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Solve:$ 4 \frac{1}{10}-\left(2 \frac{1}{2}-\left\{\frac{5}{6}-\left(\frac{2}{5}+\frac{3}{10}-\frac{4}{15}\right)\right\}\right. $
Given:
\( 4 \frac{1}{10}-\left(2 \frac{1}{2}-\left\{\frac{5}{6}-\left(\frac{2}{5}+\frac{3}{10}-\frac{4}{15}\right)\right\}\right. \)
To do:
We have to solve the given expression.
Solution:
\( 4 \frac{1}{10}-\left(2 \frac{1}{2}-\left\{\frac{5}{6}-\left(\frac{2}{5}+\frac{3}{10}-\frac{4}{15}\right)\right\}\right. \)
$ \begin{array}{l} =\frac{4\times 10+1}{10} -\left[\frac{2\times 2+1}{2} -\left\{\frac{5}{6} -\left(\frac{2\times 6+3\times 3-4\times 2}{30}\right)\right\}\right]\ \ =\frac{40+1}{10} -\left[\frac{4+1}{2} -\left\{\frac{5}{6} -\left(\frac{12+9-8}{30}\right)\right\}\right]\ \ =\frac{41}{10} -\left[\frac{5}{2} -\left\{\frac{5}{6} -\frac{13}{30}\right\}\right]\ \ =\frac{41}{10} -\left[\frac{5}{2} -\left\{\frac{5\times 5-13}{30}\right\}\right]\ \ =\frac{41}{10} -\left[\frac{5}{2} -\left\{\frac{25-13}{30}\right\}\right]\ \ =\frac{41}{10} -\left[\frac{5}{2} -\frac{12}{30}\right]\ \ =\frac{41}{10} -\left[\frac{5\times 15-12}{30}\right]\ \ =\frac{41}{10} -\left[\frac{75-12}{30}\right]\ \ =\frac{41}{10} -\frac{63}{30}\ \ =\frac{41\times 3-63}{30}\ \ =\frac{123-63}{30}\ \ =\frac{60}{30}\ \ =2 \end{array}$
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