Real Numbers : Exercise 1.2 (Mathematics NCERT Class 10th)



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

Sol.       (i) We use the division method as shown below :
6

               Therefore, 140 = 2 × 2 × 5 × 7  = {2^2} \times 5 \times 7

             (ii) We use the division method as shown below:
7

             Therefore, 156 = 2 × 2 × 3 × 13   = {2^2} \times 3 \times 13

            (iii) We use the division method as shown below :
8

             Therefore, 3825 = 3 × 3 × 5 × 5 × 17  = {3^2} \times {5^2} \times 17

            (iv) We use the division method as shown below :

 9

          Therefore, 5005 = 5 × 7 × 11 × 13

         (v) We use the division method as shown below :

 10

          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

11                             12

                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

 13                              14

             
              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

15                                 16

           
           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 = {2^2} × 3, 15 = 3 × 5 and 21 = 3 × 7
                Therefore, LCM = {2^2} × 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
 = {2^3},9 = 3\, \times 3 = {3^2}, 25 = 5 × 5 = {5^2}
                Therefore, LCM  = {2^3} \times {3^2} \times {5^2} = 8 \times 9 \times 25 = 1800
                and HCF = 1


 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
              \Rightarrow 9 × LCM (306 × 657) = 306 × 657
              \Rightarrow LCM (306,657)  = {{306 \times 657} \over 9}  = 22338


Q.5.     Check whether {6^n} can end with the digit 0 for any natureal number n.
Sol.        If the number {6^n}, for any n ends with the digit zero, then it is divisible by 5. That is, the prime factorisation of {6^n} contains the prime 5. That is, not possible as the only prime in the factorisation of {6^n} is 2 and 3 and the uniqueness of the Fundamental Theorem of Arihmetic guarantees that there are no other primes in the factorisation of {6^n}. So, there is no n \in N for which {6^n} ends with the digit zero.   


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
            \Rightarrow It is a composite number.
           Again, 7 × 6 × 5 × 4 × 3 × 1 × 1 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 1 × 1 + 1)
            \Rightarrow It is a composite number.


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

17                      18

            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.



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