By applying Euclid’s division lemma
(i) 54 = 32 × 1 + 22
Since remainder ≠ 0, apply division lemma on division of 32 and remainder 22.
32 = 22 × 1 + 10
Since remainder ≠ 0, apply division lemma on division of 22 and remainder 10.
22 = 10 × 2 + 2
Since remainder ≠ 0, apply division lemma on division of 10 and remainder 2.
10 = 2 × 5 [remainder 0]
(ii) By applying division lemma
24 = 18 × 1 + 6
Since remainder = 6, apply division lemma on divisor of 18 and remainder 6.
18 = 6 × 3 + 0
∴ Hence, HCF of 18 and 24 = 6
(iii) By applying Euclid’s division lemma
70 = 30 × 2 + 10
Since remainder ≠ 0, apply division lemma on divisor of 30 and remainder 10.
30 = 10 × 3 + 0
∴ Hence HCF of 70 and 30 is = 10.
(iv) By applying Euclid’s division lemma
88 = 56 × 1 + 32
Since remainder ≠ 0, apply division lemma on divisor of 56 and remainder 32.
56 = 32 × 1 + 24
Since remainder ≠ 0, apply division lemma on divisor of 32 and remainder 24.
32 = 24 × 1 + 8
Since remainder ≠ 0, apply division lemma on divisor of 24 and remainder 8.
24 = 8 × 3 + 0
∴ HCF of 56 and 88 is = 8.
(v) By applying Euclid’s division lemma
495 = 475 × 1 +20
Since remainder ≠ 0, apply division lemma on divisor of 475 and remainder 20.
475 = 20 × 23 + 15
Since remainder ≠ 0, apply division lemma on divisor of 20 and remainder 15.

20 = 15 × 1 + 5
Since remainder ≠ 0, apply division lemma on divisor of 15 and remainder 5.
15 = 5 × 3 + 0
∴ HCF of 475 and 495 is = 5.
(vi) By applying Euclid’s division lemma
243 = 75 × 3 + 18
Since remainder ≠ 0, apply division lemma on divisor of 75 and remainder 18.
75 = 18 × 4 + 3
Since remainder ≠ 0, apply division lemma on divisor of 18 and remainder 3.
18 = 3 × 6 + 0
∴ HCF of 243 and 75 is = 3.
(vii) By applying Euclid’s division lemma
6552 = 240 × 27 + 72
Since remainder ≠ 0, apply division lemma on divisor of 240 and remainder 72.
210 = 72 × 3 + 24
Since remainder ≠ 0, apply division lemma on divisor of 72 and remainder 24.
72 = 24 × 3 + 0
∴ HCF of 6552 and 240 is = 24.
(viii) By applying Euclid’s division lemma
1385 = 155 × 8 + 145
Since remainder ≠ 0, applying division lemma on divisor 155 and remainder 145
155 = 145 × 1 + 10
Since remainder ≠ 0, applying division lemma on divisor 10 and remainder 5
10 = 5 × 2 + 0
∴ Hence HCF of 1385 and 155 = 5.
(ix) By applying Euclid’s division lemma
190 = 100 × 1 + 90
Since remainder ≠ 0, applying division lemma on divisor 100 and remainder 90.
90 = 10 × 9 + 0
∴ HCF of 100 and 190 = 10
(x) By applying Euclid’s division lemma
120 = 105 × 1 + 15
Since remainder ≠ 0, applying division lemma on divisor 105 and remainder 15.
105 = 15 × 7 + 0
∴ HCF of 105 and 120 = 15