New 55% Off on All Courses - Enroll Now! Click Here

    Question

    If 'x' is the smallest positive integer divisible by 9,

    15, and 28, then determine the second smallest positive integer that is divisible by all three of these numbers.
    A 2400 Correct Answer Incorrect Answer
    B 2520 Correct Answer Incorrect Answer
    C 3420 Correct Answer Incorrect Answer
    D 3260 Correct Answer Incorrect Answer

    Solution

    Lowest positive integer divisible by 9, 15 and 28 = LCM (9, 15, 28) So, LCM of (9, 15, and 28) = 22 X 32 X 51 X 71 = 1260 Therefore, the next larger number which is divisible by all the three given numbers = 2 X 1260 = 2520

    Practice Next