# RD Sharma Solutions Class 10 Mathematics Solutions for Real Numbers Exercise 1.2 in Chapter 1 - Real Numbers

Question 1 Real Numbers Exercise 1.2

Use Euclid’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

(v) 136, 170 and 255

Given integers here are 225 and 135. On comparing, we find 225 > 135.

So, by applying Euclid’s division lemma to 225 and 135, we get 867 = 225 x 3 + 192

Since the remainder ≠ 0. So 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 remainder 45.

⇒ 90 = 45 x 2 + 0

Since the remainder at this stage is 0, the divisor will be the HCF. Hence, the H.C.F of 225 and 135 is 45.

(ii) 196 and 38220

Given integers 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

Given integers here are 867 and 255. On comparing, we find 867 > 255.

So, by applying Euclid’s division lemma to 867 and 225, we get,

867 = 225 x 3 + 192

Since the remainder 192 ≠ 0. So we apply the division lemma to the divisor 225 and remainder

1. We get,

225 = 192 x 1 + 33

Now we apply the division lemma to the new divisor 192 and remainder 33. We get, 192 = 33 x 5 + 27

Now we apply the division lemma to the new divisor 33 and remainder 27. We get, 33 = 27 x 1 + 6

Now we apply the division lemma to the new divisor 27 and remainder 6. We get, 27 = 6 x 4 + 3

Now we apply the division lemma to the new divisor 27 and remainder 6. We get, 6 = 3 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 3.

(iv) 184, 230 and 276

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. So we apply the division lemma to the divisor 184 and remainder 46. 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 remainder 0. Thus, 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

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. So we apply the division lemma to the divisor 136 and remainder 34.

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. So we apply the division lemma to the divisor 34 and remainder 17.

We get,

34 = 17 x 2 + 0

So, this stage has remainder 0. Thus, the HCF of the third number 255 and 34 is 17. Hence, the HCF of 136, 170, and 255 is 17.

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