Math, asked by priyakonduru0708, 8 days ago

FIND THE HCF OF THE FOLLOWING IN LONG DIVVISION METHOD:-A)216,124 B)60,84 C)51,93 D)124,224 E)119,136 F)256,78​

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Answered by as3152865
1

Answer:

Q1) Find the HCF of the following numbers :

(i) 18, 48 (ii) 30, 42 (iii) 18, 60 (iv) 27, 63

(v) 36, 84 (vi) 34, 102 (vii) 70, 105, 175

(viii) 91, 112, 49 (ix) 18, 54, 81 (x) 12, 45, 75

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SOLUTION:

i) 18,48

18=2\times3\times3

48=2\times2\times2\times2\times3

The common factor of 18 and 48 are 2,3. thus, HCF of 18 and 48 is2\times3=6

ii) 30,42

30=2\times3\times5

42=2\times3\times7

The common factor of 30 and 42 are 2 and 3

thus, HCF of 30,42 are 2\times3=6

iii) 18,60

18=2\times3\times3

60=2\times2\times3\times5

The common factor of 18 and 60 are 2,3. thus, HCF of 18 and 60 are 2\times3=6

iv) 27, 63

27=3\times3\times363=3\times3\times7

The common factor for 27 and 63 is 3

thus, HCF of 27 and 63 is 3\times3=9

v) 36,84

36=2\times2\times3\times384=2\times2\times3\times7

The common factor of 36 and 84 is 2\times3

thus, HCF of 36 and 84 is 2\times2\times3=12

vi) 34, 102

34=2\times17102=2\times3\times17

The common factor in 34 and 102 is 2, 17.

thus, HCF for 34 and 102 is 2\times17=34

vi) 70, 105, 175

70=2\times5\times7

105=3\times5\times7

175=5\times5\times7

The common factor in 70, 105 and 175 are 5, 7.

thus, HCF for 70,105 and 175 is 5\times7=35

vii) 91,112,49

91=7\times13112=2\times2\times2\times2\times749=7\times7

The common factors of 91, 112, and 49 are 7.

therefore, HCF of 91, 112 and 49 are 7.

viii) 18, 54,81

18=2\times3\times354=2\times3\times3\times381=3\times3\times3\times3

therefore common factors between 18, 54, 81 is 3X3=9

xi) 12,45,75

therefore, the coomon factor is 3.

HCF of 12,45,75 is 3.

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