Question

$50\times 5÷2+24$
$140÷\left[\right(-11)\times (-3)-(-42)÷14-\frac{3}{5}+\frac{3}{8}÷\frac{6}{4}$

Answer 1

Given that

Part (1) $50\times 5÷2+24$

Part (2) $140÷[(-11)\times (-3)-(-42)÷14]-\frac{3}{5}+\frac{3}{8}÷\frac{6}{4}$

Solve part (1).

By $BODMAS$ rule,

$50\times 5÷2+24$

$=50\times \frac{5}{2}+24$

$=25\times 5+24$

$=125+24$

$=149ans.$

Solve part (2).

By $BODMAS$ rule,

$140÷[(-11)\times (-3)-(-42)÷14]-\frac{3}{5}+\frac{3}{8}÷\frac{6}{4}$

$=140÷[33+42÷14]-\frac{3}{5}+\frac{3}{8}÷\frac{6}{4}$

$=140÷[33+3]-\frac{3}{5}+\frac{3}{8}÷\frac{6}{4}$

$=140÷36-\frac{3}{5}+\frac{3}{8}÷\frac{6}{4}$

$=\frac{140}{36}-\frac{3}{5}+\frac{3}{8}\times \frac{4}{6}$

$=\frac{140}{36}+\frac{1}{4}-\frac{3}{5}$

$=\frac{745-540}{180}$

$=\frac{41}{36}$ $ans.$