RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2: This exercise explains to students about Euclid’s Division Algorithm, which is an important property of integers. This feature is also used to calculate the HCF of two or even three numbers. If you’re facing any difficulty in understanding a problem, the RD Sharma Solutions Class 10 is the perfect resource for you. Students can get RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2 PDF for a detailed solution to these exercise questions.
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RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2 PDF
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1. Define the HCF of two positive integers and find the HCF of the following pairs of numbers:
(i) 32 and 54
Solution:
Apply Euclid’s division lemma on 54 and 32
54 = 32 x 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 x 2 + 2
Since remainder ≠ 0, apply division lemma on 10 and remainder 2
10 = 2 x 5 + 0
Therefore, the HCF of 32 and 54 is 2.
(ii) 18 and 24
Solution:
Apply Euclid’s division lemma on 24 and 18
24 = 18 x 1 + 6.
Since remainder ≠ 0, apply the division lemma on divisor 18 and remainder 6
18 = 6 x 3 + 0.
Therefore, the HCF of 18 and 24 is 6.
(iii) 70 and 30
Solution:
Apply Euclid’s division lemma on 70 and 30
70 = 30 x 2 + 10.
Since remainder ≠ 0, apply division lemma on divisor 30 and remainder 10
30 = 10 x 3 + 0.
Therefore, the HCF of 70 and 30 is 10.
(iv) 56 and 88
Solution:
Apply Euclid’s division lemma on 56 and 88
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, the HCF of 56 and 88 is 8.
(v) 475 and 495
Solution:
By applying Euclid’s division lemma on 495 and 475, we get
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, the HCF of 475 and 495 is 5.
(vi) 75 and 243
Solution:
By applying Euclid’s division lemma on 243 and 75,
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, the HCF of 75 and 243 is 3.
(vii) 240 and 6552
Solution:
By applying Euclid’s division lemma on 6552 and 240, we get
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, the HCF of 240 and 6552 is 24.
(viii) 155 and 1385
Solution:
By applying Euclid’s division lemma on 1385 and 155, we get
1385 = 155 x 8 + 145.
Since remainder ≠ 0, apply the division lemma on divisor 155 and remainder 145.
155 = 145 x 1 + 10.
Since remainder ≠ 0, apply the 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, the HCF of 155 and 1385 is 5.
(ix) 100 and 190
Solution:
By applying Euclid’s division lemma on 190 and 100, we get
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, the HCF of 100 and 190 is 10.
(x) 105 and 120
Solution:
By applying Euclid’s division lemma on 120 and 105, we get
120 = 105 x 1 + 15.
Since remainder ≠ 0, apply division lemma on divisor 105 and remainder 15
105 = 15 x 7 + 0.
Therefore, the HCF of 105 and 120 is 15.
2. Use Euclid’s division algorithm to find the HCF of
(i) 135 and 225
Solution:
The integers given here are 225 and 135. On comparing, we find 225 > 135.
So, by applying Euclid’s division lemma to 225 and 135, we get
225 = 135 x 1 + 90
Since the remainder ≠ 0, we apply the division lemma to the divisor 135 and remainder 90.
⇒ 135 = 90 x 1 + 45
Now, we apply the division lemma to the new divisor 90 and the remainder 45.
⇒ 90 = 45 x 2 + 0
Since the remainder at this stage is 0, the divisor will be the HCF.
Hence, the HCF of 225 and 135 is 45.
(ii) 196 and 38220
Solution:
The integers given here are 196 and 38220. On comparing, we find 38220 > 196.
So, by applying Euclid’s division Lemma to 38220 and 196, we get,
38220 = 196 x 195 + 0
Since the remainder at this stage is 0, the divisor will be the HCF.
Hence, the HCF of 38220 and 196 is 196.
(iii) 867 and 255
Solution:
The integers given here are 867 and 255. On comparing, we find 867 > 255.
So, by applying Euclid’s division Lemma to 867 and 255, we get,
867 = 255 x 3 + 102
Since the remainder 102 ≠ 0, we apply the division lemma to the divisor 255 and remainder 102, and we get,
255 = 102 x 2 + 51
Now, we apply the division lemma to the new divisor 102 and the remainder 51, we get,
102 = 51 x 2 + 0
Since the remainder at this stage is 0, the divisor will be the HCF.
Hence, the HCF of 867 and 255 is 51.
(iv) 184, 230 and 276
Solution:
Let’s first choose 184 and 230 to find the HCF by using Euclid’s division lemma.
Thus, we obtain
230 = 184 x 1 + 46
Since the remainder 46 ≠ 0, we apply the division lemma to the divisor 184 and the remainder 46, and we get,
184 = 46 x 4 + 0
The remainder at this stage is 0, the divisor will be the HCF, i.e., 46 for 184 and 230.
Now, we again use Euclid’s division lemma to find the HCF of 46 and 276, and we get
276 = 46 x 6 + 0
So, this stage has a remainder of 0, and the HCF of the third number 276 and 46 is 46.
Hence, the HCF of 184, 230 and 276 is 46.
(v) 136, 170 and 255
Solution:
Let’s first choose 136 and 170 to find the HCF by using Euclid’s division lemma.
Thus, we obtain
170 = 136 x 1 + 34
Since the remainder 34 ≠ 0, we apply the division lemma to the divisor 136 and remainder 34, and we get,
136 = 34 x 4 + 0
The remainder at this stage is 0, the divisor will be the HCF, i.e., 34 for 136 and 170.
Now, we again use Euclid’s division lemma to find the HCF of 34 and 255, and we get
255 = 34 x 7 + 17
Since the remainder 17 ≠ 0, we apply the division lemma to the divisor 34 and remainder 17, and we get,
34 = 17 x 2 + 0
So, this stage has a remainder of 0. Thus, the HCF of the third number, 255 and 34, is 17.
Hence, the HCF of 136, 170, and 255 is 17.
3. Find the HCF of the following pair of integers and express it as a linear combination of them,
(i) 963 and 657
Solution:
By applying Euclid’s division lemma on 963 and 657, we get
963 = 657 x 1 + 306………. (1)
As the remainder ≠ 0, apply the division lemma on divisor 657 and remainder 306
657 = 306 x 2 + 45………… (2)
As the remainder ≠ 0, apply the division lemma on divisor 306 and remainder 45
306 = 45 x 6 + 36…………. (3)
As the remainder ≠ 0, apply the division lemma on divisor 45 and remainder 36
45 = 36 x 1 + 9…………… (4)
As the remainder ≠ 0, apply the division lemma on divisor 36 and remainder 9
36 = 9 x 4 + 0……………. (5)
Thus, we can conclude that HCF = 9.
Now, in order to express the found HCF as a linear combination of 963 and 657, we perform
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:
By applying Euclid’s division lemma on 592 and 252, we get
592 = 252 x 2 + 88……… (1)
As the remainder ≠ 0, apply the division lemma on divisor 252 and remainder 88
252 = 88 x 2 + 76………. (2)
As the remainder ≠ 0, apply the division lemma on divisor 88 and remainder 76
88 = 76 x 1 + 12………… (3)
As the remainder ≠ 0, apply the division lemma on divisor 76 and remainder 12
76 = 12 x 6 + 4………….. (4)
Since the remainder ≠ 0, apply the division lemma on divisor 12 and remainder 4
12 = 4 x 3 + 0……………. (5)
Thus, we can conclude that HCF = 4.
Now, in order to express the found HCF as a linear combination of 592 and 252, we perform
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:
By applying Euclid’s division lemma on 506 and 1155, we get
1155 = 506 x 2 + 143…………. (1)
As the remainder ≠ 0, apply the division lemma on divisor 506 and remainder 143
506 = 143 x 3 + 77…………….. (2)
As the remainder ≠ 0, apply the division lemma on divisor 143 and remainder 77
143 = 77 x 1 + 66……………… (3)
Since the remainder ≠ 0, apply the division lemma on divisor 77 and remainder 66
77 = 66 x 1 + 11……………….. (4)
As the remainder ≠ 0, apply the division lemma on divisor 66 and remainder 11
66 = 11 x 6 + 0………………… (5)
Thus, we can conclude that HCF = 11.
Now, in order to express the found HCF as a linear combination of 506 and 1155, we perform
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:
By applying Euclid’s division lemma on 1288 and 575, we get
1288 = 575 x 2+ 138………… (1)
As the remainder ≠ 0, apply the division lemma on divisor 575 and remainder 138
575 = 138 x 4 + 23……………. (2)
As the remainder ≠ 0, apply the division lemma on divisor 138 and remainder 23
138 = 23 x 6 + 0……………….. (3)
Thus, we can conclude that HCF = 23.
Now, in order to express the found HCF as a linear combination of 1288 and 575, we perform
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.
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