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# Class 10 RD Sharma Solutions- Chapter 1 Real Numbers – Exercise 1.2 | Set 1

• Last Updated : 13 Jan, 2021

### (i) 32 and 54

Solution:

Applying Euclid’s Division Lemma on 54 , we get,

54 = 32 x 1 + 22

Applying Euclid’s Division Lemma on 32 and remainder 22 , we get,

32 = 22 x 1 + 10

Now, remainder ≠ 0, apply division lemma on 22 and remainder 10, we get,

22 = 10 x 2 + 2

Applying Euclid’s Division Lemma on 10 and remainder 2 , we get,

10 = 2 x 5 + 0

Therefore, the H.C.F. of 32 and 54 is 2.

### (ii) 18 and 24

Solution:

Applying Euclid’s Division Lemma on 24 and 18, we get,

24 = 18 x 1 + 6.

Applying Euclid’s Division Lemma on 18 and remainder 6., we get,

18 = 6 x 3 + 0.

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

### (iii) 70 and 30

Solution:

Applying Euclid’s Division Lemma on 70 and 30, we get,

70 = 30 x 2 + 10.

Applying Euclid’s Division Lemma on 30 and remainder 10, we get,

30 = 10 x 3 + 0.

Since, remainder is 0 now,

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

### (iv) 56 and 88

Solution:

Applying Euclid’s Division Lemma on 56 and remainder 88, we get,

88 = 56 x 1 + 32.

Applying Euclid’s Division Lemma on 56 and remainder 32, we get,

56 = 32 x 1 + 24.

Applying Euclid’s Division Lemma on 32 and remainder 24, we get,

32 = 24 x 1+ 8.

Applying Euclid’s Division Lemma on 24 and remainder 8, we get,

24 = 8 x 3 + 0.

Since, the remainder is now 0,

Therefore, H.C.F. of 56 and 88 is 8

### (v) 475 and 495

Solution:

Applying Euclid’s Division Lemma on 475 and 495, we get,

495 = 475 x 1 + 20.

Applying Euclid’s Division Lemma on 475 and remainder 20, we get,

475 = 20 x 23 + 15.

Applying Euclid’s Division Lemma on 20 and remainder 15, we get,

20 = 15 x 1 + 5.

Applying Euclid’s Division Lemma on 15 and remainder 5, we get,

15 = 5 x 3+ 0.

Since, the remainder is now 0,

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

### (vi) 75 and 243

Solution:

Applying Euclid’s Division Lemma on 243 and 75, we get,

243 = 75 x 3 + 18.

Applying Euclid’s Division Lemma on 75 and remainder 18, we get,

75 = 18 x 4 + 3.

Applying Euclid’s Division Lemma on 18 and remainder 3, we get,

18 = 3 x 6+ 0.

Since, the remainder is now 0,

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

### (vii) 240 and 6552

Solution:

Applying Euclid’s Division Lemma on 6552 and 240, we get,

6552 = 240 x 27 + 72.

Applying Euclid’s Division Lemma on 240 and remainder 72, we get,

240 = 72 x 3+ 24.

Applying Euclid’s Division Lemma on 72 and remainder 24, we get,

72 = 24 x 3 + 0.

Since, the remainder is now 0,

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

### (viii) 155 and 1385

Solution:

Applying Euclid’s Division Lemma on 1385 and 155, we get,

1385 = 155 x 8 + 145.

Applying Euclid’s Division Lemma on 155 and remainder 145, we get,

155 = 145 x 1 + 10.

Applying Euclid’s Division Lemma on 145 and remainder 10, we get,

145 = 10 x 14 + 5.

Applying Euclid’s Division Lemma on 10 and remainder 5, we get,

10 = 5 x 2 + 0.

Since, the remainder is now 0,

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

### (ix) 100 and 190

Solution:

Applying Euclid’s Division Lemma on 190 and 100, we get,

190 = 100 x 1 + 90.

Applying Euclid’s Division Lemma on 100 and remainder 90, we get,

100 = 90 x 1 + 10.

Applying Euclid’s Division Lemma on 90 and remainder 10, we get,

90 = 10 x 9 + 0.

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

### (x) 105 and 120

Solution:

Applying Euclid’s Division Lemma on 120 and 105, we get,

120 = 105 x 1 + 15.

Applying Euclid’s Division Lemma on 105 and remainder 15, we get,

105 = 15 x 7 + 0.

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

### (i) 135 and 225

Solution:

On comparing both the integers, we find 225 > 135.

Applying Euclid’s Division Lemma on 225 and 135, we get,

225 = 135 x 1 + 90

Applying Euclid’s Division Lemma on 135 and remainder 90, we get,

⇒ 135 = 90 x 1 + 45

Applying Euclid’s Division Lemma on 90 and remainder 45, we get,

⇒ 90 = 45 x 2 + 0

Since, the remainder is now 0,

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

### (ii) 196 and 38220

Solution:

On comparing both the integers, we find 38220 > 196.

Applying Euclid’s Division Lemma on 38220 and 196, we get,

38220 = 196 x 195 + 0

Since, the remainder is now 0,

Hence, the HCF of 38220 and 196 is 196

### (iii) 867 and 255

Solution:

On comparing both the integers, we find 867 > 255.

Applying Euclid’s Division Lemma on 867 and 255 , we get,

867 = 225 x 3 + 192

Applying Euclid’s Division Lemma on 225 and remainder 192 , we get,

225 = 192 x 1 + 33

Applying Euclid’s Division Lemma on 192 and remainder 33 , we get,

192 = 33 x 5 + 27

Applying Euclid’s Division Lemma on 33 and remainder 27 , we get,

33 = 27 x 1 + 6

Applying Euclid’s Division Lemma on 27 and remainder 6 , we get,

27 = 6 x 4 + 3

Applying Euclid’s Division Lemma on 6 and remainder 3 , we get,

6 = 3 x 2 + 0

Since the remainder is now 0,

Hence, the HCF of 867 and 255 is 3.

### (iv) 184, 230 and 276

Solution:

We will first choose from 184 and 230 to find the HCF by using Euclid’s division lemma.

Thus, we obtain

230 = 184 x 1 + 46

Applying Euclid’s Division Lemma on 184 and remainder 46 , we get,

184 = 46 x 4 + 0

The HCF is therefore, 230.

Applying Euclid’s Division Lemma on 46 and 276 , we get,

276 = 46 x 6 + 0

Therefore, the HCF of the third number 276 and 46 is 46.

### (v) 136, 170 and 255

Solution:

We will first choose from 136 and 170 to find the HCF by using Euclid’s division lemma.

We get,

170 = 136 x 1 + 34

Applying Euclid’s Division Lemma on 136 and remainder 34 , we get,

136 = 34 x 4 + 0

Since, the remainder is now 0, the divisor will be the HCF i.e., 34 for 136 and 170.

Applying Euclid’s Division Lemma on 34 and 255 , we get,

255 = 34 x 7 + 17

Applying Euclid’s Division Lemma on 34 and remainder 17 , we get,

34 = 17 x 2 + 0

Since, the remainder is now 0,

Therefore, the HCF of 136, 170 and 255 is 17.

### (i) 963 and 657

Solution:

Applying Euclid’s Division Lemma on 963 and 657 , we get,

963 = 657 x 1 + 306………. (1)

Applying Euclid’s Division Lemma on 657 and remainder 306 , we get,

657 = 306 x 2 + 45………… (2)

Applying Euclid’s Division Lemma on 306 and remainder 45 , we get,

306 = 45 x 6 + 36…………. (3)

Applying Euclid’s Division Lemma on 45 and remainder 36 , we get,

45 = 36 x 1 + 9…………… (4)

Applying Euclid’s Division Lemma on 36 and remainder 9 , we get,

36 = 9 x 4 + 0……………. (5)

Therefore, the HCF is 9.

We can express the HCF as a linear combination of 963 and 657, by

9 = 45 – 36 x 1 [from (4)]

= 45 – [306 – 45 x 6] x 1 = 45 – 306 x 1 + 45 x 6 [from (3)]

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

= 657 x 7 – 306 x 14 – 306 x 1

= 657 x 7 – 306 x 15

= 657 x 7 – [963 – 657 x 1] x 15 [from (1)]

= 657 x 7 – 963 x 15 + 657 x 15

= 657 x 22 – 963 x 15.

### (ii) 592 and 252

Solution:

Applying Euclid’s Division Lemma on 592 and 252 , we get,

592 = 252 x 2 + 88……… (1)

Applying Euclid’s Division Lemma on 252 and remainder 88 , we get,

252 = 88 x 2 + 76………. (2)

Applying Euclid’s Division Lemma on 88 and remainder 76 , we get,

88 = 76 x 1 + 12………… (3)

Applying Euclid’s Division Lemma on 76 and remainder 12 , we get,

76 = 12 x 6 + 4………….. (4)

Applying Euclid’s Division Lemma on 12 and remainder 4 , we get,

12 = 4 x 3 + 0……………. (5)

Therefore, H.C.F. = 4.

We can express the HCF as a linear combination of 592 and 252, by

4 = 76 – 12 x 6 [from (4)]

= 76 – [88 – 76 x 1] x 6 [from (3)]

= 76 – 88 x 6 + 76 x 6

= 76 x 7 – 88 x 6

= [252 – 88 x 2] x 7 – 88 x 6 [from (2)]

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

= 252 x 7- 88 x 20

= 252 x 7 – [592 – 252 x 2] x 20 [from (1)]

= 252 x 7 – 592 x 20 + 252 x 40

= 252 x 47 – 592 x 20

= 252 x 47 + 592 x (-20)

### (iii) 506 and 1155

Solution:

Applying Euclid’s Division Lemma on 506 and 1155 , we get,

1155 = 506 x 2 + 143…………. (1)

Applying Euclid’s Division Lemma on 506 and remainder 143 , we get,

506 = 143 x 3 + 77…………….. (2)

Applying Euclid’s Division Lemma on 143 and remainder 77 , we get,

143 = 77 x 1 + 66……………… (3)

Applying Euclid’s Division Lemma on 77 and remainder 66 , we get,

77 = 66 x 1 + 11……………….. (4)

Applying Euclid’s Division Lemma on 66 and remainder 11 , we get,

66 = 11 x 6 + 0………………… (5)

Therefore, H.C.F. = 11.

We can express the HCF as a linear combination of 506 and 1155 by,

11 = 77 – 66 x 1 [from (4)]

= 77 – [143 – 77 x 1] x 1 [from (3)]

= 77 – 143 x 1 + 77 x 1

= 77 x 2 – 143 x 1

= [506 – 143 x 3] x 2 – 143 x 1 [from (2)]

= 506 x 2 – 143 x 6 – 143 x 1

= 506 x 2 – 143 x 7

= 506 x 2 – [1155 – 506 x 2] x 7 [from (1)]

= 506 x 2 – 1155 x 7+ 506 x 14

= 506 x 16 – 1155 x 7

### (iv) 1288 and 575

Solution:

Applying Euclid’s Division Lemma on 1288 and 575 , we get,

1288 = 575 x 2+ 138………… (1)

Applying Euclid’s Division Lemma on 575 and remainder 138 , we get,

575 = 138 x 4 + 23……………. (2)

Applying Euclid’s Division Lemma on 138 and remainder 23 , we get,

138 = 23 x 6 + 0……………….. (3)

Therefore, H.C.F. = 23.

We can express the found HCF as a linear combination of 1288 and 575, by

23 = 575 – 138 x 4 [from (2)]

= 575 – [1288 – 575 x 2] x 4 [from (1)]

= 575 – 1288 x 4 + 575 x 8

= 575 x 9 – 1288 x 4

### Question 4. Find the largest number which divides 615 and 963 leaving remainder 6 in each case.

Solution:

Firstly, since 6 is required as the remainder, we subtract it from both the numbers.

So, the required numbers are 615 – 6 = 609 and 963 – 6 = 957.

The required number is the HCF of newly obtained numbers,  609 and 957.

Applying Euclid’s Division Lemma , we get,

957 = 609 x 1+ 348

609 = 348 x 1 + 261

348 = 261 x 1 + 87

261 = 87 x 3 + 0.

Since, the remainder is 0,

Therefore, the required number is 87

### Question 5. If the HCF of 408 and 1032 is expressible in the form 1032m – 408 x 5, find m.

Solution:

Firstly, the HCF of 408 and 1032 is to be found.

Applying Euclid’s Division Lemma on 408 and 1032 , we get,

1032 = 408x 2 + 216.

Applying Euclid’s Division Lemma on 408 and remainder 216 , we get,

408 = 216 x 1 + 192.

Applying Euclid’s Division Lemma on 216 and remainder 192 , we get,

216 = 192 x 1 + 24.

Applying Euclid’s Division Lemma on 192 and remainder 24 , we get,

192 = 24 x 8 + 0.

Since, the remainder is 0 ,

The H.C.F of 408 and 1032 i.e., 24

So, this HCF is expressed as a linear combination that is,

24 = 1032m – 408 x 5

1032m = 24 + 408 x 5

1032m = 24 + 2040

1032m = 2064

m = 2064/1032

We obtain,

∴ m = 2

### Question 6.  If the HCF of 657 and 963 is expressible in the form 657x + 963 x – 15, find x.

Solution:

Applying Euclid’s Division Lemma on 657 and 963 , we get,

963 = 657 x 1+ 306.

Applying Euclid’s Division Lemma on 657 and remainder 306 , we get,

657 = 306 x 2 + 45.

Applying Euclid’s Division Lemma on 306 and remainder 45 , we get,

306 = 45 x 6 + 36.

Applying Euclid’s Division Lemma on 45 and remainder 36 , we get,

45 = 36 x 1 + 9.

Applying Euclid’s Division Lemma on 36 and remainder 9 , we get,

36 = 9 x 4 + 0.

Now, the remainder = 0.

Hence, the last divisor is the H.C.F of 657 and 963 i.e., 9

by expressing the HCF as a linear combination, we get ,

9 = 657x + 963 (-15).

Finding the value of x, we get

9 = 657x —14445

9 + 14445 = 657x

14454 = 657x

⇒ x = 14454 / 657

∴ x = 22

### Question 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 march in the same number of columns. What is the maximum number of columns in which they can march?

Solution:

We need to compute the max number of columns in which the army band can march, which can be done by finding the HCF of the given two numbers.

Now, this is equal to the H.C.F of 616 and 32.

Applying Euclid’s Division Lemma on 616 and 32 , we get,

616 = 32 x 19 + 8

32 = 8 x 4 + 0.

Therefore, H.C.F. = 8

∴ The maximum number of columns in which the army band can march is 8.

### Question 8. A merchant has 120 litres of oil of one kind, 180 litres of another and 240 litres 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?

Solution:

The greatest capacity of the tin for filling three different types of oil is equivalent to the the H.C.F. of the three available quantities 120,180 and 240.

Applying Euclid’s Division Lemma on 180 and 120 , we get,

180 = 120 x 1 + 60

120 = 60 x 2 + 0

Since, the remainder is now 0 ,

The HCF = 60.

Computing the H.C.F of 60 and the third quantity 240.

Applying Euclid’s Division Lemma on 240 and 60 , we get,

240 = 60 x 4 + 0

Since the remainder is 0 ,

Therefore, the HCF = 60

Therefore, the tin should be of 60 litres.

### Question 9. During a sale, colour 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?

Solution:

We have,

Number of colour pencils in a pack = 24

Number of crayons in a pack = 32.

The least number of both colour pencils and crayons that needs to be purchased is equivalent to their LCM.

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

Now,

The number of pencil packs to be bought = 96 / 24 = 4 packs

And, the number of crayon packs to be bought = 96 / 32 = 3 packs

### Question 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 the same drink, what would be the greatest number of cartons each stack would have?

Solution:

We have,

Number of coke can cartons = 144

Number of Pepsi can cartons = 90.

Therefore, the maximum number of cartons in a stack can be found by computing the H.C.F. of(144, 90).

Applying Euclid’s Division Lemma on 144 and 90 , we get,

144 = 90 x 1 + 54

90 = 54 x 1+ 36

54 = 36 x 1 + 18

36 = 18 x 2 + 0

∴ Since, the remainder is 0 ,

Therefore, the greatest number of cartons together in one stack is 18

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