The value of ($$1 \frac{1}{3}÷2\frac{6}{7}of 5 \frac{3}{5}$$)÷($$ 6 \frac{2}{5}÷4\frac{1}{2}of 5 \frac{1}{3}$$) $$\times$$ ($$ \frac{3}{4}\times 2\frac{2}{3}÷\frac{5}{9} of $$1$$\frac{1}{5}) = 1 + k$$,where $$k$$ lies between

($$1 \frac{1}{3}÷2\frac{6}{7}of 5 \frac{3}{5}$$)÷($$ 6 \frac{2}{5}÷4\frac{1}{2}of 5 \frac{1}{3}$$) $$\times$$ ($$ \frac{3}{4}\times 2\frac{2}{3}÷\frac{5}{9} of $$1$$\frac{1}{5}) = 1 + k$$
($$ \frac{4}{3}÷\frac{20}{7}of  \frac{28}{5}$$)÷($$ \frac{32}{5}÷\frac{9}{2}of \frac{16}{3}) \times (\frac{3}{4}\times \frac{8}{3}÷\frac{2}{3} of \frac{6}{5})$$ = 1 + k
$$ \frac{1}{12}÷\frac{4}{15} \times (\frac{3}{4}\times4)$$ = 1 + k
$$ \frac{1}{12}÷\frac{4}{15} \times 3$$ = 1 + k
$$ \frac{1}{12} \times \frac{15}{4} \times 3$$ = 1 + k
15/16 = 1 + k
k = 0.9375 - 1 = -0.0625
So, k lies between -0.07 and -0.06.
$$\therefore$$ The correct answer is option A.