Question

Find $x$, if $5\frac{1}{6}-[1\frac{1}{5}+\{2\frac{3}{4}÷5\frac{1}{2}÷x-(\frac{5}{6}-\frac{2}{3})\}]=2\frac{61}{120}$

• A. $\frac{4}{13}$
• B. $2$
• C. $\frac{2}{13}$
• D. $4$

Answer 1

The correct option is A $\frac{4}{13}$

Given,

$5\frac{1}{6}-[1\frac{1}{5}+\{2\frac{3}{4}÷5\frac{1}{2}÷x-(\frac{5}{6}-\frac{2}{3})\}]=2\frac{61}{120}$

$\Rightarrow \frac{31}{6}-[\frac{6}{5}+\{\frac{11}{4}÷\frac{11}{2}÷x-(\frac{5}{6}-\frac{2}{3})\}]=\frac{301}{120}$

$\Rightarrow \frac{31}{6}-[\frac{6}{5}+\{\frac{11}{4}\times \frac{2}{11}\times \frac{1}{x}-\left(\frac{5-4}{6}\right)\}]=\frac{301}{120}$

$\Rightarrow \frac{31}{6}-[\frac{6}{5}+\{\frac{1}{2x}-\frac{1}{6}\}]=\frac{301}{120}\Rightarrow \frac{31}{6}-\frac{6}{5}-\frac{1}{2x}+\frac{1}{6}=\frac{301}{120}$

$\Rightarrow \frac{32}{6}-\frac{6}{5}-\frac{1}{2x}=\frac{301}{120}\Rightarrow \frac{160-36}{30}-\frac{1}{2x}=\frac{301}{120}$

$\Rightarrow -\frac{1}{2x}=\frac{301}{120}-\frac{124}{30}\Rightarrow -\frac{1}{2x}=\frac{301-496}{120}=-\frac{195}{120}$

$\Rightarrow x=\frac{120}{2\times 195}=\frac{4}{13}$