The value of $ \frac {33}{40} + \frac {1}{5} [\frac{4}{5} – \frac{1}{5} \times (\frac {7}{8} – \frac {5}{4})] – \frac {4}{5}$ is :

Options
a) $ \frac 15$
b) $ \frac 14$
c) $ \frac 17$
d) $ \frac 13$

Solution

As per question
$ \frac {33}{40} + \frac {1}{5} [\frac{4}{5} – \frac{1}{5} \times (\frac {7}{8} – \frac {5}{4})] – \frac {4}{5}$
$ \frac {33}{40} + \frac {1}{5} [\frac{4}{5} – \frac{1}{5} \times (\frac {7 \times 1 – 5 \times 2} {8})] – \frac {4}{5}$
$ \frac {33}{40} + \frac {1}{5} [\frac{4}{5} – \frac{1}{5} \times (\frac {7 – 10 } {8})] – \frac {4}{5}$
$ \frac {33}{40} + \frac {1}{5} [\frac{4}{5} – \frac{1}{5} \times \frac {-3} {8}] – \frac {4}{5}$
$ \frac {33}{40} + \frac {1}{5} [\frac{4}{5} + \frac{3}{40}] – \frac {4}{5}$
$ \frac {33}{40} + \frac {1}{5} [\frac{4 \times 8 + 3 \times 1}{40}] – \frac {4}{5}$
$ \frac {33}{40} + \frac {1}{5} [\frac{32 + 3}{40}] – \frac {4}{5}$
$ \frac {33}{40} + \frac {1}{5} [\frac{35}{40}] – \frac {4}{5}$
$ \frac {33}{40} + \frac {7}{40} – \frac {4}{5}$
$ \frac {33 \times 1 + 7 \times 1 – 4 \times 8}{40}$
$ \frac {33 + 7 – 32}{40}$
$ \frac {40 – 32}{40}$
$ \frac {8}{40}$
$ \frac {1}{5}$

Ans: $ \frac {1}{5}$