Question

A number is increased by $10\%$ and then decreased 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. None of these

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 $10\%.$

Obtained number $=x+\frac{x}{10}=\frac{11x}{10}$

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

Now, the number $\frac{11x}{10}$ decreased by $10\%$

i.e. 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, we get that the number is effectively decreased by $1\%.$