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Mathematics Year 5 PPSMI
3/15/2011
NCERT CBSE Math Solutions for Class X …
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Real Numbers (Math) Exercise Exercise 1.1 Pag e 7 Next»
Q1 Q2 Q3 Q4 Q5
Question 1:
Use Euclid’s division algorithm to f ind the HCF of:
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(i) 135 an and d 225 Since 225 > 135, we apply the division lemma to 225 and 135 to obtain 225 = 135 × 1 + 90 Since remainder 90 ≠ 0, we we apply the division division lemma le mma to 135 and 90 to obtain 135 = 90 × 1 + 45 We co nsider the new divisor divisor 90 and new new remainder 45, and apply apply the division lemma to o btain 90 = 2 × 45 + 0 Since Sin ce the r emain emainder der is zero, the proces s stops. Since the divisor at this stage is 45, Therefore, the HCF of 135 an and d 225 is 45. (ii)196 and 38220 Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain 38220 = 196 × 195 + 0 Since Sin ce the r emain emainder der is zero, the proces s stops. Since the divisor at this stage is 196, Therefore, HCF of 196 and 38220 is 196. (iii)867 an and d 255 Since 867 > 255, we apply the division lemma to 867 and 255 to obtain 867 = 255 × 3 + 102 Since remainder 10 2 ≠ 0, we apply apply the division division lemma le mma to 255 and 102 to o btain 255 = 102 × 2 + 51 We co nsider the new divisor divisor 10 2 and new remainder remainder 51, 5 1, and apply apply the division division le mma to obtain 102 = 51 × 2 + 0
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NCERT CBSE Math Solutions for Class X …
Since Sin ce the r emain emainder der is zero, the proces s stops. Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51 .
