EXERCISE 2 SOLUTIONS: IN 41 TO 50:

41. H.C.F. of co-prime numbers is 1. So, L.C.M. = 117/1 =117. 42. 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.

42. H.C.F. of two numbers divides their L.CM. Exactly. Clearly, 8 is not a factor of 60.

43. Clearly, the numbers are (23 x 14) = 222.

44. Since 16 is not a factor of 136, it follows that there does not exist any pair of numbers with H.C.F      16 and L.CM. 136.

45. Product of numbers = 11 x 385 = 4235.

     Let the numbers be 11 a and 111 b. The, 11 a x 11 b = 4235 a b= 35.

     Now, co-primes with product 35 are (1,35) and (5,7).

     So, the numbers are (11 x 1,11 x 35) and (11 x 5 , 11 x 7).

    Since one numbers lies between 75 and 125, the suitable pair is (55, 77).

    Hence, required number = 77.

46. Product of numbers = 29 x 4147.

    Let the numbers be 29 a and 29 b. Then, 29 a x 29 b= (24 x 4147) a b = 143.

    Now, co-primes with product 143 are (1, 143) and (11, 13)

     So, the numbers are (29 x 1, 29 x 143) and (29 x 11, 29 x 13).

   Since both be numbers are greater than 29, the suitable pair is (29 x 11, 29 x 13) i.e., (319, 377).

    Required sum = (319 + 377) = 696.

47. H.C.F. of two prime numbers is 1. Product of numbers = (1 x 161) = 161.

     Let the numbers be a and b. then, a b = 161.

     Now, co-primes with product 161 are (1,161) and (7, 23).

    Since X and y are prime numbers and X>y, we have, 23 and y =7.

    3 y-x = (3 x 7) – 23 = -2.

48. H.C.F of 2436 and 1001 is 7. Also, H.C.F. of 105 and 7 is 7.

     H.C.F. of 105, 1001 and 2436 is 7.

49. Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

50. Required measurement = (H.C.F of 496, 403, 713) litres = 31 liters.