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P Pankaj Sanodiya
(a) LCM = 2 2 5 = 20 (b) LCM = 2  3  3 = 18 (c) LCM = 2 2 2 2 3 = 48 (d) LCM = 3 3 5 = 45 Yes, it can be observed that in each case, the LCM of the given numbers is the larger number. When one number is a factor of the other number, their LCM will be the larger number.

P Pankaj Sanodiya
(a) LCM = 2 2 3 3 = 36 (b) LCM = 2 2 3 5 = 60 (c) LCM = 2  3 5 = 30 (d) LCM = 2 3 5 = 30 Yes, it can be observed that in each case, the LCM of the given numbers is the product of these numbers. When two numbers are co-prime, their LCM is the product of those numbers. Also, in each case, LCM is a multiple of 3.

P Pankaj Sanodiya
LCM of 18, 24 and 32 LCM = 2 2 2 2 2 3 3 We have to find the smallest 4-digit multiple of 288. It can be observed that 288 3 = 864 and 288 4 = 1152. Therefore, the smallest 4-digit number which is divisible by 18, 24, and 32 is 1152.

P Pankaj Sanodiya
LCM of 6, 15, 18 LCM = 2 2 3 5 = 90 Required number = 90 + 5 = 95

P Pankaj Sanodiya
Maximum capacity of the required tanker =HCF of 403,434,465 403=13 31 434=2 7 31 465=3 5 31 HCF=31  Hence the maximum capacity of a container that can measure the diesel of the three containers the exact number of times is 31.

P Pankaj Sanodiya
The time period after which these lights will change = LCM of 48, 72 and 108 LCM=2 2  2 2 3 3 3 = 432 They will change together after every 432 seconds i.e., 7 min 12 seconds. Hence, they will change simultaneously at 7:07:12 am.

P Pankaj Sanodiya
CM of 8, 10 and 12 LCM = 2   2   2   3 5 = 120 We have to find the greatest 3-digit multiple of 120. It can be seen that 120 8 = 960 and 120 9 = 1080. Hence, the greatest 3-digit number exactly divisible by 8, 10, and 12 is 960.

P Pankaj Sanodiya
Smallest number = LCM of 6, 8, 12 We have to find the smallest 3-digit multiple of 24. It can be seen that 24   4 = 96 and 24   5 = 120. Hence, the smallest 3-digit number which is exactly divisible by 6, 8, and 12 is 120.

P Pankaj Sanodiya
Length =825cm=3 5 5 11  Breadth =675cm=3 3  3 5 5  Height =450cm=2 3 3 5 5  Longest tape = HCF of 825,675, and 450=  3 5 5                       =75cm  Therefore, the longest tape is 75cm.

P Pankaj Sanodiya
Step measure of 1 st  Boy =63cm  Step measure of 2 nd  Boy =70cm  Step measure of 3 rd  Boy =77cm  LCM of 63,70,77 :

P Pankaj Sanodiya
Weight of the two bags = 75 kg and 69 kg Maximum weight : HCF (75, 69) 75 = 3 5 5 69 = 3 23 HCF = 3 Hence maximum value of weight which can measure the weight of the fertiliser exact number of times is 3.

P Pankaj Sanodiya
No, the answer is not correct. 1 is the correct HCF. As  4 = 2 2 1 15 = 3 5 1 1 is common in both so HCF is 1.

P Pankaj Sanodiya
(a) 1 e.g., HCF of 2 and 3 is 1. (b) 2 e.g., HCF of 2 and 4 is 2. (c) 1 e.g., HCF of 3 and 5 is 1.

P Pankaj Sanodiya
(a) 18, 48      18 = 2 x 3 x 3 48 = 2 x 2 x 2 x 2 x 3 HCF = 2 x 3 =6   (b) 30, 42 30 = 2 x 3 x 5 42 = 2 x 3 x 7 HCF = 2 x 3 =6   (c)  18, 60  18 = 2 x 3 x 3 60 = 2 x 2 x 3 x 5 HCF = 2 x 3 = 6   (d) 27, 63 27 = 3 x 3 x 3 63 = 3 x 3 x 7 HCF = 3 x 3 = 9   (e) 36, 84    36 = 2 x 3 x 3 x 3 84 = 2 x 2 x 3 x 7 HCF = 2 x 2 x 3 = 12   (f) 34, 102 34 = 2 x 17 102 = 2 x 3 x 17 HCF = 2 x 17 = 34   (g) 70,...

P Pankaj Sanodiya
(i) 24 and 36  24 = 2 2 2 3 36 = 2 2 3 3 HCF = 2 2 3 = 12 (ii) 15, 25 and 30 15 = 3 5  25 = 5 5  30 = 2 3 5 HCF = 5 (iii) 8 and 12  8 = 2 2 2 12 = 2 2 3 HCF = 2 2 = 4 (iv) 12, 16 and 28 12 = 2 2 3  16 = 2 2 2 2 28 = 2 2 7 HCF = 2 2 = 4

P Pankaj Sanodiya
Since it is the smallest number of such type, it will be the product of 4 smallest prime numbers. that is  2 3  5 7 = 210

P Pankaj Sanodiya
45 = 5  9 Factors of 5 = 1, 5 Factors of 9 = 1. 3, 9 Therefore, 5 and 9 are co-prime numbers. Since the last digit of 25110 is 0. it is divisible by 5. Sum of the digits of 25110 = 2 + 5 + 1 +1 + 0 = 9 As the sum of the digits of 25110 is divisible by 9, therefore. 25110 is divisible by 9. Since the number is divisible by 5 and 9 both, it is divisible by 45.

P Pankaj Sanodiya
in factorization, we don't write composite numbers. All the factors should be prime numbers in this method.  (a) 24 = 2 3 4  As we know that 4 is a composite number.  Hence This is not a prime factorization.  b) 56 = 7 2 2 2 In this factorization, all the factors of 56 are prime numbers. There is no composite number.  Hence This is prime factorization.  c) 70 = 2 5  7  In this...