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. 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 $\frac{11x}{10}$ reduced by $10\%$ i.e.
Final number of $=\frac{11x}{10}-\frac{11x}{100}=\frac{110x-11x}{100}=\frac{99x}{100}=99\%ofx$

$\therefore$ 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 $1\%$.