) -- \(\text{HCF}(306, 657) = 9\). So \(\text{LCM} = \frac{306 \times 657}{9} = \frac{201042}{9} = 22338\). Both A and R are true, and R gives the formula used to compute A.
Assertion (A): \(\sqrt{5}\) is irrational.
Reason (R): The square root of every positive integer is irrational.
Reason (R): The square root of every positive integer is irrational.
Answer: (c) -- A is true (we proved \(\sqrt{5}\) is irrational). R is false: \(\sqrt{4} = 2\), \(\sqrt{9} = 3\), \(\sqrt{16} = 4\), etc., are all rational. The square root of a perfect square is rational.
Frequently Asked Questions — Real Numbers
What is Real Numbers - Chapter Exercises in NCERT Class 10 Mathematics?
Real Numbers - Chapter Exercises is a key concept covered in NCERT Class 10 Mathematics, Chapter 1: Real Numbers. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Real Numbers - Chapter Exercises step by step?
To solve problems on Real Numbers - Chapter Exercises, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 1: Real Numbers?
The essential formulas of Chapter 1 (Real Numbers) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Real Numbers - Chapter Exercises important for the Class 10 board exam?
Real Numbers - Chapter Exercises is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Real Numbers - Chapter Exercises?
Common mistakes in Real Numbers - Chapter Exercises include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Real Numbers - Chapter Exercises?
End-of-chapter NCERT exercises for Real Numbers - Chapter Exercises cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 1, and solve at least one previous-year board paper to consolidate your understanding.
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