Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 1
Chapter Name	Real Numbers
Exercise	Ex 1.2
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

Q.1     Express each number as product of its prime factors:
(i) 140      (ii) 156      (iii) 3825      (iv) 5005      (vi) 7429

Sol. (i) 

              

Therefore, 140 = 2 × 2 × 5 × 7 =22×5×7

(ii) 

Therefore, 156 = 2 × 2 × 3 × 13  =22×3×13

(iii) 

Therefore, 3825 = 3 × 3 × 5 × 5 × 17 =32×52×17

(iv)

 

Therefore, 5005 = 5 × 7 × 11 × 13

(v)

 

Therefore, 7429 = 17 × 19 × 23

 Q.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     (ii) 510 and 92     (iii) 336 and 54

Sol.       (i) 26 and 91

                             

26 = 2 × 13 and 91 = 7 × 13

Therefore, LCM of 26 and 91 = 2 × 7 × 13 = 182

and HCF of 26 and 91 = 13
               Now, 182 × 13 = 2366 and 26 × 91 = 2366
               Since, 182 × 13 = 26 × 91
               Hence verified.

(ii) 510 and 92

                             

                              

510 = 2 × 3 × 5 × 17 and 92 = 2 × 2 × 23
              Therefore, LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460
              and HCF of 510 and 92 = 2
              Now, 23460 × 2 = 46920 and 510 × 92 = 46920
              Since 23460 × 2 = 510 × 92
              Hence verified.

           

  (iii) 336 and 54

                                 

336 = 2 × 2 × 2 × 2 × 3 × 7
           and 54 = 2 × 3 × 3 × 3
           Therefore, LCM of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
           and HCF of 336 and 54 = 2 × 3 = 6

            Now, 3024 × 6 = 18144
            and 336 × 54 = 18144
            Since, 3024 × 6 = 336 × 54
            Hence verified.

Q.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

Sol.   (i) First we write the prime factorisation of each of the given numbers.
                

12 = 2 × 2 × 3 = 22 × 3, 15 = 3 × 5 and 21 = 3 × 7
                

Therefore, LCM = 22 × 3 × 5 × 7 = 420

                and, HCF = 3

                (ii) First we write the prime factorisation of each of the given numbers.
                17 = 17, 23 = 23 and 29 = 29
                Therefore, LCM = 17 × 23 × 29 = 11339
                and HCF = 1

                (iii)  First we write the prime factorisation of each of the given numbers.
                  8 = 2 × 2 × 2 =23,9=3×3=32, 25 = 5 × 5 = 52
                Therefore, LCM =23×32×52=8×9×25=1800
                and HCF = 1

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 Q.4     Given that HCF (306, 657) = 9, find LCM (306, 657).

Sol.       We know that the product of the HCF and the LCM of two numbers is equal to the product of the given numbers.
             Therefor, HCF (306,657) × LCM (306,657) = 306 × 657
             ⇒ 9 × LCM (306 × 657) = 306 × 657
             ⇒ LCM (306,657) =306×6579  = 22338

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Q.5.     Check whether 6n can end with the digit 0 for any natureal number n.

Sol.        If the number 6n, for any n ends with the digit zero, then it is divisible by 5. That is, the prime factorisation of 6n contains the prime 5. That is, not possible as the only prime in the factorisation of 6n is 2 and 3 and the uniqueness of the Fundamental Theorem of Arihmetic guarantees that there are no other primes in the factorisation of 6n. So, there is no n∈N for which 6n ends with the digit zero.   

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Q.6    Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Sol.     Since, 7 × 11 × 13 + 13 = 13 × (7 × 11 × 1 + 1)
           = 13 × (77 + 1)
           = 13 × 78
           ⇒ It is a composite number.
           Again, 7 × 6 × 5 × 4 × 3 × 1 × 1 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 1 × 1 + 1)
           ⇒ It is a composite number.

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Q.7     There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the path, 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 again at the starting point ?

Sol.      To find the LCM of 18 and 12, we have

                      

            18 = 2 × 3 × 3 and 12 = 2 × 2 × 3
             LCM of 18 and 12 = 2 × 2 × 3 × 3 = 36
             So, Sonia and Ravi will meet again at the starting point after 36 minutes.