Question

A number is increased by $10\%$ and then reduced by $10\%$. What is the effective increase or decrease in the original number?

• A. Doesn’t change
• B. Decreases by $1\%$
• C. Increases by $1\%$
• D. Increases by 2%

Answer 1

The correct option is B Decreases by $1\%$

Let the number be $x$
$10\%ofx=\frac{x}{10}$
Given, the number $x$ is increased by $10\%$ .

Hence, the increased number
$=x+\frac{x}{10}=\frac{11x}{10}$

$Now,10\%of\frac{11x}{10}=\frac{11x}{100}$

$\text{Now, the number}\frac{11x}{10}\text{has to be reduced by}10\%$

So, the final number
$=\frac{11x}{10}-\frac{11x}{100}=\frac{110x-11x}{100}=\frac{99x}{100}=99\%ofx$
$\therefore$ Effective change
$=x-\frac{99x}{100}$ $=\frac{x}{100}$
$=\frac{1}{100}x=1\%ofx$

Thus, the number is effectively decreased by 1%.