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1. Real Numbers Mathematics class 10 in English Medium ncert book solutions Exercise 1.2
1. Real Numbers Exercise 1.2 – Complete NCERT Book Solutions for Class 10 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 1. Real Numbers Exercise 1.2 to help you master concepts and score higher.
1. Real Numbers Mathematics class 10 in English Medium ncert book solutions Exercise 1.2
NCERT Solutions for Class 10 Mathematics play an important role in helping students understand the concepts of the chapter 1. Real Numbers clearly. This chapter includes the topic Exercise 1.2 , which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 10 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 1. Real Numbers is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 1.2 . By studying these updated NCERT Solutions for Class 10 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
1. Real Numbers
Exercise 1.2
EXERCISE 1.2
Q1. Express each number as a product of its prime factors:
(i) 140
Solution:
= 22 × 5 × 7
(ii) 156
Solution:
= 22 × 3 × 13
(iii) 3825
Solution:
= 32 × 52 × 17
(iv) 5005
Solution:
= 5 × 7 × 11 × 13
(v) 7429
Solution:
Q2. 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
Solution:
26 = 2 × 13
91 = 7 × 13
Common factors = 13
∴ HCF = 13
LCM = 2 × 7 × 13 = 182
Now varification,
product of the two numbers = LCM × HCF
N1 × N2 = LCM × HCF
26 × 91 = 13 × 182
2366 = 2366
Hence Varified,
(ii) 510 and 92
Solution:
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
Common factors = 2
∴ HCF = 2
LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460
Now varification,
product of the two numbers = LCM × HCF
N1 × N2 = LCM × HCF
510 × 92 = 2 × 23460
46920 = 46920
Hence Varified,
(iii) 336 and 54
Solution:
336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
Common factors = 2 × 3
∴ HCF = 6
LCM = 2 × 2 × 2× 2 × 3 × 3 × 3 × 7 = 3024
Now varification,
product of the two numbers = LCM × HCF
N1 × N2 = LCM × HCF
336 × 54 = 6 × 3024
18144 = 18144
Hence Varified,
Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
Solution:
12 = 2 × 2 × 3
15 = 5 × 3
21 = 7 × 3
Common Factors = 3
HCF = 3
LCM = 3 × 2 × 2 × 5 × 7 = 420
(ii) 17, 23 and 29
Solution:
17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 17 × 23 × 29 = 11339
(iii) 8, 9 and 25
Solution:
8 = 2 × 2 × 2
9 = 3 × 3
25 = 5 × 5
There is no common factor except 1.
∴ HCF = 1
LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5
= 8 × 9 × 25
= 1800
Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
HCF (306, 657) = 9
LCM × HCF = N1 × N2
LCM = 22338
Q5. Check whether 6n can end with the digit 0 for any natural number n.
Solution:
Prime factorisation of 6n = (2 × 3 )n
While, Any natural number which end with digit 0 has
the prime factorisation as form of (2 × 5 )n
Therefore, 6n will not end with digit 0.
Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Let A = 7 × 11 × 13 + 13
= 13 (7 × 11 + 1)
= 13 (77 + 1)
= 13 × 78
Hence this is composite number because It has at least one positive divisor other than one.
Similarily,
Let B = 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
Hence this is also a composite number because It has at least one positive divisor other than one.
Q7. 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 same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
Sonia takes 18 minutes in one round.
Ravi takes 12 minutes in one round
they will meet again at the starting point after LCM(18, 12) minutes
18 = 2 × 3 × 3
12 = 2 × 2 × 3
HCF = 2 × 3 = 6
= 36 minutes
See other sub-topics of this chapter:
1. Exercise 1.1 2. Exercise 1.2 3. Exercise 1.3 4. Exercise 1.4