Given,

$\frac{5}{8}\text{\hspace{0.17em}of 4011.33+}\frac{7}{10}\text{\hspace{0.17em}of 3411.22=?}$

It can be approximated as

$\frac{5}{8}\text{\hspace{0.17em}of 4000+}\frac{7}{10}\text{\hspace{0.17em}of 3400=?}$

= 5×500+7×340

= 2500+2300

= 4880 ≈ 4890

Hence, Option C is correct.