DSSSB TGT CS Mock 01 – Q042

The value of \(\frac{\left(6\frac{2}{5} \div 5\frac{1}{3} \text{ of } 4\frac{1}{2}\right) – \left(1\frac{1}{3} \div 3\frac{3}{5} \text{ of } \frac{5}{6}\right) + \frac{2}{5}}{\frac{3}{4} – 2\frac{2}{3} \div 1\frac{1}{5} \text{ of } \frac{5}{9} + 4\frac{1}{4}}\) is:

• \(\frac{1}{3}\)
• \(\frac{5}{6}\)
• \(\frac{2}{3}\)
• \(\frac{2}{9}\)

Using the BODMAS rule:
**Numerator:**
– First part: \(6\frac{2}{5} \div 5\frac{1}{3} \text{ of } 4\frac{1}{2}\)
\(5\frac{1}{3} \text{ of } 4\frac{1}{2} = \frac{16}{3} \times \frac{9}{2} = 24\)
\(\frac{32}{5} \div 24 = \frac{32}{5 \times 24} = \frac{4}{15}\)
– Second part: \(1\frac{1}{3} \div 3\frac{3}{5} \text{ of } \frac{5}{6}\)
\(3\frac{3}{5} \text{ of } \frac{5}{6} = \frac{18}{5} \times \frac{5}{6} = 3\)
\(\frac{4}{3} \div 3 = \frac{4}{9}\)
– Third part: \(+\frac{2}{5}\)
– Numerator = \(\frac{4}{15} – \frac{4}{9} + \frac{2}{5} = \frac{12 – 20 + 18}{45} = \frac{10}{45} = \frac{2}{9}\)
**Denominator:**
\(\frac{3}{4} – 2\frac{2}{3} \div 1\frac{1}{5} \text{ of } \frac{5}{9} + 4\frac{1}{4}\)
– \(1\frac{1}{5} \text{ of } \frac{5}{9} = \frac{6}{5} \times \frac{5}{9} = \frac{2}{3}\)
– \(2\frac{2}{3} \div \frac{2}{3} = \frac{8}{3} \times \frac{3}{2} = 4\)
– Denominator = \(\frac{3}{4} – 4 + 4\frac{1}{4} = \left(\frac{3}{4} + \frac{17}{4}\right) – 4 = \frac{20}{4} – 4 = 5 – 4 = 1\)
**Final Result:**
\(\frac{\text{Numerator}}{\text{Denominator}} = \frac{2/9}{1} = \frac{2}{9}\).
Hence, the correct option is (d).