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Exercise - 1.2

Question-1 :-  Express each number as product of its prime factors:
(i) 140,  (ii) 156,  (iii) 3825,  (iv) 5005,  (v) 7429

Solution :-
(i)   Prime factors of 140  = 2 x 2 x 5 x 7 = 2² x 5 x 7
(ii)  Prime factors of 156  = 2 x 2 x 3 x 13 = 2² x 3 x 13
(iii) Prime factors of 3825 = 3 x 3 x 5 x 5 x 17 = 3² x 5² x 17
(iv)  Prime factors of 5005 = 5 x 7 x 11 x 13
(v)   Prime factors of 7429 = 17 x 19 x 23
 

Question-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

Solution :-
(i)  26 and 91
     Factor of 26 = 2 x 13
     Factor of 91 = 7 x 13
     HCF (26,91) = 13
     LCM (26,91) = 2 x 7 x 13 = 182
     Product of (26,91) = 26 x 91 = 2366
     HCF x LCM = 13 x 182 = 2366

(ii) 510 and 92
     Factor of 510 = 2 x 3 x 5 x 17
     Factor of 92 = 2 x 2 x 23
     HCF (510,92) = 2
     LCM (510,92) = 2 x 2 x 3 x 5 x 17 x 23 = 23460
     Product of (510,92) = 510 x 92 = 46920
     HCF x LCM = 2 x 23460 = 46920

(iii)336 and 54
     Factor of 336 = 2 x 2 x 2 x 2 x 3 x 7
     Factor of 54 = 2 x 3 x 3 x 3
     HCF (336,54) = 2 x 3 = 6
     LCM (336,54) = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7 = 3024
     Product of (336,54) = 336 x 54 = 18144
     HCF x LCM = 6 x 3024 = 18144
    

Question-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

Solution :-
(i)  12, 15 and 21
     Factor of 12 = 2 x 2 x 3
     Factor of 15 = 3 x 5
     Factor of 21 = 3 x 7
     HCF (12, 15, 21) = 3
     LCM (12, 15, 21) = 2 x 2 x 3 x 5 x 7 = 420
     Product of (12, 15, 21) = 12 x 15 x 21 = 1260
     HCF x LCM = 3 x 420 = 1260

(ii) 17, 23 and 29
     Factor of 17 = 17
     Factor of 23 = 23
     Factor of 29 = 29
     HCF (17, 23, 29) = 1
     LCM (17, 23, 29) = 17 x 23 x 29 = 11339
     Product of (17, 23, 29) = 17 x 23 x 29 = 11339
     HCF x LCM = 1 x 3055 = 11339

(iii)8, 9 and 25
     Factor of 8 = 2 x 2 x 2
     Factor of 9 = 3 x 3
     Factor of 25 = 5 x 5
     HCF (8, 9, 25) = 1
     LCM (8, 9, 25) = 2 x 2 x 2 x 3 x 3 x 5 x 5 = 1800
     Product of (8, 9, 25) = 8 x 9 x 25 = 1800
     HCF x LCM = 1 x 1800 = 1800
    

Question-4 :-  Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution :-
 HCF (306, 657) = 9
 HCF x LCM = 306 x 657
 9 x LCM = 306 x 657
 LCM = (306 x 657)/9
 LCM = 22338
 

Question-5 :-  Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution :-
 If any number ends with the digit 0, it should be divisible by 10 or in other words,
 it will also be divisible by 2 and 5 as 10 = 2 × 5
 Prime factorisation of 6ⁿ = (2 ×3)ⁿ
 It can be observed that 5 is not in the prime factorisation of 6ⁿ. 
 Hence, for any value of n, 6ⁿ will not be divisible by 5.
 Therefore, 6ⁿ cannot 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 :-
 Numbers are of two types - prime and composite. Prime numbers can be divided by 1 and only itself, 
 whereas composite numbers have factors other than 1 and itself.
 It can be observed that
 7 × 11 × 13 + 13 
 = 13 × (7 × 11 + 1) 
 = 13 × (77 + 1)
 = 13 × 78 
 = 13 ×13 × 6 
The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 
 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)
 = 5 × (1008 + 1) 
 = 5 ×1009 
 1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. 
 Hence, it 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 :-
 It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path.
 As they are going in the same direction, they will meet again at the same time when Ravi will have completed 
 1 round of that circular path with respect to Sonia. And the total time taken for completing this 
 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively 
 i.e., LCM of 18 minutes and 12 minutes.
 
 18 = 2 × 3 × 3 And, 12 = 2 × 2 × 3
 LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36
 Therefore, Ravi and Sonia will meet together at the starting pointafter 36 minutes.
 
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