Question

Using appropriate properties, find
$
  \left( i \right){\text{ }} - \dfrac{2}{3} \times \dfrac{3}{5} + \dfrac{5}{2} - \dfrac{3}{5} \times \dfrac{1}{6} \\
  \left( {ii} \right){\text{ }}\dfrac{2}{5} \times \left( { - \dfrac{3}{7}} \right) - \dfrac{1}{6} \times \dfrac{3}{2} + \dfrac{1}{{14}} \times \dfrac{2}{5} \\
$

Answer 1

Hint: Try to find out common number and then proceed further with simple mathematics.

$
  \left( i \right){\text{ }} - \dfrac{2}{3} \times \dfrac{3}{5} + \dfrac{5}{2} - \dfrac{3}{5} \times \dfrac{1}{6} \\
   = - \dfrac{2}{3} \times \dfrac{3}{5} - \dfrac{3}{5} \times \dfrac{1}{6} + \dfrac{5}{2} \\
$
Taking $\dfrac{3}{5}$common, we get
$
   = \dfrac{3}{5}\left( { - \dfrac{2}{3} - \dfrac{1}{6}} \right) + \dfrac{5}{2} \\
   = \dfrac{3}{5}\left( {\dfrac{{ - 2 \times 2 - 1}}{6}} \right) + \dfrac{5}{2} \\
   = \dfrac{3}{5}\left( {\dfrac{{ - 5}}{6}} \right) + \dfrac{5}{2} \\
   = \dfrac{{ - 1}}{2} + \dfrac{5}{2} \\
   = \dfrac{4}{2} \\
   = 2 \\
$

$
  \left( {ii} \right){\text{ }}\dfrac{2}{5} \times \left( { - \dfrac{3}{7}} \right) - \dfrac{1}{6} \times \dfrac{3}{2} + \dfrac{1}{{14}} \times \dfrac{2}{5} \\
   = \dfrac{2}{5} \times \left( { - \dfrac{3}{7}} \right) + \dfrac{1}{{14}} \times \dfrac{2}{5} - \dfrac{1}{6} \times \dfrac{3}{2} \\
 $
Taking $\dfrac{2}{5}$common from above equation, we get
$
   = \dfrac{2}{5} \times \left( { - \dfrac{3}{7} + \dfrac{1}{{14}}} \right) - \dfrac{1}{6} \times \dfrac{3}{2} \\
   = \dfrac{2}{5} \times \left( {\dfrac{{ - 3 \times 2 + 1}}{{14}}} \right) - \dfrac{1}{6} \times \dfrac{3}{2} \\
   = \dfrac{2}{5} \times \left( {\dfrac{{ - 5}}{{14}}} \right) - \dfrac{1}{6} \times \dfrac{3}{2} \\
   = \dfrac{{ - 1}}{7} - \dfrac{1}{2} \times \dfrac{1}{2} \\
   = \dfrac{{ - 1}}{7} - \dfrac{1}{4} \\
   = \dfrac{{ - 4 - 7}}{{7 \times 4}} \\
   = \dfrac{{ - 11}}{{28}} \\
$

Note: Just follow the rules of BODMAS to get the correct answer. This means that you should do what is possible within parentheses first, then exponents, then multiplication and division (from left to right), and then addition and subtraction (from left to right).