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Maths chapter 1 exercise 1.1 question 1 |
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Answer» (i) HCF of 135 and 225:Applying the Euclid’s lemma to 225 and 135, (where 225 > 135), we get225 = (135 × 1) + 90Since, 90 ≠ 0, therefore, applying the Euclid’s lemma to 135 and 90, we have:135 = (90 × 1) + 45But 45 ≠ 0∴ Applying Euclid’s Lemma to 90 and 45, we get90 = (45 × 2) + 0Here, r = 0, so our process stops. Since, the divisor at the last step is 45,∴ HCF of 225 and 135 is 45.(ii) HCF of 196 and 38220:We start dividing the larger number 38220 by 196, we get38220 = (196 × 195) + 0∵ r = 0∴ HCF of 38220 and 196 is 196.(iii) HCF of 867 and 255:Here, 867 > 255∴ Applying Euclid’s Lemma to 867 and 255, we get867 = (255 × 3) + 102Since, 102 ≠ 0, therefore, applying the Euclid’s lemma to 255 and 102, we have:255 = (102 × 2) + 51But 51 ≠ 0∴ Applying Euclid’s Lemma to 102 and 51, we get102 = (51 × 2) + 0Here, r = 0, so our process stops. Since, the divisor at the last step is 51,∴ HCF of 867 and 255 is 51. Pata nhi kya answer hai ? |
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