class 8: factors & multiples – exercise 4b · pdf filefind the greatest number that...
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Class 8: Factors & Multiples – Exercise 4B
1. Use prime factorization method to find the H.C.F. of the following:
i. 204, 1190
ii. 1445, 1785
iii. 1512, 4212
iv. 280, 315, 385
v. 576, 792, 1512
vi. 1197, 5320, 4389
Note: Prime Factorization Method:
Step 1: Express each one of the given numbers as a product of prime factors
Step 2: The product of terms containing least power of common prime factors gives the
HCF of the given numbers.
Answers:
i. 204, 1190
2 204
2 1190
2 102
5 595
3 51
7 119
17 17
17 17
1
1
204 = 22 x 3 x 17
1190 = 2 x 5 x 7 x 17
HCF = 2 x 17 = 34
ii. 1445, 1785
5 1445
5 1785
17 289
3 357
17 17
7 119
1
17 17
1
1445 = 5 x 172
1785 = 5 x 3 x 7 x 17
HCF = 5 x 17 = 85
iii. 1512, 4212
2 1512
2 4212
2 756
2 2106
2 378
3 1053
3 189
3 351
3 63
3 117
3 21
3 39
7 7
13 13
1
1
1512 = 23 x 3
3 x 7
4212 = 22 x 3
4 x 13
HCF = 22 x 3
3 = 108
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iv. 280, 315, 385
2 280
5 315
5 385
2 140
3 63
7 77
2 70
3 21
11 11
5 35
7 7
1
7 7
1
1
280 = 23 x 5 x 7
315 = 5 x 32 x 7
385 = 5 x 7 x 11
HCF = 5 x 7 = 35
v. 576, 792, 1512
2 576
2 792
2 1512
2 288
2 396
2 756
2 144
2 198
2 378
2 72
3 99
3 189
3 36
3 33
3 63
3 12
11 11
3 21
2 4
1
7 7
2 2
1
1
576 = 27 x 3
2
792 = 23 x 3
2 x 11
1512 = 23 x 3
3 x 7
HCF = 23 x 3
2 = 72
vi. 1197, 5320, 4389
3 1197
2 5320
7 4389
3 399
2 2660
3 627
7 133
7 1330
19 209
19 19
2 190
11 11
1
5 95
1
19 19
1
1197 = 32 x 7 x 19
5320 = 23 x 7 x 5 x 19
4389 = 3 x 7 x 19 x 11
HCF = 7 x 19 = 133
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2. Find the H.C.F. of the following numbers using long division method:
i. 837, 992
ii. 2923, 3239
iii. 5508, 9282
iv. 204, 1190, 1445
v. 370, 592, 1036
vi. 1701, 2106, 2754
Answer:
i. 837, 992
837 992 1
837
155 837 5
775
62 155 2
124
31 62 2
62
HCF = 31
0
ii. 2923, 3239
2923 3239 1
2923
316 2923 9
2844
79 316 4
316
HCF = 79
0
iii. 5508, 9282
5508 9282 1
5508
3774 5508 1
3774
1734 3774 2
3468
306 1734 5
1530
204 306 1
204
102 204 2
204
HCF = 102
0
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iv. 204, 1190, 1445
1190 1445 1
1190
255 1190 4
1020
170 255 1
170
85 170 2
170
HCF of 1445 and 1190 =
85
0
85 204 2
170
34 85 2
68
17 34 2
34
0
HCF = 17
v. 370, 592, 1036
592 1036 1
592
444 592 1
444
148 444 3
444
HCF of 1036 and 592 =
148 0
148 370 2
296
74 148 2
148
0
HCF = 74
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vi. 1701, 2106, 2754
2106 2754 1
21906
648 2106 3
1944
162 648 4
648
HCF of 2754 and 2106 =
162 0
162 1701 10
1620
81 162 2
162
0
HCF = 81
3. Reduce each of the following fractions to its lowest terms:
i. 682
868
ii. 777
1147
iii. 1095
1168
Answer:
Note: To reduce the fractions, first find out the HCF of the Numerator and the Denominator.
The divide the numerator by the HCF and Denominator by HCF
i. 682
868
HCF of 868 and 682 = 62
682 868 1
682
186 682 3
558
129 186 1
124
62 124 2
124
0
Therefore 682
868 =
682 ÷62
868 ÷62 =
11
14
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i. 777
1147
HCF of 1147 and 777 = 37
777 1147 1
777
370 777 2
740
37 370 10
370
0
Therefore 777
1147 =
777 ÷37
1147 ÷37 =
21
31
ii. 1095
1168
HCF of 1168 and 1095 = 73
1095 1168 1
1095
73 1095 15
1095
0
Therefore 1095
1168 =
1095 ÷73
1168 ÷73 =
15
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4. Which of the following numbers are co-primes?
i. 18, 25
ii. 62, 81
iii. 69, 92
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Answer:
Note: The natural numbers are said to be co-primes if the HCF of the numbers is 1.
You can calculate the HCF by using any of the two methods…prime factorization or long division
method.
i. 18, 25
HCF of 18, 25 = 1
Hence, 18 and 25 are co-primes.
ii. 62, 81
HCF of 62, 81 = 1
Hence, 62 and 81 are co-primes.
iii. 69, 92
HCF of 69, 92 = 23
Hence, 69 and 92 are not co-primes.
5. Find the greatest number that exactly divides 105, 1001 and 2436.
Answer:
We need to find the HCF of 105, 1001 and 2436.
1001 2436 2
2002
434 1001 2
868
133 434 3
399
35 133 3
105
28 35 1
28
7 28 4
28
0
7 105 15
105
0
Hence the HCF of the three numbers is 7. That means, 7 is the largest number that would divide all the
three numbers perfectly.
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6. Find the largest number which can divide 290, 460 and 552 leaving the remainders 4, 5 and 6
respectively.
Answer:
The largest number should divide (290-4), (460-5) and (552-6) or 286, 455 and 546 perfectly. Now find
the HCF of 286, 455 and 546.
455 546 1
455
91 455 5
455
0
91 286 3
273
13 91 7
91
0
Hence 13 is the number that can divide 290, 460 and 552 leaving the remainders 4, 5 and 6 respectively.
7. Find the largest number which can divide 1354, 1866 and 2762 leaving the same remainder 10 in each
case.
Answer:
The largest number should divide (1354 – 10), (1866 – 10) and (2762 – 10) or 1344, 1856 and 2752
perfectly. Now find the HCF of 1344, 1856 and 2752.
1856 2752 1
1856
896 1856 2
1792
64 896 14
896
0
64 1344 21
1344
0
HCF is 64. Hence 64 is the largest number which can divide 1354, 1866 and 2762 leaving the same
remainder 10 in each case.
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8. Find the greatest number that will divide 1305, 4665 and 6905 so as to leave the same remainder in each
case.
Answer:
Required number
= HCF of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
= HCF of 3360, 2240, 5600.
3360 5600 1
3360
2240 3360 1
2240
1120 2240 2
2240
0
1120 3360 3
3360
0
HCF is 1120.
The required number is 112 that will divide 1305, 4665 and 6905 so as to leave the same remainder in
each case.
9. Three pieces of timber 13m 44cm, 18m 56cm and 27m 52cm have to be divided into planks of same
lengths. What is the greatest possible length of each plank?
Answer:
The lengths are 1344cm, 1856cm and 2752cm. Find HCF.
1856 2752 1
1856
896 1856 2
1792
64 896 14
896
0
64 1344 21
1344
0
HCF is 64. Hence greatest length is 64 cm.
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10. An NGO wishes to distribute 1651 pencils and 2032 erasers among poor children in such a way that
each child gets the same number of pencils and the same number of erasers. Find the maximum number
of children among whom these pencils and erasers are distributed.
Answer:
Take HCF of 1651 and 2032.
1651 2032 1
1651
381 1651 4
1524
127 381 3
381
0
HCF = 127.
The maximum number of children among whom these pencils and erasers are distributed = 127
11. Find the greatest possible length of a rope which can be used to measure exactly the lengths 5m 13cm,
7m 83cm and 10m 80cm.
Answer:
Lengths are 513 cm, 783 cm and 1080 cm.
HCF of 513, 783 and 1080
783 1080 1
783
297 783 2
594
189 297 1
189
108 189 1
108
81 108 1
81
27 81 3
81
0
27 513 19
513
0
Hence HCF = 27. Therefore the length of the rope is 27 cm.
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12. Three different containers contain different quantity of mixture of milk and water whose measurements
are 403kg, 465kg, and 527kg. What biggest measure must be there to measure all different quantities an
exact number of times.
Answer:
Required Number is the HCF of 403, 465 and 527.
465 527 1
465
62 465 7
434
31 62 2
62
0
31 403 13
403
0
HCF = 31. Hence 31 liter measure would be required to measure the three given quantities.
13. Find the least number of square tiles required to pave the ceiling of a room 15m 17cm long and 9m 2cm
broad.
Answer:
The dimensions of the ceiling are 1517 cm and 902 cm.
The required number of times = HCF of 1517 and 902.
902 1517 1
902
615 902 1
615
287 615 2
574
41 287 7
287
0
HCF = 41
Dimensions of tiles = 41 cm.
Therefore the number of tiles = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑒𝑖𝑙𝑖𝑛𝑔
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑖𝑙𝑒=
(1517 × 902)
41 × 41 = 814.
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14. A school admitted 1190 boys and 204 girls in a year. The authorities decided to divide these students
into classes in such a way that each class gets the same number of boys and the same number of girls.
Find the maximum number of classes that can be formed.
Answer:
Required number is the HCF of 1190 and 204.
204 1190 5
1020
170 204 1
170
34 170 5
170
0
HCF = 34.
Number of classes = 34
Size of the class = 41 students
15. In a training camp, there were 195 boys and 143 girls. The coordinator instructed the pupils to form
separate teams for boys and girls such that each team includes equal number of pupils. Find the
maximum number of pupils that each team can have and also the number of teams so formed.
Answer:
Required number is HCF of 195 and 143.
143 195 1
143
52 143 2
104
39 52 1
39
13 39 3
39
0
HCF = 13.
Number of teams = 13.
Number of pupil in each team = (195 + 143) / 13 = 26.
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