Download factors and multiples

6
FACTORS AND MULTIPLES
1.(A) Find the prime factors of :
(i) 96
(ii)
Ans. (i) 96
(ii)
2 96
2
48
2
24
2
12
2
6
3
3
116
116
(iii) 245
2
116
5
245
2
58
7
49
29
7
7
1
1
(B.) If Pn means prime - factors of n, find :
(i) P6
(ii)
P24
(iii)
P36
(iv)
P42
Ans. (i) F6 = 1, 2, 3, 6 ∴ P6 (Prime factor of 6) = 2 and 3.
(ii) F24 = 1, 2, 3, 4, 6, 8, 12, 24 ∴ P24 = 2 and 3.
(iii) F36 = 1, 2, 3, 4, 6, 9, 12, 18, 36 ∴ P36 = 2 and 3.
(iv) F42 = 1, 2, 3, 6, 7, 14, 21, 42 ∴ P42 = 2, 3 and 7.
(C.) List the elements of each of M(8) and M(12). Hence, find the least element in the
set M(8) ∩ M(12) . Is it the L.C.M. of 8 and 12 ?
Ans. M(8) = { 8, 16, 24, 32, 40, 48, ...} M(12) = { 12, 24, 36, 48, 60, ...}
Set of common multiples of 8 and 12 = M(8) ∩ M(12) = { 24, 48}
The smallest element of this set is 24. Hence, the LCM of 8 and 12 = 24. Yes, it is
LCM of 8 and 12.
(D). List the elements of each of M(10) and M(15). Hence find the LCM of 10 and 15.
Ans. M(10) = { 10, 20, 30, 40, 50, ..} M(15) = { 15, 30, 45, 60 ... }
Set of commom multiples of 10 and 15 = M(10) ∩ M(15) = { 30, 60}
The smallest element of this set is 30 Hence, the LCM of 10 amd 15 = 30
2.
List the elements of P30 and P63. Hence, find : (i) P30 ∩ P63
(ii) P30 ∪ P63
Ans. We have: 30 = 2 × 3 × 5 63 = 3 × 3 × 7
Class - VI Mathematics
1
Question Bank
P30 = {2, 3, 5} and P63 = { 3, 7}
(i) P30 ∩ P63 = {2, 3, 5} ∩ {3, 7} = {3}
(ii)
P30 ∪ P63 = {2, 3, 5} ∪ {3, 7} = {2, 3, 5, 7}
3. (i) If P45 ∩ P63 = Pn , find the value of n.
(ii) If P(n) ∩ P(210) = P(25), find the n given that n < 6.
Ans. (i) We have : 45 = 3 × 3 × 5, 63 = 3 × 3 × 7
P45 = {3, 5} and P63 = {3, 7} P45 ∩ P63 = {3,5} ∩ {3, 7} = {3} = P3
∴
Hence, n = 3.
(ii) P(n) ∩ P(210) = P(25), P(n) ∩ P{2, 5, 7, 3} = P{5, 5} P(n) = 5
4. Express each one of the following as a product of prime factors :
(i) 105
(ii)
462 (iii) 1035 (iv) 1197 (v)
4641
Ans. (i) 105 = 3 × 5 × 7 (ii) 462 = 2 × 3 × 7 × 11 (iii) 1035 = 3 × 3 × 5 × 23
2 462
3 1035
3 105
3 231
3 345
5 35
7 77
5 115
7 7
11 11
23 23
1
1
1
(iv) 1197 = 3 × 3 × 7 × 19
(v) 4641 = 3 × 7 × 13 × 17
3 1197
3 4641
3
399
7
7
133
13 221
19 19
1
5. Find the H.C.F. of
(i) 16, 24
Ans. (i)
1547
17 17
1
(ii) 32, 40
(iii) 8, 20, 32
(ii) 32 and 40
(iv) 14, 28, 35
2
32
2
40
2
16
2
24
2
16
20
8
2
12
2
2
2
8
2
10
2
4
2
6
2
4
5
5
2
2
3
3
2
2
1
Class - VI Mathematics
1
1
1
2
Question Bank
∴ 16 = 2 × 2 × 2 × 2, 24
= 2 × 2 × 2 × 3,
HCF = 2 × 2 × 2 = 8.
(iii) 8, 20, 32
2
2
2
∴ 32 = 2 × 2 × 2 × 2 × 2, 40
= 2 × 2 × 2 × 5,
HCF = 2 × 2 × 2 = 8.
8
4
2
20
2
2
10
1
5
5
2
32
2
16
2
8
2
4
2
2
1
1
1
8 = 2 × 2 × 2, 20 = 2 × 2 × 5, 32 = 2 × 2 × 2 × 2 ×2, HCF = 2 × 2 = 4.
∴
(iv) 14, 28, 35
2
14
7
7
2
28
2
14
7
7
5
35
7
7
1
1
1
14 = 2 × 7, 28 = 2 × 2 × 7, 35 = 5 × 7, HCF = 7.
∴
6. Find the HCF of the following numbers using prime factorisation method :
(i) 32, 56
(ii) 72, 96
(iii) 81, 108
(iv) 135, 180
(v) 108, 216
(vi) 144, 180, 198
Ans. (i) 32, 56
(ii) 72, 96
2
56
2
28
2
14
7
7
1
2
96
2
48
3
24
9
2
8
3
2
4
1
2
2
2
32
2
72
2
16
2
36
2
8
2
18
2
4
3
2
2
3
1
1
∴ 32 = 2 × 2 × 2 × 2 × 2,
56 = 2 × 2 × 2 × 7,
HCF = 2 × 2 × 2 = 8.
Class - VI Mathematics
∴
3
72 = 2 × 2 × 2 × 3 × 3,
96 = 2 × 2 × 2 × 2 ×2 × 3,
HCF = 2 × 2 × 2 × 3 = 24
Question Bank
(iii) 81, 108
3 81
(iv) 135, 180
3 135
2
108
3
27
2
54
3
3
9
3
27
3
3
3
9
1
3
3
2
180
45
2
90
3
15
3
45
5
5
3
15
1
5
5
1
81 = 3 × 3 × 3 × 3,
108 = 2 × 2 × 3 × 3 × 3,
HCF = 3 × 3 × 3 = 27.
(v) 168, 216
∴
1
∴ 135 = 3 × 3 × 3 × 5,
180 = 2 × 2 × 3 × 3 × 5,
HCF = 3 × 3 × 5 = 45.
(vi) 144, 180, 198
2
168
2
216
2
144
2
180
2
84
2
108
2
72
2
90
2
198
2
42
2
54
2
36
3
45
3
99
3
21
3
27
2
18
3
15
3
33
7
7
3
9
3
9
5
5
11 11
1
3
3
3
3
1
1
1
1
∴ 168 = 2 × 2 × 2 × 3 × 7,
216 = 2 × 2 × 2 × 3 × 3 × 3,
HCF = 2 × 2 × 2 × 3 = 24
∴ 144 = 2 × 2 × 2 × 2 × 3 × 3,
180 = 2 × 2 × 3 × 3 × 5,
198 = 2 × 3 × 3 × 11
HCF = 2 × 3 × 3 = 18
7. Find the HCF of the following numbers using long division method :
(i) 24, 64
(ii) 144, 312
(iii) 252, 576
(iv) 576, 920
(v) 605, 935
(vi) 112, 140, 168
Ans. (i) 24, 64
(ii) 144, 312
24
64 2
48
16
24 1
16
8 16 2
16
0
∴ HCF of 24 and 64 = 8
Class - VI Mathematics
144 312 2
288
24 144
144
0
6
∴ HCF of 144 and 312 = 24
4
Question Bank
(iii) 252, 576
(iv) 575, 920
252 576 2
504
72 252 3
216
36 72 2
72
0
∴ HCF of 252 and 576 = 36
(v) 605, 935
572
345
575 1
345
230 345 1
230
115 230 5
230
0
∴ HCF of 575 and 920 = 115
(vi) 112, 140, 168
112
605 935 1
605
330
920 1
575
605 1
330
275 330 1
275
55 275 5
275
0
140 1
112
28
112 4
112
0
∴ HCF of 605 and 935 = 55.
Now, find the HCF of 28 and 168
28 168 6
168
0
HCF of 112, 140 and 168 = 28.
∴
8. Find the greatest number that exactly divides 385 and 735.
Ans. Find the HCF of 385 and 735
∴ The required number = 35.
385 735 1
385
350 385 1
350
35 350 8
350
0
Class - VI Mathematics
5
Question Bank
9. Find the greatest number that exactly divides 306, 450 and 540.
Ans.
10.
Ans.
11.
Ans.
18 540 30
306 450 1
306
540
0
144 306 2
288
18 144 8
144
0
∴ The required number = 18.
Now, find the HCF of 18 and 540
Two vessels contain 104 litres and 91 litres of milk. Find the measure of a bucket of
maximum capacity which can measure the milk of either vessel an exact number of
times :
∴ Required capacity of bucket = 13 litres.
91 104 1
91
13 91 7
91
0
Find the LCM of the following number using prime factorisation method:
(i) 16, 24 (ii) 36, 45 (iii) 72, 108 (iv) 18, 20, 30
(v) 15, 18, 24
(i) 16, 24
(ii) 36, 45
2 16
3 45
3 45
2 36
2
8
3
15
2
18
3
15
2
4
5
5
3
9
5
5
2
2
1
3
3
1
16 = 2 × 2 × 2 × 2 = 24
24 = 2 × 2 × 2 × 3 = 23 × 3
∴ LCM = 24 × 3 = 16 × 3 = 48
1
36 = 2 × 2 × 3 × 3 = 22 × 32
45 = 3 × 3 × 5 = 32 × 5
∴ LCM = 22 × 32 × 5 = 4 × 9 × 5 = 180
(iii) 72, 108
(iv) 18, 20, 30
2
72
2
108
2
36
2
54
2
18
3
27
3
9
3
9
3
3
3
3
1
Class - VI Mathematics
1
2
18
2
20
2
30
3
9
2
10
3
15
3
3
5
5
5
5
1
1
6
1
1
Question Bank
18 = 2 × 3 × 3 = 2 × 32
20 = 2 × 2 × 5 = 22 × 5
30 = 2 × 3 × 5 = 2 × 3 × 5
72 = 2 × 2 × 2 × 3 × 3
108 = 2 × 2 × 3 × 3 × 3
∴ LCM = 23 × 33 = 8 × 27 = 216
∴ LCM = 22 × 32 × 5 = 4 × 9 × 5 = 180
(v) 15, 18, 24
3
15
5
5
2
18
2
24
3
9
2
12
3
3
2
6
1
3
3
1
1
15 = 3 × 5 = 3 × 5; 18 = 2 × 3 × 3; 24 = 2 × 2 × 2 × 3
∴ LCM = 23 × 32 × 5 × = 8 × 9 × 5 = 360
12. Find the L.C.M. of the following using common division method:
(i) 8, 10, 12, 16
(ii)
12, 15, 18, 24
(iii) 15, 18, 20, 27
(iv) 20, 25, 30, 45
(v) 16, 18, 24, 32, 36
(vi) 120, 210, 225
Ans. (i) 8, 10, 12, 16
(ii)
12, 15, 18, 24
2
12 15 18
2
8
10
12
2
4
5
6
8
2
6 15
9
12
2
2
5
3
4
3
3 15
9
6
1
5
3
2
5
3
2
1
∴ LCM = 2 × 2 × 2 × 5 × 3 × 2 = 240 ∴ LCM = 2 × 2 × 3 × 5 × 3 × 2 = 360
(iii) 15, 18, 20, 27
(iv) 20, 25, 30, 45
2
15 18
20
3
15
9
10
3
5
3
3
5
1
2
20 25 30
27
3
10 25 15 45
10
9
5
1
10
3
10 25
2 5
1
2
3
5 15
1 3
∴ LCM = 2 × 3 × 3 × 5 × 2 × 3 = 540.
∴ LCM = 2 × 3 × 5 × 2 × 5 × 3 = 900
7
Question Bank
Class - VI Mathematics
(v) 16, 18, 24, 32, 36
(vi)
120, 210, 225
2
16 18
24
32
36
2
120 210 225
2
8
9
12
16
18
3
60 105 225
2
4
9
6
8
9
5
20
35
75
2
2
9
3
4
9
3
4
7
15
3
1
9
3
2
9
4
7
5
3
1
3
1
2
3
1
1
1
2
1
∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 2 = 288. ∴ LCM = 2 × 3 × 3 × 4 × 5 × 5 × 7 = 12600
13. The HCM of two numbers is 19 and their LCM. is 228. If one of the number is 57,
find the other.
Ans. Let the other number be ‘x’ ∴ Product of two numbers = HCF × LCM
⇒
57 × x = 19 × 228
19 × 228
x=
= 76.
⇒
57
∴ The other number is 76.
14. The HCF of two numbers is 23 and their LCM is 276. If one of number is 92, find
the other.
Ans. Let the other number be ‘x’. ∴ Product of two numbers = HCF × LCM
⇒
92 × x = 23 × 276 ⇒ x =
23 × 276 276
=
= 69.
92
4
∴ The other number is 69.
15. The product of two numbers is 1734 and their HCF is 17. Find their LCM.
Ans. HCF × LCM = Product of two numbers ⇒ 17 × LCM = 1734
∴ LCM = 1734 ÷ 17 = 102.
16. The product of two numbers is 3174 and their LCM is 138. Find their HCF.
Ans. HCF × LCM = Product of two numbers ⇒ HCF × 138 = 3174
∴ LCM = 3174 ÷ 138 = 23
17. The H.C.F. and the L.C.M of two numbers are 50 and 300 respectively. If one of
the numbers is 150, find the other number.
Ans. HCF = 50 and LCM = 300, One number = 150
Product of LCM and HCF = 300 × 50 = 15000
Class - VI Mathematics
8
Question Bank
∴
The other number =
Product of LCM and HCF 15000
= 100
=
150
One number
18. The product of two numbers is 432 and their LCM is 72. Find their HCF.
Ans. Product of two numbers = Product of their LCM and HCF
Here, product of two number = 432 LCM = 72
HCF = 432 ÷ 72 = 6
∴
19. Can there be two numbers with HCF 12 and LCM 64 ? Give reasons in support of
your answer.
Ans. No, because HCF of two numbers always divides their LCM.
20. An electronic device makes a beep after every 15 minutes. Another device makes
a beep after every 20 minutes. They beeped together at 10 a.m. At what time will
they make the next beep together ?
5
Ans. LCM of 15 and 20 = 5 × 3 × 4 = 60
15 20
3
4
The next beep will be after 60 minutes i.e. + 60 minutes = 11 a.m.
21. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12
minutes respectively. After what interval of time will they toll together again ?
Ans. Find the LCM of 2, 4, 6. 8, 10, 12.
2
2
4
6
8
10
12
2
1
2
3
4
4
6
3
1
1
3
2
5
3
1
1
1
2
5
1
∴ LCM = 2 × 2 × 3 × 2 × 5 = 120.
∴ The six bells toll together again after 120 minutes, i.e. after 2 hours.
22. Find the least number which when divided by 12, 15, 18, 24 and 36 leaves no
remainder.
Ans. The least number which is exactly divisible by each given numbers is their LCM.
∴ Required number = LCM of 12, 15, 18, 24 and 36. ∴ LCM = least required
number
Class - VI Mathematics
9
Question Bank
2
12, 15, 18,
24,
36
2
6, 15,
9,
12,
18
3
3, 15,
9,
6,
9
3
1,
5,
3,
2,
3
1,
5,
1,
2,
1
= 2 × 2 × 3 × 3 × 5 × 2 = 360
Hence, the least required number = 360.
23. Find the least number which when increased by one is exactly divisible by 12, 18,
24, 32 and 40.
2
Ans.
12, 18, 24,
32,
40
16,
20
2
6,
9, 12,
2
3,
9,
6,
8,
10
3
3,
9,
3,
4,
5
1,
3,
1,
4,
5
∴ LCM = 2 × 2 × 2 × 3 × 3 × 4 × 5 = 1440
∴ The required number = 1441 – 1 = 1439.
Class - VI Mathematics
10
Question Bank