EXERCISE 2 SOLUTIONS: IN 31 TO 40:

31. let the numbers be 3 X, 4 X and 5 X. The, their L.CM. =60 X. So, 60 X = 2400 or X=40.

    The numbers are (3 x 40), (4 x 40) and (5 x 40).

     Hence, required H.C.F.=40.

32. Other number (11 x 7700 /275). = 308.

33. let the numbers be x and (2000 – X). Then, their L.C. M. = X (2000 – X).

     So, X (2000 – X) = 211879    X²-2000 X + 21879 = 0

    (X – 1989) X-11) =0 X = 1989 or X=11.

     Hence, the numbers are 1989 and 11.

34. Let the numbers be X and 4 X. Then, X x 4 X = 84 x 21 x² = (84 x 21 / 4) = X= 21.

    Hence, larger number = 4 X  = 84.

35. Let the numbers be X and (100 – X).

    The, X(100 – X) = 5 x 495  X² - 100 X + 2475 = 0

    (X- 55) (X – 45) = 0               X = 55 or X = 45.

    The numbers are 45 and 55,

          Required difference = (55 – 45) = 10.

36. Let the numbers be X and (X +2).

      Then, (X (X + 2) = 24 X²+ 2 X -24 = 0 (X – 4) ( X + 6) = 0 X = 4. .

       So, the numbers are 4 and 6.

37. Let the numbers be a and b. Then, a + B = 55 and a b = 5 x 120 = 600.

      Required sum = 1/a + 1/b = a + b /a b =55 / 600 = 11 /120.

38. Let H.C.F.  Be h and, L.C.M. be I. then, I = 45 H and I + h = 1150.

    45 h + h =1150 or h = 25. So, I = (1150 – 25) = 1125.

    Hence, other number – (25 x 1125/125) = 225.

39. First number = (50 x 2) = 100. Second number = (50 x 250 /100) = 125.

40. L.C.M. product of numbers /H.C.F.  = 1320/ 6 = 220.