Class 10 Maths Chapter 1 Exercise 1.1 Solutions
If you are preparing for Class 10 board exams and looking for clear, step-by-step solutions of Exercise 1.1 from Chapter 1 (Real Numbers), this page provides complete book-style answers. Each question is followed immediately by its detailed solution for better understanding.
BRIEF SYNOPSIS
1. Euclid's Algorithm for HCF : Euclid, a Greek mathematician, devised an interesting method to find the HCF of two numbers. This method is known as Euclid's Algorithm. An algorithm is a stepwise process to solve a problem.
Steps to find HCF:
- Identify the greater number.
- Consider greater number as dividend and smaller as divisor.
- Find quotient and remainder.
- If remainder is zero, divisor is HCF.
- Otherwise repeat the process.
2. Prime Number : A number is called a prime number if it has no factor other than 1 and itself.
3. Composite Number : A number is composite if it has at least one factor other than 1 and itself.
4. Fundamental Theorem of Arithmetic : Every composite number can be expressed as a product of primes uniquely.
5. HCF and LCM Properties :
- HCF of co-prime numbers is 1.
- LCM of co-prime numbers equals their product.
EXERCISE 1.1
Question 1
Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Solution:
(i)
\[ 140 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7 \](ii)
\[ 156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13 \](iii)
\[ 3825 = 3^2 \times 5^2 \times 17 \](iv)
\[ 5005 = 5 \times 7 \times 11 \times 13 \](v)
\[ 7429 = 17 \times 19 \times 23 \]Question 2
Find the LCM and HCF of the following pairs of integers and verify that \(\text{LCM} \times \text{HCF} = \text{product of the two numbers}\).
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
(i) 26 and 91
\[ 26 = 2 \times 13 \] \[ 91 = 7 \times 13 \] \[ \therefore \text{LCM of 26 and 91} = 2 \times 7 \times 13 = 182 \] \[ \text{HCF of 26 and 91} = 13 \] \[ 182 \times 13 = 2366 \] \[ 26 \times 91 = 2366 \] \[ \therefore 182 \times 13 = 26 \times 91 \]Hence verified.
(ii) 510 and 92
\[ 510 = 2 \times 3 \times 5 \times 17 \] \[ 92 = 2 \times 2 \times 23 \] \[ \therefore \text{LCM of 510 and 92} = 2 \times 2 \times 3 \times 5 \times 17 \times 23 = 23460 \] \[ \text{HCF of 510 and 92} = 2 \] \[ 23460 \times 2 = 46920 \] \[ 510 \times 92 = 46920 \] \[ \therefore 23460 \times 2 = 510 \times 92 \]Hence verified.
(iii) 336 and 54
\[ 336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7 \] \[ 54 = 2 \times 3 \times 3 \times 3 \] \[ \therefore \text{LCM of 336 and 54} = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 7 = 3024 \] \[ \text{HCF of 336 and 54} = 2 \times 3 = 6 \] \[ 3024 \times 6 = 18144 \] \[ 336 \times 54 = 18144 \] \[ \therefore 3024 \times 6 = 336 \times 54 \]Hence verified.
Question 3
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
Solution:
(i) First we write the prime factorisation of each of the given numbers.
\[ 12 = 2 \times 2 \times 3 = 2^2 \times 3 \] \[ 15 = 3 \times 5 \] \[ 21 = 3 \times 7 \] \[ \therefore \text{LCM} = 2^2 \times 3 \times 5 \times 7 = 420 \] \[ \text{HCF} = 3 \](ii) First we write the prime factorisation of each of the given numbers.
\[ 17 = 17 \] \[ 23 = 23 \] \[ 29 = 29 \] \[ \therefore \text{LCM} = 17 \times 23 \times 29 = 11339 \] \[ \text{HCF} = 1 \](iii) First we write the prime factorisation of each of the given numbers.
\[ 8 = 2 \times 2 \times 2 = 2^3 \] \[ 9 = 3 \times 3 = 3^2 \] \[ 25 = 5 \times 5 = 5^2 \] \[ \therefore \text{LCM} = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800 \] \[ \text{HCF} = 1 \]Question 4
Given that \(\text{HCF}(306,657)=9\), find \(\text{LCM}(306,657)\).
Solution:
\[ 9 \times \text{LCM} = 306 \times 657 \] \[ \text{LCM} = \frac{306 \times 657}{9} = 22338 \]Question 5
Check whether \(6^n\) can end with the digit 0 for any natural number \(n\).
Solution:
\[ 6 = 2 \times 3 \] \[ 6^n = (2 \times 3)^n \]The prime factors are only 2 and 3. Since 5 is not present, the number cannot end with 0.
Therefore, there is no natural number \(n\) for which \(6^n\) ends with the digit 0.
Question 6
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:
\[ 7 \times 11 \times 13 + 13 = 13(77 + 1) = 13 \times 78 \]Hence, it is composite.
The second expression is divisible by 5, hence it is composite.
Question 7
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round, while Ravi takes 12 minutes. After how many minutes will they again be at the starting point?
Solution:
Prime factorisation:
\[ 18 = 2 \times 3^2 \] \[ 12 = 2^2 \times 3 \]LCM is:
\[ \text{LCM} = 2^2 \times 3^2 \] \[ = 4 \times 9 = 36 \]Therefore, they will meet again after 36 minutes.
Frequently Asked Questions (MCQ Practice – Real Numbers)
Q1. How many equivalent rational numbers can a rational number have?
A rational number has infinitely many equivalent rational numbers. Example: \( \frac{2}{3} = \frac{4}{6} = \frac{8}{12} \dots \) Therefore, the correct option is (d) Undefined/Infinite.
Q2. What is the quotient of a non-zero rational number and an irrational number?
The quotient of a non-zero rational number and an irrational number is always an irrational number. Therefore, the correct option is (b) Irrational number.
Q3. Which of the following is an irrational number?
\(5 + \sqrt{2}\) is an irrational number. Therefore, the correct option is (d).
Q4. What is the value of \((\sqrt{5} + \sqrt{7})(\sqrt{5} - \sqrt{7})\)?
\[ (\sqrt{5} + \sqrt{7})(\sqrt{5} - \sqrt{7}) = 5 - 7 = -2 \] This is a rational number. Therefore, the correct option is (c).
Q5. What is the sum of a rational number and an irrational number?
The sum of a rational and an irrational number is always irrational. Example: \(3 + (\sqrt{3} - 3) = \sqrt{3}\). Therefore, the correct option is (d) Irrational number.
Q6. What is the difference between a rational number and an irrational number?
Rational − Irrational = Always Irrational. Example: \( \frac{3}{4} - \left(\frac{3}{4} + 5\sqrt{3}\right) = -5\sqrt{3} \). Therefore, the correct option is (a).
Q7. The number \(4\sqrt{3}\) is:
\(4\sqrt{3} =\) Rational × Irrational = Irrational. Therefore, the correct option is (c).
Q8. What is the product of \(\sqrt{2}\) and \( (2 - \sqrt{2}) \)?
\[ \sqrt{2}(2 - \sqrt{2}) = 2\sqrt{2} - 2 \] This is an irrational number. Therefore, the correct option is (a).
Q9. What is the simplest form of 148 / 185?
\[ \frac{148}{185} = \frac{37 \times 4}{37 \times 5} = \frac{4}{5} \] Therefore, the correct option is (a).
Q10. Which of the following is a prime number?
2 is a prime number because it has only two factors: 1 and itself. Therefore, the correct option is (c).
Q11. What is the prime factorisation of 144?
\[ 144 = 2^4 \times 3^2 \] Therefore, the correct option is (a).
Q12. If HCF = 5 and LCM = 200, what is the product of the two numbers?
\[ a \times b = \text{HCF} \times \text{LCM} = 5 \times 200 = 1000 \] Therefore, the correct option is (b).
Q13. Is Exercise 1.1 important for board exam?
Yes, questions based on HCF, LCM and prime factorisation are frequently asked.
Q14. What is Euclid’s Algorithm?
It is a stepwise method to find HCF of two numbers.
Q15. Where can I download Class 10 Maths Ex 1.1 PDF?
You can use the download button above for printable version.
