Correct option is A533−415−485+?=815\frac{53}{3} - \frac{41}{5} - \frac{48}{5} + ? = \frac{8}{15}353−541−548+?=158?=815−533+895? = \frac{8}{15} - \frac{53}{3} + \frac{89}{5}?=158−353+589? = 8−265+26715=1015\frac{8 - 265 + 267}{15} = \frac{10}{15}158−265+267=1510?=23 ? = \frac{2}{3}?=32