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:
\[ \begin{aligned} \text{(i) }26 &= 2 \cdot 13,\;91 = 7 \cdot 13 \;\Rightarrow\; \mathrm{HCF}=13,\;\mathrm{LCM}=182,\;\text{and }13 \times 182 = 26 \times 91.\\ \text{(ii) }510 &= 2 \cdot 3 \cdot 5 \cdot 17,\;92 = 2^2 \cdot 23 \;\Rightarrow\; \mathrm{HCF}=2,\;\mathrm{LCM}=2^2 \cdot 3 \cdot 5 \cdot 17 \cdot 23.\\ \text{(iii) }336 &= 2^4 \cdot 3 \cdot 7,\;54 = 2 \cdot 3^3 \;\Rightarrow\; \mathrm{HCF}=6,\;\mathrm{LCM}=3024,\;\text{and }6 \times 3024 = 336 \times 54. \end{aligned} \]
Q3: Find the LCM and HCF of the following sets of three numbers by prime factorization:
Solution:
\[ \begin{aligned} \text{(i) }12 &= 2^2 \cdot 3,\;15 = 3 \cdot 5,\;21 = 3 \cdot 7 \;\Rightarrow\; \mathrm{HCF}=3,\;\mathrm{LCM}=420.\\ \text{(ii) }17,23,29 \;(\text{all primes})\;\Rightarrow\; \mathrm{HCF}=1,\;\mathrm{LCM}=17 \cdot 23 \cdot 29 = 11339.\\ \text{(iii) }8 &= 2^3,\;9 = 3^2,\;25 = 5^2 \;\Rightarrow\; \mathrm{HCF}=1,\;\mathrm{LCM}=2^3 \cdot 3^2 \cdot 5^2 = 1800. \end{aligned} \]
Q4: Given that HCF(306, 657) = 9, find LCM(306, 657).
Solution:
\[ \mathrm{LCM} = \frac{306 \times 657}{9} = 22338. \]
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:
\[ \begin{aligned} 7\cdot11\cdot13 + 13 &= 13\,(7\cdot11 + 1) = 13 \cdot 78,\\ 7! + 5 &= 5\,(1 + 6!) = 5 \cdot 721. \end{aligned} \] 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.