Question

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

• A. Doesn’t change
• B. Decreases by 1%
• C. Increases by $1\%$
• D. Increases by 10%

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 $\frac{11x}{10}$ reduced by $10\%$ i.e.,
Final number $=\frac{11x}{10}-\frac{11x}{100}$

$=\frac{110x-11x}{100}$

$=\frac{99x}{100}=99\%ofx$

$\therefore$ Effective change = final number - initial number
$=99\%ofx-100\%ofx$
$=-1\%ofx$

Thus, the number is effectively decreased by $1\%$.