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Tags: RD Sharma Class 6 Math Solution Chapter 2 Playing with Numbers Exercise 2.9, 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 2.9 (Lowest Common Multiple - L.C.M)
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Playing With Numbers
Exercise 2.9 (Lowest Common Multiple – L.C.M)
Question 1 (Refer Book - Mathematics Class VI R.D. Sharma)
Solution:
i. 48, 60
We have,
2
48, 60
2
24, 30
3
12, 15
2
4, 5
2
2, 5
5
1, 5
1, 5
Therefore, L.C.M of given numbers = 2x2x2x2x3x5 = 240
ii. 42, 63
We have,
21
42, 63
2
2, 3
3
1, 3
1, 1
Therefore, L.C.M of given numbers = 21x2x3 = 126
iii. 18, 17
We have,
17
18, 17
2
18, 1
3
9, 1
3
3, 1
1, 1
Therefore, L.C.M of given numbers = 17x2x3x3 = 306
iv. 15, 30, 90
We have,
3
15, 30, 90
5
5, 15, 30
3
1, 3, 6
2
1, 1, 2
1, 1, 1
Therefore, L.C.M of given numbers = 3x5x3x2 = 90
v. 56, 65, 85
We have,
5
56, 65, 85
2
56, 13, 17
2
28, 13, 17
2
14, 13, 17
7
7, 13, 17
13
1, 13, 17
17
1, 1, 17
1, 1, 1
Therefore, L.C.M of given numbers = 2x2x2x5x7x13x17 = 61880
vi. 180, 384, 144
We have,
2
180, 384, 144
2
90, 192, 72
2
45, 96, 36
2
45, 48, 18
3
45, 24, 9
3
15, 8, 3
2
5, 8, 1
2
5, 4, 1
2
5, 2, 1
5
5, 1, 1
1, 1, 1
Therefore, L.C.M of given numbers = 2x2x2x2x2x2x2x3x3x5 = 5760
vii. 108, 135, 162
We have,
3
108, 135, 162
3
36, 45, 54
3
12, 15, 18
2
4, 5, 6
2
2, 5, 3
3
1, 5, 3
5
1, 5, 1
1, 1, 1
Therefore, L.C.M of given numbers = 2x2x3x3x3x3x5 = 1620
viii. 28, 36, 45, 60
We have,
2
28, 36, 45, 60
2
14, 18, 45, 30
3
7, 9, 45, 15
3
7, 3, 15, 5
5
7, 1, 5, 5
7
7, 1, 1, 1
1, 1, 1, 1
Therefore, L.C.M of given numbers = 2x2x3x3x5x7= 1260
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