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LCM and HCF Exercise 2

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  1. Question 1 of 14
    1. Question

    The H.C.F. and L.C.M. of two numbers are 50 and 250 respectively. If the first number is divided by 2, the quotient is 50. The second number is:

    Hint

    First number = (50 × 2) = 100.

    Second number lcm-and-hcf-q-40314.png

  2. Question 2 of 14
    2. Question

    The sum of two numbers is 2000 and their L.C.M. is 21879. The two numbers are:

    Hint

    Let the numbers be x and (2000 – x).

    Then, their L.C.M. = x(2000 – x).

    So, x(2000 – x) = 21879

    ⇒ x² – 2000x + 21879 = 0

    ⇒ (x –1989)(x – 11) = 0

    ⇒ x = 1989 or 11.

    Hence, the numbers are 1989 and 11.

  3. Question 3 of 14
    3. Question

    The H.C.F. of two numbers is 8. Which one of the following can never be their L.C.M.?

    Hint

    H.C.F. of two numbers divides their L.C.M. exactly. 8 is not a factor of 60.

  4. Question 4 of 14
    4. Question

    The L.C.M. of three different numbers is 120. Which of the following cannot be their H.C.F.?

    Hint

    Since H.C.F. is always a factor of L.C.M., we cannot have three numbers with H.C.F. 35 and L.C.M. 120.

  5. Question 5 of 14
    5. Question

    Which of the following fractions is the largest?

    Hint

    L.C.M. of 8, 16, 40 and 80 = 80.

    lcm-and-hcf-q-40308.png

    lcm-and-hcf-q-40302.png

    So, 7/8 is the largest.

  6. Question 6 of 14
    6. Question

    Which is the least number that must be subtracted from 1856, so that the remainder when divided by 7, 12 and 16 will leave the same remainder 4?

    Hint

    Suppose least number be x

    1856 – x = n(LCM of 7, 12, 16) + 4

    ⇒ 1856 -x = n (336) + 4

    we should take n = 5 so that n(336) is nearest to 1856 and n (336)< 1856

    1856 – x = 1680 + 4 = 1684

    x = 1856 – 1684 = 172

  7. Question 7 of 14
    7. Question

    Find the greatest number that will divide 148, 246 and 623 leaving remainders 4, 6 and 11 respectively.

    Hint

    Required number

    = H.C.F of (148 – 4), (246 – 6) and (623 – 11)

    = H.C.F of 144, 240 and 612 = 12

  8. Question 8 of 14
    8. Question

    Monica, Veronica and Rachael begin to jog around a circular stadium. They complete their revolutions in 48 seconds, 64 seconds and 72 seconds respectively. After how many seconds will they be together at the starting point?

    Hint

    Required time = LCM of 48, 64 and 72

    lcm-and-hcf-q-40284.png

    LCM = 2 × 2 × 2 × 2 × 3 × 4 × 3 = 576 seconds.

  9. Question 9 of 14
    9. Question

    Three girls start jogging from the same point around a circular track and each one completes one round in 24 seconds, 36 seconds and 48 seconds respectively. After how much time will they meet at one point?

    Hint

    Required time

    = L.C.M of 24, 36 and 48

    = 144 seconds

    = 2 minutes 24 seconds

  10. Question 10 of 14
    10. Question

    Three friends A, B and C start running around a circular stadium and complete a single round in 24, 36 and 30 seconds respectively. After how many minutes will they meet again at the starting point?

    Hint

    Required time

    = L.C.M of 24, 36 and 30

    = 360 seconds

    = 6 minutes

  11. Question 11 of 14
    11. Question

    Swapnil, Aakash and Vinay begin to jog around a circular stadium. They complete their revolutions in 36 seconds, 48 seconds and 42 seconds respectively. After how many seconds will they be together at the starting point?

    Hint

    They will be together at the starting point after the

    L.C.M of 36, 48 and 42 = 1008 seconds

  12. Question 12 of 14
    12. Question

    The circumferences of the fore and hind-wheels of a carriage are 2⅖ and 3³⁄₇ respectively. A chalk mark is put on the point of contact of each wheel with the ground at any given moment. How far will the carriage have travelled so that its chalk marks may be again on the ground at the same time?

    Hint

    Required distance

    = L.C.M of lcm-and-hcf-q-40278.png

    lcm-and-hcf-q-40272.png

    Hence, carriage will travelled 24m so that its chalk marks may be again on the ground at the same time.

  13. Question 13 of 14
    13. Question

    Three bells chime at an interval of 18, 24 and 32 minutes respectively. At a certain time they begin to chime together. What length of time will elapse before they chime together again.

    Hint

    L.C.M of 18, 24 & 32 = 288

    Hence they would chime after every 288 min.

    ⇒ 4 hrs 48 min

  14. Question 14 of 14
    14. Question

    HCF of 3240, 3600 and a third number is 36 and their LCM is 2⁴ × 3⁵ × 5² × 7². The third number is :

    Hint

    Let the third number be x.

    Product of numbers = LCM × HCF

    Therefore,

    lcm-and-hcf-q-40266.png

    ⇒ lcm-and-hcf-q-40260.png

    ⇒ lcm-and-hcf-q-40254.png

    ⇒ lcm-and-hcf-q-40248.png

    = lcm-and-hcf-q-40242.png

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