What is the fourth proportional to 1312,934 and 74513{1\over2}, 9{3\over4} \ and\ 7{4\over5} 1321,943 and 754? A519305 {19\over30}53019 B79207 {9\over20}7209 C611246{11\over24 }62411 D417364{17\over36}43617 Show answer & explanationCorrect answer: AExplanationThe correct answer is option 1 i.e. 519305 {19\over30}53019. ⇒1312934=745x⇒x=934×7451312\frac{13\frac{1}{2}}{9\frac{3}{4}} = \frac{7\frac{4}{5}}{x} \Rightarrow x = \frac{9\frac{3}{4} \times 7\frac{4}{5}}{13\frac{1}{2}}9431321=x754⇒x=1321943×754 1312=27213\frac{1}{2} = \frac{27}{2}1321=227 934=3949\frac{3}{4} = \frac{39}{4}943=439 745=3957\frac{4}{5} = \frac{39}{5}754=539 ⇒ x=394×395÷272=39×394×5×227=1521×220×27=3042540x = \frac{39}{4} \times \frac{39}{5} \div \frac{27}{2} = \frac{39 \times 39}{4 \times 5} \times \frac{2}{27} = \frac{1521 \times 2}{20 \times 27} = \frac{3042}{540}x=439×539÷227=4×539×39×272=20×271521×2=5403042 ⇒ 3042540=50790=55790=51930\frac{3042}{540} = \frac{507}{90} = {5\frac{57}{90}} = {5\frac{19}{30}}5403042=90507=59057=53019