H.C.F and L.C.M Aptitude Test Paper 521) The LCM of two prime number x and y is 161. If x>y, find the value of 15y7x.
Answer: B Explanation: Note: The HCF of two prime numbers = 1 LCM/HCF =161/1 =161 22) Find the smallest digit that will leave same remainder 4 in each case when divided by 12, 15, 20, and 54.
Answer: C Explanation: Take LCM of 12, 15, 20, and 54. It will be = 540. ATQ, the remainder in each case = 4. Hence, the required number = 540+4 = 544. 23) What is the number nearest to 10000 which is exactly divisible by 3, 4, 5, 6, 7, and 8?
Answer: B Explanation: Take LCM of 3, 4, 5, 6, 7, and 8. It will be = 840. Now, 10000/840 = 760 (remainder). 840  760 = 80. If we add 80 in the given number, the number is exactly divisible by 840. So, the required number is 10080. H.C.F and L.C.M Aptitude Test Paper 1 H.C.F and L.C.M Aptitude Test Paper 2 H.C.F and L.C.M Aptitude Test Paper 3 H.C.F and L.C.M Aptitude Test Paper 4 H.C.F and L.C.M Aptitude Concepts
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