Chapter 1 - Real Numbers, RD Sharma Solutions - (Part - 3) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No. 1.27

Q.1. Define HCF of two positive integers and find the HCF of the following pairs of numbers:
(i) 32 and 54
Ans. We need to find H.C.F. of 32 and 54. 
By applying division lemma 
54 = 32 × 1 + 22 
Since remainder ≠ 0, apply division lemma on 32 and remainder 22 
32 = 22 x 1 + 10 
Since remainder ≠ 0, apply division lemma on 22 and remainder 10 
22 = 10 × 2 + 2 
Since remainder ≠ 0, apply division lemma on 10 and remainder 2 
10 = 2 × 5 + 0 
Therefore, H.C.F. of 32 and 54 is 2.

(ii) 18 and 24 
Ans. We need to find H.C.F. of 18 and 24. 
By applying division lemma
24 = 18 x 1 + 6
Since remainder ≠ 0, apply division lemma on divisor 18 and remainder 6
18 = 6 x 3 + 0.
Therefore, H.C.F. of 18 and 24 is 2.

(iii) 70 and 30
Ans. We need to find H.C.F. of 70 and 30. 
By applying Euclid’s Division lemma
70 = 30 x 2 + 10.
Since remainder ≠ 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) 56 and 88
Ans. We need to find H.C.F. of 56 and 88.
By applying Euclid’s Division lemma
88 = 56 x 1 + 32.
Since remainder ≠ 0, apply division lemma on 56 and remainder 32
56 = 32 x 1 + 24.
Since remainder ≠ 0, apply division lemma on 32 and remainder 24
32 = 24 x 1 + 8.
Since remainder ≠ 0, apply division lemma on 24 and remainder 8
24 = 8 x 3 + 0.
Therefore, H.C.F. of 56 and 88 = 8.

(v) 475 and 495
Ans. We need to find H.C.F. of 475 and 495.
By applying Euclid’s Division lemma
495 = 475 x 1 + 20.
Since remainder ≠ 0, apply division lemma on 475 and remainder 20
475 = 20 x 23 + 15.
Since remainder ≠ 0, apply division lemma on 20 and remainder 15
20 = 15 x 1 + 5.
Since remainder ≠ 0, apply division lemma on 15 and remainder 5
15 = 5 x 3 + 0.
Therefore, H.C.F. of 475 and 495 = 5.

(vi) 75 and 243 
Ans. We need to find H.C.F. of 75 and 243.
By applying Euclid’s Division lemma
243 = 75 x 3 + 18.
Since remainder ≠ 0, apply division lemma on 75 and remainder 18
75 = 18 x 4 + 3.
Since remainder ≠ 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) 240 and 6552
Ans.  We need to find H.C.F. of 240 and 6552.
By applying Euclid’s Division lemma
6552 = 240 x 27 + 72.
Since remainder ≠ 0, apply division lemma on divisor 240 and remainder 72
240 = 72 x 3 + 24.
Since remainder ≠ 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) 155 and 1385
Ans. We need to find H.C.F. of 155 and 1385.
By applying Euclid’s Division lemma
1385 = 155 x 8 + 145.
Since remainder ≠ 0, apply division lemma on divisor 155 and remainder 145
155 = 145 x 1 + 10.
Since remainder ≠ 0, apply division lemma on divisor 145 and remainder 10
145 = 10 x 14 + 5.
Since remainder ≠ 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) 100 and 190 
Ans. We need to find H.C.F. of 100 and 190.
By applying Euclid’s division lemma
190 = 100 x 1 + 90.
Since remainder ≠ 0, apply division lemma on divisor 100 and remainder 90
100 = 90 x 1 + 10.
Since remainder ≠ 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) 105 and 120
Ans. We need to find H.C.F. of 105 and 120.
By applying Euclid’s division lemma
120 = 105 x 1 + 15.
Since remainder ≠ 0, apply division lemma on divisor 105 and remainder 15
105 = 15 x 7 + 0.
Therefore, H.C.F. of 105 and 120 = 15.

Q.2. Use Euclid's division algorithm to find the HCF of
(i) 135 and 225
Ans. Given integers are 225 and 135. Clearly 225 > 135. So we will apply Euclid’s division lemma to 225 and 135, we get,
867 = (225)(3) + 192
Since the remainder 90 ≠ 0. So we apply the division lemma to the divisor 135 and remainder 90. We get, 135 = (90)(1) + 45
Now we apply the division lemma to the new divisor 90 and remainder 45. We get, 90 = (45)(2) + 0
The remainder at this stage is 0. So the divisor at this stage is the H.C.F.
So the H.C.F of 225 and 135 is 45.

(ii) 196 and 38220
Ans. Given integers are 38220 and 196. Clearly 38220 > 196. So we will apply Euclid’s division lemma to 38220 and 196, we get, 38220 = (196)(195) + 0
The remainder at this stage is 0. So the divisor at this stage is the H.C.F.
So the H.C.F of 38220 and 196 is 196.

(iii) 867 and 255
Ans. Given integers are 867 and 255. Clearly 867 > 225. So we will apply Euclid’s division lemma to 867 and 225, we get, 867 = (225)(3) + 192
Since the remainder 192 ≠ 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 6 and remainder 3. We get, 6 = (3)(2) + 0
The remainder at this stage is 0. So the divisor at this stage is the H.C.F.
So the H.C.F of 867 and 255 is 3.

(iv) 184, 230 and 276
Ans. Given integers are 184, 230 and 276. 
Let us first find the HCF of 184 and 230 by Euclid lemma. 
Clearly, 230 > 184. So, we will apply Euclid’s division lemma to 230 and 184. 
230 = 184 × 1 + 46 
Remainder is 46 which is a non-zero number. Now, apply Euclid’s division lemma to 184 and 46. 
184 = 46 × 4 + 0 
The remainder at this stage is zero. Therefore, 46 is the HCF of 230 and 184. 
Now, again use Euclid’s division lemma to find the HCF of 46 and 276. 
276  = 46 × 6 + 0 
The remainder at this stage is zero. Therefore, 46 is the HCF of 184, 230 and 276.

(v) 136, 170 and 255
Ans. Given integers are 136, 170 and 255. 
Let us first find the HCF of 136, 170 by Euclid lemma. 
Clearly, 170 > 136. So, we will apply Euclid’s division lemma to 136 and 170. 
170 = 136 × 1 + 34 
Remainder is 34 which is a non-zero number. 
Now, apply Euclid’s division lemma to 136 and 34. 
136 = 34 × 4 + 0 
The remainder at this stage is zero. Therefore, 34 is the HCF of 136 and 170. 
Now, again use Euclid’s division lemma to find the HCF of 34 and 255. 
255  = 34 × 7 +17 
Remainder is 17 which is a non-zero number. Now, apply Euclid’s division lemma to 34 and 17. 
34 = 17 × 2 + 0 
The remainder at this stage is zero. Therefore, 17 is the HCF of 136, 170 and 255.

(vi) 1260 and 7344
Ans. 1260 and 7344 
1260 < 7344 
Thus, we divide 7344 by 1260 by using Euclid's division lemma 
7344 = 1260 × 5 +1044 
∵ Remainder is not zero, 
∴ we divide 1260 by 1044 by using Euclid's division lemma 
1260 = 1044 × 1 + 216 
∵ Remainder is not zero, 
∴ we divide 1044 by 216 by using Euclid's division lemma 
1044 = 216 × 4 + 180 
∵ Remainder is not zero, 
∴ we divide 216 by 180 by using Euclid's division lemma 
216 = 180 × 1 + 36 
∵ Remainder is not zero, 
∴ we divide 180 by 36 by using Euclid's division lemma 
180 = 36 × 5 + 0 
Since, Remainder is zero Hence, HCF of 1260 and 7344 is 36.

(vii) 2048 and 960
Ans. 2048 and 960 2048 > 960 
Thus, we divide 2048 by 960 by using Euclid's division lemma 
2048 = 960 × 2 + 128 
∵ Remainder is not zero, 
∴ we divide 960 by 128 by using Euclid's division lemma 
960 = 128 × 7 + 64 
∵ Remainder is not zero, 
∴ we divide 128 by 64 by using Euclid's division lemma 
128 = 64 × 2 + 0 
Since, Remainder is zero 
Hence, HCF of 2048 and 960 is 64.

Q.3. Find the HCF of the following pairs of integers and express it as a linear combination of them.
(i) 963 and 657
Ans. We need to find the H.C.F. of 963 and 657 and express it as a linear combination of 963 and 657.
By applying Euclid’s division lemma 963 = 657 x 1 + 306.
Since remainder ≠ 0, apply division lemma on divisor 657 and remainder 306
657 = 306 x 2 + 45.
Since remainder ≠ 0, apply division lemma on divisor 306 and remainder 45
306 = 45 x 6 + 36.
Since remainder ≠ 0, apply division lemma on divisor 45 and remainder 36
45 = 36 x 1 + 9.
Since remainder ≠ 0, apply division lemma on divisor 36 and remainder 9
36 = 9 x 4 + 0.
Therefore, H.C.F. = 9.
Now,
9 = 45 - 36 x 1
= 45 - [306 - 45 x 6] x 1
= 45 - 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 - 657 x 1] x 15
= 657 x 7 - 963 x 15 + 657 x 15
= 657 x 22 - 963 x 15

(ii) 592 and 252
Ans. We need to 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 = 252×2+88
Since remainder ≠ 0, apply division lemma on divisor 252 and remainder 88
252 = 88×2+76
Since remainder  ≠ 0, apply division lemma on divisor 88 and remainder 76
88 = 76×1+12
Since remainder  ≠ 0, apply division lemma on divisor 76 and remainder 12
76 = 12×6+4
Since remainder  ≠ 0, apply division lemma on divisor 12 and remainder 4
12 = 4×3+0.
Therefore, H.C.F. = 4.
Now,
4 = 76 − 12 × 6
= 76 − [88 − 76 × 1] × 6
= 76 − 88 × 6 + 76 × 6
= 76 × 7 − 88 × 6
= (252 − 88 × 2) × 7 − 88 × 6
= 252 × 7 − 88 × 14 − 88 × 6
= 252 × 7 − 88 × 20
= 252 × 7 − [592 − 252 × 2] × 20
= 252 × 7 − 592 × 20 + 252 × 40
= 252 × 47 − 592 × 20
= 252 × 47 + 592 × (−20)

(iii) 506 and 1155
Ans. We need to find the 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.
Since remainder ≠ 0, apply division lemma on divisor 506 and remainder 143
506 = 143 x 3 + 77.
Since remainder ≠ 0, apply division lemma on divisor 143 and remainder 77
143 = 77 x 1 + 66.
Since remainder ≠ 0, apply division lemma on divisor 77 and remainder 66
77 = 66 x 1 + 11.
Since remainder ≠ 0, apply division lemma on divisor 66 and remainder 11
66 = 11 x 6 + 0.
Therefore, H.C.F. = 11.
Now,
11 = 77 - 66 x 1
= 77 - [143 - 77 x 1] x 1
= 77 - 143 x 1 + 77 x 1
= 77 x 2 - 143 x 1
= [506 - 143 x 3] x 2 - 143 x 1
= 506 x 2 - 143 x 6 - 143 x 1
= 506 x 2 - 143 x 7
= 506 x 2 - [1155 - 506 x 2] x 7
= 506 x 2 - 1155 x 7 + 506 x 14
= 506 x 16 - 1155 x 7.

(iv) 1288 and 575
Ans. We need to find the 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.
Since remainder ≠ 0, apply division lemma on divisor 575 and remainder 138
575 = 138 x 4 + 23.
Since remainder ≠ 0, apply division lemma on divisor 138 and remainder 23
138 = 23 x 6 + 0.
Therefore, H.C.F. = 23.
Now,
23 = 575 - 138 x 4
= 575 - [1288 - 575 x 2] x 4
= 575 - 1288 x 4 + 575 x 8
= 575 x 9 - 1288 x 4.