The simplification of left[frac{1}{2}+frac{1} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To simplify the expression $\left[\frac{1}{2}+\frac{1}{2}\left\{\frac{3}{4}-\frac{1}{2}\left(\frac{7}{8}-\frac{3}{4}\right)\right\}\right]$,we fol

Vedclass · Accepted Answer

$\frac{27}{32}$

Vedclass · Answer

(B) To simplify the expression $\left[\frac{1}{2}+\frac{1}{2}\left\{\frac{3}{4}-\frac{1}{2}\left(\frac{7}{8}-\frac{3}{4}ight)ight\}ight]$,we follow the $BODMAS$ rule.Step $1$: Solve the innermost bracket (parentheses): $\left(\frac{7}{8}-\frac{3}{4}ight) = \left(\frac{7-6}{8}ight) = \frac{1}{8}$.Step $2$: Solve the curly bracket: $\left\{\frac{3}{4}-\frac{1}{2} 	imes \frac{1}{8}ight\} = \left\{\frac{3}{4}-\frac{1}{16}ight\} = \left\{\frac{12-1}{16}ight\} = \frac{11}{16}$.Step $3$: Solve the square bracket: $\left[\frac{1}{2}+\frac{1}{2} 	imes \frac{11}{16}ight] = \left[\frac{1}{2}+\frac{11}{32}ight] = \left[\frac{16+11}{32}ight] = \frac{27}{32}$.

The simplification of $\left[\frac{1}{2}+\frac{1}{2}\left\{\frac{3}{4}-\frac{1}{2}\left(\frac{7}{8}-\frac{3}{4}\right)\right\}\right]$ yields

Similar Questions

If $3x + 7 = x^2 + P = 7x + 5$,what is the value of $P$?

$\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right) \ldots \left(1-\frac{1}{25}\right)$ is equal to

If $a=25, b=15, c=-10$; then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}$ is

If $2 = x + \frac{1}{1 + \frac{1}{3 + \frac{1}{4}}}$,then the value of $x$ is

What will come in the place of the question mark $(?)$ in the following expression?
$\frac{28}{65} \times \frac{195}{308} \div \frac{39}{44} + \frac{5}{26} = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module