**Ans . **

3

**Explanation :**Using Rule 1, Required number

= LCM X HCF

First number

= 864 x 144 = 432

288

**Ans . **

3

**Explanation :**Using Rule 1, LCM × HCF = 1st Number x 2nd Number

225 x 5 = 25 x X

X = 225 x 5 = 45

25

**Ans . **

3

**Explanation :**Given that L.C.M. of two numbers = 1820

H.C.F. of those numbers = 26

One of the number is 130

Another number = 1820 x 26 = 364

130

**Ans . **

2

**Explanation :**Using Rule 1, We have,

First number × second number

= LCM × HCF

Second number = 1920 x 16 = 240

128

**Ans . **

1

**Explanation :**Using Rule 1,

Product of two numbers = HCF × LCM

= 12906 × 14818

= LCM x 478

LCM = 12906 x 14818

478

= 400086

**Ans . **

4

**Explanation :**Using Rule 1, H.C.F. of the two 2-digit numbers = 16

Hence, the numbers can be expressed as 16x and 16y, where x and y are prime to each other.

Now, First number × second number

= H.C.F. × L.C.M.

16x × 16y = 16 × 480

xy = 16 480 = 30

16 x 16

The possible pairs of x andy, satisfying the conditionxy = 30 are : (3, 10), (5, 6), (1, 30), (2, 15) Since the numbers are of 2-digits each. Hence, admissible pair is (5, 6)

Numbers are : 16 × 5 = 80

**Ans . **

2

**Explanation :**Using Rule 1,

We know that,

First number × Second number = LCM × HCF

Second number

16 x 160 = 80

32

**Ans . **

2

**Explanation :**LCM = Product of two numbers

HCF

4107 = 111

37

Obviously, numbers are 111 and 37 which satisfy the given condition. Hence, the greater number = 111

**Ans . **

2

**Explanation :**Using Rule 1,

First number × Second number = HCF × LCM

Second number

15 x 300 = 75

60

**Ans . **

3

**Explanation :**Let the numbers be 12x and 12y where x and y are prime to each other.

LCM = 12xy

12xy = 924

xy = 77

Possible pairs = (1,77) and (7,11)

**Ans . **

3

**Explanation :**Using Rule 1,

First number × second number = LCM × HCF Let the second number be x.

10x = 30 × 5

x = 30 x 5 = 15

10

**Ans . **

1

**Explanation :**Using Rule 1,

HCF × LCM = Product of two numbers

8 × LCM = 1280

LCM = 1280 = 160

160

**Ans . **

4

**Explanation :**Using Rule 1,

First number × second number = HCF × LCM

24 × second number = 8 × 48

Second number = 8 x 48 = 16

24

**Ans . **

2

**Explanation :**Using Rule 1,

First number × second number = HCF × LCM

84 × second number = 12 × 336

Second number

= 12 x 336 = 48

84

**Ans . **

4

**Explanation :**Let the numbers be 6x and 6y where x and y are prime to each other.

6x × 6y = 216

xy = 216 = 6

6 x 6

LCM = 6xy = 6 × 6 = 36

**Ans . **

1

**Explanation :**Using Rule 1,

Second number = HCF x LCM

First number

18 x 378 = 126

54

**Ans . **

3

**Explanation :**Let the number be 15x and 15y, where x and y are co –prime.

15x × 15y = 6300

xy = 6300 = 28

15 x 15

So, two pairs are (7, 4) and (14, 2)

**Ans . **

4

**Explanation :**Using Rule 1,

First number × Second number = HCF × LCM

75 × Second number = 15 × 225

Second number = 15 x 225 = 45

75

**Ans . **

1

**Explanation :**Using Rule 1,

First number × second number = HCF × LCM

52 × second number = 4 x 520 = 40

52

**Ans . **

4

**Explanation :**Using Rule 1,

First number × Second number = HCF × LCM

864 × Second number = 96 × 1296

Second number = 96 x 1296 = 144

864

**Ans . **

2

**Explanation :**Using Rule 1,

Let LCM be L and HCF be H, then L = 4H

H + 4H = 125

5H = 125

H = 125 = 25

5

L = 4 × 25 = 100

Second number = L x H

First number

= 100 x 25 = 25

100

**Ans . **

1

**Explanation :**HCF of two-prime numbers = 1

Product of numbers = their LCM = 117

117 = 13 × 9 where 13 & 9 are co-prime. L.C.M (13,9) = 117.

**Ans . **

2

**Explanation :**HCF = 12 Numbers = 12x and 12y where x and y are prime to each other.

12x × 12y = 2160

xy = 2160

12 x 12

= 15 = 3 × 5, 1 × 15 Possible pairs = (36, 60) and (12, 180)

**Ans . **

1

**Explanation :**Using Rule 1, Second number

= H.C.F. x L.C.M.

First Number

= 27 x 2079 = 297

189

**Ans . **

2

**Explanation :**Here, HCF = 13 Let the numbers be 13x and 13y where x and y are Prime to each other.

Now, 13x × 13y = 2028

xy = 2028 = 12

13 x 13

The possible pairs are : (1, 12), (3, 4), (2, 6) But the 2 and 6 are not co-prime.

The required no. of pairs = 2

**Ans . **

2

**Explanation :**HCF = 13 Let the numbers be 13x and 13y.

Where x and y are co-prime.

LCM = 13 xy

13 xy = 45

xy = 455 = 35 = 5 x 7

13

Numbers are 13 × 5 = 65 and 13 × 7 = 91

**Ans . **

4

**Explanation :**HCF of two numbers is 8. This means 8 is a factor common to both the numbers. LCM is common multiple for the two numbers, it is divisible by the two numbers. So, the required answer = 60

**Ans . **

4

**Explanation :**Let the numbers be 23x and 23y where x and y are co-prime.

LCM = 23 xy As given, 23xy = 23 × 13 × 14

x = 13, y = 14

The larger number = 23y = 23 × 14 = 322

**Ans . **

4

**Explanation :**LCM = 2 × 2 × 2 × 3 × 5 Hence, HCF = 4, 8, 12 or 24 According to question 35 cannot be H.C.F. of 120.

**Ans . **

3

**Explanation :**Using Rule 1,

First number = 2 × 44 = 88

First number × Second number = H.C.F. × L.C.M.

88 × Second numebr = 44 × 264

Second number = 44 x 264 = 13

88

**Ans . **

3

**Explanation :**Using Rule 4, L.C.M. of 4, 6, 8, 12 and 16 = 48

Required number = 48 + 2 = 50

**Ans . **

4

**Explanation :**Using Rule 4, LCM of 15, 12, 20, 54 = 540

Then number = 540 + 4 = 544 [4 being remainder]

**Ans . **

4

**Explanation :**Using Rule 4, The greatest number of five digits is 99999. LCM of 3, 5, 8 and 12

2 3, 5, 8, 12

2 3, 5, 4, 6

3 3, 5, 2, 3

1, 5, 2, 1

LCM = 2 × 2 × 3 × 5 × 2 = 120 After dividing 99999 by 120, we get 39 as remainder 99999 – 39 = 99960 = (833 × 120) 99960 is the greatest five digit number divisible by the given divisors. In order to get 2 as remainder in each case we will simply add 2 to 99960.

Greatest number = 99960 + 2 = 99962

**Ans . **

3

**Explanation :**Using Rule 4, LCM of 4, 5, 6, 7 and 8

= 2 4, 5, 6, 7, 8

2 2, 5, 3, 7, 4

1, 5, 3, 7, 2

= 2 × 2 × 2 × 3 × 5 × 7 = 840.

let required number be 840 K + 2 which is multiple of 13. Least value of K for which (840 K + 2) is divisible by 13 is K = 3

Required number = 840 × 3 + 2 = 2520 + 2 = 2522

**Ans . **

4

**Explanation :**Required time = LCM of 252, 308 and 198 seconds

2 252, 308, 198

2 126, 154, 99

7 63, 77, 99

9 9, 11, 99

11 1, 11, 11

1, 1, 1

LCM = 2 × 2 × 7 × 9 × 11 = 2772 seconds = 46 minutes 12 seconds

**Ans . **

2

**Explanation :**15 = 3 × 5

18 = 3^{2}× 2

21 = 3 × 7

24 = 2^{3}× 3

LCM = 8 × 9 × 5 × 7 = 2520 The largest number of four digits = 9999

2520 ) 9999 ( 3

7560

2439

Required number = 9999 – 2439 – 4 = 7556 (Because 15 – 11 = 4

18 – 14 = 4

21 – 17 = 4

24 – 20 = 4)

**Ans . **

3

**Explanation :**LCM of 21, 36 and 66

LCM = 3 × 2 × 7 × 6×11 = 3 × 3 × 2 × 2 × 7 × 11

Required number = 3^{2}× 2^{2}× 7^{2}× 11^{2}= 213444

**Ans . **

1

**Explanation :**Using Rule 5, Here 4 – 1 = 3, 5 – 2 = 3, 6 – 3 = 3

The required number =

LCM of (4, 5, 6) – 3

= 60 – 3 = 57

**Ans . **

2

**Explanation :**LCM of 4, 6, 10, 15 = 60 Least number of 6 digits = 100000

The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020

Required number (N) = 100020 + 2 = 100022 Hence,

the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5

**Ans . **

3

**Explanation :**The LCM of 12, 18, 21, 30

2 12, 18, 21, 30

3 6, 9, 21, 15

2, 3, 7, 5

LCM = 2 × 3 × 2 × 3 × 7 × 5 = 1260

The required number

1260 = 630

2

**Ans . **

4

**Explanation :**2 10, 16, 24

2 5, 8, 12

2 5, 4, 6

2 5, 2, 3

3 5, 1, 3

5 5, 1, 1

1, 1, 1

LCM = 2^{2}× 2^{2}× 3 × 5

Required number = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 3600

**Ans . **

2

**Explanation :**Required number of students =

LCM of 6, 8, 10 = 120

**Ans . **

2

**Explanation :**LCM of 4, 6, 8, 9

2 4, 6, 8, 9

3 2, 3, 4, 9

3 1, 3, 2, 9

1, 1, 2, 3

LCM = 2 × 2 × 3 × 2 × 3 = 72

Required number = 72, because it is exactly divisible by 4, 6, 8 and 9 and it leaves remainder 7 when divided by 13

**Ans . **

2

**Explanation :**Using Rule 5, Here, 12 – 5 = 7, 16 – 9 = 7

Required number = (L.C.M. of 12 and 16) – 7 = 48 – 7 = 41

**Ans . **

3

Using Rule Number 5,**Explanation :**

Here, Divisor – remainder = 1 e.g., 10 – 9 = 1, 9 – 8 = 1, 8 – 7 = 1

Required number = (L.C.M. of 10, 9, 8) –1 = 360 – 1 = 359

**Ans . **

4

**Explanation :**We find LCM of 5, 6 and 8

5=5

6=3×2

8=2^{3}

= 2^{3}×3 × 5 = 8 × 15 = 120 Required number = 120K + 3

when K = 2, 120 × 2 + 3 = 243 required no. It is completely divisible by 9

**Ans . **

4

**Explanation :**LCM of 16, 18, 20 and 25 = 3600

Required number = 3600K + 4 which is exactly divisible by 7 for certain value of K. When K = 5, number = 3600 × 5 + 4 = 18004 which is exactly divisible by 7.

**Ans . **

2

**Explanation :**LCM of 3, 5, 6, 8, 10 and 12 = 120

Required number = 120x + 2, which is exactly divisible by 13. 120x + 2 = 13 × 9x + 3x + 2 Clearly 3x + 2 should be divisible by 13. For x=8,3x + 2 is divisible by 13.

Required number = 120x + 2 = 120 × 8 + 2 = 960 + 2 = 962

**Ans . **

4

**Explanation :**LCM of 6, 9, 15 and 18

2 6, 9, 15, 18

3 3, 9, 15, 9

3 1, 3, 5, 3

1, 1, 5, 1

LCM = 2 × 3 × 3 × 5 = 90

Required number = 90k + 4, which must be a multiple of 7 for some value of k. For k = 4, Number = 90 × 4 + 4 = 364, which is exactly divisible by 7.

**Ans . **

2

**Explanation :**Using Rule 9, We will find the LCM of 16, 24, 30 and 36.

2 16, 24, 30, 36

2 8, 12, 15, 18

2 4, 6, 15, 9

3 2, 3, 15, 9

2, 1, 5, 3

LCM = 2 × 2 × 2 × 3 × 2 × 5 × 3 = 720 The largest number of five digits = 99999 On dividing 99999 by 720, the remainder = 639

The largest five-digit number divisible by 720 = 99999 – 639 = 99360

Required number = 99360 + 10 = 99370

**Ans . **

2

**Explanation :**LCM of 5, 10, 12, 15

2 5, 10, 12, 15

3 5, 5, 6, 15

5 5, 5, 2, 5

1, 1, 2, 1

LCM = 2 × 3 × 5 × 2 = 60

Number = 60k + 2 Now, the required number should be divisible by 7. Now, 60k + 2 = 7 × 8k + 4k + 2 If we put k = 3, (4k + 2) is equal to 14 which is exactly divisible by 7.

Required number = 60 × 3 + 2 = 182

**Ans . **

3

**Explanation :**LCM of 9, 10 and 15 = 90 Þ The multiple of 90 are also divisible by 9, 10 or 15.

21 × 90 = 1890 will be divisible by them.

Now, 1897 will be the number that will give remainder 7. 1936 – 1897 Required number = 1936 – 1897 = 39

**Ans . **

1

**Explanation :**The difference between the divisor and the corresponding remainder is same in each case ie. 18 – 5 = 13, 27 – 14 = 13, 36 – 23 = 13

Required number = (LCM of 18, 27, and 36 ) – 13 = 108 – 13 = 95