Question

A number is increased by $10\%$ an then the increased number is decreased by $10\%.$ Find the net increase or decrease percent.

Answer 1

Let the number is increased by $10$%

$x+\frac{10x}{100}=\frac{11x}{10}$

The number is decreased by $10$%

$=\frac{11x}{10}-\frac{11x}{10}\times \frac{10}{100}$

$=\frac{11x}{10}-\frac{11x}{100}=\frac{99x}{100}$

Decrease in percentage is

Decrease $=x-\frac{99x}{100}=\frac{x}{100}$

So, Decrease %

$=\frac{\frac{x}{100}}{x}\times 100\%$

$=1$%