1. Real Numbers Mathematics Exercise - 1.2 class 10 Maths in English - CBSE Study
NCERT Solutions for Class 10 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 1. Real Numbers with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 1.2 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 10 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 10 English Medium Mathematics All Chapters:
1. Real Numbers
2. 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
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