Question

A number is increased by 10% and then it is decreased by 10%. What is the percentage change in the number?

Answer 1

Answer:

$\sf{The \ number \ reduces \ by \ 1\%.}$

Given:

$\sf{A \ number \ is \ increased \ by \ 10\%}$

$\sf{ and \ then \ it \ is \ decreased \ by \ 10\%.}$

To find:

$\sf{What \ is \ the \ percentage \ change \ in}$

$\sf{the \ number.}$

Solution:

$\sf{Let \ the \ number \ be \ x.}$

$\sf{Now, \ 10\% \ of \ x=\dfrac{10}{100}\times \ x}$

$\sf{\therefore{10\% \ of \ x=\dfrac{x}{10}}}$

$\sf{According \ to \ the \ first \ condition}$

$\sf{Number \ is \ increased \ by \ 10\%}$

$\sf{\leadsto{x+\dfrac{x}{10}}}$

$\sf{\leadsto{\dfrac{11x}{10}}}$

$\sf{Now, \ 10\% \ of \ \dfrac{11x}{10}=\dfrac{10}{100}\times\dfrac{11x}{10}}$

$\sf{10\% \ of \ \dfrac{11x}{10}=\dfrac{11x}{100}}$

$\sf{According \ to \ the \ second \ condition}$

$\sf{Number \ decreases \ by \ 10\%}$

$\sf{\leadsto{\dfrac{11x}{10}-\dfrac{11x}{100}}}$

$\sf{\leadsto{\dfrac{110x-11x}{100}}}$

$\sf{\leadsto{\dfrac{99x}{100}}}$

$\sf{Let \ \dfrac{99x}{100} \ be \ y\% \ of \ x}$

$\sf{\therefore{y=\dfrac{99x}{100}\times{100}{x}}}$

$\sf{\therefore{y=99\%}}$

$\sf\purple{\tt{\therefore{The \ number \ reduces \ by \ 1\%.}}}$