Question

Simply: $14÷\left(3of2-3+4\right)-7\left(5-3\right)$.

Answer 1

Solving the question with the help of BODMAS.

B - Bracket

O - Of

D - Division

M - Multiplication

A - Addition

S - Subtraction

Firstly, we have to solve to bracket, then of (change of into $\times$) , division , multiplication, addition and subtraction.

$\begin{array}{rcl}14÷\left(3of2-3+4\right)-7\left(5-3\right)& =& 14÷\left(3\times 2-3+4\right)-7\left(5-3\right)\\ & =& 14÷\left(6-3+4\right)-7\left(5-3\right)\\ & =& 14÷\left(10-3\right)-7\left(5-3\right)\\ & =& 14÷\left(7\right)-7\left(5-3\right)\\ & =& 14÷\left(7\right)-7\left(2\right)\\ & =& 2-14\\ & =& -12\end{array}$

Hence, the answer of this question is $-12$.