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.

class 8: factors & multiples – exercise 4b · pdf filefind the greatest number that...

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