RD Chapter 1- Real Numbers Ex-1.1 |
RD Chapter 1- Real Numbers Ex-1.3 |
RD Chapter 1- Real Numbers Ex-1.4 |
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 |

Define H.C.F. of two positive integers and find the H.C.F. of the following pairs of numbers.

(i) 32 and 54

(ii) 18 and 24

(iii) 70 and 30

(iv) 56 and 88

(v) 475 and 495

(vi) 75 and 243

(vii) 240 and 6552

(viii) 155 and 1385

(ix) 100 and 190

(x) 105 and 120

**Answer
1** :
Definition : The greatest among the common divisor oftwo or more integers is the Greatest Common Divisor (G.C.D.) or Highest CommonFactor (H.C.F.) of the given integers.

(i) HC.F. of 32 and 54

Factors 32 = 1, 2, 4, 8, 16, 32

and factors of 54 = 1, 2, 3, 6, 9, 18, 27, 54

H.C.F. = 2

(ii) H.C.F. of 18 and 24

Factors of 18 = 1, 2, 3, 6, 9, 18

and factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Highest common factor = 6

H.C.F. = 6

Factors of 18 = 1, 2, 3, 6, 9, 18

and factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Highest common factor = 6

H.C.F. = 6

(iii) H.C.F. of 70 and 30

Factors of 70 = 1, 2, 5, 7, 10, 14, 35, 70

and factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30

H.C.F. = 10

Factors of 70 = 1, 2, 5, 7, 10, 14, 35, 70

and factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30

H.C.F. = 10

(iv) H.C.F. of 56 and 88

Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56

and factors of 88 = 1, 2, 4, 8, 11, 22, 44, 88

H.C.F. = 8

Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56

and factors of 88 = 1, 2, 4, 8, 11, 22, 44, 88

H.C.F. = 8

(v) H.C.F. of 475 and 495

Factors of 475 = 1, 5, 25, 19, 95, 475

and factors of 495 = 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495

H.C.F. = 5

Factors of 475 = 1, 5, 25, 19, 95, 475

and factors of 495 = 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495

H.C.F. = 5

(vi) H.C.F. of 75 and 243

Factors of 75 = 1, 3, 5, 15, 25, 75

Factors of 243 = 1, 3, 9, 27, 81, 243

H.C.F. = 3

Factors of 75 = 1, 3, 5, 15, 25, 75

Factors of 243 = 1, 3, 9, 27, 81, 243

H.C.F. = 3

(vii) H.C.F. of 240 and 6552

Factors of 240 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 48, 60, 80, 120,240

Factors of 6552 = 1, 2, 3, 4, 6, 7, 8, 9, 12, 13, 14, 18, 21, 24, 26, 28, 36,39, 42, 52, 56, 63, 72, 91, 104, 117, 126, 156, 168, 182, 234,252, 273, 312,364, 488, 504, 546, 728, 819, 936, 1092, 1638, 2184, 3276, 6552

H.C.F. = 24

Factors of 240 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 48, 60, 80, 120,240

Factors of 6552 = 1, 2, 3, 4, 6, 7, 8, 9, 12, 13, 14, 18, 21, 24, 26, 28, 36,39, 42, 52, 56, 63, 72, 91, 104, 117, 126, 156, 168, 182, 234,252, 273, 312,364, 488, 504, 546, 728, 819, 936, 1092, 1638, 2184, 3276, 6552

H.C.F. = 24

(viii) H.C.F. of 155 and 1385

Factors of 155 = 1, 5, 31, 155

Factors of 1385 = 1, 5, 277, 1385

H.C.F. = 5

Factors of 155 = 1, 5, 31, 155

Factors of 1385 = 1, 5, 277, 1385

H.C.F. = 5

(ix) 100 and 190

Use Euclid’s division algorithm to find the H.C.F. of

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

(iv) 184, 230 and 276

(v) 136,170 and 255

**Answer
2** :

(i) H.C.F. of 135 and 225

135 < 225

225 = 135 x 1 + 90

135 = 90 x 1 +45

45 = 45 x 2 + 0

Last remainder = 0

and last divisor = 45

H.C.F. = 45

(ii) H.C.F. of 196 and 38220

196 < 38220

38220 = 196 x 195 + 0

Last remainder = 0

and last divisor = 196

H.C.F. = 196

(iii) H.C.F. 867 and 255

255 < 867

867 = 255 x 3 + 102

255 = 102 x 2 + 51

102 = 51 x 2 + 0

Last remainder = 0

and last divisor = 51

H.C.F. = 51

(iv) H.C.F. of 184, 230 and 276

Let us find the highest common factor (H.C.F.) of 184 and 230

Hence, H.C.F. of 184 and 230 = 46

Now, find the H.C.F. of 276 and 46

Hence, H.C.F. of 276 and 46 = 46

Required H.C.F. of 184, 230 and 276 = 46

(v) H.C.F. of 136, 170 and 255

Let us find the highest common factor (H.C.F.) of 136 and 70

Hence, H.C.F. of 136 and 170 = 34

Now, find the H.C.F. of 34 and 255

Hence, highest common factor of 34 and 255 = 17

Required H.C.F. of 136, 170 and 255 = 17

Find the H.C.F. of the following pairs of integers and express it as a linear combinations of them.

(i) 963 and 657

(ii) 592 and 252

(iii) 506 and 1155

(iv) 1288 and 575

**Answer
3** :

(i) 963 and 657

(ii) HCF of 592 and 252

76x + 88y

Where x = 7, y = -6

(iii) 506 and 1155

H.C.F. = 11

(iv) 1288 and 575

H.C.F. = 23

= 575 x 9 + 1288 x (-4)

= ax + by

x = 9, y = -4

**Answer
4** :

The given numbers are 615 and 963

Remainder in each case = 6

615 – 6 = 609 and 963 – 6 = 957 are divisible by the required number which is the H.C.F. of 609 and 957 = 87

Hence the required largest number = 87

**Answer
5** :

408, 1032

H.C.F. = 24

Which is in the form of 1032m – 408 x 5 comparing, we get m = 2

**Answer
6** :

657 and 963

H.C.F. = 9

**Answer
7** :

The required number of columns will be the H.C.F. of 616 and 32

Using Euclid’s division

We get H.C.F. = 4

Number of columns = 4

**Answer
8** :

Quantity of oil of one kind =120l

and quantity of second kind = 180l

and third kind of oil = 240l

Maximum capacity of oil in each tin = H.C.F. of 120l, 180l and 240l

H.C.F. of 120 and 180 = 60

**Answer
9** :

Number of pencils in each pack = 24

and number of crayons pack = 32

Highest number of pencils and crayons in packs will be = H.C.F. of 24 and 32 = 8

Number of pencil’s pack = 24/8 = 3

and number of crayon’s pack = 32/8 = 4

**Answer
10** :

Number of Coke Cans Cartons = 144

and number Pepsi Cartons = 90

Required greatest number of cartons of each = H.C.F. of 144 and 90 = 18

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