Here is the solution of NCERT Class 10 Maths Chapter 1 Real Number Exercise 1.2. There are 7 Questions in Class 10 NCERT Maths Chapter 1 Real Numbers Exercise 1.2. All 7 Questions under Exercise 1.1 are based on the fundamental theorem of arithmetic. You can go through each one of them one by one.
Question 1: Express each number as a product of its prime factors:
(i) 140
Solution 1 (i): 140 = 22 x 5 x 7
(ii) 156
Solution 1 (ii): 156 = 23 x 7
(iii) 3825
Solution 1 (iii): 3825 = 32 x 52 x 17
(iv) 5005
Solution 1 (iv): 5005 = 5 x 7 x 11 x 13
(v) 7429
Solution 1 (v): 7429 = 17 x 19 x 23
Question 2: Find the LCM and HCF of the following pairs of integers and veriry that LCM x HCF = product of the two numbers.
(i) 26 and 91
Solution 1 (i): 26 and 91
26 = 2 x 13
91 = 7 x 13
HCF = 13
LCM = 14 x 13 = 182
Product of two numbers = 26 x 91 = 2366
HCF x LCM = 13 x 182 = 2366
Hence, it is verified.
(ii) 510 and 92
Solution 1 (ii): 510 and 92
510 = 2 x 3 x 5 x 17
92 = 2 x 2 x 23
HCF = 2
LCM = 4 x 15 x 17 x 23 = 23460
Product of two numbers = 510 x 92 = 46920
HCF x LCM = 2 x 23460 = 46920
Hence, it is verified.
(iii) 336 and 54
Solution 1 (iii): 336 and 54
336 = 2 x 2 x 2 x 2 x 3 x 7
54 = 2 x 3 x 3 x 3
HCF = 6
LCM = 8 x 81 x 7 = 3024
Product of two numbers = 336 x 54 = 18144
HCF x LCM = 6 x 3024 = 18144
Hence, it is verified.
Question 3: Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
Solution 3 (i): 12, 15 and 21
12 = 2 x 2 x 3 x 1
15 = 3 x 5 x 1
21 = 3 x 7 x 1
HCF = 3
LCM = 4 x 15 x 7 = 420
(ii) 17, 23 and 29
Solution 3 (ii): 17, 23 and 29
17 = 17 x 1
23 = 23 x 1
29 = 29 x 1
HCF = 1
LCM = 17 x 23 x 29 = 11339
(iii) 8, 9 and 25
Solution 3 (iii): 8, 9 and 25
8 = 2 x 2 x 2 x 1
9 = 3 x 3 x 1
25 = 5 x 5 x 1
HCF = 1
LCM = 8 x 9 x 25 = 1800
Question 4: Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution 4: 306 = 2 x 3 x 3 x 17
657 = 3 x 3 x 73
HCF = 9
LCM = 9 x 34 x 73 = 22338
Question 5: Check whether 6n can end with the digit 0 for any natural number n.
Solution 5: In order to check whether 6n can end with the digit 0 for any natural number n, let us find the factors of 6.
It’s seen that the factors of 6 are 2 and 3.
So, 6n = (2 x 3)n
6n =2n x 3n
Since, the prime factorization of 6 does not contain 5 and 2 as its factor, together.
We can thus conclude that 6n can never end with the digit 0 for any natural number n.
Question 6: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution 6: So, basically there are two types of numbers i.e., prime numbers and composite numbers.
Understanding that,
Prime numbers are those numbers having 1 and the number itself as factors.
Composite numbers are those numbers having factors other than 1 and itself.
7 x 11 x 13 + 13 = 13[7 x 11 + 1] = 13(77 + 1) = 13 x 78 = 13 x 2 x 3 x 13 = 2 x 3 x 13 x 13
2, 3 and 13 are prime numbers.
So according to the fundamental theorem of arithmetic, every composite number can be uniquely expressed as a product of prime numbers.
So, this is a composite number
Similarly 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5[7 x 6 x 4 x 3 x 2 x 1 + 1] = 5(1008 + 1) = 5 x 1009
5 and 1009 are prime numbers.
So according to the fundamental theorem of arithmetic is a composite number.
Question 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 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 7: Sonia takes 18 minutes to drive one round of the field.
Ravi takes 12 minutes for the same.
If they both started at the same point and at the same time and in the same direction then the time taken by them to meet again at the starting point is the LCM of 18 and 12.
18 = 2 × 3 × 3 = 2 × 32
12 = 2 × 2 × 3 = 22 × 3
∴ L.C.M. = 22 × 32 = 36
∴ After 36 minutes, they both meet at the same point.
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