Correct option is AGiven:(434−313)÷(34−35)+35×57×79\left( 4 \frac{3}{4} - 3 \frac{1}{3} \right) \div \left( \frac{3}{4} - \frac{3}{5} \right) + \frac{3}{5} \times \frac{5}{7} \times \frac{7}{9}(443−331)÷(43−53)+53×75×97Solution:(434−313)÷(34−35)+35×57×79\left( 4 \frac{3}{4} - 3 \frac{1}{3} \right) \div \left( \frac{3}{4} - \frac{3}{5} \right) + \frac{3}{5} \times \frac{5}{7} \times \frac{7}{9}(443−331)÷(43−53)+53×75×97=(194−103)÷(34−35)+35×57×79 \left( \frac{19}{4} - \frac{10}{3} \right) \div \left( \frac{3}{4} - \frac{3}{5} \right) + \frac{3}{5} \times \frac{5}{7} \times \frac{7}{9}(419−310)÷(43−53)+53×75×97 =194−103=1712,34−35=320 \frac{19}{4} - \frac{10}{3} = \frac{17}{12}, \quad \frac{3}{4} - \frac{3}{5} = \frac{3}{20}419−310=1217,43−53=203 =(1712)÷(320)+35×57×79\left( \frac{17}{12} \right) \div \left( \frac{3}{20} \right) + \frac{3}{5} \times \frac{5}{7} \times \frac{7}{9}(1217)÷(203)+53×75×97 =859+13=859+39=889\frac{85}{9} + \frac{1}{3} = \frac{85}{9} + \frac{3}{9} = \frac{88}{9}985+31=985+93=988 =9799\frac{7}{9}997