NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1#
Welcome to our comprehensive NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1. This exercise introduces fundamental concepts of real numbers including prime factorization, Least Common Multiple (LCM), Highest Common Factor (HCF), and composite numbers. Our expert teachers have prepared detailed, step-by-step solutions to help CBSE Class 10 students understand every concept clearly and score full marks in their board exams.
1. Express each number as a product of its prime factors: #
(i) 140
$$= 2 × 2 × 5 × 7 = 2² × 5 × 7$$
(ii) 156
$$= 2 × 2 × 3 × 13 = 2² × 3 × 13$$
(iii) 3825
$$= 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17$$
(iv) 5005
$$= 5 × 7 × 11 × 13$$
(v) 7429
$$= 17 × 19 × 23$$
Key Concept: Prime factorization is the foundation for finding LCM and HCF of numbers. Every composite number can be uniquely expressed as a product of prime factors (Fundamental Theorem of Arithmetic).
2. 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
$26 = 2 × 13$
$91 = 7 × 13$
$HCF = 13$
$LCM = 2 × 7 × 13 = 182$
Product of the two numbers = $26 × 91 = 2366$
$HCF × LCM = 13 × 182 = 2366$
∴ $HCF × LCM$ = product of two numbers
(ii) 510 and 92
$510 = 2 × 3 × 5 × 17$
$92 = 2 × 2 × 23$
$HCF = 2$
$LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460$
Product of the two numbers = $510 × 92 = 46920$
$HCF × LCM = 2 × 23460 = 46920$
∴ $HCF × LCM$ = product of two numbers
(iii) 336 and 54
$336 = 2 × 2 × 2 × 2 × 3 × 7 = 2⁴ × 3 × 7$
$54 = 2 × 3 × 3 × 3 = 2 × 3³$
$HCF = 2 × 3 = 6$
$LCM = 2⁴ × 3³ × 7 = 3024$
Product of the two numbers = $336 × 54 = 18144$
$HCF × LCM = 6 × 3024 = 18144$
∴ $HCF × LCM$ = product of two numbers
3. Find the LCM and HCF of the following integers by applying the prime factorisation method. #
(i) 12, 15 and 21
$12 = 2 × 2 × 3 = 2² × 3$
$15 = 3 × 5$
$21 = 3 × 7$
$HCF = 3$
$LCM = 2² × 3 × 5 × 7 = 420$
(ii) 17, 23 and 29
$17 = 1 × 17$
$23 = 1 × 23$
$29 = 1 × 29$
$HCF = 1$
$LCM = 17 × 23 × 29 = 11339$
(iii) 8, 9 and 25
$8 = 2 × 2 × 2 = 2³$
$9 = 3 × 3 = 3²$
$25 = 5 × 5 = 5²$
$HCF = 1$
$LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800$
4. Given that HCF (306, 657) = 9, find LCM (306, 657). #
Answer:
We have been given the HCF of two numbers $(306, 657) = 9$,
We have to find the LCM of $(306, 657)$.
we know that $LCM × HCF =$ Product of two numbers
Substitute the values
$$\Rightarrow LCM × 9 = 306 × 657$$
$$\Rightarrow LCM = \dfrac{306 × 657}{9}$$
∴ $LCM = 22338$
Therefore, the LCM of $(306, 657) = 22338$.
5. Check whether $6^n$ can end with the digit 0 for any natural number n #
Answer
To determine whether $6^n$ can end with the digit $0$, we use the fact that any number ending in $0$ must be divisible by both $2$ and $5$.
Now, express $6^n$ in terms of its prime factors:
$$6^n = (2 × 3)^n = 2^n × 3^n$$
From this factorization, we see that the only prime factors of $6^n$ are $2$ and $3$.
Since the prime factor $5$ is not present in $6^n$ for any natural number $n$, the number $6^n$ is never divisible by $5$.
Therefore, $6^n$ cannot end with the digit $0$ for any natural number $n$.
6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.#
(i) $7 × 11 × 13 + 13$
$7 × 11 × 13 + 13$
$= 13(7 × 11 + 1)$
$= 13(77 + 1)$
$= 13 × 78$
$= 13 × 2 × 3 × 13$
Since $7 × 11 × 13 + 13$ can be expressed as a product of two numbers greater than $1$, it is a composite number.
(ii) $7 × 6 × 5 × 4 × 3 × 2 × 1 + 5$
$7 × 6 × 5 × 4 × 3 × 2 × 1 + 5$
$= 5(7 × 6 × 4 × 3 × 2 × 1 + 1)$
$= 5(1008 + 1)$
$= 5 × 1009$
Since $7 × 6 × 5 × 4 × 3 × 2 × 1 + 5$ can be expressed as a product of two numbers greater than $1$, it is a composite number.
7. 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#
Answer
$18 = 2 × 3 × 3 = 2 × 3^2$
$12 = 2 × 2 × 3 = 2^2 × 3$
$LCM (18, 12) = 2^2 × 3^2 = 4 × 9 = 36$
∴ They will meet again at the starting point after 36 minutes.