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% of $x$ =$\frac{x}{10}$

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% of $x$

$\therefore$ Effective change = Final number - Initial number
= 99% of $x$- 100% of $x$
= -1% of $x$

Thus, the number is effectively decreased by 1%.