**Exercise 1.2**

**Q.1: Define HCF of two positive integers and find the HCF of the following pairs of number:**

**(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**

**Â **

**Sol:**

**HCF Definition:Â **HCF is the Highest Common Factor i.e. it is the largest number that divides any two or more numbers exactly (without leaving any remainder).

**(i)**Â Find the H.C.F. of 32 and 54.

Now, apply Euclid’s Division Lemma 54 = 32 x 1 + 22

Since remainder â‰ 0, apply division lemma on 32 and remainder 22

32 = 22 x 1 + 10

Since remainder â‰ 0, apply division lemma on 22 and remainder 10

22 = 10 x 2 + 2

Since remainder â‰ 0, apply division lemma on 10 and 2

10 = 2 x 5 + 0

Therefore, H.C.F. of 32 and 54 is

**(ii)**Â Find the H.C.F. of 18 and 24.

Now, apply Euclid’s Division Lemma

24 = 18 x 1 + 6.

Since remainder â‰ 0, apply division lemma on divisor 18 and remainder 6

18 = 6 x 3 + 0.

Therefore, H.C.F. of 18 and 24 is 6

**(iii)**Â Find the H.C.F. of 70 and 30.

Now apply Euclid’s Division lemma

70 = 30 x 2 + 10.

Since remainder â‰ 0, apply division lemma on divisor 30 and remainder 10

30 = 10 x 3 + 0.

Therefore, H.C.F. of 70 and 30 = 10

**(iv)**Â Find the H.C.F. of 56 and 88.

Applying Euclid’s Division lemma we get,

88 = 56 x 1 + 32.

Since remainder â‰ 0, apply division lemma on 56 and remainder 32

56 = 32 x 1 + 24.

Since remainder â‰ 0, apply division lemma on 32 and remainder 24

32 = 24 x 1+ 8.

Since remainder â‰ 0, apply division lemma on 24 and remainder 8

24 = 8 x 3 + 0. Therefore, H.C.F. of 56 and 88 = 8

**(v)**Â Find the H.C.F. of 475 and 495.

By applying Euclid’s Division lemma we get,

495 = 475 x 1 + 20.

Since remainder â‰ 0, apply division lemma on 475 and remainder 20

475 = 20 x 23 + 15.

Since remainder â‰ 0, apply division lemma on 20 and remainder 15

20 = 15 x 1 + 5.

Since remainder â‰ 0, apply division lemma on 15 and remainder 5

15 = 5 x 3+ 0.

Therefore, H.C.F. of 475 and 495 = 5

**(vi)**Â Find the H.C.F. of 75 and 243.

By applying Euclid’s Division lemma

243 = 75 x 3 + 18.

Since remainder â‰ 0, apply division lemma on 75 and remainder 18

75 = 18 x 4 + 3.

Since remainder â‰ 0, apply division lemma on divisor 18 and remainder 3

18 = 3 x 6+ 0.

Therefore, H.C.F. of 75 and 243 = 3

**(vii)**Â Find the H.C.F. of 240 and 6552.

Applying Euclid’s Division lemma we get,

6552 = 240 x 27 + 72.

Since remainder â‰ 0, apply division lemma on divisor 240 and remainder 72

240 = 72 x 3+ 24.

Since remainder â‰ 0, apply division lemma on divisor 72 and remainder 24

72 = 24 x 3 + 0.

Therefore, H.C.F. of 240 and 6552 = 24

**(viii)**Â Find the H.C.F. of 155 and 1385.

By applying Euclid’s Division lemma

1385 = 155 x 8 + 145.

Since remainder â‰ 0, apply division lemma on divisor 155 and remainder 145.

155 = 145 x 1 + 10.

Since remainder â‰ 0 apply division lemma on divisor 145 and remainder 10

145 = 10 x 14 + 5.

Since remainder â‰ 0, apply division lemma on divisor 10 and remainder 5

10 = 5 x 2 + 0.

Therefore, H.C.F. of 155 and 1385 = 5

**(ix)**Â Find the H.C.F. of 100 and 190.

By applying Euclid’s division lemma

190 = 100 x 1 + 90.

Since remainder â‰ 0, apply division lemma on divisor 100 and remainder 90

100 = 90 x 1 + 10.

Since remainder â‰ 0, apply division lemma on divisor 90 and remainder 10

90 = 10 x 9 + 0.

Therefore, H.C.F. of 100 and 190 = 10

**(x)**Â Find the H.C.F. of 105 and 120.

By applying Euclid’s division lemma

120 = 105 x 1 + 15.

Since remainder â‰ 0, apply division lemma on divisor 105 and remainder 15

105 = 15 x 7 + 0.

Therefore, H.C.F. of 105 and 120 = 15.

**Â **

**Q.2: Use Euclidâ€™s division algorithm to find the HCF of**

**(i) 135 and 225**

**(ii) 196 and 38220**

**(iii) 867 and 255**

**(iv) 184, 230 and 276 [not available]**

**(v) 136, 170 and 255 [not available]**

**Â **

**Sol.**

(i) The integers here are: 225 and 135.

It is seen that 225 > 135.

So, by applying Euclid’s division lemma to 225 and 135, we get,

867 = (225) (3) + 192

Since the remainder â‰ 0. So we apply the division lemma to the divisor 135 and remainder 90.

â‡’ 135 = (90) (1) + 45

Now we apply the division lemma to the new divisor 90 and remainder 45.

â‡’ 90 = (45) (2) + 0

Since the remainder at this stage is 0, the divisor will the HCF.

Hence, the H.C.F of 225 and 135 is 45

**(ii)**Â The integers here are: 38220 and 196. It is seen that 38220 > 196.

So, apply Euclid’s division lemma to 38220 and 196.

38220 = (196) (195) + 0

Since the remainder at this stage is 0, the divisor will the HCF. Since the remainder at this stage is 0, the divisor will the HCF.

Hence, the HCF of 38220 and 196 is 196

**(iii)**Â The integers here are: 867 and 255. It is seen that 867 > 225.

So we will apply Euclid’s division lemma to 867 and 225, we get,

867 = (225) (3) + 192

Since the remainder 192 â‰ 0. So we apply the division lemma to the divisor 225 and remainder 192. We get,

225 = (192) (1) + 33

Now we apply the division lemma to the new divisor 192 and remainder 33. We get,

192 = (33) (5) + 27

Now we apply the division lemma to the new divisor 33 and remainder 27. We get,

33 = (27) (1) + 6

Now we apply the division lemma to the new divisor 27 and remainder 6. We get,

27 = (6) (4) + 3

Now we apply the division lemma to the new divisor 27 and remainder 6. We get,

6 = (3) (2) + 0

Since the remainder at this stage is 0, the divisor will the HCF.

Hence, the HCF of 867 and 255 is 3.

**Q.3: Find the HCF of the following pair of integers and express it as a linear combination of them,**

**(i) 963 and 657**

**(ii) 592 and 252**

**(iii) 506 and 1155**

**(iv) 1288 and 575**

**Sol:**

**(i)**Â Find theÂ H.C.F. of 963 and 657 and express it as a linear combination of 963 and 657.

Now, apply Euclid’s division lemma,

963 = 657 x 1 + 306.

As the remainder â‰ 0, apply division lemma on divisor 657 and remainder 306

657 = 306 x 2 + 45.

As the remainder â‰ 0, apply division lemma on divisor 306 and remainder 45

306 = 45 x 6 + 36.

As the remainder â‰ 0, apply division lemma on divisor 45 and remainder 36

45 = 36 x 1 + 9.

As the remainder â‰ 0, apply division lemma on divisor 36 and remainder 9

36 = 9 x 4 + 0.

So, H.C.F. = 9.

Now, 9 = 45 â€“ 36 x 1

= 45 – [306 â€“ 45 x 6] x 1 = 45 â€“ 306 x 1 + 45 x 6

= 45 x 7 – 306 x 1 = [657 -306 x 2] x 7 – 306 x 1

= 657 x 7 – 306 x 14 – 306 x 1

= 657 x 7 – 306 x 15

= 657 x 7 – [963 â€“ 657 x 1] x 15

= 657 x 7 – 963 x 15 + 657 x 15

=Â __657 x 22 â€“ 963 x 15__.

**(ii)**Â Find the H.C.F. of 592 and 252 and express it as a linear combination of 592 and 252.

By applying Euclid’s division lemma

592 = 252 x 2 + 88

As the remainder â‰ 0, apply division lemma on divisor 252 and remainder 88

252 = 88 x 2 + 76

As theÂ remainder â‰ 0, apply division lemma on divisor 88 and remainder 76

88 = 76 x 1 + 12

As theÂ remainder â‰ 0, apply division lemma on divisor 76 and remainder 12

76 = 12 x 6 + 4

SincAs the remainder â‰ 0, apply division lemma on divisor 12 and remainder 4

12 = 4 x 3 + 0.

So, H.C.F. = 4.

Now, 4 = 76 – 12 x 6

= 76 â€“ 88 – 76 x 1 x 6

= 76 – 88 x 6 + 76 x 6

= 76 x 7 – 88 x 6

= 252 – 88 x 2 x 7 – 88 x 6

= 252 x 7- 88 x 14- 88 x 6

= 252 x 7- 88 x 20

= 252 x 7 â€“ 592 â€“ 252 x 2 x 20

= 252 x 7 – 592 x 20 + 252 x 40

= 252 x 47 â€“ 592 x 20

=Â __252 x 47 + 592 x (-20)__

**(iii)**Â Find theÂ H.C.F. of 506 and 1155 and express it as a linear combination of 506 and 1155. By applying Euclid’s division lemma

1155 = 506 x 2 + 143.

As theremainder â‰ 0, apply division lemma on divisor 506 and remainder 143

506 = 143 x 3 + 77.

As theÂ remainder â‰ 0, apply division lemma on divisor 143 and remainder 77

143 = 77 x 1 + 66.

SincAs the remainder â‰ 0, apply division lemma on divisor 77 and remainder 66

77 = 66 x 1 + 11.

As theÂ remainder â‰ 0, apply division lemma on divisor 66 and remainder 11

66 = 11 x 6 + 0.

So, H.C.F. = 11.

Now, 11 = 77 â€“ 66 x 1 = 77â€”[143 â€“ 77 x 1] x 1

= 77 â€“ 143 x 1 + 77 x 1

= 77 x 2 â€“ 143 x 1

= [506 â€“ 143 x 3] x 2 â€“ 143 x 1

= 506 x 2 â€“ 143 x 6 â€“ 143 x 1

= 506 x 2 â€“ 143 x 7 = 506 x 2 â€“ [1155 â€“ 506 x 2] x 7 = 506 x 2 â€“ 1155 x 7+ 506 x 14

=Â __506 x 16 â€“ 1155 x 7__

**(iv)**Â Find theÂ H.C.F. of 1288 and 575 and express it as a linear combination of 1288 and 575. By applying Euclid’s division lemma

1288 = 575 x 2+ 138.

As theremainder â‰ 0, apply division lemma on divisor 506 and remainder 143

575 = 138 x 4 + 23.

As the remainder â‰ 0, apply division lemma on divisor 143 and remainder 77

138 = 23 x 6 + 0.

So, H.C.F. = 23.

Now, 23 = 575 â€“ 138 x 4 = 575â€”[1288 â€“ 575 x 2] x 4

__= 575 â€“ 1288 x 4 + 575 x 8__

**Â **

**Q.4: Find the largest number which divides 615 and 963 leaving remainder 6 in each case.**

**Sol:**

First, find the largest number which divides 615 and 963 without leaving the remainder 6 in each case.

The required number when divides 615 and 963, leaves remainder 6.

This means 615 â€” 6 = 609 and 963â€” 6 = 957 are exactly divisible by the number.

So, the required number = H.C.F. (609, 957).

By applying Euclid’s division lemma

957 = 609 x 1+ 348

609 = 348 x 1 + 261

348 = 216 x 1 + 87

261 = 87 x 3 + 0.

â‡’Â H.C.F. = 87.

Therefore, the required number is 87

**Q.5: If the HCF of 408 and 1032 is expressible in the form 1032m â€“ 408 x 5, find m.**

**Sol:**

First, find “m” if the H.C.F of 408 and 1032 is expressible in the form 1032 m â€” 408 x 5

The integers here are: 408 and 1032

It is clearly seen that 408 < 1032

By applying Euclid’s division lemma, we get 1032 = 408x 2 + 216.

Here, the remainder â‰ 0.

So apply Euclid’s division lemma on divisor 408 and remainder 216

408 = 216 x 1 + 192.

As the remainder â‰ 0, again apply division lemma on divisor 216 and remainder 192

216 = 192 x 1 + 24.

Again the remainder â‰ 0. So, apply division lemma again on divisor 192 and remainder 24

192 = 24 x 8 + 0.

Now, it is seen that the remainder is 0.

Hence, the last divisor is the H.C.F of 408 and 1032.

So,

24 = 1032m â€” 408 x 5

1032m = 24 + 408 x 5

1032m = 24 + 2040

1032m = 2064

\(m = \frac{ 2064 }{ 1032 }\)

m = 2

**Â **

**Q.6: If the HCF of 657 and 963 is expressible in the form 657x + 963x – 15, find x.**

**Sol:**

Here, find “x” if the H.C.F of 657 and 963 is expressible in the form 657x + 963y (-15).

The integers here are: 657 and 963.

By applying Euclid’s division lemma, we get,

963 = 657 x 1+ 306.

Here, the remainder â‰ 0 and so we apply Euclid’s division lemma on divisor 657 and remainder 306

657 = 306 x 2 + 45.

Now, continue applying division lemma till the remainder=0.

306 = 45 x 6 + 36.

Agian,

45 = 36 x 1 + 9.

Again,

36 = 9 x 4 + 0.

Now, the remainder = 0.

So, H.C.F. = 9.

It is given that H.C.F = 657x + 936 (-15).

Hence, 9 = 657x â€”14445

9 + 14445 = 657x

14454 = 657x

\(x = \frac{ 14454 }{ 657 }\)

Now, solve the above equation,

â‡’ x = 22.

**Â **

**Q.7: An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to in the same number of columns. What is the maximum number of columns in which they can march?**

**Â ****Sol.**

It is given that an army contingent of 616 members is to march behind an army band of 32 members in a parade. Also, the two groups are to march in the same number of columns. We need to find the maximum number of columns in which they can march.

So,

Members in army = 616

Members in band = 32.

Therefore, the maximum number of columns = H.C.F of 616 and 32.

By applying Euclid’s division lemma

616 = 32 x 19 + 8

32 = 8 x 4 + 0.

So, H.C.F. = 8

âˆ´ The maximum number of columns in which they can march is 8

**Q.8: A merchant has 120 liters of oil of one kind, 180 liters of another and 240 liters of the third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?**

**Sol:**

In the question, it is given that the merchant has 3 different oils of 120 liters, 180 liters and 240 liters respectively.

So, the greatest capacity of the tin for filling three different types of oil can be obtained by finding the H.C.F. of 120,180 and 240.

Now, apply Euclid’s division lemma.

180 = (120) (1) + 60

120 = (60) (2) + 0

Since the divisor at the last step is 60, the HCF (120, 180) = 60.

Now, find the H.C.F. (60 and 240)

240 = (60) (4) + 0

Here, since the divisor at the last step is 60, the HCF (240, 60) is 60.

So, the tin should be of 160 liters.

**Â **

**Q.9: During a sale, color pencils were being sold in packs of 24 each and crayons in packs of 32 each. If you want full packs of both and the same number of pencils and crayons, how many of each would you need to buy?**

**Sol:**

To get full packs of both color pencils and crayons and of the same number, we need to find the number of each we need to buy.

It is given,

Number of color pencils in one pack = 24

Number of crayons in pack = 32.

So, the least number of both colors has to be purchased

L.C.M of 24 and 32 = 2 x 2 x 2 x 2 x 2 x 3 = 96

âˆ´ The number of packs of pencils to be bought

\(\frac{ 96 }{ 24 } = 4\),

And,

The number of packs of crayon to be bought

\(\frac{ 96 }{ 32 } = 3\)

**Q.10: 144 cartons of coke cans and 90 cartons of Pepsi cans are to be stacked in a canteen. If each stack is of the same height and is to contain cartons of same drink, what would be the greatest number of cartons each stack would have?**

**Sol:**

It is given,

Number of cartons of coke cans = 144

Number of cartons of Pepsi cans = 90.

So, the greatest number of cartons in one stack will be H.C.F. (144, 90).

Now, by applying Euclid’s division lemma:

144 = 90 x 1 + 54

90 = 54 x 1+ 36

54 = 36 x 1 + 18

36 = 18 x 2 + 0

âˆ´ H.C.F. =18.

Hence, the greatest number cartons in one stack = 18

**Q.11: Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively.**

**Sol:**

The statement “the required number when divides 285 and 1249, leaves remainder 9 and 7” means

285 – 9 = 276 and 1249 -7 = 1242 are exactly divisible by the number.

So, the required number = H.C.F. of 276 and 1242.

Now, apply Euclid’s division lemma,

1242 = 276 x 4 + 138

276 = 138 x 2 + 0.

So, H.C.F. = 138

âˆ´ The required number is 138

**Q.12: Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.**

**Sol:**

For the question statement to be true,

280 – 4 = 276 and 1245 â€“ 3 = 1242 has to be exactly divisible by the number.

So, the required number = H.C.F. of 276 and 1242.

Bow, apply division lemma

1242 = 276 x 4 + 138

276 = 138 x 2 + 0.

So, H.C.F. = 138.

âˆ´ The required number is 138

**Q.13: What is the largest number which that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively?**

**Sol:**

For this question statement to be true,

626 â€“ 1 = 625,

3127 â€“ 2 = 3125,

And 15628 – 3 = 15625 has to be exactly divisible by the number.

So, the required number will be H.C.F. (625, 3125 and 15625).

Now consider 625 and 3125 andÂ apply division lemma

3125 = 625 x 5 + 0.

âˆ´ H.C.F. (625, 3125) = 625

Now, consider 625 and 15625 andÂ apply division lemma

15625 = 625 x 25 + O.

âˆ´ H.C.F. (625, 3125, 15625) = 625

So, the required number is 625

**Q.14: Find the greatest number that will divide 445,572 and 699 leaving remainders 4,5 and 6 respectively.**

**Sol:**

For this question,

445 – 4 = 441, 572 – 5 = 567 and 699 – 6 = 693 has to be exactly divisible by the number.

So, the required number = H.C.F. of 441, 567 and 693.

Now,Â consider 441 and 567 andÂ apply division lemma

567 = 441 x 1 + 126

441 = 126 x 3 + 63

126 = 63 x 2 + 0.

âˆ´Â H.C.F. of 441 and 567 = 63

Now, consider 63 and 693 andÂ apply Euclid’s division lemma

693 = 63 x 11 + 0.

âˆ´ H.C.F. of 441, 567 and 693 = 63

So, the required number is 63

**Q.15: Find the greatest number which divides 2011 and 2623 leaving remainders 9 and 5 respectively.**

**Sol:**

For this,

2011 – 9 = 2002 and 2623 â€“ 5 = 2618 has to be exactly divisible by the number.

Now, the required number will be the H.C.F. of 2002 and 2618

Applying Euclid’s division lemma

2618 = 2002 x 1 + 616

2002 = 616 x 3 + 154

616 = 154 x 4+ 0.

So, the H.C.F. of 2002 and 2618 = 154

âˆ´ The required number is 154

**Â **

**Q.16: Two brands of chocolates are available in packs of 24 and 15 respectively. If I need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind I would need to buy?**

**Sol:**

In the question, it is given that,

Number of chocolates of 1st brand in one pack = 24

Number of chocolates of 2nd brand in one pack = 15.

So, the least number of chocolates he need to purchase is

L.C.M. of 24 and 15 = 2 x 2 x 2 x 3 x 5 = 120

Hence, the number of packet of 1^{st}Â brand is

\(\frac{ 120 }{ 24 }\)Â = 5,

And the number of packet of 2^{nd}Â brand is

\(\frac{ 120 }{ 15 } = 8\)

**Q.17: A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10ft by 8ft. what would be the size in inches of the tile required that has to be cut and how many such tiles are required?**

**Sol:**

In the question it is given that.

Size of bathroom = 10 ft by 8 ft

= (10 x 12) inch by (8 x 12) inch

= 120 inch by 96 inch

Now, the largest size of tile required will be the HCF (120 and 96)

Apply Euclid’s division lemma

120 = 96 x 1 + 24

96 = 24 x 4 + 0

â‡’Â HCF = 24

â‡’ The Largest size of tile that is required = 24 inches and,

\(no. of tiles required = \frac{ area of bathroom }{ area of 2 tile } = \frac{ 120 x 96 }{ 24 x 24 } = 5 x 4 = 20 tiles\)

**Q.18: 15 pastries and 12 biscuit packets have been donated for a school fete. These are to be packed in several smaller identical boxes with the same number of pastries and biscuits packets in each. How many biscuit packets and how many pastries will each box contain?**

**Sol:**

Given:

Number of pastries = 15

Number of biscuit packets = 12

So, the required no of boxes to contain equal number will be tha HCF of 15 and 13

Now, apply Euclid’s division lemma

15 = 12 x 13

12 = 2 x 9 = 0

So, No. of boxes required = 3

âˆ´ Each box will containÂ \(\frac{ 15 }{ 3 } = 5\)Â pastries andÂ \(\frac{ 2 }{ 3 }\)Â biscuit packs.

**Q.19: 105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals on every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went on each trip?**

**Sol:**

Given:

Number of goats = 205

Number of donkey = 140

Number of cows = 175

So, the largest number of animals in one trip will be the HCF (105. 140 and 175).

Now, first consider the numbers 105 and 140 andÂ apply division lemma

140 = 105 x 1 + 35

105 = 35 x 3 + 0 4

So, HCF (105 and 140) = 35

Now consider 35 and 175 andÂ apply Euclid’s division lemma

175 = 35 x 5 +0

So, HCF (105, 140, 175) 35.

**Q.20: The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm and 4 m 50 cm, respectively. Determine the longest rod which can measure the three dimensions of the room exactly.**

**Sol:**

Length of room = 8m 25 cm = 825 cm

Breadth of room = 6m 75m = 675 cm

Height of room = 4m 50m = 450 cm

The required longest rod = HCF of 825, 675 and 450

First consider 675 and 450

By applying Euclid’s division lemma

675 = 450 x 1 + 225

450 = 225 x 2 + 0

Therefore, HCF of 675 and 450 = 825

Now consider 625 and 825

By applying Euclid’s division Lemma:

825 = 225 x 3 + 150

225 = 150 x 1+75

150 = 75 x 2 + 0

Therefore, HCF of 825, 675 and 450 = 75

**Â **

**Q.21: Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two different ways.**

**Sol:**

We need to express the H.C.F. of 468 and 222 as 468x + 222y

Where x, y are integers in two different ways.

The integers here are: 468 and 222, where 468 > 222

By applying Euclid’s division lemma, we get 468 = 222 x 2 + 24.

Since the remainder â‰ 0, so apply division lemma on divisor 222 and remainder 24

222 = 24 x 9 + 6.

Since the remainder â‰ 0, so apply division lemma on divisor 24 and remainder 6

24 = 6 x 4 + 0.

We observe that remainder is 0. So the last divisor 6 is the H.C.F. of 468 and 222 from we have

6 = 222 â€“ 24 x 9

6 = 222 – (468 â€“ 222 x 2) x 9

6 = 222 â€“ 468 x 9 + 222 x 18

6 = 222 x 19 â€“ 468 x 9 [Substituting 24 = 468 â€” 222 x 2]

16 = 222y + 468x, where x = – 9 and y = 19.