A number is increased by $40\%$ and then the increased number is decreased by $40\%$. Find the net increase or decrease percent.

$16\%$

Step by Step Explanation:
1. Let the number be $x$.

We are given that a number is increased by $40\%$.

So, increased percentage = $(100+40)\%$ = $140\%$

Then, \begin{aligned} \text{ Increased number } =& \space 140\% \text{ of } x = \bigg( \dfrac { 140 } { 100 } \times x \bigg) = \dfrac { 7 } { 5 } x \\ \end{aligned} Now, the number is decreased by $40 \%$. So, the decreased number will be $(100-40)\%$ = $60\%$ of the increased number. \begin{aligned} \text{ Decreased number } =& \space 60\% \text{ of } \dfrac { 7 } { 5 } x = \bigg( \dfrac { 60 } { 100 } \times \dfrac { 7 } { 5 } x \bigg) = \dfrac { 420 } { 500 } x \\ \therefore \text{ Net decrease } =& \text{ Original Number - Decreased Number } = \bigg( x - \dfrac { 420 } { 500 } x \bigg) = \dfrac { 80 } { 500 } x \\ \end{aligned}
2. Therefore, \begin{aligned} \text{ Net decrease }\% =& \dfrac { \text{ Net decrease } } { \text{ Original number } } \times 100\% \\ =& \bigg( \dfrac { 80 x } { 500 } \div { x } \times 100 \bigg)\% \\ = & \bigg( \dfrac { 80 x } { 500 } \times \dfrac { 1 } { x } \times 100 \bigg)\% = 16\% \end{aligned} Hence, the net decrease is $16\%$.