H.C.F. AND L.C.M. OF NUMBERS- EXERCISE 2 ( SOLUTIONS: 71 TO 75:)

71. L.C.M. of 5, 6, 7, 8= 840.

       Required number = (840 x  2 + 3 ) = 1683.

72. L.C.M. of 16, 18, 20, 25 =3600. Required number is of the form 3600 k + 4.

       Least value of k for which (3600 k + 4) is divisible by 7 is k = 5.

       Required number = (3600 x 5 + 4) = 18004.

73. L.C.M. of 2, 4, 8, 10, and 12 is 120.

         So, the bells will toll together after every 120 seconds, i.e., 2 minutes.

         In 30 minutes, they will toll together [(30/2) + 1)] =16 times.

74. Interval after which the devices will beep together

      = (L.C.M. of 30, 60, 90, 105) mi. = 1260 min. = 21  hrs.

      So, the devices will again beep together 21 hrs. after 12 noon I;e., at 9 a.m.

75. L.C.M. of 252, 308 and 198 = 2772.

        So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

H.C.F. AND L.C.M. OF NUMBERS -(EXERCISE 2 SOLUTIONS 61 TO 70:)

61. Greatest number of 4 digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600.

         On dividing 9999 by, 600, the remainder is 399.

         Required number = (9999 – 399) = 9600.

61. L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37.

        Number to be added = (60 – 37) = 23.

62. The least number divisible by 16, 20, 24

         = L.C.M. of 16, 20, 24 = 240 = 2 x 2 x 2 x 2 x 3 x 5.

         To make it a perfect square, it must be multiplied by 3 x 5.

         Required number =240 x 3 x 5 =360.

63. Required number = (L.C.M. of 12, 16, 18, 21, 28) + 7 = 1008 + 7 =1015.

64. Required number = (L.C.M. of 24, 32, 36, 54) – 5 864 – 5 = 859.

65. Required number = (L.C.M. of 12, 15, 20, 54) + 8 540 + 8 = 548

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66. Greatest number of 4 digits is 9999. L.C.M.  Of 4, 7 and 13 = 364.

         On dividing 9999 by 363, remainder obtained is 171.

         Greatest number of 4 digits divisible by 4, 7 and 13 = (9999 -171) = 9828.

         Hence, required number = (9828 + 3) 9831.

67. least number of 6 digits is 100000. L.C.M. of 4, 6, 10 and 15= 60

H.C.F. AND L.C.M. OF NUMBERS (EXERCISE 2 SOLUTIONS 451 TO 60:)

51. Required number of students = H.C.F. of 1001 and 910 = 91 

52. Largest size of the tile = H.C.F. of 378 cm and 525 cm = 21 cm.

53. REQUIRED NUMBER = H.C.F.  Of (91 – 43), (183 – 91) and (183 – 43)

                                         = H.C.F. of 48, 92 and 140 = 4

54. N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)

     = H.C.F of 3360, 2240 and 5600 = 1120.

      Sum of digits in N= (1+ 1 + 2 + 0) = 4.

55.  Required number = H.C.F of (1356 – 12), (1868 – 12) and (2764 -12)

                                   H.C.F. of 1344, 1856 and 2752 = 64.

56. Required number = H.C.F. of (1657 – 6) and (2037 – 5)

                                  H.C.F. of 1651 and 2032 =127.

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

       8/8 = 70/80; 13/16 = 65/80; 31/40 = 62/80.

       Since, 70/80 > 63/80> 65/80 > so 7/8 > 63/80 > 13/16> 31/40.

       So, 7/8 is the largest.

58. L.C.M.  of 12, 18, 21, 30

H.C.F. AND L.C.M. OF NUMBERS. IMPORTANT FACTS AND FORMULA

       1. Factors and multiples: if a number a divides another number b exactly, we say that a is a factor of b. in this case, b is called multiple of a. 

        2.  Highest common factor (H.C.F.) or greatest common measure (G.C.M.) or greatest common divisor (G.C.D.): the H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly. 

       There are two methods of finding the H.C.F. of a given  set of numbers. 

         I. Factorization method: express each one of the given numbers as the product of prime factors. the product of least powers of common prime factors given H.C.F. 

         II. Finding the H.C.F. of more than two numbers: suppose we have to find the H.C.F of three numbers. then, H.C.F. of [(H.C.F. of any two) and (the third number ] gives the H.C.F of three given numbers. 

     3. least common multiple (L.C.M):  the least number which is exactly divisible by each one of the given numbers is called their L.C.M. 

          i. Factorization method of finding L L.C.M.: resolve each one of the given numbers into a product of prime factors. the, L.C.M. is the product of highest powers of all the factors. 

         II.common division method (short - cut method ) of finding L.C.M: arrange the given numbers in a row in any order. divide by a number which divides exactly at least two of the given numbers and  carry forward the numbers which are not divisible. repeat the above process till no tow of the numbers are divisible by the same number except 1. the product of the divisors and the divided numbers is the required L.C.M. of the given numbers. 

   4. product of two numbers = product of their H.C.F. and L.C.M. 

   5. co-primes: two numbers are said to be co-primes if their H.C.F is 1. 

   6. H.C.F. and L.C.M. of fractions:

   

      I.H.C.F.= H.C.F. of numerators         II. L.C.M. = L.C.M. of numerators 

                                   /       /        /                                            /       /          /

                        L.C.M. of denominators                           H.C.F. of denominators

    7. H.C.F. and L.C.M. of decimal fractions:  in given numbers,make the same numbers of decimal places  by annexing zeros is some numbers, if necessary. . considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. now, in the result, mark off as many decimal places as are there in each of the given numbers. 

    8. comparison of fractions: find the L.C.M. of the denominators of the given fractions. convert each of the fractions into an equivalent fraction with L.C.M as the denominator, by multiplying both the numerator and denominator by the same number. the resultant fraction with the greatest numerator is the greatest.