The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is

Product of numbers = 11 x 385 = 4235

 

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235  =>  ab = 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 number lies 75 and 125, the suitable pair is  (55,77)

 

Hence , required number = 77