Welcome to StudyMats. This comprehensive educational guide for Class 10 Mathematics Chapter 1 Real Number Solutions is prepared to assist students in preparing effectively for their school, board, and competitive examinations. All questions, exercise solutions, and concept notes strictly adhere to the latest NCERT curriculum and CBSE guidelines.
Exercise 1.1 – Chapter 1 (Real Numbers)
Q1: Express each number as a product of its prime factors:
Solution:
\[ \begin{aligned} 140 &= 2^2 \cdot 5 \cdot 7,\\ 156 &= 2^2 \cdot 3 \cdot 13,\\ 3825 &= 3^2 \cdot 5^2 \cdot 17,\\ 5005 &= 5 \cdot 7 \cdot 11 \cdot 13,\\ 7429 &= 17 \cdot 19 \cdot 23. \end{aligned} \]
Q2: Find the LCM and HCF of the following pairs and verify that HCF × LCM = product of the two numbers:
Solution:
\[ \begin{aligned} 140 &= 2^2 \cdot 5 \cdot 7,\\ 156 &= 2^2 \cdot 3 \cdot 13,\\ 3825 &= 3^2 \cdot 5^2 \cdot 17,\\ 5005 &= 5 \cdot 7 \cdot 11 \cdot 13,\\ 7429 &= 17 \cdot 19 \cdot 23. \end{aligned} \]Q3: Find the LCM and HCF of the following sets of three numbers by prime factorization:
Solution:
\[ \begin{aligned} \text{(i)} & \quad 12 = 2^2 \cdot 3, \; 15 = 3 \cdot 5, \; 21 = 3 \cdot 7 \\ & \quad \Rightarrow\; \mathrm{HCF}=3, \;\mathrm{LCM}=420. \\[1em] \text{(ii)} & \quad 17, 23, 29 \;(\text{all primes}) \\ & \quad \Rightarrow\; \mathrm{HCF}=1, \;\mathrm{LCM}=11339. \\[1em] \text{(iii)} & \quad 8 = 2^3, \; 9 = 3^2, \; 25 = 5^2 \\ & \quad \Rightarrow\; \mathrm{HCF}=1, \;\mathrm{LCM}=1800. \end{aligned} \]Q4: Given that HCF(306, 657) = 9, find LCM(306, 657).
Solution:
\begin{align*} \mathrm{LCM} &= \frac{306 \times 657}{9} \\ &= 22338. \end{align*}Q5: Check whether \(6^n\) can end with the digit 0 for any natural number \(n\).
Solution:
No — because \(6^n\) does not have a factor of 5, it cannot end in zero.
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.
Solution:
\[ \begin{aligned} 7 \cdot 11 \cdot 13 + 13 &= 13(7 \cdot 11 + 1) \\ &= 13 \cdot 78. \\[2em] 7! + 5 &= 5(1 + 6!) \\ &= 5 \cdot 721. \end{aligned} \] Both expressions are factorable into integers greater than 1, so they are composite.
Q7: Sonia takes 18 minutes and Ravi takes 12 minutes to complete one round of a circular path. Suppose they start together. After how much time will they meet again at the starting point?
Solution:
They will meet again after \( \mathrm{LCM}(18,\,12) = 36 \) minutes.
💡 Important Exam & Study Tips
- Master NCERT Fundamentals: Ensure thorough familiarity with standard textbook definitions, formulas, and in-text examples before solving advanced problems.
- Step-by-Step Presentation: In board examinations, neat presentation, correct formula application, and labeled steps earn maximum step-marking.
- Practice Under Timed Conditions: Regular practice with previous years' question papers (PYQs) builds exam speed and accuracy.
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❓ Frequently Asked Questions (FAQs)
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