Question : The value of $\frac{\left(1 \frac{1}{9} × 1 \frac{1}{20} ÷ \frac{21}{38}-\frac{1}{3}\right) ÷\left(2 \frac{4}{9} ÷ 1 \frac{7}{15} \text { of } \frac{3}{5}\right)}{\frac{1}{5} \text { of } \frac{1}{5} ÷ \frac{1}{125}-\frac{1}{25} ÷ \frac{1}{5} \text { of } \frac{1}{5}}$ lies between ____.
Option 1: 0.1 and 0.15
Option 2: 0.2 and 0.25
Option 3: 0.15 and 0.2
Option 4: 0.25 and 0.3
Correct Answer: 0.15 and 0.2
Solution :
Given: $\frac{\left(1 \frac{1}{9} × 1 \frac{1}{20} ÷ \frac{21}{38}-\frac{1}{3}\right) ÷\left(2 \frac{4}{9} ÷ 1 \frac{7}{15} \text { of } \frac{3}{5}\right)}{\frac{1}{5} \text { of } \frac{1}{5} ÷ \frac{1}{125}-\frac{1}{25} ÷ \frac{1}{5} \text { of } \frac{1}{5}}$
= $\frac{(\frac{10}{9} × \frac{21}{20}\div\frac{21}{38} – \frac{1}{3})\div(\frac{22}{9}\div(\frac{22}{15}\times\frac{3}{5}))}{(\frac{1}{5} \times \frac{1}{5}) ÷ \frac{1}{125}-\frac{1}{25} ÷ (\frac{1}{5} \times \frac{1}{5})}$
= $\frac{(\frac{10}{9} × \frac{21}{20}\times\frac{38}{21} – \frac{1}{3})\div(\frac{22}{9}\div\frac{22}{25})}{\frac{1}{25} ÷ \frac{1}{125}-\frac{1}{25} ÷ \frac{1}{25}}$
= $\frac{(\frac{19}{9} – \frac{1}{3})\div(\frac{22}{9}\times\frac{25}{22})}{\frac{1}{25} \times \frac{125}{1}-\frac{1}{25} \times \frac{25}{1}}$
= $\frac{(\frac{19-3}{9}\div\frac{25}{9})}{5-1}$
= $\frac{(\frac{16}{9}\times\frac{9}{25})}{5-1}$
= $\frac{\frac{16}{25}}{4}$
= $\frac{4}{25}$
= 0.16
Hence, the correct answer is 0.15 and 0.2.
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