Exercise 1.1(cbse 10th)

 Question 1:

Use Euclid's division algorithm to find the HCF of:

(i) 135 and 225 (ii) 196 and 38220

Answer 1:

(i) 135 and 225

Since 225 135, we apply the division lemma to 225 and 135 to obtain 

225= 135 x 1 +90

Since remainder 90not equal to 0 , we apply the division lemma to 135 and 90 to

obtain

135 =90 x 1 + 45

We consider the new divisor 90 and new remainder 45, and apply the

division lemma to obtain

90 =2 x 45 +0

Since the remainder is zero, the process stops. Since the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220

Since 38220 > 196, we apply the division lemma to 38220 and 196 to

obtain

38220= 196 x 195 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 196, Therefore, HCF of 196 and 38220 is 196.

(iii) 867 and 255

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain

 867= 255 x 3 + 102

Since remainder 102 is equal to 0, we apply the division lemma to 255 and 102

to obtain

255=102×2+51

We consider the new divisor and new remainder 51 apply the division lemma to obtain 

102=51×2+0

Since the remainder is 0 the process stops

Since the divisor at this stage is 51

Therefore HCF of 867 and 255 is 51.