Friday, 20 February 2015

EXERCISE 2 SOLUTIONS: IN 21 TO 30:

21. Let the required numbers be x, 2 x and 3 x. Then, their H.C.F. = X. so, X =12.

     The numbers are 12, 24 and 36.

22. Let the numbers be 3 X and 4 X. Then, their H.C.F = X. so, X= 4
       So, the numbers are 12 and 16.
       L.C.M.of 12 and 16 = 48.

23. Let the required numbers be 27 a and 27 b. Then, 27 a + 27 b = 216 a+ b = 8.
      Now, co-primes with sum 8 are (1,7) and (3,5).
      Required numbers are (27 x 1 , 27 x 7 ) and (27 x 3, 27 x 5) i.e., (27, 189) and (91,135).
      Out of these, the given one in the answer is the pair (27,189).

24. Let the required numbers be 33 a and 33 b. then 33 a +33 b =528 a + b = 16.
    Now, co-prime with sum 16 are (1, 15), (3, 13) , (5,11) and (7, 9).
    Required number is (33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11).
   (33 x 7, 33 x 9).
   The number of such pairs is 4.

25. Numbers with H.C.F. 15 must contain 15 as a factor.
    Now, multiples of 15 between 40 and 100 are 45, 60, 75 and 90.
    Number-pairs with H.C.F. 15 are (45, 60),(45,75), (60,75) and (75,90).
    H.C.F. of (60, 90) is 30 and that of (45,90) is 45]
    Clearly, there are 4 such pairs.

26. out of the given numbers, the two with H.C.F 12 AND DIFFERENCE 12 ARE 84 AND 96.

27. Let the numbers be 37 a and 37 b. Then, 37 a x 37 b =4107 a b=3.

    No, co-primes with product 3 are (1, 3).
    So, the required numbers are (37 x 1, 37) i.e., (1.111).
   Greater number = 111.

28. let the numbers be 13 a and 13 b. Then, 13 a x 13 b =2028 a b = 12.
    Now, co-primes with product 12 are (1, 12)0 and (3, 4).
    So, the required numbers are (13 x 1, 12) and (13 x 3, 13 x 4).
    Clearly, there are 2 such pairs.

29. Since the numbers are co-prime, they contain only 1 as the common factors also; the given two    products have the middle number in common.
   So, middle number = H.C.F. of 551 and 1073 = 29;
   First number = (551/29) = 19; third number = (1073/29) = 37.
   Required sum = (19 +29 + 37) = 85.

30. let the numbers be 2 X and 3 X. Then, their L.C.M. =6 X = 48 or X =8.
   The numbers are 16 and 24.
   Hence, required sum = (16 + 24) =40.

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