Question

A number is increased by 10% and the increased number is decreased by 10%. Show that the net decrease is 1%.

Answer 1

Let x be the number.
The number is increased by 10%.
∴ Increased number = 110% of x = $\left(x\times \frac{110}{100}\right)=\left(\frac{11x}{10}\right)$
The number is, then, decreased by 10%.

∴ Decreased number = 90% of $\left(\frac{11x}{10}\right)$ = $\left(\frac{11x}{10}\times \frac{90}{100}\right)=\left(\frac{99x}{100}\right)$

Net decrease = $\left(x-\frac{99x}{100}\right)=\frac{\left(100x-99x\right)}{100}=\frac{x}{100}$
Net decrease percentage = $\left(\frac{x}{100}\times \frac{1}{x}\times 100\right)=1$