1. Hints:- (a) L.C.M. of 5 and 20 = 2 x 2 x 5 = 20 (b) L.C.M. of 6 and 18 = 2 x 3 x 3 = 18 (c) L.C.M. of 12 and 48 = 2 x 2 x 2 x 2 x 3 = 48 (d) L.C.M. of 9 and 45 = 3 x 3 x 5 = 45 From these all cases, we can conclude that if the smallest number if the factor of largest number, then the L.C.M. of theseRead more

    Hints:-
    (a) L.C.M. of 5 and 20
    = 2 x 2 x 5
    = 20

    (b) L.C.M. of 6 and 18
    = 2 x 3 x 3
    = 18

    (c) L.C.M. of 12 and 48
    = 2 x 2 x 2 x 2 x 3
    = 48

    (d) L.C.M. of 9 and 45
    = 3 x 3 x 5
    = 45
    From these all cases, we can conclude that if the smallest number if the factor of largest number, then the L.C.M. of these two numbers is equal to that of larger number.

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

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  2. Hints:- (a) L.C.M. of 9 and 4 = 2 x 2 x 3 x 3 = 36 (b) L.C.M. of 12 and 5 = 2 x 2 x 3 x 5 = 60 (c) L.C.M. of 6 and 5 = 2 x 3 x 5 = 30 (d) L.C.M. of 15 and 4 = 2 x 2 x 3 x 5 = 60 Yes, the L.C.M. is equal to the product of two numbers in each case. And L.C.M. is also the multiple of 3. https://www.tiwRead more

    Hints:-
    (a) L.C.M. of 9 and 4
    = 2 x 2 x 3 x 3
    = 36

    (b) L.C.M. of 12 and 5
    = 2 x 2 x 3 x 5
    = 60

    (c) L.C.M. of 6 and 5
    = 2 x 3 x 5
    = 30

    (d) L.C.M. of 15 and 4
    = 2 x 2 x 3 x 5
    = 60

    Yes, the L.C.M. is equal to the product of two numbers in each case.
    And L.C.M. is also the multiple of 3.

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

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  3. Hints:- L.C.M. of 18, 24 and 32 = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288 The smallest four-digit number = 1000 Therefore, the required number is 1000 + (288 – 136) = 1152. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

    Hints:-

    L.C.M. of 18, 24 and 32 = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288
    The smallest four-digit number = 1000

    Therefore, the required number is 1000 + (288 – 136) = 1152.

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

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  4. L.C.M. of 6, 15 and 18 = 2 x 3 x 3 x 5 = 90 Therefore, the required number = 90 + 5 = 95 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

    L.C.M. of 6, 15 and 18 = 2 x 3 x 3 x 5 = 90
    Therefore, the required number = 90 + 5 = 95

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

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    • 6
  5. Hints:- The maximum capacity of container = H.C.F. (403, 434, 465) Factors of 403 = 13 x 31 Factors of 434 = 2 x 7 x 31 Factors of 465 = 3 x 5 x 31 H.C.F. = 31 Therefore, 31 litres of container is required to measure the quantity.   https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chRead more

    Hints:-

    The maximum capacity of container = H.C.F. (403, 434, 465)
    Factors of 403 = 13 x 31
    Factors of 434 = 2 x 7 x 31
    Factors of 465 = 3 x 5 x 31
    H.C.F. = 31
    Therefore, 31 litres of container is required to measure the quantity.

     

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

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