>>11031334first we simplify the problem, by dividing 5 by 11.
5/11 is x
50/11 is 10x
subtract equations
45/11 is 9x
45/9 is 11x
subtract first equation
45/9-5/11 is 10x
450/99 is 10x
divide by 10
45/99 is x
now we know that since the denominator is form of 2 9s, the repeating decimal must be,
x is .454545454545454545....... and so on forever.
now we take the reciprocal.
1/x is 1/.4545454545454545....
so we do long division
_______________
.454545454545...|1.000000000....
is equivalent to:
____________
.4545454545...|.99999999999....
so we find an upper and lower bound, given upper and lower bounds for 1 (one).
45 goes into 99 twice, and 100 twice with remainders 9 and 10, respectively.
Thus, the first digit in our answer must be 2. Going back to the original problem, we see that the denominator (divisor) is 5. Thus we can be sure that the quotient (answer) must be a terminating decimal. For this reason, I won't bore you with the rigmarole of finding the rest of the answer rigorously. Instead, you can guess and check the possibilities of answers by multiplying : 5* 2 is 10, (too small) 5*2.1 is 10.5 (still too small) 5*2.2 is.... you get the idea, so on and so forth until you find the answer.
