RD Chapter 1- Real Numbers Ex-1.1 |
RD Chapter 1- Real Numbers Ex-1.2 |
RD Chapter 1- Real Numbers Ex-1.3 |
RD Chapter 1- Real Numbers Ex-1.5 |
RD Chapter 1- Real Numbers Ex-1.6 |
RD Chapter 1- Real Numbers Ex-VSAQS |
RD Chapter 1- Real Numbers Ex-MCQS |

Find the L.C.M. and H.C.F. of the following pairs of integers and verify that L.C.M. x H.C.F. = Product of the integers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

**Answer
1** :

(i) 26 and 91

26 = 2 x 13

91 = 7 x 13

H.C.F. = 13

and L.C.M. = 2 x 7 x 13 = 182

Now, L.C.M. x H.C.F. = 182 x 13 = 2366

and 26 x 91 = 2366

L.C.M. x H.C.F. = Product of integers

Find the L.C.M. and H.C.F. of the following integers by applying the prime factorisation method :

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

(iv) 40, 36 and 126

(v) 84, 90 and 120

(vi) 24,15 and 36

**Answer
2** :

**Answer
3** :
HCF of 306, 657 = 9

**Answer
4** :

H.C.F. of two numbers = 16

and their L.C.M. = 380

We know the H.C.F. of two numbers is a factor of their L.C.M. but 16 is not a factor of 380 or 380 is not divisible by 16

It can not be possible.

**Answer
5** :

First number = 725

Let second number = x

Their H.C.F. = 145

and L.C.M. = 2175

Second number = 435

**Answer
6** :

H.C.F. of two numbers = 16

and product of two numbers = 3072

**Answer
7** :

First number = 30

Let x be the second number

Their L.C.M. = 180 and H.C.F. = 6

We know that

first number x second number = L.C.M. x H.C.F.

30 x x = 180 x 6

⇒ x = = 36

Second number = 36

**Answer
8** :

L.C.M. of 520 and 468

= 2 x 2 x 9 x 10 x 13 = 4680

The number which is increased = 17

Required number 4680 – 17 = 4663

**Answer
9** :

Dividing by 28 and 32, the remainders are 8 and 12 respectively

28 – 8 = 20

32 – 12 = 20

Common difference = 20

Now, L.C.M. of 28 and 32

= 2 x 2 x 7 x 8 = 224

Required smallest number = 224 – 20 = 204

**Answer
10** :

L.C.M. of 35, 56, 91

= 5 x 7 x 8 x 13 = 3640

Remainder in each case = 7

The required smallest number = 3640 + 7 = 3647

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