**Ans . **

1

**Explanation :**

2 28, 42

2 14, 21

7 7, 21

1, 3

= 2 × 2 × 7 × 3 = 84 H.C. F. of 28 and 42

28 ) 42 ( 1

28

14 ) 28 ( 2

28

0

H.C. F = 14

Required ratio = 84 = 6:1

14

**Ans . **

3

**Explanation :**Let the two numbers are 2x and 3x respectively. According to question, LCM = 54 x ( ) 3 2 54 ´ = Þ x = 9 Numbers = 2x = 2 × 9 = 18 and, 3x = 3 × 9 = 27 Sum of the two numbers = 18 + 27 = 45

**Ans . **

3

**Explanation :**Suppose the numbers are 4x and 5x respectively According to question x × 4 × 5 = 120 Þ x = 6 Required numbers = 4 × 6 = 24 and = 5 × 6 = 30

**Ans . **

1

**Explanation :**Let the numbers be 2x, 3x and 4x respectively. HCF = x = 12 Numbers are : 2 ×12 = 24 3 ×12 = 36, 4 ×12 = 48 LCM of 24, 36, 48 = 2 × 2 × 2 × 3 × 3 × 2 = 144

**Ans . **

3

**Explanation :**Let the number be 3x and 4x. Their LCM = 12x According to the question, 12x = 240

x = 240 = 20

12

Smaller number = 3x = 3 × 20 = 60

**Ans . **

3

**Explanation :**Let the numbers be 3x and 4x. Their LCM = 12x 12x = 84

x = 84 = 7

12

Larger number = 4x = 4 × 7 = 28

**Ans . **

1

**Explanation :**Numbers = 3x and 4x HCF = x = 4

LCM = 12x = 12 × 4 = 48

**Ans . **

3

**Explanation :**Let the numbers be 4x and 4y where x and y are prime to each other. LCM = 4xy

(4x + 4y) = 7

4xy 12

12 (x + y) = 7 xy

x = 3, y = 4

Smaller number = 4 × 3 = 12

**Ans . **

4

**Explanation :**Using Rule 1, Let the numbers be 3x and 4x respectively First number × second number

= HCF × LCM

3x × 4x = 2028

x^{2}= 2028 = 169

3 x 4

x = √169 = 13 Sum of the numbers = 3x + 4x = 7x = 7 × 13 = 91

**Ans . **

3

**Explanation :**If the numbers be 2x and 3x, then LCM = 6x 6x = 48

x = 8 Required sum = 2x + 3x = 5x = 5 × 8 = 40

**Ans . **

4

**Explanation :**Let the numbers be 4x and 5x. H.C.F. = x = 8 Numbers = 32 and 40 Their LCM = 160

**Ans . **

2

**Explanation :**If the numbers be 3x and 4x, then HCF = x = 5 Numbers = 15 and 20 LCM = 60

**Ans . **

1

**Explanation :**Numbers = x , 2 x and 3 x (let) Their H.C.F. = x = 12 Numbers = 12, 24 and 36

**Ans . **

4

**Explanation :**Using Rule 1, Product of two numbers = HCF × LCM

Numbers = zx and zy

zx × zy = z × LCM

LCM = xyz

**Ans . **

4

**Explanation :**HCF of numbers = 21 Numbers = 21x and 21y Where x and y are prime to each other. Ratio of numbers = 1 : 4

Larger number = 21 × 4 = 84

**Ans . **

4

**Explanation :**Using Rule 1, Let the numbers be x and (x + 2). Product of numbers = HCF × LCM x (x + 2) = 24

x^{2}+ 2x – 24 = 0

x^{2}+ 6x – 4x – 24 = 0

x (x + 6) – 4 (x + 6) = 0

(x – 4) (x + 6) = 0

x = 4, as x ¹ – 6 = 0

Numbers are 4 and 6.

**Ans . **

1

**Explanation :**Using Rule 1, Suppose 1st number is x then, 2nd number = 100 – x

LCM × HCF = 1st number × 2nd number

495 × 5 = x × (100 – x)

495 × 5 = 100x – x^{2}

x^{2}– 55x – 45x – 2475 = 0

(x – 45) (x – 55) = 0

x = 45 or x = 55

Then, difference = 55 – 45 = 10

**Ans . **

2

**Explanation :**Let the number be 29x and 29y respectively where x and y are prime to each other.

LCM of 29x and 29y = 29xy

Now, 29xy = 4147

xy = 4147 = 143

29

Thus xy = 11 × 13

Numbers are 29 × 11

= 319 and 29 × 13 = 377

Required sum

= 377 + 319 = 696

**Ans . **

4

**Explanation :**Let HCF be h and LCM be l. Then, l = 84h and l + h = 680 84h + h = 680 h = 680 = 8

85 l = 680 – 8 = 672

Other number

= 672 x 8

56

= 96

**Ans . **

2

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

12x + 12y = 84

12 (x + y) = 84

x + y = 84 = 7

12

Possible pairs of numbers satisfying this condition = (1,6), (2,5) and (3,4). Hence three pairs are of required numbers

**Ans . **

3

**Explanation :**Let the numbers be 21x and 21y where x and y are prime to each other. 21x + 21y = 336

21 (x + y) = 336

x + y = 336 =16 21 Possible pairs = (1, 15), (5, 11), (7, 9), (3, 13)

**Ans . **

3

**Explanation :**Let the number be x and y. According to the question, x + y = 45 ......... (i)

Again, x – y = 1( x + y)

9

or x – y = 145

9 or x – y = 5 ..... (ii)

By (i) + (ii) we have,

x + y = 45

x – y = 5

= 50 2x

or, x = 25

y = 45 – 25 = 20.

Now, LCM of 25 and 20 = 100

**Ans . **

3

**Explanation :**Let the numbers be 17x and 17y where x and y are co-prime.

LCM of 17x and 17y = 17 xy

According to the question,

17xy = 714

xy = 714 = 42 = 6 x 7

17

x = 6 and y = 7

or, x = 7 and y =6.

First number = 17x

= 17 × 6 = 102

Second number = 17y

=17 × 7 = 119

Sum of the numbers

= 102 + 119 = 221

**Ans . **

3

**Explanation :**Using Rule 1, Let the larger number be x. Smaller number = x – 2

First number × Second number = HCF × LCM x (x – 2) = 24

x^{2}– 2x – 24 = 0

x^{2}– 6x + 4x – 24 = 0

x (x – 6) + 4 (x – 6) = 0

(x – 6) (x + 4) = 0

x = 6 because x not equal to 4

**Ans . **

2

**Explanation :**HCF of two numbers = 27 Let the numbers be 27x and 27y

where x and y are prime to each other.

According to the question,

27x + 27y = 216

27 (x + y) = 216

x + y = 216 = 8

27

Possible pairs of x and y = (1, 7) and (3, 5)

Numbers =(27, 189) and (81, 135)

**Ans . **

1

**Explanation :**Using Rule 1, Let the HCF of numbers = H

Their LCM = 12H

According to the question,

12H +H = 403

13H = 403

H =403 = 31

13

LCM = 12 × 31

Now,

First number × second number

= HCF × LCM

= 93 × Second Number

= 31 × 31 × 12

Second number =

31 x 31 x 12

93

= 124

**Ans . **

3

**Explanation :**Let the numbers be 48x and 48y where x and y are co-primes.

48x + 48y = 384

48 ( x + y) = 384

x + y =384 =8

48

Possible and acceptable pairs of x and y satisfying this condition are : (1, 7) and (3, 5).

Numbers are : 48 × 1 = 48

and 48 × 7 = 336

and 48 × 3 = 144 and 48 × 5 = 240

Required difference

= 336 – 48 = 288

**Ans . **

2

**Explanation :**Let the numbers be 3x and 3y.

3x + 3y = 36

x + y = 12 ... (i)

and 3xy = 105 ... (ii)

Dividing equation (i) by (ii), we have

x + y =12

3xy 3xy 105

1 + 1 =4

3xy 3xy 35

**Ans . **

4

**Explanation :**Let the numbers be 10x and 10y where x and y are prime to each other.

LCM = 10 xy

10xy = 120

xy = 12

Possible pairs = (3, 4) or (1, 12)

Sum of the numbers

= 30 + 40 = 70

**Ans . **

3

**Explanation :**Let the numbers be x, y and z which are prime to one another. Now, xy = 551

yz = 1073

y = HCF of 551 and 1073

y = 29

x = 551 =19

29

and

z = 1073 =37

29

Sum = 19 + 29 + 37 = 85

**Ans . **

3

**Explanation :**HCF of two numbers = 4.

Hence, the numbers can be given by 4x and 4y where x and y are co-prime. Then,

4x + 4y = 36

4 (x + y) = 36

x + y = 9

Possible pairs satisfying this condition are : (1, 8), (4, 5), (2, 7)

**Ans . **

2

**Explanation :**Let the numbers be 2x and 2y where x and y are prime to each other.

LCM = 2xy

2xy = 84

xy = 42 = 6 × 7

Numbers are 12 and 14.

Hence Sum = 12 + 14 = 26

**Ans . **

3

**Explanation :**Let the numbers be x H and yH where H is the HCF and yH x H.

LCM = xy H

xyH = 2yH

x = 2 Again, x H – H = 4

2H – H = 4

H = 4

Smaller number = x H = 8

**Ans . **

4

**Explanation :**Using Rule 1, Let the H.C.F. be H.

L.C.M. = 20H

Then, H + 20H = 2520

21 H = 2520

H = 2520 = 120

21

L.C.M. = 20H = 20×120= 2400

As,

First number × Second number

= L.C.M. × H.C.F.

480 × Second number = 2400 × 120

Second number = 2400 x 120 = 600

480

**Ans . **

1

**Explanation :**Using Rule 1, If the HCF = H, then

LCM = 44 H

44 H + H = 1125

45 H = 1125

H = 1125 = 25

45

LCM = 44 × 25 = 1100

Now

First number × Second number

= LCM × HCF

25 × Second number

= 1100 × 25

Second number

= 1100 x 25 = 1100

25

**Ans . **

4

**Explanation :**Let no. are x and y and HCF = A, LCM = B

Using Rule, we have

xy = AB

x + y = A + B (given) ...(i)

(x–y)^{2}= (x + y)^{2}– 4xy

or, (x–y)^{2}= (A + B)^{2}– 4 AB

or, (x–y)^{2}= (A – B)2

or, (x–y) = A – B ...(ii)

Using (i) and (ii), we get

x = A and y = B

A^{3}+ B^{3}= x^{3}+ y^{3}

**Ans . **

3

**Explanation :**Let the numbers be 7x and 7y where x and y are co-prime.

Now, LCM of 7x and 7y = 7xy

7xy = 140

xy = 140 = 20

7

Now, required values of x and y whose product is 50 and are coprime, will be 4 and 5.

Numbers are 28 and 35 which lie between 20 and 45.

Required sum = 28 + 35 = 63

**Ans . **

4

**Explanation :**

Firstly, we find the LCM of 30, 36 and 80.

2 30, 36, 80

2 15, 18, 40

3 15, 9, 20

5 5, 3, 20

1, 3, 4

LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720

Required number = Multiple of 720 = 720 × 5 = 3600; because 3000 < 3600 < 4000

**Ans . **

3

**Explanation :**LCM of 5, 6, 7 and 8 = 840

2 5, 6, 7, 8

5, 3, 7, 4

LCM = 2 × 5 × 3 × 7 × 4 = 840

Required number = 840x + 3

which is divisible by 9 for a certain least value of x.

Now,

840x + 3 = 93x × 9 + 3x + 3

3x + 3, is divisible by 9 for x = 2

Required number = 840 × 2 + 3

= 1680 + 3 = 1683

Sum of digits = 1 + 6 + 8 + 3

= 18

**Ans . **

1

**Explanation :**

2 12, 18, 21, 28

2 6, 9, 21, 14

3 3, 9, 21, 7

7 1, 3, 7, 7

1, 3, 1, 1

LCM = 2 × 2 × 3 × 3 × 7= 252 The largest 4-digit number = 9999

252 ) 9999 ( 39

756

2439

2268

171

Required number = 9999 – 171 = 9828

**Ans . **

4

**Explanation :**LCM of 8, 12 and 16 = 48 Required number = 48a + 3 which is divisible by 7.

x = 48a + 3

= (7 × 6a) + (6a + 3) which is divisible by 7.

i.e. 6a + 3 is divisible by 7.

When a = 3, 6a + 3 = 18 + 3

= 21 which is divisible by 7.

x = 48 × 3 + 3 = 144 + 3 = 147

**Ans . **

1

**Explanation :**

2 12, 16, 18, 21

2 6, 8, 9, 21

3 3, 4, 9, 21

1, 4, 3, 7

LCM = 2 × 2 × 3 × 4 × 3 × 7 = 1008

Multiple of 1008 = 2016

Required number = 2016 – 2000 = 16 = x

Sum of digits of x = 1 + 6 = 7

**Ans . **

3

**Explanation :**

2 12, 18, 21

3 6, 9, 21

2, 3, 7

LCM of 12, 18 and 21

= 2 × 3 × 2 × 3 × 7 = 252

Of the options, 10080 ÷ 252 = 40

**Ans . **

1

**Explanation :**2 30, 36, 80

2 15, 18, 40

3 15, 9, 20

5 5, 3, 20

1, 3, 4

LCM = 2 × 2 × 3 × 3 × 4 × 5 = 720 Required number = 2 × 720 + 11 = 1440 + 11 = 1451

**Ans . **

2

**Explanation :**2 12, 18, 21, 32

2 6, 9, 21, 16

3 3, 9, 21, 8

1, 3, 7, 8

LCM = 2 × 2 × 3 × 3 × 7 × 8 = 2016 Required number = 2016 × 2 = 4032

**Ans . **

4

**Explanation :**2 210

3 105

5 35

7

210 = 2 × 3 × 5 × 7 = 5 × 6 × 7 Required answer = 5 + 6 = 11

**Ans . **

4

**Explanation :**Let the numbers be 12x and 12y.

Their LCM = 12xy when x and y are prime to each other.

y = 1056 = 8

132 Other number = 12y

= 12 × 8 = 96

**Ans . **

2

**Explanation :**When 36798 is divided by 78, remainder = 60

The least number to be subtracted = 60

**Ans . **

1

**Explanation :**LCM of 18, 21 and 24

2 18, 21, 24

3 9, 21, 12

3, 7, 4

LCM = 2 × 3 × 3 × 7 × 4 = 504 Now compare the divisors with their respective remainders. We observe that in all the cases the remainder is just 11 less than their respective divisor. So the number can be given by 504 K – 11. Where K is a positive integer Since 23 × 21 = 483 We can write 504 K – 11 = (483 + 21) K – 11 = 483 K + (21K – 11) 483 K is multiple of 23, since 483 is divisible by 23.

So, for (504K – 11) to be multiple of 23, the remainder (21K – 11) must be divisible by 23.

Put the value of K = 1, 2, 3, 4, 5, 6, ..... and so on successively.

We find that the minimum value of K for which (21K – 11) is divisible by 23. is 6, (21 × 6 – 11) = 115 which is divisible by 23.

Therefore, the required least number = 504 × 6 – 11 = 3013

**Ans . **

4

**Explanation :**Using Rule 7, Clearly, 122 – 2 = 120 and 243 – 3 = 240 are exactly divisible by the required number.

Required number = HCF of 120 and 240 = 120

**Ans . **

2

**Explanation :**P = 2

^{3}× 3^{10}× 5 Q = 2^{5}× 3 × 7 HCF = 2^{3}× 3

**Ans . **

4

**Explanation :**Let the original fraction be x.

y

x+2 = 1

y+1 6

6x – 24 = y + 1

6x – y = 25 .......(i) Again, x+2 = 1

y+1 3

3x + 6 = y + 1

3x – y = –5 .......(ii) By equation (i) – (ii),

6x – y – 3x + y = 25 + 5

= 3x = 30

x = 10 From equation (i), 60 – y = 25

= y = 35 LCM of 10 and 35 = 70

**Ans . **

4

**Explanation :**HCF of a and b = 12

Numbers = 12x and 12y where x and y are prime to each other. Q a > b > 12

a = 36; b = 24

**Ans . **

4

**Explanation :**Let the numbers be 9x and 9y where x and y are prime to each other. According to the question, 9x + 9y = 99 Þ 9(x + y) = 99

x + y = 11 Possible pairs = (1, 10) (2, 9), (3, 8), (4, 7), (5, 6