| **Answers & Solutions ** | | 1 . | Answer : Option B | Explanation : | Required numbers = (LCM of 12, 16, 18, 21, 28) + 7 = 1008 + 7 = 1015 | | | 2 . | Answer : Option C | Explanation : | Other number = [11 ×$7700\over275$ ] = 308 | | | 3 . | Answer : Option D | Explanation : | Let the numbers be 37a and 37b. Then, 37a × 37b = 4107 ab = 3 Now, co-primes with product 3 are (1, 3) So, the required numbers are (37 × 1, 37 × 3) i.e., (1, 111) Therefore greater number = 111 | | | 4 . | Answer : Option C | Explanation : | Required length = HCF of 700 cm, 385 cm and 1295 cm = 35 cm | | | 5 . | Answer : Option A | Explanation : | Required number = HCF of (91 - 43), (183 - 91) and (183 - 43) = HCF of 48, 92 and 140 = 4 | | | 6 . | Answer : Option C | Explanation : | Given numbers are 1.08, 0.36 and 0.90 HCF of 108, 36 and 90 is 18 HCF of given numbers – 0.18 | | | 7 . | Answer : Option D | Explanation : | Let the required numbers be x, 2x and 3x The, their HCF = x, So x = 12 The numbers are 12, 24, 36 | | | 8 . | Answer : Option B | Explanation : | Largest size of the tile. HCF of 378 cm and 525 cm = 21 cms. | | | 9 . | Answer : Option A | Explanation : | LCM = $Product\,\, of \,\,numbers \over HCF$ =$1320\over6$ =220 | | | 10 . | Answer : Option B | Explanation : | HCF of co-prime numbers is 1 So, LCM = $117\over1$= 117 | | | | |

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