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Tags: RD Sharma Class 6 Math Solution Chapter 2 Playing with Numbers Exercise 2.6, Solution of RD Sharma for class 6, RD Sharma solutions Grade VI, Solution of RD Sharma exercises for sixth standard, R.D. Sharma Mathematics solutions and practice pages, Playing With Numbers Exercise 2.6 (Higest Common Factor-H.C.F)
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Playing With Numbers
Exercise 2.6 (Highest Common Factor – H.C.F)
Question 1 (Refer Book - Mathematics Class VI R.D. Sharma)
Solution:
i. 144, 198
First, we find the prime factorization of each of the given numbers by division method.
2
144
2
72
2
36
2
18
3
9
3
2
198
3
99
3
33
3
11
Thus 144 = 2x2x2x2x3x3
198 = 2x3x3x3x11
Therefore, HCF of 144 and 198 is 2x3x3 (Common factors)= 18
ii. 81, 117
First, we find the prime factorization of each of the given numbers by division method.
3
81
3
27
3
9
3
3
117
3
39
13
Thus 81 = 3x3x3x3
117 = 3x3x13
Therefore, HCF of 81 and 117 is 3x3 (Common factors)= 9
iii. 84, 98
First, we find the prime factorization of each of the given numbers by division method.
2
84
2
42
3
21
7
2
98
7
49
7
Thus 84= 2x2x3x7
98 = 2x7x7
Therefore, HCF of 84 and 98 is 2x7 (Common factors)= 14
iv. 225, 450
First, we find the prime factorization of each of the given numbers by division method.
3
225
3
75
5
25
5
2
450
3
225
3
75
5
25
5
Thus 225 = 3x3x5x5
450 = 2x3x3x5x5
Therefore, HCF of 225 and 450 is 3x3x5x5 (Common factors)= 225
v. 170, 238
First, we find the prime factorization of each of the given numbers by division method.
2
170
5
85
17
2
238
7
119
17
Thus 170 = 2x5x17
238 = 2x7x17
Therefore, HCF of 170 and 238 is 2x17 (Common factors)= 34
vi. 504, 980
First, we find the prime factorization of each of the given numbers by division method.
2
504
2
252
2
126
3
63
3
21
7
2
980
2
490
5
245
7
49
7
Thus 504 = 2x2x2x3x3x7
980 = 2x2x5x7x7
Therefore, HCF of 504 and 980 is 2x2x7 (Common factors)= 28
vii.150, 140, 210
First, we find the prime factorization of each of the given numbers by division method.
2
150
3
75
5
25
5
2
140
2
70
5
35
7
2
210
3
105
5
35
7
Thus 150 = 2x3x5x5
140 = 2x2x5x7
210 = 2x3x5x7
Therefore, HCF of 150, 140 and 210 is 2x5 (Common factors)= 10
viii. 84, 120, 138
First, we find the prime factorization of each of the given numbers by division method.
2
84
2
42
3
21
7
2
120
2
60
2
30
3
15
5
2
138
3
69
23
Thus 84 = 2x2x3x7
120 = 2x2x2x3x5
138 = 2x3x23
Therefore, HCF of 84, 120 and 138 is 2x3 (Common factors)= 6
ix. 106, 159, 265
First, we find the prime factorization of each of the given numbers by division method.
2
106
53
3
159
53
5
265
53
Thus 106 = 2x53
159 = 3x53
265 = 5x53
Therefore, HCF of 106, 159 and 265 is 53 (Common factors)= 53
Question 2 (Refer Book - Mathematics Class VI R.D. Sharma)
Solution:
i. H.C.F of two consecutive numbers is 1.
ii. H.C.F of two consecutive even numbers is 2.
iii. H.C.F of two consecutive odd numbers is 1
Question 3 (Refer Book - Mathematics Class VI R.D. Sharma)
Solution:
No, the answer is not correct. The correct H.C.F is 1.
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