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.