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

**Answer:**

^@16\%^@

**Step by Step Explanation:**

- 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} @^ - 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\%^@.