DRDO Multi tasking staff Model paper 2

Before we proceed to solve this, we need to understand BODMAS: B – Brackets O – OF D – Division M – Multiplication A – Addition S – Subtraction This is the order in which the operators are to be solved. We have $\frac{1}{3}+\left(\frac{5}{8}\times\frac{4}{5}\right)-\left(\frac{2}{3}\div\frac{4}{5}\right)+\frac{3}{4}\text{ of }\frac{2}{5}$ We have two brackets in this case. Hence, we solve them first. $\frac{5}{8}\times\frac{4}{5}=\frac{1}{2}$ And, $\frac{2}{3}\div\frac{4}{5}=\frac{2}{3}\times\frac{5}{4}=\frac{5}{6}$ Substituting them back, we have $\frac{1}{3}+\frac{1}{2}-\frac{5}{6}+\frac{3}{4}\text{ of }\frac{2}{5}$ Next we solve that operator containing “OF”. Hence, $\frac{3}{4}\text{ of }\frac{2}{5}=\frac{3}{4}\times\frac{2}{5}=\frac{3}{10}$ Substituting them back, we have $\frac{1}{3}+\frac{1}{2}-\frac{5}{6}+\frac{3}{10}$ This is the simplified version. The LCM of the denominators is 30. Hence, we have $\frac{10}{30}+\frac{15}{30}-\frac{25}{30}+\frac{9}{30}=\frac{9}{30}=\frac{3}{10}$